Page 3 :
Herb Silverman, College of Charleston, Department of Mathematics, Charleston, SC 29424, U.S.A., , S. Ponnusamy, Indian Institute of Technology, Madras, Department of Mathematics, Chennai, 600 036, India, , Cover design by Alex Gerasev., Mathematics Subject Classification (2000): 11A06, 11M41, 30-XX, 32-XX (primary); 26Axx, 40Axx,, 26Bxx, 33Bxx, 26Cxx, 28Cxx, 31Axx, 35Axx, 37F10, 45E05, 76M40 (secondary), Library of Congress Control Number: 2006927602, ISBN-10: 0-8176-4457-1, ISBN-13: 978-0-8176-4457-4, , eISBN: 0-8176-4513-6, , Printed on acid-free paper., c 2006 Birkhäuser Boston, , All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Birkhäuser Boston, c/o Springer Science+Business Media LLC, 233, Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or, scholarly analysis. Use in connection with any form of information storage and retrieval, electronic, adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden., The use in this publication of trade names, trademarks, service marks and similar terms, even if they, are not identified as such, is not to be taken as an expression of opinion as to whether or not they are, subject to proprietary rights., , Printed in the United States of America., 987654321, www.birkhauser.com, , (TXQ/MP)
Page 5 :
Preface, , The student, who seems to be engulfed in our culture of specialization, too, quickly feels the necessity to establish an “area” of special interest. In keeping, with this spirit, academic bureaucracy has often forced us into a compartmentalization of courses, which pretend that linear algebra is disjoint from, modern algebra, that probability and statistics can easily be separated, and, even that advanced calculus does not build from elementary calculus., This book is written from the point of view that there is an interdependence between real and complex variables that should be explored at every opportunity. Sometimes we will discuss a concept in real variables and, then generalize to one in complex variables. Other times we will begin with, a problem in complex variables and reduce it to one in real variables. Both, methods—generalization and specialization—are worthy of careful consideration., We expect “complex” numbers to be difficult to comprehend and “imaginary” units to be shrouded in mystery. Hopefully, by staying close to the, real field, we shall overcome this regrettable terminology that has been thrust, upon us. The authors wish to create a spiraling effect that will first enable, the reader to draw from his or her knowledge of advanced calculus in order to, demystify complex variables, and then use this newly acquired understanding, of complex variables to master some of the elements of advanced calculus., We will also compare, whenever possible, the analytic and geometric character of a concept. This naturally leads us to a discussion of “rigor”. The, current trend seems to be that anything analytic is rigorous and anything, geometric is not. This dichotomy moves some authors to strive for “rigor” at, the expense of rich geometric meaning, and other authors to endeavor to be, “intuitive” by discussing a concept geometrically without shedding any analytic light on it. Rigor, as the authors see it, is useful only insofar as it clarifies, rather than confounds. For this reason, geometry will be utilized to illustrate, analytic concepts, and analysis will be employed to unravel geometric notions,, without regard to which approach is the more rigorous.
Page 6 :
viii, , Preface, , Sometimes, in an attempt to motivate, a discussion precedes a theorem., Sometimes, in an attempt to illuminate, remarks about key steps and possible, implications follow a theorem. No apologies are made for this lack of terseness, surrounding difficult theorems. While brevity may be the soul of wit, it is not, the soul of insight into delicate mathematical concepts. In recognition of the, primary importance of observing relationships between different approaches,, some theorems are proved in several different ways. In this book, traveling, quickly to the frontiers of mathematical knowledge plays a secondary role to, the careful examination of the road taken and alternative routes that lead to, the same destination., A word should be said about the questions at the end of each section. The, authors feel deeply that mathematics should be questioned—not only for its, internal logic and consistency, but for the reasons we are led where we are., Does the conclusion seem “reasonable”? Did we expect it? Did the steps seem, natural or artificial? Can we re-prove the result a different way? Can we state, intuitively what we have proved? Can we draw a picture?1, “Questions”, as used at the end of each section, cannot easily be categorized. Some questions are simple and some are quite challenging; some are, specific and some are vague; some have one possible answer and some have, many; some are concerned with what has been proved and some foreshadow, what will be proved. Do all these questions have anything in common? Yes., They are all meant to help the student think, understand, create, and question. It is hoped that the questions will also be helpful to the teacher, who, may want to incorporate some of them into his or her lectures., Less need be said about the exercises at the end of each section because, exercises have always received more favorable publicity than have questions., Very often the difference between a question and an exercise is a matter of, terminology. The abundance of exercises should help to give the student a, good indication of how well the material in the section has been understood., The prerequisite is at least a shaky knowledge of advanced calculus. The, first nine chapters present a solid foundation for an introduction to complex, variables. The last four chapters go into more advanced topics in some detail,, in order to provide the groundwork necessary for students who wish to pursue, further the general theory of complex analysis., If this book is to be used as a one-semester course, Chapters 5, 6, 7,, 8, and 9 should constitute the core. Chapter 1 can be covered rapidly, and, the concepts in Chapter 2 need be introduced only when applicable in latter, chapters. Chapter 3 may be omitted entirely, and the mapping properties in, Chapter 4 may be omitted., We wanted to write a mathematics book that omitted the word “trivial”., Unfortunately, the Riemann hypothesis, stated on the last page of the text,, 1, , For an excellent little book elaborating on the relationship between questioning, and creative thinking, see G. Polya, How to Solve It, second edition, Princeton, University press, Princeton, New Jersey, 1957.
Page 7 :
Preface, , ix, , could not have been mentioned without invoking the standard terminology, dealing with the trivial zeros of the Riemann zeta function. But the spirit, if, not the letter, of this desire has been fulfilled. Detailed explanations, remarks,, worked-out examples and insights are plentiful. The teacher should be able to, leave sections for the student to read on his/her own; in fact, this book might, serve as a self-study text., A teacher’s manual containing more detailed hints and solutions to questions and exercises is available. The interested teacher may contact us by, e-mail and receive a pdf version., We wish to express our thanks to the Center for Continuing Education, at the Indian Institute of Technology Madras, India, for its support in the, preparation of the manuscript., Finally, we thank Ann Kostant, Executive Editor, Birkhäuser, who has, been most helpful to the authors through her quick and efficient responses, throughout the preparation of this manuscript., , S. Ponnusamy, IIT Madras, India, June 2005, , Herb Silverman, College of Charleston, USA
Page 8 :
Contents, , Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii, 1, , Algebraic and Geometric Preliminaries . . . . . . . . . . . . . . . . . . . . 1, 1.1 The Complex Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1, 1.2 Rectangular Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5, 1.3 Polar Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15, , 2, , Topological and Analytic Preliminaries . . . . . . . . . . . . . . . . . . . ., 2.1 Point Sets in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 2.2 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 2.3 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 2.4 Stereographic Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 2.5 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., , 25, 25, 32, 39, 44, 48, , 3, , Bilinear Transformations and Mappings . . . . . . . . . . . . . . . . . . ., 3.1 Basic Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 3.2 Linear Fractional Transformations . . . . . . . . . . . . . . . . . . . . . . . . ., 3.3 Other Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., , 61, 61, 66, 85, , 4, , Elementary Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91, 4.1 The Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91, 4.2 Mapping Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100, 4.3 The Logarithmic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108, 4.4 Complex Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114, , 5, , Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121, 5.1 Cauchy–Riemann Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121, 5.2 Analyticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130, 5.3 Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
Page 9 :
xii, , Contents, , 6, , Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153, 6.1 Sequences Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153, 6.2 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164, 6.3 Maclaurin and Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173, 6.4 Operations on Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186, , 7, , Complex Integration and Cauchy’s Theorem . . . . . . . . . . . . . . . 195, 7.1 Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195, 7.2 Parameterizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207, 7.3 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217, 7.4 Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226, , 8, , Applications of Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 243, 8.1 Cauchy’s Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243, 8.2 Cauchy’s Inequality and Applications . . . . . . . . . . . . . . . . . . . . . . 263, 8.3 Maximum Modulus Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275, , 9, , Laurent Series and the Residue Theorem . . . . . . . . . . . . . . . . . . 285, 9.1 Laurent Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285, 9.2 Classification of Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293, 9.3 Evaluation of Real Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308, 9.4 Argument Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331, , 10 Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349, 10.1 Comparison with Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . 349, 10.2 Poisson Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358, 10.3 Positive Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371, 11 Conformal Mapping and the Riemann Mapping Theorem . . 379, 11.1 Conformal Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379, 11.2 Normal Families . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390, 11.3 Riemann Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395, 11.4 The Class S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405, 12 Entire and Meromorphic Functions . . . . . . . . . . . . . . . . . . . . . . . . 411, 12.1 Infinite Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411, 12.2 Weierstrass’ Product Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422, 12.3 Mittag-Leffler Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437, 13 Analytic Continuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445, 13.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445, 13.2 Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 458, References and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473, Index of Special Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475
Page 10 :
Contents, , xiii, , Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479, Hints for Selected Questions and Exercises . . . . . . . . . . . . . . . . . . . . 485
Page 11 :
1, Algebraic and Geometric Preliminaries, , The mathematician Euler once said, “God made integers, all else is the work, of man.” In this chapter, we have advanced in the evolutionary process to, the real number system. We partially characterize the real numbers and then,, alas, find an imperfection. The quadratic equation x2 + 1 = 0 has no solution., A new day arrives, the complex number system is born. We view a complex, number in several ways: as an element in a field, as a point in the plane, and, as a two-dimensional vector. Each way is useful and in each way we see an, unmistakable resemblance of the complex number system to its parent, the, real number system. The child seems superior to its parent in every way except, one—it has no order. This sobering realization creates a new respect for the, almost discarded parent., The moral of this chapter is clear. As long as the child follows certain, guidelines set down by its parent, it can move in new directions and teach us, many things that the parent never knew., , 1.1 The Complex Field, We begin our study by giving a very brief motivation for the origin of complex, numbers. If all we knew were positive integers, then we could not solve the, equation x + 2 = 1. The introduction of negative integers enables us to obtain, a solution. However, knowledge of every integer is not sufficient for solving, the equation 2x − 1 = 2. A solution to this equation requires the study of, rational numbers., While all linear equations with integers coefficients have rational solutions,, there are some quadratics that do not. For instance, irrational numbers are, needed to solve x2 − 2 = 0. Going one step further, we can find quadratic, equations that have no real (rational or irrational) solutions. The equation, x2 + 1 = 0 has no real solutions because the square of any real number, is nonnegative. In order to solve this equation, we must “invent” a number
Page 12 :
2, , 1 Algebraic and Geometric Preliminaries, , √, whose square is −1. This number, which we shall denote by i = −1, is called, an imaginary unit., Our sense of logic rebels against just “making up” a number that solves a, particular equation. In order to place this whole discussion in a more rigorous, setting, we will define operations involving combinations of real numbers and, imaginary units. These operations will be shown to conform, as much as possible, to the usual rules for the addition and multiplication of real numbers. We, may express any ordered pair of real numbers (a, b) as the “complex number”, a + bi, , or, , a + ib., , (1.1), , The set of complex numbers is thus defined as the set of all ordered pairs, of real numbers. The notion of equality and the operations of addition and, multiplication are defined as follows:1, (a1 , b1 ) = (a2 , b2 ) ⇐⇒ a1 = a2 , b1 = b2 ,, (a1 , b1 ) + (a2 , b2 ) = (a1 + a2 , b1 + b2 ),, (a1 , b1 )(a2 , b2 ) = (a1 a2 − b1 b2 , a1 b2 + a2 b1 )., The definition for the multiplication is more natural than it appears to be,, for if we denote the complex numbers of the form (1.1), multiply as we would, real numbers, and use the relation i2 = −1, we obtain, (a1 + ib1 )(a2 + ib2 ) = a1 a2 − b1 b2 + i(a1 b2 + a2 b1 )., Several observations should be made at this point. First, note that the formal, operations for addition and multiplication of complex numbers do not depend, on an imaginary number i. For instance, the relation i2 = −1 can be expressed, as (0, 1)(0, 1) = (−1, 0). The symbol i has been introduced purely as a matter, of notational convenience. Also, note that the order pair (a, 0) represents the, real number a, and that the relations, (a, 0) + (b, 0) = (a + b, 0), , and, , (a, 0)(b, 0) = (ab, 0), , are, respectively, addition and multiplication of real numbers. Some of the, essential properties of real numbers are as follows: Both the sum and product, of real numbers are real numbers, and the order in which either operation is, performed may be reversed. That is, for real numbers a and b, we have the, commutative laws, a+b=b+a, , and a · b = b · a., , (1.2), , The associative laws, a + (b + c) = (a + b) + c, 1, , and a · (b · c) = (a · b) · c,, , The symbol ⇐⇒ stands for “if and only if” or “equivalent to.”, , (1.3)
Page 13 :
1.1 The Complex Field, , 3, , and the distributive law, a · (b + c) = a · b + a · c, , (1.4), , also holds for all real numbers a, b, and c. The numbers 0 and 1 are, respectively, the additive and multiplicative identities. The additive inverse of a is, −a, and the multiplicative inverse of a (= 0) is the real number a−1 = 1/a., Stated more concisely, the real numbers form a field under the operations of, addition and multiplication., Of course, the real numbers are not the only system that forms a field., The rational numbers are easily seen to satisfy the above conditions for a, field. What is important in this chapter is that the complex numbers also, form a field. The additive identity is (0, 0), and the additive inverse of (a, b), is (−a, −b). The multiplicative inverse of (a, b) = (0, 0) is, , , b, a, ., ,− 2, a2 + b2, a + b2, We leave the confirmation that the complex numbers satisfy all the axioms, for a field as an exercise for the reader., The discerning math student should not be satisfied with the mere verification of a proof. He/she should also have a “feeling” as to why the proof, works. Did the reader ask why the multiplicative inverse of (a, b) might be, expected to be, , , a, b, ,− 2, ?, a2 + b2, a + b2, Let us go through a possible line of reasoning. If we write the inverse of, (a, b) = a + bi as, 1, ,, (a + ib)−1 =, a + ib, then we want to find a complex number c + di such that, 1, = c + id., a + ib, By cross multiplying, we obtain ac + i2 bd + i(ad + bc) = 1, or, , ac − bd = 1,, ad + bc = 0., The solution to these simultaneous equation is, c=, , a2, , a, ,, + b2, , d=−, , a2, , b, ., + b2, , Can the reader think of other reasons to suspect that
Page 14 :
4, , 1 Algebraic and Geometric Preliminaries, , (a, b)−1 =, , , , , a, b, ,, −, ?, a2 + b2, a2 + b2, , Let z = (x, y) be a complex number. Then x and y are called the real part of, z, Re z, and the imaginary part of z, Im z, respectively. Denote the set of real, numbers by R and the set of complex numbers by C. There is a one-to-one, correspondence between R and a subset of C, represented by x ↔ (x, 0) for, x ∈ R, which preserves the operations of addition and multiplication. Hence, we will use the real number x and the ordered pair (x, 0) interchangeably., We will also denote the ordered pair (0, 1) by i. Because a complex number, is an ordered pair of real numbers, we use the terms C = R2 or C = R × R, interchangeably. Thus R × 0 is a subset of C consisting of the real numbers., As noted earlier, an advantage of the field C is that it contains a root, of z 2 + 1 = 0. In Chapter 8 we will show that any polynomial equation, a0 + a1 z + · · · + an z n = 0 has a solution in C. But this extension from R to, C is not without drawbacks. There is an important property of the real field, that the complex field lacks. If a ∈ R, then exactly one of the following is, true:, a = 0, a > 0, −a > 0 (trichotomy)., Furthermore, the sum and the product of two positive real numbers is positive, (closure)., A field with an order relation < that satisfies the trichotomy law and these, two additional conditions is said to be ordered. In an ordered field, like the, real or rational numbers, we are furnished with a natural way to compare any, two elements a and b. Either a is less than b (a < b), or a is equal to b (a = b),, or a is greater than b (a > b). Unfortunately, no such relation can be imposed, on the complex numbers, for suppose the complex numbers are ordered; then, either i or −i is positive. According to the closure rule, i2 = (−i)2 = −1 is, also positive. But 1 must be negative if −1 is positive. However, this violates, the closure rule because (−1)2 = 1., To sum up, there is a complex field that contains a real field that contains a, rational field. There are advantages and disadvantages to studying each field., It is not our purpose here to state properties that uniquely determine each, field, although this most certainly can be done., Questions 1.1., 1. Can a field be finite?, 2. Can an ordered field be finite?, 3. Are there fields that properly contain the rationals and are properly, contained in the reals?, 4. When are two complex numbers z1 and z2 equal?, 5. What complex numbers may be added to or multiplied by the complex, number a + ib to obtain a real number?, 6. How can we separate the quotient of two complex numbers into its real, and imaginary parts?
Page 15 :
1.2 Rectangular Representation, , 5, , 7. What can we say about the real part of the sum of the two complex, numbers? What about the product?, 8. What kind of implications are there in defining a complex number as, an ordered pair?, 9. If a polynomial of degree n has at least one solution, can we say more?, 10. If we try to define an ordering of the complex numbers by saying that, (a, b) > (c, d) if a > b and c > d, what order properties are violated?, 11. Can any ordered field have a solution to x2 + 1 = 0?, Exercises 1.2., , √, 1. Show that the set of real numbers of the form a + b 2, where a and b, are rational, is an ordered field., 2. If a and b are elements in a field, show that ab = 0 if and only if either, a = 0 or b = 0., 3. Suppose a and b are elements in an ordered field, with a < b. Show that, there are infinitely many elements between a and b., 4. Find the values of, (a) (−2, 3)(4, −1), (b) (1 + 2i){3(2 + i) − 2(3 + 6i)}, (d) (1 + i)4, (c) (1 + i)3, (e) (1 + i)n − (1 − i)n ., 5. Express the following in the form x + iy:, (a) (1 + i)−5, √, (c) eiπ/2 + 2eiπ/4, a + ib a − ib, −, (e), a − ib a + ib, (g) (2 + i)2 + (2 − i)2, , (b) (3 − 2i)/(1 − i), (d) (1 + i)eiπ/6, 3 + 5i, 1+i, (f), +, 7 + i √4 + 3i, (4 + 3i) 3 + 4i, (h), 3+i, √, (j) (−1 + i 3)60, √, ( 3 − i)2 (1 + i)5, √, (l), ., ( 3 + i)4, , (ai40 − i17 ), , (a−real), √ −1 + i, 1 + a2 + ia, √, (k), , (a−real), a − i 1 + a2, 6. Show that, , , √ 6, √ 3, ±1 ± i 3, −1 ± 3, = 1 and, =1, 2, 2, (i), , for all combinations of signs., 7. For any integers k and n, show that in = in+4k . How many distinct, values can be assumed by in ?, , 1.2 Rectangular Representation, Just as a real number x may be represented by a point on a line, so may a, complex number z = (x, y) be represented by a point in the plane (Figure 1.1).
Page 16 :
6, , 1 Algebraic and Geometric Preliminaries, , y, y, , 0, , z = (x, y )= x +iy, , x, , x, , Figure 1.1. Cartesian representation of z in plane, , Each complex number corresponds to one and only one point. Thus the, terms complex number and point in the plane are used interchangeably. The, x and y axes are referred to as the real axis and the imaginary axis, while the, xy plane is called the complex plane or the z plane., There is yet another interpretation of the complex numbers. Each point, (x, y) of the complex plane determines a two-dimensional vector (directed line, segment) from (0, 0), the initial point, to (x, y), the terminal point. Thus the, complex number may be represented by a vector. This seems natural in that, the definition chosen for addition of complex numbers corresponds to vector, addition; that is,, (x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 )., Geometrically, vector addition follows the so-called parallelogram rule, which, we illustrate in Figure 1.2. From the point z1 , construct a vector equal in, magnitude and direction to the vector z2 . The terminal point is the vector, z1 + z2 . Alternatively, if a vector equal in magnitude and direction to z1 is, joined to the vector z2 , the same terminal point is reached. This illustrates, the commutative property of vector addition. Note that the vector z1 + z2, is a diagonal of the parallelogram formed. What would the other diagonal, represent?, , Figure 1.2. Illustration for parallelogram law
Page 17 :
1.2 Rectangular Representation, , 7, , Figure 1.3. Modulus of a complex number z, , By the magnitude (length) of the vector (x, y) we mean the distance of, the point z = (x, y) from the origin. This distance is called the modulus, or absolute value of the complex number z, and denoted by |z|; its value is, , x2 + y 2 . For each positive real number r, there are infinitely many distinct, values (x, y) whose absolute value is r = |z|, namely the points on the circle, x2 + y 2 = r2 . Two of these points, (r, 0) and (−r, 0), are real numbers so that, this definition agrees with the definition for the absolute value in the real field, (see Figure 1.3)., Note that, for z = (x, y),, , |x| = |Re z| ≤ |z|,, |y| = |Im z| ≤ |z|., The distance between any two points z1 = (x1 , y1 ) and z2 = (x2 , y2 ) is, , |z2 − z1 | = (x2 − x1 )2 + (y2 − y1 )2 ., The triangle inequalities, , , |z1 + z2 | ≤ |z1 | + |z2 |,, |z1 − z2 | ≥ | |z1 | − |z2 | |, , say, geometrically, that no side of a triangle is greater in length than the, sum of the lengths of the other two sides, or less than the difference of the, lengths of the other two sides (Figure 1.2). The algebraic verification of these, inequalities is left to the reader., Among all points whose absolute value is the same as that of z = (x, y),, there is one which plays a special role. The point (x, −y) is called the conjugate, of z and is denoted by z. If we view the real axis as a two-way mirror, then z, is the mirror image of z (Figure 1.4)., From the definitions we obtain the following properties of the conjugate:
Page 18 :
8, , 1 Algebraic and Geometric Preliminaries, , Figure 1.4. Mirror image of complex numbers, , , , z1 + z2 = z 1 + z 2 ,, z1 z2 = z 1 z 2 ., , (1.5), , Some of the important relationships between a complex number z = (x, y), and its conjugates are, ⎧, z + z = (2x, 0) = 2Re z,, ⎪, ⎪, ⎨ z − z = (0, 2y) = 2iIm z,, , (1.6), |z| = |z| = x2 + y 2 ,, ⎪, ⎪, ⎩, zz = |z|2 ., The squared form of the absolute value in (1.6) is often the most workable., For example, to prove that the absolute value of the product of two complex, numbers is equal to the product of their absolute values, we write, |z1 z2 |2 = (z1 z2 )(z1 z2 ) = (z1 z2 )(z 1 z 2 ) = (z1 z 1 )(z2 z 2 ) = (|z1 | |z2 |)2 ., Moreover, the conjugate furnishes us with a method of separating the inverse, of a complex number into its real and imaginary parts:, (a + bi)−1 =, , 1, a, b, a − bi, a + bi, ·, = 2, − 2, i., = 2, 2, 2, a + bi a + bi, a +b, a +b, a + b2, , Equation of a line in C. Now we may rewrite the equation of a straight, line in the plane, with the real and imaginary axes as axes of coordinates, as, , , , , z−z, z+z, +b, + c = 0,, ax + by + c = 0, a, b, c ∈ R; i.e., a, 2, 2i
Page 20 :
10, , 1 Algebraic and Geometric Preliminaries, , |z1 − z2 | < |1 − z1 z 2 |, and if either |z1 | = 1 or |z2 | = 1, then, |z1 − z2 | = |1 − z 1 z2 |., (ii) |z1 + z2 |2 + |z1 − z2 |2 = 2(|z1 |2 + |z2 |2 ) ( Parallelogram identity ); for,, L.H.S = (z1 + z2 )(z 1 + z 2 ) + (z1 − z2 )(z 1 − z 2 ), = [|z1 |2 + (z1 z 2 + z 1 z2 ) + |z2 |2 ], + [|z1 |2 − (z1 z 2 + z 1 z2 ) + |z2 |2 ], = R.H.S., Example 1.3. Let us use the triangle inequality to find upper and lower, bounds for |z 4 − 3z + 1|−1 whenever |z| = 2. To do this, we need to find m, and M so that m ≤ |z 4 − 3z + 1|−1 ≤ M for |z| = 2. As |3z − 1| ≤ 3|z| + 1 = 7, for |z| = 2, we have, |z 4 − 3z + 1| ≥ | |z 4 | − |3z − 1| | ≥ 24 − 7 = 9, and |z 4 − 3z + 1| ≤ |z|4 + |3z − 1| = 24 + 7 = 23. Thus, for |z| = 2, we have, 1, 1, ≤ |z 4 − 3z + 1|−1 ≤ ., 23, 9, , •, , Example 1.4. Suppose that we wish to find all circles that are orthogonal to, both |z| = 1 and |z − 1| = 4. To do this, we consider two circles:, C1 = {z : |z − α1 | = r1 },, , C2 = {z : |z − α2 | = r2 }., , These two circles are orthogonal to each other if (see Figure 1.5), r12 + r22 = |α1 − α2 |2 ., In view of this observation, the conditions for which a circle |z − α| = R is, orthogonal to both |z| = 1 and |z − 1| = 4 are given by, 1 + R2 = |α − 0|2 and 42 + R2 = |α − 1|2 = 1 + |α|2 − 2Re α, which give R = (|α|2 − 1)1/2 and Re α = −7. Consequently,, α = −7 + ib and R = (49 + b2 − 1)1/2 = (48 + b2 )1/2, and the desired circles are given by, Cb : |z − (−7 + ib)| = (48 + b2 )1/2 ,, , b ∈ R., , •
Page 22 :
12, , 1 Algebraic and Geometric Preliminaries, , Figure 1.6. Equilateral triangle ABC, , (αα)(ββ)(γγ) = (αα)3 and (αα)(ββ)(γγ) = (ββ)3 ., Thus,, 1, 1, 1, + + = 0 =⇒ |α|3 = |β|3 = |γ|3 =⇒ |α| = |β| = |γ|,, α β, γ, showing that ABC is equilateral., Here is an alternate proof. First we remark that equilateral triangles are, preserved under linear transformations f (z) = az + b, which can be easily, verified by replacing zj by azj + b (j = 1, 2, 3) in (1.10). By a suitable transformation, we can reduce the problem to a simpler one. If z1 , z2 , z3 are the, vertices of a degenerated equilateral triangle (i.e., z1 = z2 = z3 ), then (1.10), holds. If two of the vertices are distinct, then, by a suitable transformation,, we can take z1 = 0 and z2 = 1. Then (1.10) takes the form 1 + z32 = z3 , which, gives, √, √, 1−i 3, 1+i 3, z3 =, or, ., 2, 2, •, In either case {0, 1, z3 } forms vertices of an equilateral triangle., Example 1.6. Suppose we wish to describe geometrically the set S given by, S = {z : |z − a| − |z + a| = 2c}, , (0 = a ∈ C, c ≥ 0),, , (1.11), , for the following situations:, (i) c > |a|, , (ii) c = 0, , (iii) 0 < c < a, , (iv) c = a > 0., , The triangle inequality gives that, |2a| = |z − a − (z + a)| ≥ |z − a| − |z + a| = 2c, i.e., c ≤ |a|., Thus, there are no complex numbers satisfying (1.11) if c > |a|. Hence, S = ∅, whenever c > |a|.
Page 23 :
1.2 Rectangular Representation, , 13, , If c = 0, we have |z − a| = |z + a| which shows that S is the line that is, the perpendicular bisector of the line joining a and −a., Next, we consider the case a > c > 0. Then, writing z = x + iy,, |z − a| − |z + a| = 2c ⇐⇒ |z − a|2 = (2c + |z + a|)2, ⇐⇒ |z − a|2 = 4c2 + |z + a|2 + 4c|z + a|, ⇐⇒ c|z + a| + c2 = −aRe z (Re z < 0), ⇐⇒ c2 [|z|2 + a2 + 2aRe z] = (c2 + aRe z)2, ⇐⇒ c2 |z|2 − a2 (Re z)2 = c2 (c2 − a2 ), y2, x2, = 1., ⇐⇒ 2 − 2, c, a − c2, Further, we observe that for |z − a| − |z + a| to be positive, we must have, Re z < 0. Thus, if a > c > 0 we have, , y2, x2, =1, S = x + iy : 2 − 2, c, a − c2, and so S describes a hyperbola with focii at a, −a., Finally, if c = a then, |z − a| − |z + a| = 2a ⇐⇒ |z + a| = −Re (z + a) =⇒ Re (z + a) < 0, and therefore, S in this case is the interval (−∞, −a]., , •, , Questions 1.7., 1. In Figure 1.2, would we still have a parallelogram if the vector z2 were, in the same or the opposite direction as that of z1 ?, 2. Geometrically, can we predict the quadrant of z1 +z2 from our knowledge, of z1 and z2 ?, 3. Why don’t we define multiplication of complex numbers as vector multiplication?, 4. When does the triangle inequality become an equality?, 5. What would be the geometric interpretation of the inequality for the, sum of n complex numbers?, 6. Name some interesting relationships between the points (x, y) and, (−x, y)., 7. If a and b√are positive, rational, √, √numbers, why might we want to call the, √, numbers a + b and a − b real conjugates?, √, √, 8. Is every rational number algebraic? Are 3 and 5 5 − 3i algebraic?, Note: A number is algebraic if it is a solution of a polynomial (in z), with integer coefficients. Numbers which are not algebraic are called, transcendental numbers., 9. What does |z|2 + βz +√βz + γ = 0 represent if |β|2 ≥ γ?, 10. Is |z + 1| + |z − 1| ≤ 2 2 if |z| ≤ 1?
Page 24 :
14, , 1 Algebraic and Geometric Preliminaries, , Exercises 1.8., 1. If z1 = 3 − 4i and z2 = −2 + 3i, obtain graphically and analytically, (b) 3z1 − 2z 2, (c) z1 − z 2 − 4, (a) 2z1 + 4z2, (d) |z1 + z2 |, (e) |z1 − z2 |, (f) |2z 1 + 3z 2 − 1|., 2. Let z1 = x1 + iay1 and z2 = x2 − ib/y1 , where a, b are real. Determine, a condition on y1 so that z1−1 + z2−1 is real., 3. Identify all the points in the complex plane that satisfy the following, relations., (a) 1 < |z| ≤ 3, (b) |(z − 3)/(z + 3)| < 2, (c) |z − 1| + |z + 1| = 2, (d) Re (z − 5) = |z| + 5, (e) Re z 2 > 0, (f) Im z 2 > 0, (g) Re ((1 − i)z) = 2, (h) |z − i| = Re z, (i) Re (z) = |z|, (j) Re (z 2 ) = 1, (l) [Im (iz)]2 = 1., (k) z = 5/(z − 1) (z = 1), 4. Let |(z − a)/(z − b)| = M , where a and b are complex constants and, M > 0. Describe this curve and explain what happens as M → 0 and, as M → ∞., 5. Find a complex form for the hyperbola with real equation 9x2 −4y 2 = 36., 6. If |z| < 1, prove that, , , , , , , z, 1+z, 1, 1, 1, (b) Re, (c) Re, (a) Re, >, >−, > 0., 1−z, 2, 1−z, 2, 1−z, 7. If P (z) is a polynomial equation with real coefficients, show that z1 is a, root if and only if z 1 is a root. Conclude that any polynomial equation, of odd degree with real coefficients must have at least one real root. Can, you prove this using elementary calculus?, 8. Prove that, for every n ≥ 1,, |z1 + z2 + · · · + zn | ≤ |z1 | + |z2 | + · · · + |zn |., 9. Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be complex numbers. Prove the, Schwarz inequality,, n, , k=1, , , , 2, , ak bk, , ≤, , n, , k=1, , , 2, , |ak |, , n, , , , 2, , |bk | | ., , k=1, , When will equality hold?, 10. Define e(α) = cos α + i sin α, for α real. Prove the following., (a) e(0) = 1, (b) |e(α)| = 1, (c) e(α1 + α2 ) = e(α1 )e(α2 ), (d) e(nα) = [e(α)]n ., Which of these properties does the real-valued function f (x) = ex, satisfy?
Page 25 :
1.3 Polar Representation, , 15, , 11. Show that the line connecting the complex numbers z1 and z2 is perpendicular to the line connecting z3 and z4 if and only if, Re {(z1 − z2 )(z 3 − z 4 )} = 0., 12. If a, b are real numbers in the unit interval (0, 1), then when do the three, points z1 = a + i, z2 = 1 + ib and z3 = 0 form an equilateral triangle?, 13. If |zj | = 1 (j = 1, 2, 3) such that z1 + z2 + z3 = 0, then show that zj ’s, are the vertices of an equilateral triangle., , 1.3 Polar Representation, In Section 1.2, the magnitude of the vector z = x + iy was discussed. What, about its direction? A measurement of the angle θ that the vector z (= 0), makes with the positive real axis is called an argument of z (see Figure 1.7)., Thus, we may express the point z = (x, y) in the “new” form, (r cos θ, r sin θ)., This, of course, is just the polar coordinate representation for the complex, number z. We have the familiar relations, , y, r = |z| = x2 + y 2 and tan θ = ., x, The real numbers r and θ, like x and y, uniquely determine the complex number z. Unfortunately, the converse isn’t completely true. While z uniquely, determines the x and y, hence r, the value of θ is determined up to a multiple of 2π. There are infinitely many distinct arguments for a given complex, number z, and the symbol arg z is used to indicate any one of them. Thus the, arguments of the complex number (2, 2) are, π, + 2kπ, 4, , (k = 0, ±1, ±2, . . . )., , This inconvenience can sometimes (although not always) be ignored by distinguishing (arbitrarily) one particular value of arg z. We use the symbol Arg z, to stand for the unique determination of θ for which −π < arg z ≤ π. This θ, is called the principal value of the argument. To illustrate,, Arg (2, 2) =, , π, ,, 4, , π, Arg (0, −5) = − ,, 2, , Arg (−1,, , √, , 3) =, , 2π, ., 3, , Note that Re z > 0 is equivalent to |Arg z| < π/2. If x = y = 0, the expression, tan θ = y/x has no meaning. For this reason, arg z is not defined when z = 0., Suppose that z1 and z2 have the polar representations, z1 = r1 (cos θ1 + i sin θ1 ), , and z2 = r2 (cos θ2 + i sin θ2 ).
Page 28 :
18, , 1 Algebraic and Geometric Preliminaries, , Example 1.11. Let z = sin θ + i cos 2θ and w = cos θ + i sin 2θ. We wish to, show that there exists no value of θ for which z = w. To do this, we first note, that, z = w ⇐⇒ sin θ = cos θ and cos 2θ = sin 2θ., There exists no values of θ satisfying both conditions, because sin θ = cos θ, implies that cos 2θ = cos2 θ − sin2 θ = 0, and so the second condition reduces, to sin 2θ = 2 sin θ cos θ = 0, i.e., sin θ = 0 = cos θ., •, Remark 1.12. Geometric considerations (Figures 1.2 and 1.7) indicate that, the rectangular representation will frequently be more useful for problems, involving sums of complex numbers, with polar representation being more, •, useful for problems involving products., If we let z1 = z2 = · · · = zn in (1.13), we obtain, z n = rn (cos nθ + i sin nθ)., , (1.14), , For |z| = 1 (the unit circle), (1.14) reduces to, (cos θ + i sin θ)n = cos nθ + i sin nθ,, , (1.15), , a theorem of DeMoivre., The possibility of finding nth roots of the complex number is suggested by, 1/n, (1.14). A complex number z is an nth root of z0 if z n = z0 , written z = z0 ., The problem is to reverse the multiplicative operation and determine a, number which, when multiplied by itself n times, furnishes us with the original, number. Given a complex number z0 = r0 (cos θ0 + i sin θ0 ), how do you find a, complex number z = r(cos θ + i sin θ) such that z n = z0 ? By (1.14), we must, have, rn (cos nθ + i sin nθ) = r0 (cos θ0 + i sin θ0 )., , (1.16), , Since | cos α + i sin α| = 1 for all real α, (1.16) yields the relations, r n = r0 ,, , cos nθ + i sin nθ = cos θ0 + i sin θ0 ., , (1.17), , 1/n, , The first relation in (1.17) shows that |z| = r0 , which we already knew, (why)? But the second gives important information about the argument of, z, namely, that n arg z differs from arg z0 by a multiple of 2π (that is, nθ =, θ0 + 2kπ, k = 0, ±1, ±2, . . . ):, θ=, , θ0 + 2kπ, ., n, , How many integers k in (1.18) produce distinct solutions? We have, , , , , , θ0 + 2kπ, θ0 + 2kπ, 1/n, 1/n, cos, + i sin, ., z = z0 = r0, n, n, , (1.18), , (1.19)
Page 29 :
1.3 Polar Representation, , 19, , For each k (k = 0, 1, 2, . . . , n − 1), there is a different value for z. We leave it, for the reader to verify that there are no more solutions. Thus, given z0 = 0,, there are exactly n distinct complex numbers z such that z n = z0 ., By letting z0 = 1 in (1.19), we may find the nth roots of unity. If z n = 1,, then, , , , , 2kπ, 2kπ, z = cos, + i sin, (k = 0, 1, 2, . . . , n − 1)., (1.20), n, n, Geometrically, the solutions represent the n vertices of a regular polygon of, n sides inscribed in a circle with center at the origin and radius equal to one., See Figures 1.9 and 1.10 for the inscribed square and pentagon., By (1.20), the difference in the arguments of any two successive nth roots, of unity is constant (2π/n). If we let, ω = cos, , 2π, 2π, + i sin, ,, n, n, , then each root of unity may be expressed as a multiple of ω; that is,, ω, ω 2 , ω 3 , . . . , ω n−1 ,, , ω n = ω 0 = 1., , This gives interesting information about the sums and products of the roots of, the unity, namely, that the product of any two roots of unity is also a root of, unity, and that the sum of all nth roots of unity is zero. The latter statement, follows from the identify, 1 + ω + ω 2 + · · · + ω n−1 =, , 1 − ωn, ., 1−ω, , Using (1.19), we easily see, for instance, the following:, √, (a) ∗√3 + 4i = ±(2 + i), (b) ∗ −3 + 4i = ±(1 + 2i), , Figure 1.9. Illustration for the 4th roots of unity
Page 30 :
20, , 1 Algebraic and Geometric Preliminaries, , Figure 1.10. Illustration for the 5th roots of unity, , ⎛, √, , √, , (c) ∗ 1 + i = ± ⎝, , √, , 2+1, +i, 2, , ⎞, 2 − 1⎠, 2, , √, (d) ∗2i = ±(1 + i), √, , √, 1−i 3, 3−i, =±, (e) ∗, 2, 2, √, , , √, 3+i, √, (f) ∗ 1 + i 3 = ±, 2, √, (g) ∗√−5 − 12i = ±(−2 + 3i), (h) ∗√5 + 12i = ±(3 + 2i), (i) ∗ −5 + 12i = ±(2 + 3i)., √, Here ∗ a + ib denotes the two 2th roots of the complex number a + ib., Since the n nth roots of unity are given by (1.20), we have, z n − 1 = (z − 1)(z − ω1 )(z − ω2 ) · · · (z − ωn−1 ),, , ωk = ω k = e2πki/n ., , Dividing both sides by z − 1, using the identity, 1 + z + z 2 + · · · + z n−1 =, , 1 − zn, (z = 1),, 1−z, , and letting z → 1, we have, n = (1 − ω1 )(1 − ω2 ) · · · (1 − ωn−1 ), and, n = (1 − ω 1 )(1 − ω 2 ) · · · (1 − ω n−1 )., As (1 − e−iθ )(1 − eiθ ) = 2(1 − cos θ) = 4 sin2 (θ/2), it follows that, n2 =, , n−1, , k=1, , |1 − ωk |2 =, , n−1, , , 4 sin2, , k=1, , , , kπ, n, , , , = 22(n−1), , n−1, , k=1, , sin2, , , , kπ, n, , , .
Page 31 :
1.3 Polar Representation, , 21, , Taking the positive square root on both sides we have, n = 2n−1, , n−1, , , , sin, , k=1, , kπ, n, , , , n > 1., , (1.21), , We can make the following generalization: Consider the equation, Ma (z) = z 2n − 2z n an cos nφ + a2n = 0 (n ∈ N, a ∈ R+ , φ ∈ R)., Solving this for z n , we find z n = an e±inφ so that, Ma (z) = [z n − an einφ ][z n − an e−inφ ]., Therefore, using the concept of nth root of a complex number, we can write, Ma (z) =, =, , n , , , z − aei(φ+2kπ/n), , , , z − ae−i(φ+2kπ/n), , , , k=1, n , , k=1, , , , , 2kπ, + a2 ., z 2 − 2za cos φ +, n, , Some special cases of (1.22) follow:, (a) Taking φ = 0, we have, (z n − an )2 =, , n , , , z 2 − 2za cos, , k=1, , , , 2kπ, n, , , , , + a2 ., , (b) Taking φ = π/n, we have, (z n + an )2 =, , , , , n , , (2k + 1)π, + a2 ., z 2 − 2za cos, n, , k=1, , (c) If a = 1 then, on dividing (1.22) by z n , z = 0, we have, , , n , , 2kπ, z + z −1 − 2 cos φ +, z n + z −n − 2 cos(nφ) =, n, k=1, , and so, if z = eiθ , this becomes, cos(nθ) − cos(nφ) = 2n−1, , , , n , , 2kπ, cos θ − cos φ +, n, , k=1, , which is, for cos θ = cos φ, equivalent to, , , n−1, , 2kπ, cos(nθ) − cos(nφ), n−1, =2, cos θ − cos φ +, ., cos θ − cos φ, n, k=1, , (1.22)
Page 32 :
22, , 1 Algebraic and Geometric Preliminaries, , In the limiting case when θ, φ → 0, the above reduces to, n−1, , n=2, , n−1, , k=1, , , sin, , kπ, n, , , ,, , which is nothing but (1.21)., Questions 1.13., 1. What problem would be created by defining the argument of z = 0 to, be zero?, 2. Loosely speaking, for complex numbers z1 and z2 we have, arg(z1 z2 ) = arg z1 + arg z2 ., What real-valued functions have the property that, f (x1 x2 ) = f (x1 ) + f (x2 )?, 3., 4., 5., 6., 7., 8., , 9., 10., 11., 12., 13., 14., 15., 16., 17., 18., 19., 20., 21., , When does Arg (z1 z2 ) = Arg z1 + Arg z2 ?, How are the complex numbers z1 and z2 related if arg(z1 ) = arg z2 ?, How are the arguments arg(z1 ) and arg z2 related if z1 = z2 ?, How are the arguments arg(z1 ) and arg z2 related if Re (z1 z 2 ) = |z1 z2 |?, How are the arguments arg(z1 ) and arg z2 related if |z1 +z2 | = |z1 |+|z2 |?, As the complex number z approaches the negative real axis from above, and below, what is happening to Arg z? What if z approaches the positive real axis from above and below?, How do the arguments of z and 1/z compare?, How do the arguments of z and z compare?, How do the arguments of z and 1/z compare?, What is the position of the complex number (cos α + i sin α)z relative, to the position of z?, What are some differences between the terms angle, real number, and, argument?, Of what use might the binomial theorem be in this section?, For which integers n does z n = 1 have, only real solutions?, For which complex numbers z does z/z = z/|z|?, Is it always √, the case that for any given nonzero complex number, either, √, z 2 = z or z 2 = −z?, Which postulates for a field are satisfied by the roots of unity under, ordinary addition and multiplication of complex numbers?, What can you say about the nth roots of an arbitrary complex number?, For α an arbitrary real number, how many solutions might you expect, z α = 1 to have?, If z = eiα (α ∈ (0, 2π)), is (1 + z)/(1 − z) equal to i cot(α/2)?
Page 33 :
1.3 Polar Representation, , 23, , Exercises 1.14., , √, 1. For a fixed positive integer n, determine the real part of (1 + i 3)n ., 2. Find two complex numbers z1 and z2 so that, Arg (z1 z2 ) = Arg z1 + Arg z2 ., , 3. Find two complex numbers z1 and z2 so that, Arg (z1 z2 ) = Arg z1 + Arg z2 ., 4. Describe the following regions geometrically., (a) Arg z = π/6, |z| > 1, (b) π/4 < Arg z < π/2, (c) −π < Arg z < 0, |z + i| > 2, (d) 1 < |z − 1| < 5., 5. If |1 − z| < 1, show that |Arg z| < π/2., 6. If |z| < 1, show that |Arg ((1 + z)/(1 − z))| < π/2., 7. If Re z > 0, show that Re (1/z) > 0. If Re z > a > 0, what can you say, about Re (1/z)?, 8. If |z| = 1, z = −1, show that z may be expressed in the form, z=, , 1 + it, ,, 1 − it, , where t is a real number., 9. Write the polar form of the following:, 1 + cos φ + i sin φ, (a), (0 < φ < π/2), 1 + cos φ − i sin φ, 1 + cos φ + i sin φ, (b), 1 − cos φ − i sin φ, (c) 1 − sin φ + i cos φ (0 < φ < π/2), (d) − sin φ − i cos φ, (e) (1 + i)√n (n ∈ N) √, (f) (1 + i 3)n + (1 − i 3)n (n ∈ N)., 10. Find all values of the following and simplify the expressions as much as, possible., √, (a) i1/2, (b) i1/4, (c) (−i)1/3, (d) 1 + i, √, √, √, (e) 6 8, (f) 4 + 3i, (g) (4 − 3i)1/3, (h) 2 + i, √, 11. If ω = (−1 + i 3)/2 is a cube root of unity and if, Sn = 1 − ω + ω 2 + · · · + (−1)n−1 ω n−1 ,, then find a formula for Sn ., 12. Let ω be a cube root of unity and let a, b, c be real. Determine a condition, on a, b, c so that (a + bω + cω 2 )3 is real., 13. Let ω be a cube root of unity. Determine the value of, (a) (1 + ω)3, , (b) (1 + 2ω + ω 2 )(1 + ω + 2ω 2 ), , (c) (1 + ω + 2ω 2 )9, , (d) (1 + 3ω + 2ω 2 )(1 + 4ω + 3ω 2 ).
Page 34 :
24, , 1 Algebraic and Geometric Preliminaries, , 14. Let ω = 1 be an nth root of unity. Show that, n, ., 1−ω, n, 15. Let ωk = cos(2kπ/n) + i sin(2kπ/n). Show that k=1 |ωk − ωk−1 | < 2π, for all values of n. What happens as n approaches ∞?, 16. Find the roots of the equation (1 + z)5 = (1 − z)5 ., 17. Find α, β, γ and δ such that the roots of the equation, 1 + 2ω + 3ω 2 + · · · + nω n−1 = −, , z 5 + αz 4 + βz 3 + γz 2 + δz + η = 0, lie on a regular pentagon centered at 1., 18. Prove that for any real x and a natural number n,, n, , −1, ix + 1, = 1., ei2n cot (x), ix − 1, 19. Find a positive integer n such that, √, (i) ( 3 + i)n = 2n, (ii) (−1 + i)n = 2n/2 .
Page 35 :
2, Topological and Analytic Preliminaries, , The neighborhood of a young child consists of the people very close on the left, and right. As we get older we think in terms of two-dimensional neighborhoods, (the people around the corner) or even three-dimensional neighborhoods (the, people in the world). In this chapter we do likewise. We develop numerous, methods for accurately describing sets in the real line (one-dimensional) and, the plane (two-dimensional). In order to track down the elusive point at infinity, it becomes necessary to introduce the sphere (three-dimensional)., When a set is described in a satisfactory manner, we become concerned, about its image. We investigate conditions under which properties of a set are, preserved when the set is transformed into a new set. A remarkable outcome, of our investigation is that the removal of a single point from one set may, entirely change its character, whereas the removal of infinitely many points, from a different set may be insignificant. The removal of two points from a set, on the line may give it more affinity to a set in the plane than to its former, self. In this chapter we learn that in a sense all points are equal but some, points are more equal than others., , 2.1 Point Sets in the Plane, A neighborhood of a real number x0 is an interval in the form (x0 − , x0 + ),, where is any positive real number. Thus we may say that an neighborhood, of x0 is the set of points x ∈ R for which |x − x0 | < . There are different ways, to extend this one-dimensional neighborhood concept to include points in the, plane. A square neighborhood of a point (x0 , y0 ) is the set of all points (x, y), whose coordinates satisfy the two inequalities, |x − x0 | < ,, , |y − y0 | < ., , It consists of all points inside a square centered at (x0 , y0 ). The sides of the, square are parallel to the coordinate axes and have length 2 . A circular
Page 36 :
26, , 2 Topological and Analytic Preliminaries, , Figure 2.1. Illustration for open sets in the plane, , neighborhood of (x0 , y0 ) is the set of all points (x, y) whose distance from, (x0 , y0 ) is less than . It consists of all points (x, y) such that, , (x − x0 )2 + (y − y0 )2 < ,, i.e., points inside a circle centered at (x0 , y0 ) whose radius is . Observe that, every square neighborhood of a point contains a circular neighborhood of the, point, and every circular neighborhood of a point contains a square neighborhood of the point (for a smaller , of course). This is illustrated in Figure 2.1., From our point of view (that a point in the plane represents a complex number), it will be more convenient to deal with circular neighborhoods, for then, an neighborhood of the complex number z0 consists of all points z ∈ C, satisfying the inequality |z − z0 | < . Such a neighborhood is denoted by, N (z0 ; )., Care must be taken to distinguish between a neighborhood on the real line, and a neighborhood in the plane. For example, {x ∈ R : −1 < x < 1} is a, neighborhood of 0, a point on the line; it is not a neighborhood of (0, 0), a point, in the plane. A point in the plane is not permitted to have a one-dimensional, neighborhood., The definitions and theorems in this section are valid simultaneously for, points on the line and points in the plane, when the concepts of neighborhood, are suitably interpreted. A set is said to be bounded if it is contained in some, disk centered at the origin. A point is said to be an interior point of a set if, there is some neighborhood of the point contained in the set. An important, distinction between the bounded sets, A = {z ∈ C : |z − z0 | < } and B = {z ∈ C : |z − z0 | ≤ }, is that every point in A is an interior point. To see this, let z1 be any point, in A. Then |z − z0 | = δ for some δ, 0 ≤ δ < . But for η = ( − δ)/2, we, have N (z1 ; η) ⊂ N (z0 ; δ) (see Figure 2.2). Of course, no point on the circle, |z − z0 | = is an interior point of B. A set A is called an open set if every, point in A is an interior point. We have shown that a neighborhood of a point, in the plane is an open set. Other simple examples of open sets in the plane, are
Page 37 :
2.1 Point Sets in the Plane, , 27, , Figure 2.2. Description for an interior point, , (a), (b), (c), (d), (e), (f), , the empty set,, the set of all complex numbers,, {z : |z| > r}, r ≥ 0,, {z : r1 < |z| < r2 }, 0 ≤ r1 < r2 ,, the intersection of any two open sets,, the union of any collection of open sets., , Remark 2.1. An open interval on the real line is not an open set in the plane,, since any neighborhood of a point will contain points in the plane that are, •, not real., A deleted neighborhood of z0 , denoted by N (z0 ; ), is the set of all points, z such that 0 < |z − z0 | < . That is, the point z0 is “punched out” from the, set. A point z0 is called a limit point of a set A if every deleted neighborhood, of z0 contains a point of A. Note that a limit point z0 may or may not be in, the set A., Examples 2.2., (i) The limit points of the open set |z| < 1 are |z| ≤ 1;, that is, all the points of the set and all the points on the unit circle, |z| = 1. If ∂Δ = {z : |z| = 1} and Δ = {z : |z| ≤ 1}, then all points, of Δ are its limit points and no other point is a limit point of Δ. The, same is true for ∂Δ. On the other hand, all points of Δ \ {0} together, with 0 and the points of ∂Δ are limit points of Δ \ {0}. Note that 0 and, the points of ∂Δ are not in Δ \ {0}., (ii) The set A = {1/n : n ∈ N}, where N = {1, 2, 3, . . . , n, . . . }, has 0 as, a limit point (regardless of whether the set is considered a subset of, the line or the plane) and 0 is not in the set. Similarly, the set A =, {eiπ/n : n ∈ N} has 1 as its only limit point, see Figure 2.3., (iii) If A consists of the set of points that have both coordinates rational,, •, then every point in the plane is a limit point of A., A set is said to be closed if it contains all of its limit points. The union of, a set A and its limit points is called the closure of A, and is denoted by A., Some examples of closed sets in the plane are, (a) the empty set,, (b) the set of all complex numbers,
Page 38 :
28, , 2 Topological and Analytic Preliminaries, , Figure 2.3. Description of limit point 1 of {eiπ/n : n ∈ N}, , (c), (d), (e), (f), (g), , {z : |z| ≥ r}, r ≥ 0,, {z : r1 ≤ |z| ≤ r2 }, 0 ≤ r1 < r2 ,, the union of any two closed sets,, the intersections of any collection of closed sets,, {z : |z| ≤ 1}., , Some examples of sets that are not closed in the complex plane C are Δ,, Δ \ {0}, Δ \ {0}. Finally, we remark that the set Δ \ {0} is neither closed nor, open., Theorem 2.3. If z0 is a limit point of A, then every neighborhood of z0 contains infinitely many points of A., Proof. Assume that some deleted neighborhood of z0 contains only a fi=, nite number of points of A. Let the points be z1 , z2 , . . . , zn and, mini=1,2, ... ,n |z0 − zi |. Then N (z0 ; ) contains no points of A, and z0 can’t be, a limit point of A., Corollary 2.4. Every finite set is closed., Proof. The set contains all of its limit points—all “none” of them., For the set |z| ≤ 1, we would like to distinguish the interior points from, the points on the unit circle. A point z0 is called a boundary point of A if, every neighborhood of z0 contains points in A and points not in A (in the, complement of A). The set of all boundary points of A is called boundary of, A. For example, the circle |z| = 1 is the boundary for both the bounded set, |z| < 1 and the unbounded set |z| > 1., Remark 2.5. The boundary points determine the “openness” or “closedness”, of a set. An open set cannot contain any of its boundary points, whereas a, closed set must contain all of its boundary points (why?). Clearly, an interior, point of a set A is a limit point of A but a limit point may or may not be an, •, interior point of A.
Page 39 :
2.1 Point Sets in the Plane, , 29, , Figure 2.4. Description between open and connected sets, , We would also like to distinguish between the two sets, A = {z : |z| < 1} and B = {z : |z| < 1} ∪ {z : |z − 3| < 1}., Set A is “all one piece”, while set B consists of two pieces (Figure 2.4). A, set S is said to be connected if there do not exist disjoint open sets U and V, satisfying the following conditions:, (i) U ∪ V ⊃ S,, , (ii) U ∩ S = φ,, , V ∩ S = φ., , In particular, if an open connected set can be expressed as the disjoint, union of two open sets U and V , then either U = φ or V = φ. Set A above is, connected and set B is not., An open connected set is called a domain. A region is a domain together, with some, none, or all of its boundary points.1 We might think that the, counterpart of a real-valued function of a real variable being defined on an, open set is a complex-valued function of a complex variable being defined on, an open set. But this is not the case. Actually, the counterpart of an open, interval in R is a domain. Note that an open interval in R is a connected, subset of R. Likewise a domain is open as well as connected. The “negative”, definition for connectedness is sometimes difficult to visualize. But when the, connected set is a domain, we have the following useful property., Theorem 2.6. Any two points in a domain can be joined by a polygonal line, that lies in the domain., Proof. Choose a point z0 in the domain D. It suffices to show that every point, in D can be joined to z0 by a polygonal line that lies in D. Let A denote the, set of all points in D that can be so joined to z0 and let B denote all those, points that cannot. Note that A ∪ B = D and A ∩ B = φ. We wish to show, that B is empty., 1, , The reader is warned that some authors use the term “region” for what we call, a domain (following the modern terminology), and others make no distinction, between the terms.
Page 40 :
30, , 2 Topological and Analytic Preliminaries, , If a point z1 is in A, then z1 is in D. Since D is open, there exists an 1 > 0, such that N (z1 ; 1 ) ⊂ D. But all the points in N (z1 ; 1 ) can be joined to z1, by a straight line segment. Therefore, each point in N (z1 ; 1 ) must be in A,, which means that A is an open set., Similarly, if a point z2 is in B, then there exists an 2 > 0 such that, N (z2 ; 2 ) ⊂ D. All the points in this neighborhood must also lie in B, for if, some point b ∈ N (z2 ; 2 ) could be joined to z0 by a polygonal line, then the, straight line segment from z2 to b could be connected to the polygonal line, from z0 to b in order to form a polygonal line from z0 to z2 . Thus B is an, open set. Consequently, neither A nor B can contain any boundary points., Since D is connected, either A or B must be empty. But z0 ∈ A, so that B is, empty. This completes the proof., Note that a domain may contain two points that cannot be joined by a, single straight line segment, as is illustrated in Figure 2.5., , Figure 2.5. Connected domains, , Remark 2.7. We could have required that the polygonal line of Theorem 2.6, be parallel to the coordinate axes. The only modification in the proof is the, observation that any point in a disk can be joined to the center by combining, •, a line segment parallel to the x axis with one parallel to the y axis., The converse of Theorem 2.6 is also true: if any two points of an open set, can be joined by a polygonal line, then the set is connected. The proof is left, for the exercises. Also, in the exercises an example is given of a connected set,, two of whose points cannot be joined by a polygonal line that lies in the set., With the above definitions, we are furnished with a method for adequately, characterizing most sets on either the line or the plane., Examples 2.8., (i) Let A = {z ∈ C : |z| ≤ 1, excluding the points zn =, 1/n (n ∈ N)}. Then the set A is not open because the points on the, unit circle have been included and is not closed because the limit points, zn = 1/n (n ∈ N) have been excluded. The set is bounded, connected, and has a boundary consisting of the unit circle, the points zn = 1/n,, and the origin.
Page 41 :
2.1 Point Sets in the Plane, , 31, , (ii) Let A = {z ∈ C : Re z > 0} ∪ {z : Re z < −2}. This set is open, not, closed, not bounded, and not connected. Its boundary consists of all, points on the lines Re z = 0 and Re z = −2., (iii) Let A = {z ∈ C : −π/4 ≤ Arg z ≤ π/4}. This set is connected, closed,, not open, and not bounded. Its boundary consists of the origin together, •, with the rays Arg z = π/4 and Arg z = −π/4., Questions 2.9., 1. What alternative definitions of “bounded” might we have used?, 2. What can we say about unions and intersections of open and closed, sets?, 3. What can we say about the complements of open and closed sets?, 4. What sets are open (closed) in both the plane and the line?, 5. What sets are both open and closed?, 6. Can a set have infinitely many points without having a limit point?, 7. What is the relation between the boundary points and limit points?, 8. How does the closure of the intersection of two sets compare with the, intersection of their closures?, 9. What can you say about intersections and unions of connected sets?, 10. What can you say about a set in which every pair of points can be joined, by a straight line segment lying in the set?, 11. How does the set described in the previous question compare to a set, in which there exists a point that can be joined to any other point by, a straight line segment lying in the set? What is an example of such a, set?, 12. What are the boundary points of a deleted neighborhood of z0 ?, 13. What are the boundary points of the complex plane?, Exercises 2.10., 1. Prove that a neighborhood of a point on the real line (an open interval), is an open set in R., 2. Show that a set A of complex numbers is bounded if and only if, given, z0 ∈ C, there exists a real number M such that z ∈ N (z0 ; M ) for every, z ∈ A. Can M be chosen independent of z0 ?, 3. Show that a set of complex numbers is bounded if and only if both the, sets of its real and imaginary parts are bounded., 4. Describe the following sets., (a) {z ∈ C : 1 < |z| < 2, excluding points for which z ∈ R}, (b) {z ∈ C : z = (x, y), x and y are rational}, (c) {x ∈ R : x − irrational}, (d) {x ∈ R : x, Z}, ∈, ∞, (e) {n ∈ N : n=1 [1/n, n]}, (f) {z ∈ C : |z| > 2, |Arg z| < π/6}, (g) {z ∈ C : |z + 1| < |z − i|}
Page 42 :
32, , 2 Topological and Analytic Preliminaries, , 5., , 6., , 7., 8., 9., 10., 11., 12., 13., , (h) {z ∈ C : |z + 1| = |z − i|}, (i) {z ∈ C : |Re z| + |Im z| = 1}., Which of the following subsets are connected?, (a) D = {z ∈ C : |z| < 1} ∪ {z ∈ C : |z + 2| ≤ 1}, (b) D = [0, 2) ∪ {2 + 1/n : n ∈ N}., Prove that the union of an arbitrary collection of open sets is open and, that the intersection of a finite number of open sets is open. Also, show, that ∩∞, n=1 {z : |z| < 1/n} is not an open set., Show that a set is open if and only if its complement is closed., Show that the intersection of an arbitrary collection of closed sets is, closed and the union of a finite number of closed sets is closed., Show that the limit points of a set form a closed set., Show that A, the closure of A, is the smallest closed set containing A., Show that a set is connected if any two of its points can be joined by a, polygonal line., Show that if a set A is connected, then A is connected. Is the converse, true?, Show that the union of two domains is a domain if and only if they have, a point in common., , 2.2 Sequences, A sequence {zn } of complex numbers is formed by assigning to each positive, integer n a complex number zn . The point zn is called the nth term of the, sequence. Care must be taken to distinguish between the terms of the sequence, and the set whose elements are the term of the sequence. For example, the, sequence {2, 2, 2, . . . } has infinitely many terms (as do all sequences), but, the set {2, 2, 2, . . . } contains only one point. In general, when we discuss settheoretic properties of a sequence, we will mean the set associated with the, terms of the sequence., A sequence {zn } is said to have a limit z0 (converge to z0 ), written, lim zn = z0, , n→∞, , or, , zn → z0 ,, , if for every > 0, there exists an integer N (depending on ) such that |z−z0 | <, whenever n > N . Geometrically, this means that every neighborhood of z0, contains all but a finite number of terms of sequence (see Figure 2.6). We, must point out that zn → z0 is equivalent to zn − z0 → 0. To illustrate, the, sequence {1/n} converges to 0; but the sequence {(−1)n }, which oscillates, between 1 and −1, does not converge. Examples of convergent sequences that, appear frequently are, 1, = 0 (p > 0), npn, (b) lim |z| = 0 (|z| < 1), (a) lim, , n→∞, n→∞, , (c) lim n1/n = 1., n→∞
Page 43 :
2.2 Sequences, , 33, , Figure 2.6. Geometric meaning of a convergence of a sequence, , Example 2.11. We can easily see that {1+in }n≥1 does not converge. Indeed,, if zn = 1 + in then, for each fixed k = 0, 1, 2, 3,, ⎧, 2 if k = 0, ⎪, ⎪, ⎨, 1, +, i if k = 1, z4n+k = 1 + i4n+k = 1 + ik =, 0, if k = 2, ⎪, ⎪, ⎩, 1 − i if k = 3, and so {1 + in } diverges. Also we remark that {1 + in } and {in } diverge or, converge together and so it suffices to deal with {in } which is easier than the, original sequence., The convergence of the sequence {in /n}n≥1 is easier to convince yourself, •, of if you draw a figure representing these points., There is a nice relationship between the convergence of a sequence of, complex numbers and the convergence of its real and imaginary parts., Theorem 2.12. Let zn = xn + iyn be a sequence of complex numbers. Then, {zn } converges to a complex number z0 = x0 +iy0 if and only if {xn } converges, to x0 and {yn } converges to y0 ., Proof. The proof is simply a consequence of the inequalities, |Re z|, |Im z| ≤ |z| ≤ |Re z| + |Im z|., To provide a detailed proof, we assume limn→∞ xn = x0 and limn→∞ yn = y0 ., Then given > 0, there exist an integer N such that n > N implies, |xn − x0 | < /2,, , |yn − y0 | < /2., , (2.1), , From (2.1) we obtain, |zn − z0 | = |xn − x0 + i(yn − y0 )| ≤ |xn − x0 | + |yn − y0 | < ,, and {zn } converges to z0 . Conversely, if we assume that limn→∞ zn = z0 , the, inequalities
Page 44 :
34, , 2 Topological and Analytic Preliminaries, , |xn − x0 | ≤ |zn − z0 |,, , |yn − y0 | ≤ |zn − z0 |, , show that {xn } and {yn } converge to x0 and y0 , respectively., Theorem 2.12 essentially says that many properties of the complex sequences may be deduced from corresponding properties of real sequences. For, example, the uniqueness of the limit of a complex sequence can be derived, either directly or from the uniqueness property of real sequences., A sequence of complex numbers {zn } is said to be bounded if there exists, an R > 0 such that |zn | < R for all n. In other words, a sequence is said to, be bounded if it is contained in some disk., Since a convergent sequence eventually clusters about its limit, the next, theorem is not too surprising., Theorem 2.13. A convergent sequence is bounded., Proof. If limn→∞ zn = z0 , then zn ∈ N (z0 ; 1) for n > N . Let, M = max{|z1 |, |z2 |, . . . , |zN |}., Then, |zn | < M + |z0 | + 1 for every n., The converse of Theorem 2.13 is not true. The sequence {1, 2, 1, 2, . . . } is, bounded and not convergent, although the odd terms and even terms both, form convergent sequences., A subsequence of a sequence {zn } is a sequence {zn k } whose terms are, selected from the terms of the original sequence and arranged in the same, order. For the sequence zn = (−1)n , we have subsequence {z2k } converging, to 1 and subsequence {z2k−1 } converging to −1., The next theorem shows that a subsequence must be at least as “well, behaved” as the original sequence., Theorem 2.14. If a sequence {zn } converges to z0 , then every subsequence, {zn k } also converges to z0 ., Proof. Given > 0, we have zn ∈ N (z0 ; ) for n > N . Hence zn k ∈ N (z0 ; ), for nk > N . Since nk ≥ n (why?), and there can be at most N terms of the, subsequence for which |zn k − z0 | ≥ ., We know that not all sets are bounded. However, if a set of real numbers, is bounded, it has a “smallest” bound. A real number M is said to be the, least upper bound (lub) of a nonempty set A of real numbers if, (i) x ≤ M for every x ∈ A. That is A is bounded above by M and M is an, upper bound for A., (ii) For any > 0, there exists a y ∈ A such that y > M − . That is, M is, the smallest among all the upper bounds of A., Similarly, the real number m is said to be the greatest lower bound (glb) of a, nonempty set A if:
Page 45 :
2.2 Sequences, , 35, , (i) x ≥ m for every x ∈ A; That is A is bounded below by m and m is a, lower bound of A., (ii) For any > 0, there exists a y ∈ A such that y < m + . That is, m is, the largest among all the lower bounds of A., The Dedekind property states that every nonempty bounded set of real numbers, has a least upper bound and a greatest lower bound. This is an amplified version, of the result that R is complete. For a proof of this, see [R1]., As we have seen, the converse of Theorem 2.13 (even for real sequences), is not true. Bounded oscillating sequences need not converge. Eliminating the, oscillation, however, will produce convergence. A real sequence {xn } is said to, be monotonically increasing (decreasing) if xn+1 ≥ xn (xn+1 ≤ xn ) for every, n. A sequence will be called monotonic if it is either monotonically increasing, or monotonically decreasing., Theorem 2.15. Every bounded monotonic sequence of real numbers converges., Proof. Let the bounded sequence {xn } be monotonically increasing. According to the Dedekind property, there exists a least upper bound of {xn }, call, it x. By the definition of lub, given > 0 there exists an integer N such that, xN > x − . Since {xn } is monotonically increasing,, x − < xn ≤ x for n > N ., Hence |xn −x| < for n > N , and {xn } converges to its least upper bound. The, proof for monotonically decreasing sequences is identical, using the greatest, lower bound instead of the least upper bound., The examples we have seen of bounded sequences that did not converge, did have convergent subsequences. To show that this is true in general, we, need the following, Lemma 2.16. Every sequence of real numbers contains a monotonic subsequence., Proof. Assume that the real sequence {xn } has the property that there are, infinitely many n such that xk ≤ xn for every k ≥ n. Let n1 be the first, such n with this property, n2 the second, etc. Then xn1 , xn2 , xn3 , . . . is a, monotonically decreasing subsequence of {xn }., On the other hand, if there are only finitely many n such that xk ≤ xn, for every k ≥ n, choose an integer m1 such that no terms of the sequence, xm1 , xm1 +1 , xm1 +2 , . . . have this property. Let m2 be the first integer greater, than m1 for which xm2 > xm1 . Continuing the process, we obtain a sequence, xm1 , xm2 , xm3 , . . . which is a monotonically increasing subsequence of {xn }., This completes the proof., Although the converse of Theorem 2.13 is not true, here is slightly a weaker, version of it.
Page 46 :
36, , 2 Topological and Analytic Preliminaries, , Theorem 2.17. Every bounded sequence of complex numbers contains a convergent subsequence., Proof. Let zn = xn + iyn , with |zn | ≤ M . Then |xn | ≤ M and |yn | ≤ M ., By Lemma 2.16, {xn } contains a monotonic subsequence {xn k }. By Theorem, 2.15, {xn k } converges., Now consider the corresponding subsequence {yn k } of {yn }. This may not, converge, but by Theorem 2.15, it does contain a convergent subsequence, {yn k(l) }. By Theorem 2.14, {xn k(l) } also converges. Applying Theorem 2.12,, the sequence, zn k(l) = xn k(l) + iyn k(l), is a convergent subsequence of {zn }, and this completes the proof., What are the relationships between the limit of a sequence, the limit points, of a sequence, and lub or glb of a sequence? The lub and glb are meaningless, in the complex number system, although (as we have just seen) these notions, for real numbers may be used to prove theorems about complex numbers. For, the sequence {n/(n + 1)}, 1 is the lub, the limit, and the unique limit point., If a convergent sequence has only finitely many distinct elements, it will have, no limit points; however, we do have the following theorem., Theorem 2.18. A point z0 is a limit point of a set A if and only if there is, a sequence of distinct points in A converging to z0 ., Proof. If a sequence {zn } of distinct points in A converges to z0 , then every, neighborhood of z0 contains all but a finite number (hence infinitely many), of points of {zn }. Therefore, z0 is a limit point of A., To prove the converse, let z0 be a limit point of A. For every integer n,, choose a point zn ∈ N (z0 ; 1/n) ∩ A. Since every neighborhood of A contains, infinitely many distinct points, we may assume the points of the sequence, {zn } to be distinct., Given > 0, choose N such that 1/N < . Then zn ∈ N (z0 ; ) for n > N ,, and the sequence {zn } converges to z0 ., Combining the previous two theorems, we obtain, Theorem 2.19. (Bolzano–Weierstrass)Every bounded infinite set in the complex plane has a limit point., Proof. Choose any sequence of distinct points in the set. By Theorem 2.17,, this sequence contains a convergent subsequence; and by Theorem 2.18, the, limit of this convergent subsequence is a limit point of the set. This completes, the proof., A sequence {zn } of complex numbers is said to be a Cauchy sequence if, for every > 0, there exists an integer N (depending on ) such that
Page 47 :
2.2 Sequences, , |zm − zn | <, , whenever, , 37, , m, n > N., , What is the difference between a Cauchy sequence and a convergent sequence?, Geometrically, for a convergent sequence, all but a finite number of points are, close to a fixed point (the limit of the sequence), while for a Cauchy sequence, all but a finite number of points are close to each other. We will show, for, complex sequences, that these concepts are equivalent. Moreover, from our, exercises, we see that the algebra of complex sequences is essentially the same, as that for the real sequences studied in real-variable theory., Theorem 2.20. (Cauchy Criterion)The sequence {zn } converges if and only, if {zn } is a Cauchy sequence., Proof. Assume {zn } converges to z0 . By the triangle inequality,, |zm − zn | = |zm − z0 + z0 − zn | ≤ |zm − z0 | + |zn − z0 |., , (2.2), , Given > 0, both terms on the right side of (2.2) can be made less than /2, for m, n > N . Hence {zn } is a Cauchy sequence., Conversely, assume {zn } is a Cauchy sequence. Then for n > N , we have, |zn − zN | < 1. That is,, |zn | < |zN | + 1 for n > N ., Thus {zn } is a bounded sequence. By Theorem 2.17, {zn } contains a subsequence {zn k } that converges to a point (say z0 )., We will show that {zn } also converges to z0 . Once again using the triangle, inequality, we obtain, |zn − z0 | = |zn − zn k + zn k − z0 | ≤ |zn − zn k | + |zn k − z0 |., Given, , > 0, there exists an integer N such that, for n > N ,, , |zn − zn k | < /2 (because {zn } is Cauchy),, |zn k − z0 | < /2 (because {zn k } converges to z0 )., , Combining (2.3) and (2.4), we see that |zn − z0 | <, converges to z0 , and the proof is complete., , (2.3), , (2.4), , for n > N . Hence, {zn }, , Theorem 2.20 furnishes us with a general method for determining the, convergence of a sequence of complex numbers even though we may not know, in advance what its limit is. There are some systems in which not every Cauchy, sequence converges. For instance, in the field of rational√numbers, the Cauchy, sequence 1, 1.41, 1.414, . . . does not converge (because 2 is not rational). A, system in which every Cauchy sequence converges is said to be complete. In, Sprecher [S], it is shown that the real number system forms the only complete, ordered field.
Page 48 :
38, , 2 Topological and Analytic Preliminaries, , Example 2.21. Suppose that z = 1, but |z| = 1. Then, 1 k, z = 0., n→∞ n, n, , lim, , k=1, , Indeed, as (1 − z), , n, , n, , k=1, , k=1, , z k = z(1 − z n ), we have, , zk =, , |z|(1 + |z|n ), 2|z|, z(1 − z n ), ≤, ≤, 1−z, |1 − z|, |1 − z|, , so that, , n, 1 k, |z|, 2, z ≤, n, n |1 − z|, , → 0 as n → ∞., , •, , k=1, , Questions 2.22., 1. When a sequence {zn } converges to z0 , is the limit z0 unique?, 2. Let {xn } and {yn } be real sequences. If {(xn + yn )} converges, does, this mean that both {xn } and {yn } converge? How does this question, compare with Theorem 2.12?, 3. How many subsequences are there for a given sequence?, 4. Can unbounded sequences have limit points? What about monotonic, unbounded sequences?, 5. When will the least upper bound of a set be an element of the set?, 6. Can a real sequence converge to a value other than lub or glb of the, sequence?, 7. Can a sequence have infinitely many limit points?, 8. Can you think of a sequence that converges without knowing what its, limit is?, 9. How could Theorem 2.18 have been proved without appealing to Theorem 2.3?, 10. What can be said of the sequence bn = glb {an , an+1 , an+2 , . . . }, where, {an } is a real sequence? What if {an } is bounded?, 11. Suppose that {zn } converges. Does {|zn |} converge? Does {arg zn } converge? Does {Arg zn } converge?, 12. Suppose that both {Arg zn } and {|zn |} converge. Does {zn } converge?, Exercises 2.23., 1. Let {zn } converge to z0 and wn converge to w0 . Show that, (a) lim (zn + wn ) = z0 + w0 ,, n→∞, , (b) lim zn wn = z0 w0 ,, n→∞, z0, zn, (c) lim, =, provided w0 = 0., n→∞ wn, w0
Page 49 :
2.3 Compactness, , 39, , In particular, if, zn =, , 1 + n + 2i(n − 1), n1/2 + 2i(3 + 4n3 ), and wn =, ,, n, n3, , find z0 , w0 and z0 /w0 ., 2. Show that no sequence having more than one limit point can converge., 3. If {zn } converges, show that {|zn |} converges. Is the converse true?, 4. Which of the following sequences are convergent?, (a) {in }, (b) , {z0n }, where |z0 | < 1, cos n + i sin n, (c), n, , 1 + 12 + 13 + · · · + n1, (d), , n, , √, , n, , enπi/3 + − 12 − i 23, , , √ n , (g) enπi/6 + − 12 + i 23, , , (f) n cos(nπ), 2n+1, , !", (i) sin nπ, ., 8, n, If {zn }n≥1 converges to 0, prove that { n1 k=1 zk, } converges to 0. Then, n, show that {zn } converging to z0 implies that { n1 k=1 zk } converges to, z0 ., Give an example of a sequence that, (a) does not converge, but has exactly one limit point;, (b) has n limit points, for any given integer n;, (c) has infinitely many limit points., Prove that the subsequential limits (the limits of all possible subsequences) of a sequence {zn } form a closed set., Let {zn } be a sequence having the following property: Given > 0,, there exists an integer N such that for n > N , |zn+1 − zn | < . Give an, example to, that {zn } need not be a Cauchy sequence., show, n, Let sn = k=1 1/k!. Use the Cauchy criterion to show that {sn } converges., (f), , 5., , 6., , 7., 8., , 9., , 2.3 Compactness, The union of the open intervals (n − 12 , n + 12 ) for n = 1, 2, 3, . . . contains, the set of positive integers. Each interval is important in that the removal, of any one of them will leave a positive integer uncovered. For the bounded, set S = {x ∈ R : 0 < x < 1}, the union of open intervals (1/n, 1) for, n = 2, 3, 4, . . . contains S. While the removal of any one of these intervals, will prevent the union of the remaining intervals from covering S, the set S is, not contained in any finite subcollection of the intervals.
Page 50 :
40, , 2 Topological and Analytic Preliminaries, , A set is said to be countable if its elements can be put in a one-to-one, correspondence with a subset of positive integers., A collection {Oα } of open, , sets is called an open cover of a set S if S ⊂ α Oα . Note that the collection, {Oα } may contain uncountably many sets. A set S is compact if every open, cover of S contains a finite subcover., We have seen that neither the set of positive integers nor the open interval, (0, 1) is compact. However, any finite set is compact because for any open, cover we have a finite subcover formed by associating with each point one of, the open sets containing the point., The definition of compactness is not always easy to apply. We would like, to work with a more geometrically intuitive method for determining compactness. To this end we will need the following., Lemma 2.24. Let {In } be a sequence of closed and bounded intervals on the, real line. If In+1 ⊂ In for every n and the length of In approaches 0 as n → ∞,, then there is exactly one point in common to all In ., Proof. Let In = {x : an ≤ x ≤ bn }. By hypothesis,, an ≤ an+1 , bn+1 ≤ bn, , (n = 1, 2, 3, . . . ), , and, lim (bn − an ) = 0., , n→∞, , (2.5), , The sequences {an } and {bn } are both monotonic and bounded (an , bn ) ∈, [a1 , b1 ] for every n. By Theorem 2.15 both sequences must converge; and by, (2.5), they must converge to the same point, call it x0 . Since x0 = lub {an } =, glb {bn },, x0 ∈ [an , bn ] for every n, (see Figure 2.7). There cannot be another point x1 in all the In . For, if x1 (=, x0 ) were less than (resp. greater than) x0 , then x0 would not be the lub {an }, (resp. glb {bn })., Note that Lemma 2.24 is not true if closed is replaced by open. The, collection, of intervals {(0, 1/n) : n ∈ N} satisfy the hypotheses, although, #∞, (0,, 1/n), = φ., n=1, , Figure 2.7.
Page 51 :
2.3 Compactness, , 41, , Lemma 2.25. Let {Sn } be a sequence of closed and bounded rectangles in the, plane. If Sn+1 ⊂ Sn for every n and the length of the sides of Sn approaches, 0 as n → ∞, then there is exactly one point in common to all the Sn ., Proof. Let {In } and {Jn } be the projections of {Sn } into real and imaginary, axes respectively. Then {In } and {Jn } satisfy the conditions of Lemma 2.24., If, and {y0 } = ∩∞, {x0 } = ∩∞, n=1 In, n=1 Jn ,, then (see Figure 2.8) {z0 } = {(x0 , y0 )} = ∩∞, n=1 Sn ., , Figure 2.8., , Theorem 2.26. (Heine–Borel) Every closed and bounded set is compact., Proof. Let S be a closed and bounded set. Assume {Oα } is an open cover of S, that has no finite subcover. Since S is bounded, it is contained in some square, S0 whose vertices are z = ±a ± ai. The coordinate axes divide S0 into four, equal subsquares. At least one of these squares (call it S1 ) has the property, that S ∩ S1 cannot be covered by a finite subfamily of {Oα } (why?). We now, divide S1 into four more equal closed subsquares (see Figure 2.9). Again, for, at least one of these squares, denoted by S2 , there is no finite subfamily of, {Oα } that covers S ∩ S2 ., We can continue the process indefinitely, forming a sequence {Sn } of closed, squares for which there is no finite subfamily of {Oα } that covers S ∩ Sn . Note, that the length of any side of Sn is a/(2n−1 ). By Lemma 2.25, there is exactly, one point, denoted by z0 , common to all squares Sn . This point z0 must be a, limit point of S, and hence an element of S., Let Oα 0 be an element of the cover {Oα } that contains z0 . Since Oα 0 is an, open set, N (z0 ; ) ⊂ Oα 0 for some > 0. But Sn ⊂ N (z0 ; ) for n sufficiently, large. Thus Sn ∩ S has a finite subcover of {Oα }, namely, one element: Oα 0 ., This contradiction, concludes the proof.
Page 52 :
42, , 2 Topological and Analytic Preliminaries, , Figure 2.9. Illustration for Heine–Borel theorem, , The gist of the above argument is that if no finite subcollection of {Oα }, covers S, then no finite subcollection covers a carefully chosen sequence of, subsets of S. On the other hand, this sequence of subsets can be made small, enough to be contained in one of the open sets of the cover., We are now ready to collect some of the important results of the last two, sections to obtain the following major theorem., Theorem 2.27. Let S be a subset of the complex plane C. The following, statements are equivalent:, (i) S is closed and bounded., (ii) S is compact., (iii) Every infinite subset of S has a limit point in S., (iv) Every sequence in S has a subsequence that converges to a point in S., Proof. The Heine–Borel theorem states that (i) implies (ii). We will show that, (ii) implies (iii), (iii) implies (iv), and (iv) implies (i). Since each statement is, clearly correct if S is a finite set, we may suppose that S is infinite., Assume that (ii) holds. If A is an infinite subset of S having no limit point, in S, then for every point in S \A we can find a neighborhood containing no, points of A. Furthermore, for every point in A we can find a neighborhood, containing no other points of A. The collection of all such neighborhoods is, an open cover of S for which there is no finite subcover, contradicting the, compactness of S., Assume (iii) holds. Let {zn } be a sequence of distinct points in S. (Why, is it sufficient to consider only such sequences?) By hypothesis, there exists a, limit point z0 of {zn } with z0 ∈ S. By Theorem 2.18, there is some subsequence, of {zn } converging to z0 ., Assume (iv) holds. If S is unbounded, then there exists a sequence of, points {zn } in S such that |zn | > n for every n. Let {zn k } be an arbitrary, subsequence of {zn }. For any point z0 ∈ S, N (z0 ; 1) can contain no points of, {zn k } for which nk > |z0 | + 1. Hence {zn k } cannot converge to any point in
Page 53 :
2.3 Compactness, , 43, , S, contradicting our assumption: To show that S is closed, let z0 be a limit, point of S. By Theorem 2.18, there is sequence of distinct points {zn } of S, converges to z0 . By Theorem 2.14, every subsequence of {zn } converges to z0 ., According to (iv), z0 must therefore be in S. This completes the proof., Compactness is a nice property because reducing an open cover to a finite, subcover often means that only a finite number of points need be considered, in proving that a set has a certain property. For this reason, when we have, compactness, many local properties (properties that hold in a neighborhood, of each point in a set) can be shown to be global or uniform (a property of, the set as a whole)., For example, from the fact that each point may be covered by a bounded, neighborhood, we deduced that a compact set is bounded. Also, if each point, in a compact set is a positive distance from a fixed point, the set itself is a, positive distance from the point (see Exercise 2.29(3)). This, of course, is not, true in general. Each point of the open interval (0, 1) is a positive distance, from 0, but we can not find a positive real number between 0 and the set., What makes the addition of one or two points so important? Let us compare, ∞ the open interval (0, 1) with the closed interval [0, 1]. As we saw earlier,, n=2 (1/n, 1) is an open cover of (0, 1), that has no finite subcover. This cover, does not contain the points {0} and {1}. If these points were added to the set,, intervals like (− , ) and (1 − , 1 + ) would also have to be added to obtain, N, a cover. But then (− , ), (1 − , 1 + ) and n=2 (1/n, 1) for N > 1/ would, be a finite subcover., Questions 2.28., 1., 2., 3., 4., 5., 6., , What can we say about the finite union (intersection) of compact sets?, What can we say about the infinite union (intersection) of compact sets?, What can we say about the complement of a compact set?, What can we say about Cauchy sequences in compact sets?, When can we say that every subset of a compact set is compact?, We have seen that the removal of one point from a set may destroy, the compactness. How many points may be added to a set to destroy, compactness?, 7. What kind of generalizations to Lemma 2.25 might we have for compact, sets?, 8. Can we talk about “infinity” being a limit point?, , Exercises 2.29., 1. Show that the union of any bounded set and its limit points is a compact, set., 2. Show that a compact set of real numbers contains its greatest lower, bound and its least upper bound. Can this occur for a set of real numbers, that is not compact?
Page 54 :
44, , 2 Topological and Analytic Preliminaries, , 3. If S is compact and z0 ∈ S, prove that glb z∈S |z − z0 | > 0., 4. If {Sn } is a sequence, #∞of nonempty compact sets with Sn+1 ⊂ Sn for, every n, show that n=1 Sn = φ., 5. In Theorem 2.27, prove as many different implications as you can., 6. Show that the set of rational numbers are countable., 7. Show that any open cover of a subset of the plane has a countable, subcover., , 2.4 Stereographic Projection, Thus far, infinite limits have been carefully avoided. Consider the three real, sequences:, , n if n is odd, , cn = (−1)n n., an = n, bn =, 1 if n is even, Even though all three sequences grow arbitrarily large, we do not want to, say they all approach infinity. From our knowledge of finite limits, it seems, appropriate that {an } should approach infinity and that {bn } should not, since, a subsequence of {bn } converges to 1. A case for {cn } can be made either way., The standard approach is to introduce the symbols ±∞, and adjoin them to, the real numbers. The set R∞ := R ∪ {+∞, −∞} is known as the extended, real number system. In the extended real number system, we use the following, conventions:, ⎧, ±∞ + a = ±∞ = a ± ∞ for a ∈ R, ⎪, ⎪, ⎪, ⎪, ∞, ⎨ · a = a · ∞ = ∞ for a ∈ R∞ \{0}, a, = 0 for a ∈ R \{0}, ⎪, ∞, ⎪, ⎪, ⎪, a, ⎩ = ∞ for a ∈ R \{0}., ∞, 0, The expressions ∞ + ∞ = ∞, −∞ − ∞ = −∞ hold while ∞ − ∞ is not, defined. In the extended real number system, {cn } does not converge because, {c2n } approaches +∞ and {c2n+1 } approaches to −∞., A perfectly logical, if somewhat unusual, approach is to adjoin only one, point, ∞, to R. We then say that a sequence {an } approaches ∞, written, limn→∞ an = ∞, if, for any preassigned real number M , all but a finite number, of terms lie outside the interval (−M, M ). According to this definition, the, sequence {(−1)n n} does approach ∞., This latter approach can be thought to arise from the former by grabbing, the two points −∞ and +∞ (with two very long arms) and bringing them, together. The real number line is then transformed into a circle. We now make, this geometric notion more precise. Consider the unit circle x2 + y 2 = 1. For, any real number a, draw the straight line joining the points (a, 0) and (0, 1)., This line intersects the unit circle at (0, 1) and one other point (x1 , y1 ), which, we identify with the real number a. For example, points in the open interval
Page 55 :
2.4 Stereographic Projection, , 45, , (−1, 1) are identified with points in the lower half of the circle, the points −1, and 1 are identified with themselves, and the points outside the interval [−1, 1], are identified with points in the upper half of the circle (see Figure 2.10)., , Figure 2.10. Illustration for the existence of +∞ and −∞ in R, , Observe that points close to one another on the real line are always identified with points close to one another on the circle. The converse is not true., Points “far out” in the positive and negative directions are identified with, points close to one another on the unit circle. In fact, the greater the absolute, value of a real number the closer is its identification with a point near (0, 1),, the only point on the unit circle not identified with a real number. For this, reason, we identify the point (0, 1) with the point ∞. This provide us with a, one-to-one correspondence between points in the set R ∪ {∞} and the points, on the unit circle. Since the set of real numbers is not compact, the identification of R ∪ {∞} with (compact) circle is called a one-point compactification, of the real numbers., Was the elimination of −∞ worth all this effort? Not really. In fact, it, is actually useful for −∞ to mean “less than any real number”. The set, R ∪ {∞} was introduced in order to properly motivate our study of the, extended complex plane. Consider the complex sequence {zn } defined by, zn = n(cos θ + i sin θ), where 0 ≤ θ ≤ 2π. For each different value of θ,, {zn } approaches ∞ along a different ray. Furthermore, since the complex, numbers are not ordered, the symbol −∞ would have no more meaning than, the symbol i∞., In the case of complex numbers, by an M neighborhood of ∞, denoted by, N (∞; M ), we mean the set of all points whose absolute value is greater than, M . That is the exterior of the disk with radius M and center at the origin., The sequence {zn } is said to approach ∞ if for any M > 0, zn ∈ N (∞; M ), for all but a finite number of n., If we adjoin the point at ∞ to the set of complex numbers, we obtain, the extended complex number system. Sometimes C is referred to as the finite, complex plane and is designated also by |z| < ∞. Then C ∪ {∞} := C∞ is, called the extended complex plane. Note that the extended complex number
Page 56 :
46, , 2 Topological and Analytic Preliminaries, , system is conceptually different from the extended real number system, in, which two points (+∞ and −∞) are added. We first make the following, algebraic rules as definitions:, ⎧, ∞ ± z = ∞ = z ± ∞ for z ∈ C, ⎪, ⎪, ⎪, ⎪, ⎪, ⎨ ∞ · z = z · ∞ = ∞ for z ∈ C∞ \{0}, z, = 0 for z ∈ C \{0}, ⎪, ∞, ⎪, ⎪, ⎪, ⎪, ⎩ z = ∞ for z ∈ C∞ \{0}., 0, There is a difficulty in assigning meaning to the expressions ∞ + ∞, ∞ − ∞, ∞/∞, ∞ · 0 and 0/0 and so none of these expressions has meaning in C∞ ., The one-point compactification, C∞ := C ∪ {∞}, of the plane has geometric, model similar to that of the one-point compactification of the line, with the, unit circle being replaced by the unit sphere, S = {(x, y, z) : x2 + y 2 + u2 = 1}, in the 3-dimensional Euclidean sphere in R3 ., Identify the complex number a + ib with the point (a, b, 0) in R3 . By doing, so, we are free to imagine C as an object sitting inside R3 as xy plane. Having, made this identification, for every number a+ib in the complex plane, draw the, straight line in R3 connecting the points (a, b, 0) and (0, 0, 1). This line intersects the sphere x2 + y 2 + u2 = 1 at (0, 0, 1) and at one other point (x1 , y1 , u1 )., The projection from the point (0, 0, 1) on the sphere to the point (a, b, 0) in, the complex plane is called a stereographic projection (see Figure 2.11). The, sphere S is called the Riemann sphere., , Figure 2.11. Stereographic projection, , This one-to-one correspondence covers all points in the finite complex, plane and all points in the sphere except (0, 0, 1). The point at ∞ in the, extended complex- number system is identified with the point (0, 0, 1), sometimes called the north pole. Note that a neighborhood of ∞ in the complex
Page 57 :
2.4 Stereographic Projection, , 47, , plane corresponds to the interior of an arctic circle whose center is the north, pole., To find specifically the point (x1 , y1 , u1 ) on the sphere identified with the, point (a, b, 0), observe that the three points, (0, 0, 1),, , (x1 , y1 , u1 ),, , and, , (a, b, 0), , are collinear. Hence,, y1 − 0, u1 − 1, x1 − 0, =, =, =t, a, b, −1, , (2.6), , for some real scaler t. But, x21 + y12 + u21 = (at)2 + (bt)2 + (1 − t)2 = 1,, , i.e., (a2 + b2 + 1)t2 = 2t., , Solving for t, we obtain, t=, , 2, = 1 − u1, a2 + b2 + 1, , as t = 0 corresponds to (0, 0, 1), the north pole. In view of (2.6), the complex, number a + ib is then identified with the point, , , 2a, 2b, a2 + b 2 − 1, (x1 , y1 , u1 ) =, ,, ,, ., (2.7), a2 + b2 + 1 a2 + b2 + 1 a2 + b2 + 1, Rewriting (2.7), we identify the complex number z = x + iy with the point, on the sphere, , , 2y, |z|2 − 1, 2x, ,, ,, ., |z|2 + 1 |z|2 + 1 |z|2 + 1, From the second formula for t and (2.6), we conclude that, a=, , y1, x1, and b =, ., 1 − u1, 1 − u1, , Consequently, we identify the point (x, y, u) in S \ {(0, 0, 1)} with the complex, number in the plane, , , , , y, x, +i, ., 1−u, 1−u, For instance the points, z = 0 and z = 1−i correspond to the points (0, 0, −1), and (2/3, −2/3, 1/3), respectively., Questions 2.30., 1. Which theorems for finite limit remain true for infinite limits?, 2. What is the relationship between unbounded sets and neighborhoods of, ∞?
Page 58 :
48, , 2 Topological and Analytic Preliminaries, , 3., 4., 5., 6., 7., 8., 9., 10., 11., , How might we define ∞ to be a limit point of a sequence?, What might the symbol “i∞” mean?, What might the symbol “−i∞” mean?, What happens to the points on the unit circle in the complex plane, under stereographic projection?, Could we have identified the complex plane with a different sphere?, What would be a one-point compactification of Rn ?, How are the images on the Riemann sphere of z and z related?, How are the images on the Riemann sphere of z and −z related? How, about for z and −z?, What is the image of the line x + y = 1 in the complex plane, on the, Riemann sphere?, , Exercises 2.31., 1. Show that a sequence having a finite limit point cannot approach ∞., 2. If {zn } approaches ∞ and {wn } is bounded, show that {(zn + wn )}, approaches ∞., 3. Show that {zn } approaches ∞ if and only if {|zn |} approaches ∞., 4. Given a point (x1 , y1 , u1 ) on the unit sphere, find its corresponding point, in the complex plane., 5. Show that a circle on the sphere that does not pass through the north, pole corresponds to a circle in the complex plane., 6. Show that a circle on the sphere passing through the north pole corresponds to a straight line in the complex plane., 7. Show that we may identify, by stereographic projection, the complex, plane with the sphere x2 + y 2 + (u − 12 )2 = ( 12 )2 ., 8. Consider two antipodal points (x, y, u) and (−x, −y, −u) on the Riemann sphere. Show that their stereographic projections z and z are, related by zz = −1. Give a geometric interpretation., √, 9. Show that the image of the circle |z| = 3 under the stereographic, projection is the set of all points (x1 , y1 , u1 ) in the sphere described by, x21 + y12 = 3/4 and u1 = 1/2., , 2.5 Continuity, A (single-valued) function or mapping f from a set A into a set B, written, f : A → B, is a rule that associates with each element x of A a unique, element f (x), the value of f at x, of B. The set A is called the preimage (or, the domain set) of f and the subset of B associated with the element of A, is called the image of f and is denoted by f (A), i.e. f (A) = {f (x) : x ∈ A}., If the set B, called the range of the function, is equal to f (A), the function is, said to be onto. If no two elements of A are mapped onto the same element, in B, the function is said to be one-to-one on A. By f (a) = b, we will mean, that the element a ∈ A is mapped onto the element of b ∈ B.
Page 59 :
2.5 Continuity, , 49, , For each b ∈ B, we define f −1 (b) to be the set of elements in A whose, image is b. Note that f −1 (b) may be empty if f is not onto. However, if f is, one-to-one and onto, f −1 : B → A is also a one-to-one and onto function,, called the inverse function of f ., Example 2.32. The function w = f (z) = az + b, a = 0, is one-to-one in, C and the inverse function is defined by z = (w − b)/a. Note that both are, defined in the whole plane C., On the other hand, the function f defined by f (z) = z + 3z 2 is not oneto-one in |z| < 1. For,, z1 + 3z12 = z2 + 3z22 =⇒ (z1 − z2 ) = 3(z2 − z1 )(z1 + z2 ), which implies (z1 − z2 )[1 + 3(z1 + z2 )] = 0 and we see that the last equality, is true when z1 + z2 = −1/3. But there are many points z1 , z2 ∈ Δ such that, •, z1 + z2 = −1/3. However, this function is one-to-one in |z| < 1/6., We have tacitly been dealing with functions. For example, a sequence of, real numbers is a function f : N → R and a sequence of complex numbers is, a function f : N → C, where N is the set of positive integers. In stereographic, projection, a one-to-one function was found that mapped the extended complex plane onto the unit sphere. The reader (hopefully) is familiar with some, of the properties of real-valued functions of a real variable, i.e., functions mapping sets of real numbers onto sets of real numbers. For example, the function, y = f (x) = x2 , mapping the real variable x onto the real variable x2 , takes, the set of real numbers onto the set of nonnegative real numbers, the closed, interval [0, 1] onto itself, and so on., Remark 2.33. Strictly speaking, f stands for the function and f (x) for the, value of the function at the point x. However, when there is no ambiguity, we, will sometimes use the time-honored notational abuse of referring to f (x) as, a function., For z = x + iy, the complex-valued function f (z) can be viewed as a, function of the complex variable z or as a function of two real variables x and, •, y., For example, the function f (z) = z 2 may be expressed as, w = f (z) = f (x, y) = (x + iy)2 = x2 − y 2 + i(2xy),, where, Re f (z) = u(x, y) = x2 − y 2 and Im f (z) = v(x, y) = 2xy., For this function, the points (2, 1), (1, 2), and (3, −1) are mapped onto the, points (3, 4), (−3, 4), and (8, −6) respectively., Just as a real-valued function of a real variable may be viewed as a mapping, from the x axis to the y axis, so may a complex-valued function of a complex
Page 60 :
50, , 2 Topological and Analytic Preliminaries, , Figure 2.12. Concept of continuity at z0, , variable be viewed as a mapping from the xy plane (z plane) to the uv plane, (w plane). While the y axis may be placed vertically on the x axis to obtain a, complete two-dimensional picture of the real-valued function y = f (x), the z, plane and w plane must stay apart, at least in this three-dimensional world., In this book, we mostly deal with functions f : A → C where A is a subset, of C., In Chapter 3, we will be concerned with functions that map certain regions, in the z plane onto certain regions in the w plane. Right now we have the, more modest task of determining a class of functions that map points near, one another in the z plane onto points near one another in the w plane., A function f (z), defined in a domain D, is said to be continuous at a point, z0 ∈ D if for every > 0, there exists a δ > 0 (δ depending on and z0 ) such, that, |f (z) − f (z0 )| < ,, , whenever, , |z − z0 | < δ., , (2.8), , Geometrically, this means that, for every neighborhood of f (z0 ) in the w, plane, there corresponds a neighborhood of z0 in the z plane whose image is, contained in the neighborhood of f (z0 ). More formally, for every > 0, there, exists a δ > 0 such that, f (N (z0 ; δ)) ⊂ N (f (z0 ); ), , (2.9), , (see Figure 2.12). If a function is continuous at every point of D, the function is, said to be continuous in the domain D. A function f : A → C is discontinuous, (or has a discontinuity) at z0 if z0 ∈ A, yet f is not continuous at z = z0 ., Remark 2.34. We will use (2.8) and (2.9) interchangeably. The reader should, convince himself of their equivalence and strive to be equally proficient with, both., Also, we will have occasion to discuss the continuity of a function in a, region R that includes boundary points. By an neighborhood of a boundary, point z0 ∈ R, we will mean N (z0 ; ) ∩ R, and will call this an open set relative
Page 61 :
2.5 Continuity, , 51, , Figure 2.13. -neighborhood of a boundary point, , to the region R. See Figure 2.13 for an, of the closed unit disk |z| ≤ 1., , neighborhood of a boundary point, , •, , If f (z) is continuous at z0 , we write limz→z0 f (z) = f (z0 ). A function may, have a limit at a point without being continuous at the point. We say that, limz→z0 f (z) = L if for every neighborhood of L, there is a deleted neighborhood of z0 whose image is contained in the neighborhood of L. If L = f (z0 ),, the function is continuous at z0 and the word “deleted” may be deleted from, this definition., Examples 2.35., , (i) Let, , f (z) =, , z 2 if z =, 2,, 5 if z = 2., , For this function, limz→2 f (z) = 4 although the function is not continuous at z = 2., (ii) Let, ⎧, ⎨ z − 2 if z = 2,, f (z) = z 2 − 4, ⎩4, if z = 2., Then limz→2 f (z) = limz→2 1/(z + 2) = 1/4 = L. Here L = f (2). Hence, f has a limit as z → 2 but is not continuous at z = 2., (iii) If limz→a f (z) = L, then for a given > 0 there exists δ > 0 such that, | |f (z)| − |L| | ≤ |f (z) − L| <, , whenever 0 < |z − a| < δ, , and therefore,, lim |f (z)| = |L|., , z→a, , Clearly, if L = 0, limz→a |f (z)| = |L| iff limz→a f (z) = L. What happens, if L = 0? More precisely, if limz→a |f (z)| = L , then is it always the, case that limz→a f (z) exists? Remember that if limz→a f (z) = L, then, |L| = L and therefore, we have to examine when equality holds in
Page 62 :
52, , 2 Topological and Analytic Preliminaries, , | |f (z)| − L | = | |f (z)| − |L| | ≤ |f (z) − L|., Equality would imply that, , Re (f (z)L) = |f (z)| |L| or |f (z)| = Re, , L, f (z), |L|, , , = Re (eiθ f (z)), , where θ = Arg (L/|L|), or equivalently,, |eiθ f (z)| = Re (eiθ f (z)), so that eiθ f (z) is real and nonnegative which is impossible for a general, complex-valued function f (z). However, this is possible when f (z) = L, or f (z) is a real-valued function with constant sign., (iv) The signum function sgn on C is defined by, ⎧, ⎧, ⎨ |z| for z = 0, ⎨ z for z = 0, z, sgn (z) :=, = |z|, ⎩, ⎩, 0 for z = 0., 0 for z = 0, This function is clearly continuous on C \{0} and, , 1 for z = 0, |sgn (z)| =, 0 for z = 0., , •, , A point z0 in a set D ⊆ C that is not a limit point of D is called an, isolated point of D. Clearly, at an isolated point z0 , there exists a δ > 0 such, that N (z0 ; δ) ∩ D = {z0 }. A function f : D → C is obviously continuous at, all isolated points of D. For example, consider, , z for z ∈ {1 − 1/n : n = 1, 2, . . . }, f (z) =, 1 for z = 1, and let D = {1 − n1 : n = 1, 2, . . . } ∪ {1}. The only limit point of D is 1 and, so all other points of D are isolated. Since, lim f (z) = f (1) = 1,, , z→1, , f is continuous at z = 1. By definition, f is obviously continuous at the, isolated points z = 1 − 1/n, n = 1, 2, . . . . Thus, f is continuous on D., What is the relationship between limits of sequences and limits of more, general functions? A complex sequences {zn }n≥1 , which defines a mapping, f : N → C, converges to z0 if for every > 0, there exists an M > 0 such that, f (N (∞; M ) ∩ N) ⊂ N (z0 ; )., Recall that a real M neighborhood of ∞ is the set of points outside the interval, (−M, M ).
Page 63 :
2.5 Continuity, , 53, , If the preimage of f is an unbounded region instead of the set of positive, integers, we have the following analog: Let f : C → C. Then limz→∞ f (z) = L, if for > 0, there exists an M > 0 with f (N (∞; M )) ⊂ N (L; )., Even if our region is bounded, there are important similarities between, limits of the sequences and limits of more general functions. A sequence has, a limit if eventually its points are “close” to one another, while a function, of a complex variable has a limit if closeness of points in different planes, is preserved. Our next theorem shows that continuity may be viewed as an, operation that preserves convergence of sequences., Theorem 2.36. The function f (z), defined in a region R, is continuous at a, point z0 ∈ R if and only if, for every sequence {zn } in R converging to z0 , the, sequence {f (zn )} converges to f (z0 )., Proof. Let f (z) be continuous at z0 . Then, for every > 0, there exists a δ > 0, such that |f (z) − f (z0 )| < whenever |z − z0 | < δ (z ∈ R). If {zn } converges, to z0 , then |zn − z0 | < δ for n > N . By continuity, |f (zn ) − f (z0 )| < for, n > N . Since was arbitrary, the sequence {f (zn )} converges to f (z0 )., Conversely, suppose that f (z) is not continuous at z0 . Now discontinuity, of f at z0 means that (see (2.9)) for some > 0, N (f (z0 ); ) does not contain, the image of any neighborhood of z0 . This means that we can find a sequence, of points {zn } such that zn ∈ N (z0 ; 1/n) ∩ R and f (zn ) ∈ N (f (z0 ); ). As, |zn − z0 | < 1/n for all n, the sequence {zn } converges to z0 although the, sequence {f (zn )} does not converge to f (z0 ). This contradiction completes, the proof., Remark 2.37. Theorem 2.36 is equally valid for real-valued functions of a, •, real variable., Let f be a continuous function defined in a region A. What properties, of A are inherited by its image f (A)? Theorem 2.36 states that convergent, sequences in A give rise to convergent sequences in f (A). But many properties,, even for real-valued functions of a real variable, are not preserved under a, continuous map., Examples 2.38., (i) The function f (z) = |z| maps the plane onto the real, interval [0, ∞). This shows that the continuous image of an open set, need not be open. We then say that f is not an open map., (ii) The function f (x) = tan−1 x maps the real line onto (−π/2, π/2). This, shows that the continuous image of a closed set need not be closed., (iii) The function f (z) = 1/z maps the punctured disk 0 < |z| < 1 onto, the exterior of the unit disk. This shows that the continuous image of a, •, bounded set need not be bounded., But all is not lost. If we combine the “nice” properties of the last two, examples, the image must also be “nice”.
Page 64 :
54, , 2 Topological and Analytic Preliminaries, , Theorem 2.39. The continuous image of a compact set is compact., Proof. Let f : A → f (A) be continuous on the compact set A. For any, sequence {wn } in f (A), we can find a corresponding sequence {zn } in A such, that f (zn ) = wn . By Theorem 2.27, there exists a subsequence {zn k } that, converges to a point z0 ∈ A. By Theorem 2.36, f (zn k ) = wn k converges to a, point f (z0 ) ∈ f (A). Since {wn } was arbitrary, every sequence in f (A) has a, subsequence that converges in f (A). Hence f (A) must be a compact set., A function f is said to be locally constant if for each a ∈ D there exists a, neighborhood N (a; δ) of a on which f (z) = f (a) for all z., Theorem 2.40. If a continuous function on a connected set D is locally constant, then f is constant throughout., Proof. Let a be such that f (a) = b. Define, S = {z : f (z) = b} = f −1 (b)., Now S is open because f is locally continuous. But S is closed because the, singleton set {b} is closed. Since S is not empty, we must have D = S. This, completes the proof., Because the complex field is not ordered, it makes no sense to talk about, maximum and minimum values for a complex-valued function f (z). However,, the next best thing is a discussion of maxima and minima for the related realvalued function |f (z)|. It will be helpful to observe that |f (z)| is continuous, in any region where f (z) is continuous. This follows from the inequality, | |f (z2 )| − |f (z1 )| | ≤ |f (z2 ) − f (z1 )| (z1 , z2 ∈ C)., Theorem 2.41. If f (z) is continuous on a compact set E, then |f (z)| attains, a maximum and minimum on E., Proof. According to Theorem 2.39, the image of E under |f (z)|, which we shall, denote by E , is a compact set. Since E is a bounded set of real numbers, it, has a least upper bound b. As a consequence of Exercise 2.29(2), the point b, is in the set E . But this means that |f (z0 )| = b for some z0 ∈ E., The proof that |f (z)| attains its minimum is similar, with greatest lower, bound being substituted for least upper bound., A function f (z) is said to uniformly continuous in a region R if for every, > 0, there exists a δ > 0 (δ depending only on ) such that if z1 , z2 ∈ R and, |z1 − z2 | < δ, then |f (z1 ) − f (z2 )| < . This differs from continuity in a region, in that the same δ may be used for every point in the region., For example, the function f (z) = z is uniformly continuous in every region,, since we may always choose δ = .
Page 65 :
2.5 Continuity, , 55, , Examples 2.42. The function f (z) = 1/z, although continuous, is not uniformly continuous in the region 0 < |z| < 1., To see this, assume that f (z) is uniformly continuous. Then for > 0 there, exists a δ, 0 < δ < 1, to satisfy the conditions of the definition. We exploit, the sensitivity of this function near the origin. Let z1 = δ and z2 = δ/(1 + )., Then |z1 − z2 | = δ /(1 + ) < δ, but, |f (z1 ) − f (z2 )| =, , 1 1+, −, δ, δ, , =, , δ, , > ,, , showing that f is not uniformly continuous on the punctured unit disk., Here is another example. The function f (z) = z 2 is not uniformly continuous in the complex plane C., Again, assume the contrary and let > 0 be given. Then for any δ > 0,, choose, z1 = 1/δ and z2 = 1/δ + δ/(1 + )., Then, we have |z1 − z2 | = δ/(1 + ) < δ and, |z12 − z22 | = 2/(1 + ) + δ 2 /(1 + )2 > 2/(1 + )., Note that this function is uniformly continuous in any bounded region., , •, , Example 2.43. Consider f (z) = x2 − iy 2 . Clearly f is continuous on C. But, f is not uniformly continuous on C, whereas it is uniformly continuous for, |z| < R. To verify the second part we first note that, for z = x + iy and, z0 = x0 + iy0 ,, |f (z) − f (z0 )| = |(x + x0 )(x − x0 ) − i(y + y0 )(y − y0 )|, ≤ |x + x0 | |x − x0 | + |y + y0 | |y − y0 |., If z, z0 are in the disk |z| < R, then |x + x0 | < 2R and |y + y0 | < 2R. This, implies that, √, |f (z) − f (z0 )| ≤ 2R[|x − x0 | + |y − y0 |] ≤ 2 2R|z − z0 |, √, √, (since |x| + |y| ≤ 2|z|). Now, given any > 0, there exists a δ = /(2 2R), such that, |f (z) − f (z0 )| <, , whenever |z − z0 | < δ = √ ., 2 2R, , So, f is uniformly continuous on ΔR ., The first part may now be verified as in the previous two examples, and, •, so we leave this part as a simple exercise., Theorem 2.44. If f (z) is continuous on a compact set A, then f (z) is uniformly continuous on A.
Page 66 :
56, , 2 Topological and Analytic Preliminaries, , Proof. Let > 0 be given. Then, for each point zα ∈ A, there is a neighborhood (depending on and zα ) such that, |f (z) − f (zα )| <, , (2.10), , 2, , whenever |z − zα | < δα , z ∈ A. The collection of all neighborhoods of the form, N (zα ; δα /2) is a cover of A. By the compactness of A, there exists a finite, subcover, say, A⊂, , n, $, k=1, , , , Choose, δ = min, , , , δk, ., N zk ;, 2, , δn, δ1 δ2, , , ... ,, 2 2, 2, , (2.11), , ., , We wish to show that this δ will work for the whole set A., Let w1 and w2 be any two points in A such that |w1 − w2 | < δ. By (2.11),, w1 ∈ N (zk ; δk /2) for some k. According to (2.10), it follows that, |f (w1 ) − f (zk )| <, , 2, , ., , (2.12), , But we also have, δk, δk, δk, <, +, = δk ., 2, 2, 2, , |w2 − zk | ≤ |w2 − w1 | + |w1 − zk | < δ +, Hence w2 ∈ N (zk ; δk ) ∩ A and, |f (w2 ) − f (zk )| <, , 2, , ., , (2.13), , Combining (2.12)and (2.13) we obtain, |f (w1 ) − f (w2 )| ≤ |f (w1 ) − f (zk )| + |f (zk ) − f (w2 )| <, , 2, , +, , 2, , = ,, , and this completes the proof., We end the section with a remark on stereographic projection discussed in, the previous section. If π : S \ {(0, 0, 1)} → C is a function, then, according, to the rule of correspondence,, , , , , , x, y, y, x, π(x, y, u) =, ,, ,0 =, +i, 1−u 1−u, 1−u, 1−u, and π has an inverse function π −1 : C → S \ {(0, 0, 1)} with the rule of, correspondence
Page 68 :
58, , 2 Topological and Analytic Preliminaries, , there exists an index N = N (R) such that |z| > R for all n ≥ N . Similarly, if, f : A → C and z0 ∈ C is a limit point of A, then limz→z0 f (z) = ∞ iff given, R > 0 there exists a δ > 0 such that |f (z)| > R whenever 0 < |z − z0 | < δ, and z ∈ A. If ∞ is a limit point of A, then limz→∞ f (z) = L iff given > 0, there exists an R > 0 such that |f (z) − L| < whenever |z| > R and z ∈ A., Questions 2.45., 1. What ambiguities might there be if we called the preimage the domain, of the function?, 2. What is the geometric significance of a complex-valued function of a, real variable? A real-valued function of a complex variable?, 3. What properties do functions and their inverses have in common?, 4. For what kinds of functions will we have points closer (more distant) in, the w plane than in the z plane?, 5. What can we say about the continuity of sequences?, 6. Can we talk about a function being continuous at ∞?, 7. What can we say about the continuous image of a limit point of a set?, 8. How do the proofs of Theorem 2.36 and Theorem 2.18 compare?, 9. What is the largest region on which f (z) = 1/z is uniformly continuous?, 10. Can discontinuous functions map compact sets onto compact sets?, 11. If a function is uniformly continuous on a set A, is it also uniformly, continuous on every subset of A?, 12. How can you define a piecewise continuous real-valued function of a real, variable defined on an interval [a, b]?, 13. How can you define a piecewise continuous complex function of a real, variable defined on an interval [a, b]?, 14. Are piecewise continuous real-valued functions of a real variable defined, on an interval [a, b] integrable and bounded?, 15. Does f (z) = arg z define a complex function? How about, f (z) = cos(arg z) + i sin(arg z)?, Exercises 2.46., 1. Find the following limits when they exist:, z2 + 9, z→3i z − 3i, z+1, (c) lim, z→∞ z 2, z 3 + 27i, (e) lim 2, z→3i z + 9, , (a) lim, , z + z2, z→2i 1 − z, z 2 + 10z + 2, (d) lim, z→∞ 2z 2 − 11z − 6, 1 − zn, (f) lim 2, (n ≥ 1), z→1 z + 5z − 6, , (b) lim
Page 69 :
2.5 Continuity, , 59, , 2. Discuss continuity and uniform continuity for the following functions., 1, 1, (|z| < 1), (b) f (z) =, (|z| ≥ 1), 1, −, z, z, %, %, Re z, |z|, if 0 < |z| < 1, if 0 < |z| ≤ 1 (d) f (z) =, (c) f (z) =, z, z, 1, if z = 0., 0 if z = 0, , (a) f (z) =, , 3. Prove that f (z) = 1/(1 − z) is not uniformly continuous for |z| < 1., 4. Show that the function f (z) = 1/z 2 is not uniformly continuous for, 0 < Re z < 1/2 but is uniformly continuous for 1/2 < Re z < 1., 5. Let f (z) be one of the following functions each being defined in the, punctured plane C \ {0}:, z, z |z| z, Re z Im z, ,, ,, , ,, ,, ., z, z, |z| z, z, z, Is it possible to suitably define any one of the these functions at z = 0, so that the resulting function will become continuous at z = 0. Answer, the same question for the functions, zIm z, zRe z, and, ., |z|, |z|, How about for the functions, z, z, and, Re z, Im z, when it is defined for C \ {x + iy : x = 0} and C \ {x + iy : y = 0},, respectively?, 6. Discuss continuity of, ⎧, 2, ⎨ (Re z) (Im z) if z = 0, |z|2, f (z) =, ⎩, 0 if z = 0, at the all points of C., 7. Find the following limits:, (a) lim f (z), where f (z) =, , xy, + 2xi,, x2 + y 2, x, xy, + 2 i,, (b) lim f (z), where f (z) = 2, z→0, x +y, y, 3, xy, x8, (c) lim f (z), where f (z) = 3, i., + 2, 3, z→0, x +y, y +1, 8. If limz→∞ f (z) = a, and f (z) is defined for every positive integer n,, prove that limn→∞ f (n) = a. Give an example to show that the converse, is false., z→0
Page 70 :
60, , 2 Topological and Analytic Preliminaries, , 9. Show that a monotonic real-valued function of a real variable cannot, have uncountably many discontinuities., 10. Show that f : A → B is continuous if and only if for every open set O, relative to B, f −1 (O) is an open set relative to A., 11. Using Exercise 2.46(10), prove that the continuous image of a compact, set is compact., 12. Show that f : A → B is continuous if and only if for every closed set F, relative to B, f −1 (F ) is a closed set relative to A., 13. Prove that continuous image of a connected set is connected., 14. If a function, defined on a compact set, is continuous, one-to-one, and, onto, show that the inverse function also has these properties. Can compactness be omitted?, 15. Let f and g be continuous on a set A. Show that f + g, f · g, and, f /g (g = 0) are also continuous on A. What can we say if f and g are, uniformly continuous on A?, 16. Show that f (z) is continuous in a region R if and only if both Re f (z), and Im f (z) are continuous in R., 17. Show that every polynomial is continuous in the complex plane., 18. Let f (z) be continuous in the complex plane. Let A = {z ∈ C : f (z) =, 0}. Show that A is a closed set., 1, 1, = ∞ and lim 2, = 0., 19. Show that lim, z→∞ z + 2, z→4 z − 4, 20. Suppose that J : C∞ → C∞ is defined by J(z) = 1/z, z ∈ C∞ . Do our, conventions imply J(0) = ∞ and J(∞) = ∞? Does, χ(J(z), J(w)) = χ(z, w), hold in C∞ ?
Page 71 :
3, Bilinear Transformations and Mappings, , In the previous chapter we saw that a complex function of a complex variable, maps points in the z plane onto points in the w plane. After the initial excitement of this discovery wore off, it became rather tiresome to map points, onto points in computer like fashion. In this chapter we will see, for some special functions, what happens to regions in the z plane when mapped onto the, regions in the w plane. We will show that bilinear transformations map circles and straight lines onto circles and straight lines. In fact, we will discover, that—contrary to popular belief—a circle is very similar to a straight line,, at least in the extended complex plane. We also determine the most general, form of bilinear transformation which maps, •, •, •, •, , the, the, the, the, , real line R onto the unit circle |z| = 1, unit circle |z| = 1 onto itself, unit circle |z| = 1 onto R, real line R onto itself., , 3.1 Basic Mappings, The function w = f (z) = z + b, where b is a complex constant, maps sets, in the z plane onto sets in the w plane displaced through a vector b. This, mapping is known as a translation. Note that the set in the w plane will have, the same shape and size as the set in the z plane. For instance, the function, w = z + (1 + 2i) maps the square having vertices ±1 ± i onto a square having, vertices i, 2 + i, 3i, and 2 + 3i (see Figure 3.1). To show this, let z = x + iy, and w = u + iv. Then, u + iv = (x + iy) + (1 + 2i), i.e. , u = x + 1, v = y + 2., As x describes the interval [−1, 1], u describes the interval [0, 2]; as y describes, the interval [−1, 1], v describes the interval [1, 3].
Page 72 :
62, , 3 Bilinear Transformations and Mappings, , Figure 3.1. Image of a square under w = z + 1 + 2i, , The function w = az, where a = cos α + i sin α, maps a point in the z, plane onto a point in the w plane whose distance from the origin is the same, but whose argument is increased by α, the argument of a. This mapping is, called a rotation. For instance, the function w = iz maps the right half-plane, (Re z > 0) onto the upper half-plane (Im z > 0). Observing that Arg i = π/2,, we may view this geometrically as a mapping of the points in the z plane, satisfying −π/2 < Arg z < π/2 onto points in the w plane satisfying 0 <, Arg w < π. Analytically,, w = u + iv = i(x + iy) = −y + ix, i.e. u = −y, v = x., Thus x > 0 is mapped onto v > 0., For a > 0, a = 1 the function w = az is known as a magnification (although, for a < 1 it is really a contraction). This function takes regions in the z plane, and either stretches or shrinks them, depending on whether a > 1 or a < 1., For instance, the function w = 5z maps the disk |z| ≤ 1 onto the disk |w| ≤ 5., More generally, for complex values of a, the function w = az represents, both a rotation and a magnification; the expression arg a is the rotation part,, and |a| is the magnification part. Indeed, we can combine a translation, rotation, and magnification to obtain the linear function, w = f (z) = az + b,, where a and b are complex constants. Note that, |w1 − w2 | = |f (z1 ) − f (z2 )| = |a| |z1 − z2 |, so that the distance between any two points is multiplied by |a|. For instance,, the function, w = (1 − i)z + (2 + i), maps the rectangle in the z plane shown in Figure 3.2 onto the rectangle in the, w plane that has, √ twice the area, with the length of each side being increased, by a factor of 2., There is a relationship between a complex linear function and the more, familiar real-valued linear function y = ax + b, a straight line. The complexvalued function w = az +b, with a and b are complex constants, maps straight
Page 73 :
3.1 Basic Mappings, , 63, , Figure 3.2. Image of a rectangle under w = (1 − i)z + (2 + i), , lines in the z plane onto straight lines in the w plane. Note that the complex, linear functions (a = 0) always map ∞ to ∞. We leave the determination of, the effect of the constants a and b on the slope of the image line as an exercise, for the reader. Observe that w = az + b, like its real-valued counterpart, is a, one-to-one function., The mapping w = 1/z, called an inversion, takes points close to the origin, in the z plane onto points far from the origin in the w plane and points far, from the origin in the z plane onto points close to the origin in the w plane., Indeed if z = reiθ , then, 1, 1, w = = e−iθ ., z, r, In particular, as z approaches the origin, w approaches the point at ∞ in the, extended complex plane; i.e., given M > 0, there exists a δ > 0 such that, |z| < δ implies |w| > M . We thus have a one-to-one map from the extended, plane onto itself with the origin being mapped onto the point at ∞. However,, it is wrong to conclude that inversion always maps lines into lines, and circles, into circles (see Theorem 3.1), There is also a certain symmetry with respect to both the unit circle and, the real axis. Points inside (outside) the unit circle are mapped onto points, outside (inside) the unit circle, and points above (below) the real axis are, mapped onto points below (above) the real axis (see Figure 3.3)., The inversion w = 1/z is sometimes called a reflection with respect to, both the unit circle and the real axis. To see what happens to sets in the z, plane when transformed into sets in the w plane by this reflection, we solve, w = u + iv =, , 1, 1, =, z, x + iy, , for a given variable in one plane in terms of the variables in the other plane., This gives the relations, u=, , x, ,, x2 + y 2, , v=−, , y, x2 + y 2, , (3.1)
Page 74 :
64, , 3 Bilinear Transformations and Mappings, , Figure 3.3. Illustration for w = 1/z, , and, x=, , u, ,, u2 + v 2, , y=−, , v, ., u2 + v 2, , (3.2), , From (3.1) or (3.2) we obtain, x2 + y 2 =, , 1, ., u2 + v 2, , (3.3), , Now consider the equation, a(x2 + y 2 ) + bx + cy + d = 0,, , (3.4), , where a, b, c, and d are real constants. This equation represents a circle if, a = 0 and a straight line if a = 0. From (3.1), (3.2), and (3.3), we see that the, function w = 1/z maps (3.4) onto the set, d(u2 + v 2 ) + bu − cv + a = 0,, , (3.5), , which describes a circle for d = 0 and a straight line if d = 0., We can now, in view of (3.4) and (3.5), draw several conclusions about the, mapping properties of w = 1/z:, (a) Circles not passing through the origin (that is, with a = 0 and d = 0)., are mapped onto circles not passing through the origin., (b) Circles passing through the origin (that is, with a = 0 and d = 0) are, mapped onto straight lines not passing through the origin., (c) Straight lines not passing through the origin (that is, with a = 0 and, d = 0) are mapped onto circles passing through the origin., (d) Straight line passing through the origin (that is, with a = 0 and d = 0), are mapped onto straight lines passing through the origin., (e) The circle |z| = 1 maps onto the circle |w| = 1., (f) The punctured disk Δ \ {0} maps onto C \ Δ, and conversely., (g) All points on C \ Δ map onto Δ \ {0}.
Page 75 :
3.1 Basic Mappings, , 65, , In short, we have, Theorem 3.1. The function w = 1/z maps circles and straight lines onto, circles and straight lines., Is there a way to remember which maps onto which? The key lies in the, fact that the origin maps onto the point at ∞. Every straight line (and no, circle) passes through the point at ∞. Hence a straight line or circle maps, onto a straight line if it passes through the origin, and onto a circle if it does, not. Also, we note that the interior of a circle containing the origin maps onto, the exterior of a circle, and the interior of a circle not containing the origin, (nor having the origin as a boundary point) maps onto the interior of a circle., Finally, we present some precise mapping properties of w = 1/z. Consider the, circle |z − a| = R, a = 0. If w = 1/z, then we obtain that, |z − a| < R ⇐⇒, , 1, 2, − a < R ⇐⇒ |1 − aw| < R2 |w|2, w, , ⇐⇒ |w|2 (|a|2 − R2 ) − 2Re (aw) + 1 < 0., ⎧, Re (aw) > 1/2 for R = |a|, ⎪, ⎪, ⎪, ⎪, ⎪, R, a, ⎨, w− 2, < 2, for R < |a|, 2, ⇐⇒, |a| − R, |a| − R2, ⎪, ⎪, ⎪, R, a, ⎪, ⎪, > 2, for R > |a|., ⎩ w− 2, |a| − R2, R − |a|2, For example, if R = |a|, then under the inversion w = 1/z we have, •, •, •, , |z − a| < |a| is mapped onto the half-plane Re (aw) > 1/2, |z − a| = |a| is mapped onto the straight line Re (aw) = 1/2, |z − a| > |a| is mapped onto the half-plane Re (aw) < 1/2., , When |a|2 − R2 = 0, there exist two possibilities |a| > R and |a| < R., In each of these cases, mapping properties may be stated with the help of, the above discussion. For example, under the inversion w = 1/z, we have the, following:, •, •, •, , R, 3, <, for R < 3, 9 − R2, 9 − R2, R, 3, > 2, for R > 3, |z − 3| < R is mapped onto the disk w + 2, R −9, R −9, R, 3, =, for R =, 3., |z − 3| = R is mapped onto the disk w −, 2, 9−R, |9 − R2 |, |z − 3| < R is mapped onto the disk w −, , Questions 3.2., 1. The functions w = 1/z, w = z, and w = −z all map the upper half of, the unit circle onto lower half. What are their differences?
Page 76 :
66, , 3 Bilinear Transformations and Mappings, , 2. How does the area of a region compare with the area of its image for a, linear function? For an inversion?, 3. For w = 1/z, what is the image of the interior of a circle having the, origin as boundary point?, 4. What happens to conic sections other than circles under an inversion?, 5. Is there a difference between performing a translation followed by an, inversion and an inversion followed by a translation?, 6. For the four operations of translation, rotation, magnification, and inversion, which pairs may be interchanged without affecting the mapping, properties?, Exercises 3.3., 1. For the mapping w = (1 + i)z + 2, find the image of, (a) the line y = 2x, (b) the line y = 3x + 2, (c) the circle |z| = 3, (d) the circle |z − 1| = 2., 2. Find the image of the half-plane Re z > 0 under the transformation, (a) w = 2iz − i, (b) w = i/z − 1., 3. Find the image of the semi-infinite strip {z : 0 < Re z < 2, Im z > 1}, for the transformation w = (1 − i)z + (2 − i), and sketch., 4. Find a linear transformation f that maps the circle |z + 1| = 2 onto the, circle |w + i| = 3. Find also the image of |z + 1| < 2 under f ., 5. Prove that the linear transformation w = az + b maps a circle having, radius r and center z0 onto a circle having radius |a|r and center az0 +b., 6. Given a triangle with vertices at 3 + 4i, −3 + 4i, and −5i, find its image, for the transformation, (a) w = z + 5i, (b) w = iz + (2 − i), (c) w = (2 + i)z − 3., 7. Find the image of the line y = 2x + 1 under the following transformations., (a) w = 1/z, (b) w = i/z, (c) w = 1/(z − 2i)., 8. For the transformation w = 1/z, find the image of, (b) the circle |z − 1| = 2, (a) the circle |z − 2| = 1, (d) the domain Re z > 1., (c) the circle |z − 1| = 1, (e) the infinite strip 14 < Re z < 12 ., 9. Find the images of the strips 0 ≤ Re z ≤ 2 and 0 ≤ Im z ≤ 2 under the, map w = 1/z., , 3.2 Linear Fractional Transformations, By considering quotient of two linear transformations, we get a very important, class of mappings of the form, known as a linear fractional transformation:, w = T (z) =, , az + b, ,, cz + d, , (3.6)
Page 77 :
3.2 Linear Fractional Transformations, , 67, , where a, b, c and d are the complex numbers such that ad − bc = 0. This, transformation contains, as special cases, the mappings of the previous section., For c = 0 we have a linear transformation, and for a = d = 0, b = c we have, an inversion. The condition ad − bc = 0 ensures that the mapping is not a, constant. To see this, suppose that ad − bc = 0. If c = 0, we may solve for, b = ad/c and write (3.6) as, ad, ad, a, (cz + d) + b −, a b− c, a, c, c, = +, = ,, w=, cz + d, c, cz + d, c, a constant. Similarly if a = 0, then ad − bc = 0 gives, becomes, , , a z + b/a, az + b, w=, =, =, cz + bc/a, c z + b/a, , (3.7), , d = bc/a so that (3.6), a, ,, c, , again a constant. If ad − bc = 0 and a = 0, then either b = 0 or c = 0. When, a = b = 0, it follows that w = 0, and when a = c = 0, we see that w = b/d, a, constant. We will henceforth assume that ad − bc = 0. Thus, we write, , , ⎧, ad − bc, 1, a, ⎪, ⎨, −, if c = 0, az + b, 2, c, c, z, +, d/c, (3.8), = , w = T (z) =, cz + d ⎪, ⎩ a z+ b, if c = 0., d, d, The domain of the definition of T (z) is C \ {−d/c}. Clearly, T (z) is a oneto-one function on its domain. Since T is well defined for all points in the, extended complex plane except at z = −d/c and the point at ∞, we may, extend the definition of T to the extended complex plane by including these, points. Indeed, as, lim, z→−d/c, , 0, 1, cz + d, = lim, = −d, = 0,, T (z) z→−d/c az + b, a( c ) + b, , we find that limz→−d/c T (z) = ∞. Further, we have, az + b, = lim, z→∞ cz + d, z→0, lim, , a, z, c, z, , +b, a, a + bz, = lim, = ,, + d z→0 c + dz, c, , and hence for c = 0, we may define, ⎧, az + b, ⎪, ⎪, ⎨ cz + d if z = −d/c, z = ∞, T (z) = ∞, if z = −d/c, ⎪, ⎪, ⎩a, if z = ∞, d, and T defined in this way is then one-to-one onto the extended complex plane, and has an inverse that is also a linear fractional transformation. Solving for, z, in terms of w, we obtain
Page 78 :
68, , 3 Bilinear Transformations and Mappings, , ⎧, dw − b, ⎪, ⎪, ⎪, ⎨ −cw + a if w = a/c, w = ∞, z = T −1 (w) = ∞, if w = a/c, ⎪, ⎪, d, ⎪, ⎩−, if w = ∞., c, Thus regions may be mapped with equal facility from the extended z plane to, the extended w plane or from the extended w plane to the extended z plane., When cleared of fractions, (3.6) assumes the form, Azw + Bz + Cw + D = 0,, an equation that is linear in both z and w. For this reason, a linear fractional, transformation is often called a bilinear transformation., A bilinear transformation represents a one-to-one continuous mapping of, the extended complex plane onto itself with the point z = −d/c mapping onto, w = ∞ and the point z = ∞ mapping onto w = a/c. Recall that, •, •, , a linear transformation maps circles onto circles and straight lines onto, straight lines., an inversion maps circles and straight lines onto circles and straight, lines, see Theorem 3.1., , We will use these facts to deduce that bilinear transformations have similar, mapping properties. We consider w = T (z) defined by (3.8) for c = 0. Then, we have that, w = T (z) = (f3 ◦ f2 ◦ f1 )(z),, where, , and, , d, w1 = f1 (z) = z + ,, c, a, w = f3 (w2 ) = −, c, , w2 = f2 (w1 ) =, , , ad − bc, c2, , 1, w1, , , w2 ., , That is z → w = T (z) is given by the composition, z → w1 → w2 → w., The first is linear and maps circles in the z plane onto circles in the w1, plane and straight line in the z plane onto straight lines in the w1 plane; the, second is an inversion, mapping circles and straight lines in the w1 plane onto, circles and straight lines in the w2 plane; the third is again linear and maps, circles in the w2 plane onto circles in the w plane and straight lines in the, w2 plane onto straight lines in the w plane. If c = 0, the transformation is a, linear. The above results may be summarized as, Theorem 3.4. The bilinear transformation maps circles and straight lines, onto circles and straight lines.
Page 79 :
3.2 Linear Fractional Transformations, , 69, , In the foregoing proof, we have also shown that a linear fractional transformation (3.8) can be written as a composition of three types of elementary, transformations namely,, •, •, •, , the translation T1 (z) = z + B, the inversion T2 (z) = 1/z, the dilation T3 (z) = Az., , Note that a dilation is a composition of a magnification (or contraction) and, rotation. The order in which these transformations are performed is immaterial as they commute., Remark 3.5. The point z = −d/c plays the same role in the bilinear transformation as does the point z = 0 in the inversion transformation. Thus a, straight line or a circle maps onto a straight line if it passes through the point, z = −d/c, and onto a circle if it does not., •, Theorem 3.4 may often be used to simplify computation. For example, in, Exercise 1.8(6), the reader was asked to show that, , 1, z, Re, for |z| < 1., >−, 1−z, 2, Presumably, the reader separated z/(1 − z) into its real and imaginary parts,, substituted in points inside the unit disk, and then marveled at the result. We, will now apply more sophisticated techniques that give some insight into the, solution. The bilinear transformation w = z/(1 − z) maps the unit circle onto, a straight line (since z = 1 maps onto w = ∞). By choosing any two distinct, points on the unit circle, we can determine this straight line. The point −1, and i map onto the points − 12 and − 12 + 2i , respectively. Thus the image of, the circle |z| = 1 is the line Re w = − 12 (see Figure 3.4). The continuity of a, bilinear map reveals that connected sets are mapped onto connected sets (see, Exercise 2.46(13)). Since a bilinear map is also a one-to-one mapping of the, extended plane onto itself, the image of |z| < 1 is either, , Figure 3.4. Illustration for the image of the unit disk under z/(1 − z)
Page 80 :
70, , 3 Bilinear Transformations and Mappings, , Re w > −, , 1, 1, or Re w < − ., 2, 2, , To determine the correct image, we need to test only one point. The origin, mapping onto itself assures us that |z| < 1 maps onto Re w > − 12 . Alternatively, we simply note that, w=, , w, w(1 + w), w + |w|2, z, ⇐⇒ z =, =, =, ., 2, 1−z, 1+w, |1 + w|, |1 + w|2, , Thus, |z| = 1 gives, |w|2 = |1 + w|2 , i.e., Re w = −1/2, and, since 0 is mapped onto w = 0, it follows that |z| < 1 is mapped onto, Re w > −1/2 under the map w = z/(1 − z)., Example 3.6. Suppose the reader was asked to find the image of the closed, half disk {z : |z| ≤ 1, Re z ≥ 0} under the bilinear transformation, w=, , z, ., 1−z, , We know that |z| < 1 is mapped onto Re w > −1/2. Moreover, Re z ≥ 0 is, mapped onto points, Re w + |w|2 = |w + (1/2)|2 − 1/4 ≥ 0,, , i.e., |w + 1/2| ≥ 1/2., , Consequently, the image of the closed half disk {z : |z| ≤ 1, Re z ≥ 0} under, the bilinear transformation w = z/(1 − z) is, {w : Re w ≥ −1/2} ∩ {w : |w + 1/2| ≥ 1/2}., Note also that Im z ≤ 0 is mapped onto Im w ≤ 0, showing that the, image of the closed half disk {z : |z| ≤ 1, Im z ≤ 0} under w = z/(1 − z), is {w : Re w ≥ −1/2 and Im w ≤ 0}. What is the image of one-quarter disk, {z : |z| ≤ 1, Im z ≤ 0, Re z ≥ 0}?, •, Example 3.7. Suppose the reader is asked to find the image of the annulus, {z : 1 < |z| < 2} under w = z/(1 − z). To do this, we first note that, z=, and so, , w, 1+w, , |z| > 1 ⇐⇒ |w|2 > |1 + w|2 , i.e., 0 > 1 + 2Re w, , and, |z| < 2 ⇐⇒ |w|2 < 4|1 + w|2 , i.e. 0 < 3[|w + (4/3)|2 − (4/9)]., So we see easily that
Page 81 :
3.2 Linear Fractional Transformations, , •, •, , 71, , |z| > 1 is mapped into Re w < −1/2, |z| < 2 is mapped onto |w + 4/3| > 2/3, , and the required image is {w : Re w < −1/2} ∩ {w : |w + 4/3| > 2/3}., , •, , So far, we have been concerned with determining the images of sets under, a fixed bilinear transformation. We now reverse the problem and consider the, following., Problem 3.8. Given two sets, under what circumstances will there exist a, bilinear map from one set onto the other?, From elementary geometry, we know that any three points determine either, a circle or a straight line depending on whether the three points are collinear., Any bilinear transformation maps a circle or a straight line determined by, these points onto either a circle or a straight line, depending on the coefficients, of the bilinear transformation., For our next result, we need the notion of fixed points. A point z in C∞, that satisfies the equation, z = T (z), is called a fixed point of T . The identity transformation I(z) = z has every z, in C∞ as its fixed points. Concerning other bilinear transformations, we have, the following result which has many important consequences., Theorem 3.9. A bilinear transformation w = T (z) with more than two fixed, points in C∞ must be the identity transformation., Proof. Suppose that c = 0 in (3.8). Then, T is of the form, T (z) = αz + β,, , α = 0., , The solution of z = αz + β are the fixed points of T . Clearly, the solution set, is given by, (i) z = ∞ and z = β/(1 − α) whenever α = 1, (ii) z = ∞ whenever α = 1, β = 0, (iii) all z whenever α = 1, β = 0., Suppose that c = 0. Then ∞ cannot be a fixed point and the fixed point, equation z = T (z) gives the quadratic equation, cz 2 + (d − a)z − b = 0, which has at most two complex roots. Evidently, the only situation which, provides more than two fixed points is the one in which T = I, the identity, transformation., We will now show that for A (a circle or a straight line in the z plane) and, B (a circle or a straight line in the w plane), there exists a bilinear map from, A onto B.
Page 83 :
3.2 Linear Fractional Transformations, , 73, , Remark 3.12. The right side of (3.12) is called the cross ratio of the points, z1 , z2 , z3 , z and is denoted by (z1 , z2 , z3 , z). Observe that (3.12) asserts the, invariance of the cross ratio under a bilinear transformation. That is, for any, four distinct pairs of points (z1 , w1 ), (z2 , w2 ), (z3 , w3 ), and (z, w) of a bilinear, transformation, we must have, (z1 , z2 , z3 , z) = (w1 , w2 , w3 , w)., , •, , Remark 3.13. It is not surprising that three points determine a bilinear, transformation. If we divide (3.6) by one of the nonzero constants (assume, a = 0), then (3.6) may be rewritten as, w=, , z+B, ,, Cz + D, , and elementary algebra may be used to solve for three equations with three, unknowns. Moreover, the proof of Theorem 3.10 suggests a method of finding, w = T (z) satisfying the condition wj = T (zj ) whenever zj and wj are given,, j = 1, 2, 3. In many cases, T can be found with even less trouble as we can, •, seen in some of the examples of this section., Example 3.14. Let us now find a bilinear transformation that maps the, points z = i, 2, −2 onto w = i, 1, −1, respectively., To do this we may simply use (3.12) and obtain, (z − i)(2 + 2), (w − i)(1 + 1), =, ., (w + 1)(1 − i), (z + 2)(2 − i), Solving this equation, we obtain, w=, , 3z + 2i, ., iz + 6, , Similarly, it is easy to find the bilinear transformation that maps the points, z = 1 − i, 1 + i, −1 + i onto 0, 1, ∞, respectively., Indeed, by Corollary 3.11, we see that the desired transformation is, w=, , z − (1 − i), ., iz + (1 + i), , •, , Example 3.15. Let us find a bilinear transformation which maps the disk, |z + i| < 1 onto the exterior disk |w| > 4. To do this, we consider, f (z) =, , az + b, ., cz + d, , Without loss of generality we may assume that f (−i) = ∞. Then f (z) takes, the form, az + b, f (z) =, ., z+i
Page 84 :
74, , 3 Bilinear Transformations and Mappings, , Note that f (0) = −ib and f (−2i) = 2a + ib. According to our requirements,, these two points must lie on the circle |w| = 4. This gives, , |b| = 4 and |2a + ib| = 4a2 + b2 = 4., A choice satisfying these two conditions are b = 4 and a = 0. This shows that, f (z) = 4/(z + i) is a bilinear transformation which maps the circle |z + i| = 1, onto the circle |w| = 4. Since −i → ∞,, f (z) =, , 4, z+i, , is exactly a desired transformation. Note also that this is not unique as there, •, are many bilinear transformations which do the same job., Thus far we have seen that there is a unique bilinear transformation mapping three distinct points in the z plane onto three distinct points in the w, plane. This has given rise to ways of mapping circles and straight lines onto, circles and straight lines, although not uniquely. For example, to find a function mapping a line onto a circle, we may choose any three points on the line, and make them correspond with any three points on the circle. Let Im z0 > 0., Then, by Theorem 3.10, the bilinear transformation mapping such that, z0 → 0, z 0 → ∞, 0 →, is given by, w=, , z0, ,, z0, , z − z0, ., z − z0, , Note that this function maps the real line R onto the unit circle |w| = 1., Moreover, as z0 → 0, it must map the upper half-plane {z : Im z > 0} onto, the unit disk |w| < 1 and lower half-plane {z : Im z < 0} onto the exterior, |w| > 1. On the other hand, the bilinear transformation mapping the points, z = z 0 , z0 , 0 onto the points w = 0, ∞, z 0 /z0 respectively, given by, w=, , z − z0, ,, z − z0, , maps the lower half-plane onto the unit disk |w| < 1 and upper half-plane, onto the exterior |w| > 1. How about the bilinear transformation such that, z0 → 0, z 0 → ∞, 0 → eiα, , z0, ?, z0, , We will now attempt to solve the following., Problem 3.16. Characterize all bilinear transformations that map the upper, half-plane onto the interior of the unit circle.
Page 85 :
3.2 Linear Fractional Transformations, , 75, , These transformations in question, of course, must map the real line R, onto the unit circle |w| = 1. Letting, w=, , az + b, ,, cz + d, , we choose specific points to determine conditions for our coefficients. Since, |w| = 1 when z = 0 and z = ∞, we obtain, b, =1, d, , and, , a, = 1., c, , (3.13), , In view of (3.13), we have, d, b, =, ., a, c, Hence the transformation may be written as, , , b, d, a z + b/a, iα z − z0, w=, =e, z0 = − , z1 = −, ,, c z + d/c, z − z1, a, c, , (3.14), , where α ∈ R and |z0 | = |z1 |. Can we obtain additional information about the, relationship between z0 and z1 ? By letting z = 1 in (3.14), we have, |w| =, , 1 − z0, = 1,, 1 − z1, , or, |1 − z0 | = |1 − z1 |., , (3.15), , Upon simplifying (3.15), we see that Re z0 = Re z1 . Since |z1 | = |z0 |, either, z1 = z0 or z1 = z 0 . If z1 = z0 , then ad − bc = 0 and so, (3.14) reduces to a, constant. Thus z1 = z 0 , and we have, Theorem 3.17. The most general bilinear transformation of the real line R, onto the unit circle |w| = 1 is given by, w = T (z) = eiα, , z − z0, ,, z − z0, , (3.16), , where α ∈ R., Since the point z0 maps onto the origin and z 0 maps onto ∞, (3.16) maps, the upper half-plane onto the interior of the unit circle if Im z0 > 0 and onto, its exterior if Im z0 < 0. How do we characterize all bilinear transformations, that map the right half-plane {z : Re z > 0} onto the unit disk |w| < 1?, We wish to determine the most general set of coefficients such that, w=, , az + b, cz + d
Page 86 :
76, , 3 Bilinear Transformations and Mappings, , will map Im z > 0 onto Im w > 0. As in the previous example, we will first, find all bilinear mappings from the boundary of the region in the z plane onto, the boundary of the region in the w plane and then determine the subset, of these mappings that satisfy our additional criterion. Bilinear mappings, from Im z = 0 onto Im w = 0 are found by mapping any three points on, the real axis of the z plane onto any three points in the real axis of the w, plane. Let z = z1 , z2 , z3 map onto the points w = 0, 1, ∞, respectively. Since, w = [0, 1, ∞, w], the invariance of cross ratio shows that (see Corollary 3.11), w = (z1 , z2 , z3 , z) =, , (z − z1 )(z2 − z3 ), (z − z3 )(z2 − z1 ), , for any z., , Since z1 , z2 , z3 are all real, the coefficients a, b, c, d of, w=, , az + b, cz + d, , must all be real. To see what further constraints are necessary, we rewrite, w=, , (az + b)(cz + d), ac|z|2 + bd + adz + bcz, az + b, =, =, ., cz + d, |cz + d|2, |cz + d|2, , Then, whenever Im z > 0, we have, Im w =, , (ad − bc)Im z, > 0 if and only if ad − bc > 0., |cz + d|2, , Moreover, Im w < 0 if and only if ad − bc < 0. Recall that the map is onto., Hence we have the following., Theorem 3.18. The most general bilinear map of the upper half-plane, {z : Im z > 0} onto itself is given by, w=, , az + b, ,, cz + d, , where a, b, c, d are real and ad − bc > 0., Corollary 3.19. The most general bilinear map of the upper half-plane, {z : Im z > 0} onto the lower half-plane {w : Im w < 0} is given by, w=, , az + b, ,, cz + d, , where a, b, c, d are real and ad − bc < 0., Next we ask, Problem 3.20. Determine all bilinear transformations that map the unit disk, {z : |z| < 1} onto itself.
Page 87 :
3.2 Linear Fractional Transformations, , 77, , The answer to this problem will necessitate mapping the unit circle onto, itself. Any rotation of the identity function (that is, w = az, |a| = 1) is a, solution to our problem. But there are more general transformations that, take the unit disk onto itself., Theorem 3.21. The most general bilinear transformation that takes the unit, disk Δ = {z : |z| < 1} onto itself is given by, , , z − z0, f (z) = eiα, ,, (3.17), 1 − z0z, where α ∈ R and z0 ∈ Δ., If, in Theorem 3.21, we take z0 with |z0 | > 1, then f maps |z| > 1 onto, |w| < 1 so that f in this choice maps |z| < 1 onto |w| > 1., There are several proofs of this result. The easiest proof follows from the, principle of inverse points. Let us now discuss an important concept concerning, inverse/symmetric points. Let L be a line in C. Two points z and z ∗ are, called the inverse points (symmetric) with respect to the line L if L is the, perpendicular bisector of [z, z ∗ ], the line segment connecting z and z ∗ , see, Figure 3.5., , Figure 3.5. Inverse points with respect to a line L, , Then it is easy to see that every line or circle passing through both z and, z ∗ intersect L at right angles. For instance,, (i) z and z ∗ are inverse points with respect to the real axis whenever z ∗ = z,, (ii) z and z ∗ are inverse points with respect to the imaginary axis whenever, z ∗ = −z., Consider w = 1/z, z ∈ Δ = {z : |z| < 1}. Then the point z = reiθ (0 < r < 1), in Δ maps onto the point (1/r)e−iθ which lies outside the unit circle |z| = 1., Let L be the line from the center “O” through z = reiθ . Draw a line S, perpendicular to the line L through the point z. The line S intersects the, unit circle |z| = 1 at two points. Draw tangents at which S intersects the unit
Page 88 :
78, , 3 Bilinear Transformations and Mappings, , circle. It is easy to see these two tangents intersect the line L at the point, 1/z. Note that, 1, =1, |z|, z, and 0, z, 1/z lie on the same ray., Using the above discussion, we may define the inverse points with respect, to an arbitrary circle as follows: we say that two points z and z ∗ in C are, the inverse points with respect to a circle in C if every line or circle passing, through both z and z ∗ intersect at right angles. It is easy to see the following:, “Let C = {ζ : |ζ − z0 | = R} be a circle in C with center at z0 and radius R., Two points z and z ∗ are inverse points with respect to the circle C if, (i) z and z ∗ are collinear with center z0, (ii) |z − z0 | |z − z ∗ | = R2 .”, We remark the following:, •, •, , •, , If z moves close to the boundary of C, the point z ∗ also moves closer, to the boundary. In other words, every point on the circle is the inverse, point of itself., If z moves towards the center z0 , then |z −z0 | → 0 whereas |z −z ∗ | → ∞., This fact is expressed by saying that the center “z0 ” and the point at, “∞” are the inverse points with respect to the circle C. Since R is, arbitrary, the center and the point ∞ are inverse points with respect to, any circle centered at z0 and any finite radius., Let z be a point inside the circle. Then z = z0 + reiθ (r < R). If z ∗ is, the inverse point of z with respect to the circle C, then, since z and z ∗, lie on the same ray through z0 , we have, Arg (z ∗ − z0 ) = Arg (z − z0 ) = θ and |z − z0 | |z ∗ − z0 | = R2 ., This gives, z ∗ − z0 =, , , , R2, |z − z0 |, , , , R2, R2, =, ., re−iθ, z − z0, , eiθ =, , The fact discussed above may be formulated as, Corollary 3.22. Two points z and z ∗ are inverse points with respect to the, circle C = {ζ : |ζ − z0 | = R} if and only if, (z ∗ − z0 )(z − z0 ) = R2 , i.e., z ∗ = z0 +, , R2, ., z − z0, , (3.18), , Thus, one may define the inversion in C = {ζ : |ζ − z0 | = R} as a map, JC : C∞ → C∞ defined by, JC (z) = z0 +, , R2, ,, z − z0, , z = z0 .
Page 90 :
80, , 3 Bilinear Transformations and Mappings, , or equivalently,, βz + βz ∗ + c = βz + βz ∗ + c and (βz + βz ∗ + c) + (βz + βz ∗ + c) = 0., Note that this is of the form A = A and A + A = 0 which imply that A = 0., In conclusion, we have, Theorem 3.24. Two points z and z ∗ are inverse points with respect to the, line βZ + βZ + c = 0 if and only if βz + βz ∗ + c = 0., Let us state and prove a general result which covers the case of a circle., Theorem 3.25. Two points z and z ∗ are inverse points with respect to the, circle in C∞ ,, αZZ + βZ + βZ + γ = 0, , (3.19), , αzz ∗ + βz + βz ∗ + γ = 0., , (3.20), , if and only if, , (Note that line is considered as a circle of infinite radius; in this case α = 0)., Proof. Without loss of generality, we may assume that α = 1 as α = 0 has, been dealt with in Theorem 3.24. For α = 1, (3.19) is equivalent to, , |Z + β| = |β 2 | − γ., Thus, by (3.18), z and z ∗ are symmetric with respect to the circle, , , |Z + β| = |β 2 | − γ (z0 = −β and R = |β 2 | − γ), if and only if, [z ∗ − (−β)][(z − (−β))] = |β|2 − γ, i.e., zz ∗ + βz + βz ∗ + γ = 0, and the proof is complete., If we choose β = 0 and γ = −1 in Theorem 3.25, then we see that z and, z ∗ are the inverse points with respect to the unit circle |z| = 1 if and only if, zz ∗ = 1, i.e., z ∗ =, , 1, ., z, , The concept of inverse points is useful in solving mapping problems that, involve bilinear transformations because of the remarkable property which, asserts that “bilinear transformation preserve inverse points”. More precisely,, we have
Page 91 :
3.2 Linear Fractional Transformations, , 81, , Theorem 3.26. Let C be a circle in C∞ . Suppose further that w = T (z) is a, bilinear transformation and C = T (C) is the transformed circle in C∞ . The, two points z and z ∗ in C∞ are inverse points with respect to C if and only if, w = T (z) and w∗ = T (z ∗ ) are inverse points with respect to C ., Proof. Let the equation of the circle C in C∞ be given by, αZZ + βZ + βZ + γ = 0 (α ∈ R, β ∈ C, γ ∈ R)., , (3.21), , Suppose that z and z ∗ are a pair of inverse points with respect to the circle, C. By Theorem 3.25, z and z ∗ must satisfy, αzz ∗ + βz + βz ∗ + γ = 0., , (3.22), , Let W = T (Z) = (aZ + b)/(cZ + d) be a bilinear transformation. Then, Z=, , dW − b, ., −cW + a, , We wish to show that T (z) = w and T (z ∗ ) = w∗ are inverse points with, respect to the transformation circle C = T (C). First we see from (3.21) that, the image of the circle C under W = T (Z) is given by, , , , , , , , dW − b, dW − b, dW − b, dW − b, α, +β, +β, + γ = 0., −cW + a, −cW + a, −cW + a, −cW + a, The image of (3.22) under w = T (z) is the same as above except that w and, w are replaced by w and w∗ , respectively. Therefore, by Theorem 3.25, w and, w∗ must be inverse points with respect to the transformed circle C described, by the above equation. The converse follows similarly., Using this theorem, it is easy to characterize all bilinear transformations, which map a circle in C∞ to another given circle in C∞ ., Let us first find all bilinear transformations which map the unit disk, Δ = {z : |z| < 1} onto itself. To do this let f (z) to be a general bilinear, transformation which takes Δ onto itself. Clearly, there exists a z0 in Δ such, that f (z0 ) = 0. We know that z0 and 1/z 0 are inverse points with respect to, the unit circle |z| = 1 (Recall that, as z0 → 0, 1/z 0 → ∞ and so 0 and ∞, are inverses with respect to the any circle centered at the origin). As z0 → 0,, 1/z 0 → ∞, and so f must be of the form, , , , , , , z − z0, z − z0, z − z0, w = f (z) = k, = −kz 0, =A, z − 1/z 0, 1 − zz 0, 1 − zz 0, where A is a constant chosen so that |w| = 1. As |z| = 1 implies that |w| = 1,, we in particular have, , , 1 − z0, = |A| = 1,, |f (1)| = 1, i.e., A, 1 − z0
Page 92 :
82, , 3 Bilinear Transformations and Mappings, , which gives A = eiα for some real α. Thus, f has the desired form, namely, (3.17). Theorem 3.21 follows., Let us next use the symmetry principle to obtain all bilinear transformations that map the upper half-plane {z : Im z > 0} onto the unit disk |z| < 1., To see this we suppose that z0 (Im z0 > 0) is mapped to w = 0. Note that, z 0 is symmetric to z0 with respect to the real axis of the z-plane. As w = 0, and w = ∞ are symmetric with respect to the unit circle, the desired bilinear, transformation w = T (z) must carry z 0 to w = ∞. Therefore,, , , z − z0, w = T (z) = A, z − z0, for some complex constant A. When z is real, we have |w| = 1 which gives, |A| = 1, i.e., A = eiα for some α ∈ R, and T (z) is of the form (3.16). Thus,, we have provided an alternate proof of Theorem 3.17., Questions 3.27., 1. In choosing three points, why is it often convenient to pick 0, 1, and ∞?, 2. When will the sum of bilinear transformations be a bilinear transformation? The product?, 3. What kind of bilinear transformation maps ∞ onto itself?, 4. What kind of bilinear transformation maps ∞ onto the origin?, 5. How many bilinear transformations map more than two points onto, themselves?, 6. What is the form of a bilinear transformation which has one fixed point, z1 ∈ C and the other fixed point at ∞? How about, in particular, z1 = 0?, 7. Is there a bilinear transformation having no fixed point?, 8. Is there a bilinear transformation having exactly one fixed point? How, about f (z) = z/(2z + 1)?, 9. What can we say about a transformation which has ∞ as the only fixed, point? Is it simply the transformation of the form f (z) = z +β, for some, β ∈ C?, 10. Why don’t we say that f (z) = z is a bilinear transformation? Does it, have infinitely many fixed points? Is it true that the only bilinear transformation, having more than two fixed points, is the identity transformation?, 11. Does the isogonal map f (z) = z preserve cross ratio?, 12. Without substituting any points, is there a way to determine whether, a region maps inside or outside another region?, 13. What can you say about the existence of bilinear transformations from, a triangle to a square? Triangle to a circle? Triangle to a straight line?, Square to a square?, 14. From (3.16) we see that a bilinear mapping from the real line to the, unit circle is uniquely determined by finding the point that maps onto, the origin and one other point. Does this contradict the fact that three, points determine a bilinear transformation?
Page 93 :
3.2 Linear Fractional Transformations, , 83, , 15. Are there any one-to-one mappings of the extended plane onto itself,, other than bilinear transformations?, Exercises 3.28., 1. Find the cross ratio of the four roots of i1/4 and 11/4 ., 2. Find a bilinear transformation mapping the points, (a) z1 = 0, z2 = i, z3 = −i onto w1 = i, w2 = −i, w3 = 0, (b) z1 = 1, z2 = i, z3 = ∞ onto w1 = i, w2 = −1, w3 = 0, (c) z1 = 2, z2 = −1, z3 = −i onto w1 = ∞, w2 = −1, w3 = −i, (d) z1 = 2, z2 = ∞, z3 = i onto w1 = ∞, w2 = −1, w3 = 1., 3. Using the invariance property of the cross-ratio, find a bilinear transformation f in each of the following cases:, (a) {1, i, −1} onto {1, 0, i}, (b) {∞, i, 0} onto {0, i, ∞}, (c) {−i, −2 + i, 3i} onto {4, 1 + 3i, −2}, (d) {0, 1, ∞} onto {−i, 1, i}., 4. Under the transformation w = iz/(z − 1), find the image of, (a) the closed unit disk |z| ≤ 1., (b) the closed right half-plane Re z ≥ 0., (c) the closed upper half-plane Im z ≥ 0., (d) the open infinite sector π/4 < Arg z < π/2., 5. Under the transformation w = (z − 1)/(z + 1), find the image of, (b) |z| ≤ r (r > 1), (a) |z| ≤ r < 1, (c) Im z > 1, (d) Im z > Re z., 6. Find conditions for a bilinear transformation to carry a straight line in, the z-plane onto the unit circle |w| = 1., 7. Let w be a bilinear transformation from the real line onto the unit circle., If z1 is mapped onto w1 , show that z 1 is mapped onto 1/w1 ., 8. Let w be a bilinear transformation from the unit circle onto itself. If z1, is mapped onto w1 , show that 1/z 1 is mapped onto 1/w1 ., 9. Prove that the cross ratio of four distinct points is real if and only if the, four points lie on a circle or on a straight line., 10. If z1 and z2 are distinct fixed points of a bilinear transformation w =, T (z), show that the transformation may be expressed as, z − z1, w − z1, =K, ,, w − z2, z − z2, where K is a complex constant., 11. If z1 ∈ C and z2 = ∞ are two fixed points of a bilinear transformation, w = T (z), show that the transformation may be expressed as, w − z1 = K(z − z1 ),, for some complex constant K.
Page 95 :
3.3 Other Mappings, , 85, , 26. Determine the inverse point of 1 + i with respect to the circle, |z + 1 − 2i| = 2., , 3.3 Other Mappings, In this section we examine the mapping properties of functions other than, bilinear transformations. Consider the function w = z 2 . Separating this into, its real and imaginary parts we obtain, w = u + iv = (x + iy)2 = x2 − y 2 + i(2xy)., This function maps the point (a, a) in the z plane onto the point (0, 2a2 ) in, the w plane. That is, the ray y = x, with x > 0, is mapped onto the ray, (0, v), with v > 0; and the ray y = x, x < 0, is also mapped onto the ray, (0, v), v > 0. In other words, the line y = x is twice mapped onto the ray, (0, v), v ≥ 0 (see Figure 3.6). Observe that, unlike bilinear transformations,, the function w = z 2 is not one-to-one., , Figure 3.6. Image of the line y = x under w = z 2, , In general the point (x, y) = (x, mx) is mapped onto the point, (u, v) = ((1 − m2 )x2 , 2mx2 )., Since, , v, 2m, =, (m = 1),, u, 1 − m2, the straight line y = mx is mapped twice onto the ray, , , 2m, u,, v=, 1 − m2, where u assumes all the nonnegative real numbers if |m| < 1 and all nonpositive real numbers if |m| > 1. Note that the region 0 < Arg z < π/4 is mapped, onto the first quadrant, 0 < Arg z < π/2 .
Page 96 :
86, , 3 Bilinear Transformations and Mappings, , To determine the preimage of a circle for the function w = z 2 , we write, u2 + v 2 = (x2 − y 2 )2 + (2xy)2 = (x2 + y 2 )2 ., Hence a circle in the z plane, having its center at the origin and radius r, is, mapped onto a circle in the w plane having the center at the origin and radius, r2 . It is, perhaps, more natural to discuss this function in terms of its polar, coordinate representation. We have, w = z 2 = [r(cos θ + i sin θ)]2 = r2 (cos 2θ + i sin 2θ)., Thus a point with polar coordinates (r, θ) in the z plane is mapped onto the, point with polar coordinates (r2 , 2θ) in the w plane, a point whose distance, from the origin is squared and whose argument is doubled. For instance, we, have, •, , the function f (z) = z 2 maps the right half-plane, {z : Re z > 0} := {z = reiθ : 0 < r < ∞, |θ| < π/2}, , •, , onto the slit plane C \(−∞, 0]., for each fixed θ0 with 0 < θ0 ≤ π/2, the function f (z) = z 2 maps the, sector |Arg z| < θ0 onto the sector |Arg w| < 2θ0 ., , The function w = z and w = z 2 both map the unit circle onto itself; but, these mappings can no more be considered identical than can the real-valued, functions y = x and y = x2 , both mapping the closed interval [0, 1] onto, itself. The function w = z 2 describes the unit circle twice; in fact, it maps any, semicircle centered at the origin onto a circle., We should not leave the function w = z 2 without comparing it with its, real-valued counterpart, the parabola y = x2 . The line y = c in the z plane is, transformed into u = x2 − c2 and v = 2xc, from which we obtain, u=, , v 2, v2, − c2 = 2 − c2 ., 2c, 4c, , Hence the horizontal line y = c = 0 is mapped onto the parabola, u=, , v2, − c2 ., 4c2, , If c = 0, the parabola degenerates into the ray (u, 0), u ≥ 0. In a similar, fashion, we can show that the vertical line x = a = 0 maps onto the parabola, (see Figure 3.7), , 2, v, 2, −, a, ., u=−, 4a2, For n a positive integer, the function, w = z n = rn (cos nθ + i sin nθ)
Page 97 :
3.3 Other Mappings, , 87, , Figure 3.7. Image of lines parallel to coordinate axes under w = z 2, , maps the point whose distance from the origin is r onto points of distance rn ,, and points whose argument is θ onto points having argument nθ. The function, f (z) = z n maps the arc, (r, θ),, , θ0 ≤ θ < θ0 + (2π/n),, , onto a circle of radius rn centered at the origin (see Figure 3.8). Moreover,, for each fixed θ0 with 0 < θ0 ≤ π/n, we see that the function f (z) = z n, maps the sector {z : |Arg z| < θ0 } onto the sector {w : |Arg w| < nθ0 }., In particular, f (z) = z n maps the sector {z : |Arg z| < π/(2n)} onto the, half-plane {w : Re w > 0}., The function, w = z = x − iy = r(cos θ − i sin θ), is yet another function mapping the unit circle onto itself. It maps the point, (r, θ) onto the point (r, −θ). Thus the image of the unit circle described counterclockwise is the unit circle described clockwise (see Figure 3.9). Note that, the upper half-plane is mapped onto the lower half-plane., While the composition of bilinear transformations is again a bilinear transformation (Exercise 3.28(15)), the sum of bilinear transformations need not, , Figure 3.8. Image of an unbounded sector under w = z n
Page 98 :
88, , 3 Bilinear Transformations and Mappings, , Figure 3.9. Image of a circle under w = z, , be. The last function we will examine in this chapter is, , , 1, 1, z+, ., w=, 2, z, Set z = reiθ . Then, separating this into real and imaginary parts, we have, , , 1, 1, z+, w = u + iv =, 2, z, , , 1, 1, =, r(cos θ + i sin θ) +, 2, r(cos θ + i sin θ), , , , , 1, 1, 1, 1, =, r+, cos θ + i, r−, sin θ., 2, r, 2, r, The unit circle |z| = 1 is mapped onto w = u = cos θ. As θ describes the, interval [0, π], u decreases continuously from 1 to −1; as θ describes the interval, [π, 2π], u describes the interval [−1, 1]. Hence the upper and lower halves of, the unit circle are both mapped onto the closed interval [−1, 1]., It is interesting to note that the points z and 1/z both map onto the same, points under this transformation. Since 1/z lies outside the unit circle if and, only if z lies inside, it suffices to study the mapping properties for |z| > 1., From the relations, , , , , 1, 1, 1, 1, r+, cos θ, v =, r−, sin θ,, (3.23), u=, 2, r, 2, r, we see that, for r > 1,, , , u, 1, 2 (r + 1/r), , 2, , , +, , v, 1, 2 (r − 1/r), , 2, , = cos2 θ + sin2 θ = 1., , That is, the circle |z| = r > 1 is mapped onto an ellipse with major axis, along the u axis (Note also that f (z) maps the circle |z| = 1/r onto the same
Page 99 :
3.3 Other Mappings, , 89, , 1, , Figure 3.10. Image of circles under the mapping w = (z + 1/z)/2, , ellipse). As r increases, the ellipse becomes more circular, and as r decreases, to 1, the ellipse degenerates to the interval [−1, 1] (see Figure 3.10). We next, determine what happens to the ray Arg z = θ. For r ≥ 1, we see from (3.23), that the rays Arg z = 0 and Arg z = π are mapped onto themselves, although, (excluding the point at ∞) only the points (1, 0) and (−1, 0) remain fixed., Similarly, Arg z = π/2 (r > 1) is mapped onto Arg w = π/2 (r > 0) and, Arg z = −π/2 (r > 1) is mapped onto Arg w = −π/2 (r > 0). For all other, values of θ we have, according to (3.23),, &, 2 , 2 ', u 2 v 2, 1, 1, 1, r+, −, =, − r−, = 1,, (3.24), cos θ, cos θ, 4, r, r, which is the equation for a hyperbola. For r > 1, each arc of this hyperbola, is located in the same quadrant as the ray Arg z = θ (see Figure 3.11). To, summarize, the function, , , 1, 1, z+, w=, 2, z, , Figure 3.11. Image of lines under the mapping w = (z + 1/z)/2
Page 100 :
90, , 3 Bilinear Transformations and Mappings, , maps the unit circle onto the closed interval [−1, 1] twice, and all other circles, onto ellipses. It maps both the interior and exterior of the unit circle onto the, extended complex plane, excluding the real interval [−1, 1]. Finally, it maps, rays having constant arguments onto arcs of the hyperbolas., Questions 3.29., 1. For the function w = f (z) = z 2 , how does the image of y = c differ from, that of y = −c?, 2. What is the largest domain in which the function w = f (z) = z 2 is, one-to-one?, 3. Why was it is more convenient to discuss the function w = 12 (z + 1/z), than w = z + 1/z?, 4. When does a ray have constant argument?, 5. What is the largest domain for which the function f (z) = 12 (z + 1/z), will be one-to-one? Is f (z) one-to-one on the exterior domain |z| > 1?, 6. How might the mapping properties of the last two sections be combined?, Exercises 3.30., 1. Show that the function w = z 2 maps the hyperbolas x2 − y 2 = C and, xy = K onto straight lines., 2. Find the image of the region bounded by straight lines x = 1, y = 1 and, x + y = 1 under the mapping. f (z) = z 2 ., 3. Show that w = ((1 + z)/(1 − z))2 maps the disk |z| < 1 onto the plane,, excluding the ray (u, 0), u ≤ 0., 4. Show that w = z/(1−z)2 maps the disk |z| < 1 onto the plane, excluding, the ray (u, 0), u ≤ − 14 ., 5. Show that the function w = z 2 maps the disk |z − 1| ≤ 1 onto the, cardioid R = 2(1 + cos θ), 6. Discuss the mapping properties of w = z −n , n a positive integer., 7. Find a transformation which maps Ωn = {z : 0 < Arg z < π/n} (n ∈ N), onto the unit disk |w| < 1., 8. Find the image of the sector |z| < 1, 0 < Arg z < π/n, for the function, n, 2, z +1, zn + 1, (b) w =, ., (a) w = n, z −1, zn − 1, 9. Find the image of the unit disk |z| ≤ 1 for the function, , , n, , |zk | zk − z, w = f (z) =, ,, zk 1 − z k z, k=1, , where 0 < |zk | < 1 for every k.
Page 101 :
4, Elementary Functions, , Many high school students are puzzled by the following “proof”: Let a = b., Then, a2 = ab, a2 − b2 = ab − b2 , and (a + b)(a − b) = b(a − b)., Dividing by a − b, we have, a + b = b, 2b = b, and 2 = 1., The reader, of course, is not fooled by the invalid division by zero. So let us, produce an absurdity without dividing by zero. Since 1/(−1) = (−1)/1, we, take square roots to obtain, , , √ √, √, √, (1/ − 1) = (−1/1), 1/ −1 = −1/ 1, and 1/i = i/1., Cross multiplying, we have 12 = i2 or 1 = −1., In this chapter, we will show that 1 does not really equal −1. We will also, see that the complex exponential and trigonometric functions have much in, common, and that a function having a complex exponent must be defined in, terms of a logarithm., , 4.1 The Exponential Function, Recall that the real-valued function f (x) = ex has the following properties:, 1., 2., 3., 4., , ex, ex, ex, ex, , is continuous on R, ex > 0, e−x = 1/ex > 0,, → +∞ as x → +∞, e−x → 0 as x → +∞,, is equal to its derivative,, has the power series expansion, ex = 1 + x +, , x3, x2, +, + · · · for x ∈ R,, 2!, 3!
Page 102 :
92, , 4 Elementary Functions, , 5. the rule of exponents,, ex1 ex2 = ex1 +x2 for x1 , x2 ∈ R., The function f (x) = ex maps the set of real numbers one-to-one onto the set, of positive reals. Therefore, it has a continuously strictly increasing inverse, function called natural logarithm:, ln : R+ → R;, , i.e., x = ln y ⇐⇒ y = ex ., , Now we raise the following., Problem 4.1. Can we extend the definition of the real exponential function, to the complex case? If so, what properties remain the same?, In defining the complex-valued function, w = f (z) = ez = ex+iy ,, we would like to preserve the important properties of the corresponding realvalued function. If the rule of the exponents is to hold, we must have, ez = ex+iy = ex eiy ., It remains to give a “reasonable” definition for eiy ., If we could expand eiy in a power series similar to that of ex , we would, have, eiy = 1 + iy +, , (iy)3, (iy)4, (iy)5, (iy)2, +, +, +, + ··· ., 2!, 3!, 4!, 5!, , Separating (4.1) into its real and imaginary parts, we would obtain, , , , , y4, y3, y5, y2, +, + ··· + i y −, +, + ··· ., eiy = 1 −, 2!, 4!, 3!, 5!, , (4.1), , (4.2), , The power series expansion in (4.2) represents the functions cos y and sin y,, respectively. This leads to the following definition:, eiy = cos y + i sin y, , (y real)., , (4.3), , We emphasize that (4.3) is a definition, and that the above argument was, introduced only to make this definition seem plausible. In Chapter 8, we will, formally prove the validity of the complex power series expansion, thus justifying our definition. The familiar De Moivre law,, (cos y + i sin y)n = cos ny + i sin ny,, may now be expressed as (eiy )n = einy . Note that, (, |eiy | = cos2 y + sin2 y = 1, for any real number y.
Page 104 :
94, , 4 Elementary Functions, , ez e−z = e0 = 1., Consequently, the inverse of ez is e−z ., Thus, most of the important properties of ex are preserved for ez . There, is, however, a notable exception. The function ez is not one-to-one. In fact for, any complex number z,, ez+2πi = ez e2πi = ez (cos 2π + i sin 2π) = ez ., Next, suppose ez = ex+iy = 1. Then,, ex cos y = 1 and ex sin y = 0., Since ex = 0, the second relation gives, sin y = 0;, , that is, y = nπ, n ∈ Z., , But if we substitute y = nπ into the first equation, we get, ex cos nπ = 1,, so that n must be an even integer and in this case, x must be equal to 0., Hence, z is an integral multiple of 2πi. That is,, ez = 1 ⇐⇒ z = 2kπi,, , k ∈ Z., , To summarize the above discussion we need to introduce the definition of, periodic function. A function f : C → C is called periodic if there exists a, complex number such that f (z + ω) = f (z) for all z ∈ C. The number ω is, then called a “period” of f ., Theorem 4.3. The exponential function f (z) = ez is periodic with the pure, imaginary period 2πi. That is, ez+2πi = ez for all z ∈ C., We have ez+2kπi = ez for k ∈ Z. In view of this, for any two complex, numbers z1 and z2 for which ez1 = ez2 , we have ez1 −z2 = 1. Consequently,, z1 − z2 = 2kπi, k is an integer. Hence, we have, Theorem 4.4. The equality ez1 = ez2 , for z1 , z2 ∈ C, holds if and only if, z1 = z2 + 2kπi for some k ∈ Z., If the exponential function ez assumes a value once, it must—by its, periodicity—assume the value infinitely many times. We now show that ez, assumes every finite, nonzero complex number infinitely often. If ez = a + ib,, a and b both not 0, then, ex cos y = a,, Squaring both terms in (4.5), we obtain, , ex sin y = b., , (4.5)
Page 105 :
4.1 The Exponential Function, , 95, , e2x (cos2 y + sin2 y) = e2x = a2 + b2 ., Since the logarithm is well defined for positive real numbers, we have, x=, , 1, ln(a2 + b2 )., 2, , When a = 0, we divide the second expression by the first in (4.5) to obtain, , b, b, ., tan y = , i.e., y = tan−1, a, a, Hence, , 1, z = ln(a2 + b2 ) + i tan−1, 2, , , b, a, , is a solution to the equation ez = a + bi. If one of the values of tan−1 (b/a), is y0 , then y0 + 2kπ, for any integer k, must also be a value. If a = 0, from, ez = 0 + ib = ib, it follows that, ex = |b|, y = arg(ez ) = arg(ib),, and therefore, π, , ⎧, ⎨ ln |b| + i, + 2kπ if b > 0, , 2, z=, ,, ⎩ ln |b| + i − π + 2kπ if b < 0, 2, , k ∈ Z., , Example 4.5. Let ez = 5 − 5i. Then,, , , −5, 1, ln[52 + (−5)2 ] + i tan−1, 2, 5, , , π, ln 50, + i − + 2kπ , k ∈ Z., =, 2, 4, , z=, , Suppose ez = −5 + 5i. Then,, , , 5, 1, ln[(−5)2 + 52 ] + i tan−1, 2, −5, , , 3π, ln 50, +i, + 2kπ , k ∈ N., =, 2, 4, , z=, , Note that, tan, , −1, , , , −b, a, , , , −1, , = tan, , , , b, −a, , , ., , •, , The definition of the exponential in terms of the trigonometric functions, suggests that the process may be reversed. From (4.3) we obtain
Page 106 :
96, , 4 Elementary Functions, , e−iy = cos(−y) + i sin(−y) = cos y − i sin y., , (4.6), , Subtracting or adding (4.3) and (4.6) leads to, sin y =, , eiy − e−iy, ,, 2i, , cos y =, , eiy + e−iy, ., 2, , It therefore seems natural to define the complex trigonometric functions by, sin z =, , eiz − e−iz, ,, 2i, , cos z =, , eiz + e−iz, ., 2, , (4.7), , Replacing z by −z shows that, sin z = − sin(−z),, , cos z = cos(−z)., , Note also that ez = ez . Having extended the definition of sin x and cos x, it, is of interest to note that these complex trigonometric functions do not have, any additional zeros. More precisely, we see that, sin z = 0 ⇐⇒ z = kπ for some k ∈ Z., Indeed, eiz − e−iz, =0, 2i, ⇐⇒ eiz = e−iz, ⇐⇒ iz − (−iz) = 2kπi for some k ∈ Z, ⇐⇒ z = kπ for some k ∈ Z,, , sin z = 0 ⇐⇒, , as claimed. A similar argument shows that, , , 1, π for some k ∈ Z., cos z = 0 ⇐⇒ z = k +, 2, Also, this result follows from the former, because cos z = sin(z + π/2). In fact,, as, π, π, ei(z+ 2 ) = eiz eiπ/2 = ieiz and e−i(z+ 2 ) = −ie−iz ,, (4.7) gives, , ieiz − (−ie−iz ), eiz + e−iz, π ei(z+π/2) − e−i(z+π/2), sin z +, =, =, =, 2, 2i, 2i, 2, so that, replacing z by z − π2 and z by −z, respectively, gives, , π, , π , − z = cos z., and sin, sin z = cos z −, 2, 2, Similarly, we have sin 2z = 2 sin z cos z because
Page 107 :
4.1 The Exponential Function, , , 2 sin z cos z = 2, , eiz − e−iz, 2i, , , , eiz + e−iz, 2, , , =, , 97, , e2iz − e−2iz, = sin 2z., 2i, , The remaining trigonometric functions are defined by the usual relations, tan z =, , sin z, ,, cos z, , cot z =, , cos z, ,, sin z, , sec z =, , 1, ,, cos z, , csc z =, , 1, ., sin z, , With these definitions, most of the familiar real-valued trigonometric properties can be extended to the complex plane. Now for the exception. The, real-valued sine and cosine functions are bounded by 1. However, neither sin z, nor cos z is bounded in the complex plane. From the triangle inequality, we, have, | |e−iz | − |eiz | |, |ey − e−y |, eiz − e−iz, ≥, =, ., | sin z| =, 2i, 2, 2, As z approaches infinity along the ray Arg z = π/2 or Arg z = −π/2, the expression on the right grows arbitrarily large, showing that sin z is unbounded., Similarly,, eiz + e−iz, |ey − e−y |, | cos z| =, ≥, ,, 2, 2, and | cos z| also approaches ∞ as z approaches ∞ along the ray Arg z = ±π/2., In fact, to show that sin z and cos z are not bounded in C, it suffices to observe, that, e−y + ey, e−y − ey, and cos(iy) =, sin(iy) =, 2i, 2, showing that each of | sin(iy)| and cos(iy) is large whenever y is large., The identities in (4.7) may be used to find solutions for equations involving, the trigonometric functions., Example 4.6. Let us find all the complex numbers for which cos z = 2. To, do this, by the definition of cos z, we must have, (eiz + e−iz )/2 = 2,, which leads to e2iz − 4eiz + 1 = 0, a quadratic in eiz . Solving this for eiz , we, obtain, √, √, 4 ± 16 − 4, iz, = 2 ± 3., e =, 2, √, iz, i(x+iy), −y, But e = e, = e (cos x + i sin x) = 2 ± 3, which gives the relations, √, e−y cos x = 2 ± 3, e−y sin x = 0., The second relation shows that x = nπ, n ∈ Z; and the first shows that n, must be even, so that the last relation reduces to, √, e−y = 2 ± 3.
Page 108 :
98, , 4 Elementary Functions, , This gives y = − ln(2 ±, , √, , 3). Hence cos z = 2 if and only if, √, √, z = 2kπ − i ln(2 ± 3) = 2kπ ± i ln(2 + 3), k ∈ Z., , We leave it as an exercise for the reader to show that both sin z and cos z, assume every value in the complex plane., •, Finally, as in the real case, we define the hyperbolic sine and hyperbolic, cosine functions by the formulas,, sinh z =, , ez − e−z, ,, 2, , cosh z =, , ez + e−z, ,, 2, , z ∈ C., , (4.8), , As an immediate consequence of (4.8), we have the relations, sinh z = −i sin iz,, , cosh z = cos iz., , Observe that, sinh(−z) = − sinh z and cosh(−z) = cosh z., Note also that both sinh z and cosh z are periodic, with period 2πi. We may, also define, sech z =, , 1, 1, sinh z, 1, , csch z =, , tanh z =, , coth z =, ., cosh z, sinh z, cosh z, tanh z, , Questions 4.7., 1., 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., 12., 13., 14., 15., , For what functions f (z) will ef (z) be periodic?, What is the largest region in which ez is one-to-one?, What is the largest region in which ez is bounded?, What is the largest region in which sin z is bounded?, When does ef (z) = ef (z) ? Does sin(iz) = sin z? Does cos(iz) = cos z?, When does ef (z) = ef (z) ? Does sin(iz) = sin z? Does cos(iz) = cos z?, When does cos z1 = cos z2 ? When does cos z1 + cos z2 = 0? When does, sin z1 = sin z2 ? When does sin z1 + sin z2 = 0? When does ez1 + ez2 = 0?, Are there any trigonometric identities, valid for real variables, that are, not valid in the complex plane?, How do | sin z| and sin |z| compare?, How do | sin z| and | sinh z| compare?, What happens to ez as z → ∞ along different rays? What about ez + z?, Are the zero sets of sin z in C, and sin x in R the same?, Are the zero sets of cos z in C, and cos x in R the same?, Does the equation tan z = i have a solution in C?, For what values of z is | sin z| ≤ 1?
Page 109 :
4.1 The Exponential Function, , 99, , Exercises 4.8., 1. Find all values of z for which, 2, z, (b) ez = 1, (c) ee = 1., (a) e3z = 1, 2. Show that all the zeros of sin z and cos z are real., 3. (a) Show that both sin z and cos z are unbounded on the ray Arg z = θ,, 0 < |θ| < π., (b) Show that sin z is bounded only on sets contained in a horizontal, strip., 4. For |z| = r, prove that, (a) e−r ≤ |ez | ≤ er, (b) e−rn ≤ |ezn | ≤ ern , n a positive integer., When will equality hold?, 5. Prove the following identities:, (a) sin2 z + cos2 z = 1, (b) sin(z1 + z2 ) = sin z1 cos z2 + cos z1 sin z2, (c) cos(z1 + z2 ) = cos z1 cos z2 − sin z1 sin z2 ., Note: From (a) it appears that both sin z and cos z are bounded. But, we have already shown that this is not the case!, 6. (a) Separate e1/z , z = 0 into its real and imaginary parts., (b) Show that |e1/z | is bounded in the region |z| ≥ , > 0., 7. (a) Prove that eiz is periodic, with period 2π., (b) For an arbitrary nonzero complex number a, show that eaz is periodic, and find its period., 8. Prove the following inequalities:, 2, 2, 2, (a) |ez + ez | ≤ ex + ex −y, 2, (b) |eiz + eiz | ≤ e−y + e−2xy, 2, (c) | sin z| + | cos z|2 ≥ 1., 9. Prove the following hyperbolic identities:1, (a) cosh2 z − sinh2 z = 1, (b) sinh(z1 ± z2 ) = sinh z1 cosh z2 ± cosh z1 sinh z2, (c) cosh(z1 ± z2 ) = cosh z1 cosh z2 ± sinh z1 sinh z2, (d) sinh z = sinh x cos y + i cosh x sin y, (e) cosh z = cosh x cos y + i sinh x sin y, (f) | sinh z|2 = sinh2 x + cos2 y, (g) | cosh z|2 = sinh2 x + cos2 y., 10. Show that, (a) | sin z|2 = sin2 x + sinh2 y, (b) | cos z|2 = cos2 x + sinh2 y., 11. Prove that tanh z = (sinh z)/(cosh z) is periodic, with period πi., , 1, , The hyperbolic identity (a) clarifies somewhat the adjective “hyperbolic” if one, recalls that x2 − y 2 = 1 is the equation of a hyperbola in R2 .
Page 110 :
100, , 4 Elementary Functions, , 4.2 Mapping Properties, The real and imaginary parts of the nonzero complex number z = x + iy are, “equally” important in determining its position in the plane. We have, , y, |z| = x2 + y 2 and tan(arg z) = ., x, The real and imaginary parts of z = x + iy play independent roles, however,, in determining the position of the point ez in the w plane. Separating the, function w = ez into its real and imaginary components, we have, w = ez = u + iv = ex eiy ,, from which we obtain, |ez | = ex ,, , tan(arg(ez )) =, , v, = tan y., u, , These relations show that the modulus of ez depends only on the real part of, z, while the argument of ez depends only on the imaginary part of z. Indeed,, as ez = ex eiy , one has, arg(ez ) = y + 2kπ, k ∈ Z., It will therefore be of some interest to determine the image of the lines parallel, to our coordinate axes, but first we will make use of the periodicity of the, exponential. We have already seen that, e2kπi = 1, ez ew = ez+w and ez+2kπi = ez for every z ∈ C and k ∈ Z,, so that the points x0 + i(y0 + 2kπ) have the same image for every integer k., Hence we may examine the mapping properties by restricting ourselves to the, infinite strip −π < Im z ≤ π. Whatever occurs in this strip will also occur in, the strip −π + 2kπ < Im z ≤ π + 2kπ. With this restriction, Arg (ez ) = y,, −π < y ≤ π. We have the following:, •, , •, , •, , Since ez has constant modulus, all the points on the line x = x0 are, mapped onto the points equidistant from the origin. In particular, the, line segment x = x0 , −π < y ≤ π, is mapped one-to-one onto the circle, in the w plane having center at the origin and radius ex0 . As y increases, from −π to π, the circle is described in a counterclockwise direction., Since |ez | = ex > 1 if and only if x > 0, the semi-infinite-strip, {z : Re z > 0, −π < Im z ≤ π} is mapped one-to-one onto {w : |w| >, 1}, while the strip {z : Re z < 0, −π < Im z ≤ π} is mapped onto the, punctured unit disk {w : 0 < |w| < 1} (see Figure 4.1)., As |ez | = ex < 1 if and only if x < 0, the semi-infinite strip, {z : Re z < 0, 0 ≤ Im z ≤ π}, is mapped one-to-one onto the upper semi-disk {w : Im w ≥ 0, |w| < 1}, excluding the origin.
Page 111 :
4.2 Mapping Properties, , 101, , Figure 4.1. Image of line segments parallel to coordinate axes under ez, , As already noted, Re z plays no role in determining the argument of ez ., Hence the points with identical imaginary parts will map onto the points, having the same argument. For the line y = y0 , −π < y0 ≤ π, we have, w = ez = ex+iy0 = ex (cos y0 + i sin y0 )., •, , Since ex describes the positive reals, the line y = y0 is mapped one-toone onto the ray Arg w = y0 . Therefore, the infinite strip, {z : 0 < Im z < π}, is mapped one-to-one onto the upper half-plane {z : Im z > 0}, while, the strip, {z : −π < Im z < 0}, is mapped onto the lower half-plane {z : Im z < 0} (see Figure 4.2)., , Figure 4.2. Image of lines parallel to the real axis under ez, , •, , Note that the x axis, y = 0, is mapped onto the positive real axis and, the line y = π is mapped onto the negative real axis. Hence, under the, exponential function ez , the strip, {z : −π < Im z ≤ π}, is mapped one-to-one onto the punctured w plane, C \ {0}.
Page 112 :
102, , 4 Elementary Functions, , Figure 4.3. Image of a rectangle under ez, , We can combine the two previous mappings to determine the image of the, rectangles for the function w = ez . Writing the image in the polar form, we, have the rectangle, {z : A ≤ x ≤ B, −π < C ≤ y ≤ D ≤ π}, being mapped onto the region, {Reiθ : eA ≤ R ≤ eB , C ≤ θ ≤ D},, bounded by arcs and rays (see Figure 4.3)., Next consider a straight line not parallel to either of the coordinate axes., The image of this line will have neither constant modulus nor constant argument, yet it must grow arbitrarily large as x grows arbitrarily large, and, must make a complete revolution each time y increases by 2π, thus producing, a spiraling effect. If y = mx + b, m = 0, then, w = ez = ex+i(mx+b) ., Hence |ez | = ex and arg(ez ) = mx + b + 2kπ, k an integer. In polar form, we, may write w = Reiθ , with, , R = |ez | = ex, (4.9), θ = Arg (ez ) = mx + b + 2kπ,, where k = k(x) is an integer chosen so that θ always satisfies the inequality, −π < θ ≤ π. Since x describes the set of real numbers, k must describe the, set of integers. Eliminating x from the relations in (4.9), we obtain, R = e(θ−b−2kπ)/m = e−b/m e(θ−2kπ)/m ., Letting α = θ − 2kπ in (4.10), we have, , (4.10)
Page 113 :
4.2 Mapping Properties, , 103, , Figure 4.4. Logarithmic spiral, , R = Keα/m ,, , (4.11), , where K is a positive constant and α describes the set of real numbers., Equation (4.11) represents what is known as a logarithmic spiral. In Figure 4.4 we show the image of one segment of a line, and in Figure 4.5 we show, a more complete picture., , Figure 4.5., , Since the argument of iz and the argument of z differ by π/2, we expect, the function w = eiz to maps lines parallel to the y axis (x axis) onto the, same kind of figure as the function w = ez maps lines parallel to the x axis, (y axis). Setting, w = eiz = ei(x+iy) = e−y+ix ,, we see that, , |eiz | = e−y and arg(eiz ) = x + 2kπ., , Hence the line segment −π < x ≤ π, y = y0 , is mapped onto the circle having, center at the origin and radius e−y0 . The semi-infinite strip,, {z : −π < x ≤ π, y > 0},
Page 114 :
104, , 4 Elementary Functions, , Figure 4.6. Image of the line x = a under eiz, , is mapped onto the interior of the punctured unit disk, while the strip, {z : −π < x ≤ π, y < 0},, is mapped onto its exterior (see Figure 4.6). Also, the line x = x0 , −π < x0 ≤, π, is mapped onto the ray Arg w = x0 ., We will use these mapping properties of the exponential to determine those, of the trigonometric functions. We will now discuss the complex mapping w =, cos z of the z-plane onto the w-plane. As was the case with the exponential,, we wish to restrict cos z to a region where the function is one-to-one. Because, cos z is periodic with a real period of 2π, this function assumes all values in, any infinite vertical strip {z : α < Re z ≤ α + 2π}. Therefore, it suffices to, study the mapping w = cos z on the strip where α is fixed to be −π. That, is on the strip {z : −π < Re z ≤ π}. Note that cos (π/2) = 0 = cos (−π/2), showing that cos z is not one-to-one on this region. Moreover, cos z is an even, function; that is, cos z = cos(−z). This means that the points in the first and, fourth quadrants (second and third quadrants) have identical images. Also, it, follows that the image of the strip {z : −π < Re z < 0} is the same as that, of the strip {z : 0 < Re z < π}, under w = cos z. Hence the image of any set, contained in the semi-infinite strip, {z : −π < Re z ≤ π, Im z > 0}, will be duplicated in any semi-infinite strip of the form, {z : (k − 1)π < Re z ≤ (k + 1)π, ±Im z > 0}., Since cos z is real if and only if z is real, it suffices to consider the mapping, defined in the region, {z : −π < Re z ≤ π, Im z > 0} ∪ {z : 0 ≤ Re z ≤ π, Im z = 0},
Page 115 :
4.2 Mapping Properties, , 105, , where the function w = cos z is one-to-one., Recall that the function w = (1/2)(z + 1/z) maps circles onto ellipses and, rays onto arcs of hyperbolas. We may view the transformation, , , 1, 1 iz, w = cos z =, e + iz, 2, e, as successive mappings from the z plane to the ζ plane and the ζ plane to the, w plane, where, ζ = eiz = e−y eix ., For any y0 > 0, the line segment −π < x ≤ π, y = y0 , in the z plane is, mapped onto the circle |ζ| = e−y0 in the ζ plane. Then the function, , , !, 1, 1 −y0 ix, 1, e e + ey0 e−ix, ζ+, :=, w=, 2, ζ, 2, maps the circle |ζ| = e−y0 in the ζ plane onto the ellipse, , , u, 1 −y0, + ey0 ), 2 (e, , 2, , , +, , 2, , v, 1 −y0, 2 (e, , − ey0 ), , =1, , in the w plane. Hence, w = cos z maps the line segment −π < x ≤ π, y0 > 0, onto an ellipse (see Figure 4.7)., , Figure 4.7. Image of the line segment −π < x ≤ π, y0 > 0 under cos z, , Similarly, for y > 0, the half-line {z = x0 +iy : y > 0} (where x0 ∈ (−π, π), is fixed), is mapped onto the line segment, Arg ζ = Arg (e−y eix0 ) = x0 ,, , 0 < |ζ| < 1,, , which, in turn, is mapped onto an arc of the hyperbola (see Figure 4.8), , , u, cos x0, , , , 2, −, , v, sin x0, , 2, = 1.
Page 116 :
106, , 4 Elementary Functions, , Figure 4.8. Illustration for mapping properties of cos z, , Remark 4.9. For y > 0, the above mappings are not valid for the half-lines, x = 0, x = π, or x = ±π/2. If x = 0,, cos z =, , 1 −y, (e + ey ), 2, , is mapped onto the real interval u > 1, while the half-line x = π gives, 1, cos(π + iy) = − (e−y + ey ), 2, so that under w = cos z, x = π is mapped onto the real interval u < −1., Similarly, for y > 0, as, cos, , π, , , i, + iy = (e−y − ey ), 2, 2, , and, , π, , i, cos − + iy = − (e−y − ey ),, 2, 2, the half-line x = π/2 is mapped onto the negative imaginary axis, and the, half-line x = −π/2 is mapped onto the positive imaginary axis. Finally, as, cos(x + i0) =, , eix + e−ix, ,, 2, , the interval 0 ≤ x ≤ π, y = 0, is mapped onto the real interval −1 ≤ u ≤ 1,, v = 0., •, The identity, , π , sin z = cos z −, 2, , (4.12)
Page 117 :
4.2 Mapping Properties, , 107, , enables us to deduce mapping properties of sin z from those of cos z. Equation, (4.12) shows that we may view the transformation w = sin z as the translation, from the z plane to the ζ plane, where ζ = z − π/2, followed by the mapping, w = cos ζ. Thus the function, w = sin z, maps points in the region −π/2 < Re z ≤ 3π/2 in the same manner as, w = cos z, maps the point in the region −π < Re z ≤ π., For instance, the function w = cos z maps the line segment, −π < x ≤ π, y = 1,, onto the ellipse, , u2, v2, +, = 1., 1, 1, 2, 2, 4 (e + 1/e), 4 (e − 1/e), , The function w = sin z maps the line segment −π/2 < x ≤ 3π/2, y = 1, onto, the same ellipse., Finally, we remark that the relationship between the complex trigonometric and hyperbolic functions, for example,, cos(iz) = cosh z and sin(iz) = i sinh z,, allow us to discuss the action of hyperbolic functions as complex mappings., Questions 4.10., 1. What kind of function might map the complex plane, excluding the, origin, onto the strip {w : −π < Im w ≤ π}?, 2. Given a point in the z plane, does there always exist a neighborhood of, that point in which the function ez is one-to-one?, 3. How would you describe the behavior of ez as z approaches ∞?, 4. For the function ez , how does the area of a rectangle compare with the, area of its image?, 5. For the function cos z, how does the area of a rectangle compare with, the area of its image?, 6. For the function ez , how does the slope of a straight line affect the, logarithmic spiral onto which it is mapped?, 7. What functions, other than ez , are never zero in the plane?, 8. What is the largest region in which sin z is one-to-one?, 9. Given a point in the z plane, does there always exist a neighborhood of, that point in which the function sin z is one-to-one?, 10. What are the differences between the functions w = cos(z − π/2) and, w = cos z − π/2?
Page 118 :
108, , 4 Elementary Functions, , 11. What is the image of the infinite strip {z : 0 ≤ Im z ≤ π} under the, mapping w = ez ?, 12. What is the image of the infinite strip {z : 0 ≤ Im z ≤ π/2} under the, mapping w = ez ?, 13. What is the image of the disk {z : |z| ≤ π} under the mapping w = ez ?, Exercises 4.11., 1. Find the image of the following sets under the transformation w = ez ,, and sketch:, (a) −5 ≤ x ≤ 5, y = π/4, (b) x = 3, −π/2 < y < π/2, (c) −2 < x < 1, 0 < y < π, (d) x < 1, −π/3 < y < 2π/3., 2. Find the image of the region 0 ≤ x ≤ π, y ≥ 0, for the transformation, (b) w = ieiz, (c) w = ie−iz ., (a) w = eiz, 3. Find the images of the straight lines for the transformation w = ecz , c, a complex constant., 4. Show that the image of the disk |z| ≤ 1 under the transformation w = ez, is contained in the annulus 1/e ≤ |w| ≤ e., 5. Show that the image of the disk |z| ≤ 1 under the transformations, w = cos z and w = sin z are contained in the disk |w| ≤ (e2 + 1)/2e., 6. Find the image of the following sets under the transformation w = cos z., (b) − π4 < x < π4 , y = −5, (a) x = π2 , y ≥ 0, (c) 0 ≤ x < π, −2 < y < 2, (d) − π2 < x < π4 , y > 0., , 4.3 The Logarithmic Function, Before defining the logarithm of a complex number, we review some properties, of the real-valued logarithm. For every positive real number x, there exists a, unique real number y such that ey = x. We write y = ln x, and observe that, for x1 , x2 > 0, we have, ln(x1 x2 ) = ln x1 + ln x2 ., The function y = ln x maps the positive real numbers onto the set of reals,, and is the inverse of the function y = ex , which maps the real numbers onto, the positive reals (see Figure 4.9). Since ex is one-to-one, its inverse is also a, one-to-one function., There is a problem in defining the logarithm of a complex number z as, the value w for which, ew = z., We know that for z = 0 this equation has no solution in C because if w = u+iv,, then one has, |ew | = |eu+iv | = eu > 0., Thus, ew never assumes zero in C. Further, the expression
Page 119 :
4.3 The Logarithmic Function, , 109, , Figure 4.9. Mappings of ex and log x for x real, , ew = eu eiv, clearly shows that the range of ew is C \ {0}. The periodicity of the complex, exponential precludes the existence of a unique complex logarithm. For, if, ew = z, then ew+2kπi = z for any integer k. We thus define the logarithm of, a complex number z, denoted by log z, as the set of all values w = log z for, which ew = z. Thus, as in the real case, for z = 0, w = log z ⇐⇒ ew = z;, , or log z ∈ {w : ew = z}., , Since the exponential function never vanishes, there is no logarithm associated with the complex number zero; and since the exponential assumes every, nonzero complex number infinitely often, there are infinitely many values of, the logarithm associated with each nonzero complex number. More precisely,, we have, Proposition 4.12. Given z = 0, the most general solution of ew = z is given, by, w = log z = ln |z| + i(Arg z + 2kπi) := ln |z| + i arg z,, , k ∈ Z., , (4.13), , (Remember that there is no solution to ew = 0)., Proof. Setting z = reiθ (r > 0, θ = Arg z), we conclude from, ew := eu eiv = z = reiθ, that eu = r and eiv = eiθ , or equivalently, u = ln r and v = θ + 2kπ,, , k ∈ Z,, , where ln r is the natural logarithm, to the base e, of a positive real number., Therefore, we have the expression, w = u + iv = log z := ln |z| + i(Arg z + 2kπ),, which has infinitely many values at each point z = 0., , k∈Z
Page 120 :
110, , 4 Elementary Functions, , Remark 4.13. The exponential function ez has one more important special, property. Recall that, ex → +∞ as x → +∞., Is this true if we replace the real x by a complex z? The answer is clearly no!, Indeed, given 0 = z ∈ C, there exists a w such that ew = z. But then, ew+2kπi = z, holds for every k ∈ Z. Hence we can obtain w having arbitrarily large modulus, |w| such that ew = z. As a consequence, we conclude that limw→∞ ew does, not exist. Note that, lim ew = ∞,, , w=u, u→∞, , lim ew = 0, , w=u<0,, u→−∞, , whereas the limit, lim ew = lim eiv, , w=iv, v→∞, , v→∞, , does not exist. (For instance, both vn = nπ and vn = nπ + π/2 approach ∞, , •, as n → ∞ but eivn = (−1)n and eivn = 0)., Since log z is not a uniquely defined function of z, it is appropriate to, introduce the principle value of log z for z = 0. For z = 0,, ln |z| + iArg z, is called the principle value of log z and is denoted by Log z:, Log z = ln |z| + iArg z., Using this we can rewrite (4.13) in the form, log z = Log z + 2kπi,, , k ∈ Z., , We remark that the expression w = log z, z = 0, is our first example of a, multiple-valued function, a relation in which there is more than one image, associated with a complex value. Note that the multiple-valuedness of the, logarithm is related to the many values connected with the argument of a, complex number., Our methods of investigating continuity and other properties for singlevalued functions cannot be used for multiple-valued functions. Fortunately,, a multiple-valued function can quite naturally be replaced by many different, single-valued functions. The nature of multiple-valued function may then be, examined from the point of view of its single-valued counterparts., We define a branch of log z to be any single-valued function log∗ z that, ∗, satisfies the identity elog z = z for all nonzero complex values of z. There are
Page 122 :
112, , 4 Elementary Functions, , or, log(z1 z2 ) = log z1 + log z2, , (mod 2πi)., , When the cut is along the negative real axis, the branch of the logarithm under, consideration is completely determined by specifying one particular value of, the function. For instance, the principal branch is the only one for which, log 1 = 0. The branch for which log 1 = 10πi is given by, log z = Log z + 10πi., Each of the functions w = Log z + 2kπi maps the plane, excluding the origin,, onto the infinite strip (2k−1)π < Im w ≤ (2k+1)π. Recall that the exponential, function maps each strip, (2k − 1)π < Im w ≤ (2k + 1)π, onto the punctured plane. Since the behavior of each function defined in (4.14), is essentially the same, we will—unless otherwise stated—assume k = 0 and, confine ourselves to the principal branch of the logarithm., The function w = Log z is not continuous at any point on the negative, real axis. For any such point may be expressed as, z0 = r0 eπi , r0 > 0,, with Log z0 = ln r0 + iπ. But as the point z0 is approached through values, below the real axis, we have, lim Arg z = −π., , z→z0, , Hence,, Log z → ln r0 − iπ = Log z0 as z → z0, through such values., This does not mean that the logarithm function is not continuous on the, negative real axis. All we have seen is that Log z, the principal branch, is not, continuous at these points. By making our cut along a different ray, we can, find a branch of the logarithm that is continuous for negative real values. For, instance, the single-valued function, w = log z = ln |z| + i arg z, , (−π/2 < arg z ≤ 3π/2), , is continuous at all points on the negative real axis, but not on the ray arg z =, 3π/2., In other words, the logarithm function is continuous for all nonzero complex values in the following sense: Given z0 = 0, there exists a branch for which, limz→z0 log z = log z0 . However, there does not exist a branch for which log z, is continuous for all nonzero complex numbers., In view of (4.15), we can easily determine some mapping properties of the, logarithm function. The image of the circle |z| = r for the function w = Log z, is the line segment u = ln r, −π < v ≤ π (see Figure 4.10). We also have the, ray Arg z = θ mapping onto the line v = θ (see Figure 4.11).
Page 123 :
4.3 The Logarithmic Function, , 113, , Figure 4.10. Image of an annulus region under Log z, , Figure 4.11. Image of segment of rays under Log z, , Questions 4.14., 1. What is the relationship between the argument and the logarithm of a, complex number?, 2. For fixed θ0 , what changes will occur if we define θ0 − π < arg z ≤ θ0 + π, to be the principal value?, 3. What would be the consequences of defining log 0 = ∞?, 4. What is the image of spirals under the function w = log z?, 5. In what regions is log z bounded?, 6. Does log(z1 /z2 ) = log z1 − log z2 ?, 7. Does Log (z1 /z2 ) = Log z1 − Log z2 ?, 8. Does limz→∞ exp (−z 2 ) exist? Does limz→∞ exp (−z 4 ) exist?, 9. Does limz→0 exp (−1/z) exist? Does limz→0 exp (−1/z 2 ) exist? How, about limz→0 exp (−1/z 3 )? Does limz→0 exp(−1/z 4 ) exist?, Exercises 4.15., 1. Find all the values of, (a) log(1 − i), (b) log(3 − 2i), (c) log(x + iy)., 2. For any nonzero complex number z1 and z2 , prove that, Log (z1 z2 ) = Log z1 + Log z2 + 2kπi,
Page 124 :
114, , 3., 4., 5., , 6., , 4 Elementary Functions, , where k = 0, 1, −1. Give examples to show that each value of k is possible., For z = 0, prove that ln |z| ≤ | Log z| ≤ ln |z| + |Arg z|., Let f (z) be defined in a domain D with f (z) = 0 in D. Prove that, Arg f (z) = Im Log f (z) for every point z ∈ D., Find the image of straight lines parallel to the coordinate axes for the, function, (a) w = Log (iz), (b) w = Log (−iz) + 1, (c) w = Log z 2 ., iθ, Set z = re , where 0 < a < r < b and θ ∈ (−π, π). Show that under the, mapping Log z the region Ω = {z : a < |z| < b} \ [−b, −a] is mapped, onto the rectangle (ln a, ln b) × (−π, π)., , 4.4 Complex Exponents, As we have seen in Section 1.3 there are n distinct complex values associated, with z 1/n , z = 0. If we write z = reiθ , then, zk = r1/n ei[(θ+2kπ)/n], is a distinct nth root for k = 0, 1, 2, . . . , n − 1. The values of z 1/n vary as, the argument of z takes on the values θ, θ + 2π, θ + 4π, . . . , θ + 2(n − 1)π., A unique value for z 1/n may be obtained by restricting arg z to a particular, branch. This seems to indicate a link between the nth roots and logarithm of, a nonzero complex number., Indeed, we may define the function z 1/n by, w = z 1/n = e(1/n) log z = e(1/n)(ln |z|+i arg z) ., , (4.16), , This function, like the logarithm function, is multiple-valued. Upon setting, arg z = Arg z + 2kπ, k an integer, we see that (4.16) assumes different values, for k = 0, 1, . . . , n − 1; for any other integer k, one of these n values will be, repeated. More generally, if m and n are positive integers with no common, factors, we define, (z 1/n )m = e(m/n) log z , z = 0., This, too, has n distinct values. We are thus led quite naturally to define z α, for complex values of α by, z α = eα log z ., , (4.17), , If α is not a rational number, then there are infinitely many values associated, with the expression in (4.17). To see this, we first suppose that α is irrational., Then for z = reiθ = 0, we have, z α = eα log z = eα[ln r+i(θ+2kπ)] = rα eiαθ ei(2kπα) .
Page 126 :
116, , 4 Elementary Functions, , equality for each α and β need not hold. In general, z α+β assumes every, value of z α z β , but the converse is not true. For example 51/2+1/2 = 5 but, 51/2 51/2 = ±5. We leave it for the reader to show this containment for α and, β complex numbers., Recall the “proof” in the introduction that 1 = −1, where we made the, false assumption that, , √ √, 1/ − 1 = 1/ −1., Had we used the preceding results, we would have been able to reach only the, much less interesting conclusion that ±1 = ±1., As was the case with the logarithmic function, we may replace multiplevalued functions that have fractional exponents with (single-valued) branches., To illustrate, for z = reiθ (r = 0, −π < θ ≤ π) the principal branch of z 1/2 is, √, w0 = z 1/2 = e(1/2)(ln r+iθ) = rei(θ/2) (−π < θ ≤ π)., Another branch of the function is, √, w1 = z 1/2 = e(1/2)(ln r+i(θ+2π)) = − rei(θ/2), , (π < θ + 2π ≤ 3π)., , Both w0 and w1 are continuous functions, except on the negative real axis., This ray is called a branch cut for both w0 and w1 . Each of these single-valued, functions is called a determination or branch of the multiple-valued function, w = z 1/2 ., We now establish some mapping properties for the functions w0 and w1 ., The punctured plane (z = 0) is mapped by w0 onto the right half-plane,, including the positive imaginary axis, and by w1 onto the left half-plane,, including the negative imaginary axis. These functions also map circles onto, semicircles, excluding the end point (see Figure 4.12)., , Figure 4.12. Mapping properties of square root function, , We may similarly analyze other multiple-valued functions with rational, exponents. The function w = z 1/3 has three branches. We write, √, wk = 3 rei(θ+2kπ)/3 (k = 0, 1, 2; −π < θ ≤ π).
Page 127 :
4.4 Complex Exponents, , 117, , Each of these three single-valued functions is continuous except on the negative real axis, and maps the circle |z| = r onto the arc, |wk | =, , √, 3, , r,, , (2k − 1)π, (2k + 1)π, < Arg wk ≤, ., 3, 3, , The next example illustrates some of the surprising properties of complex, exponents. Consider the function w = 1z . We have, w = 1z = ez log 1 = e2kπiz = e2kπi(x+iy) = e−2kπy e2kπix ., For each integer k, the function w = 1z is defined in the whole plane. If k = 0,, the principal branch of the logarithm, then w ≡ 1. This is what we expect., But consider a different determination of the logarithm, and assume that, k = k0 , k0 > 0. The function w = 1z is then periodic, with period 1/k0 . If z, is a positive integer, then 1z = 1. If z is real, then 1z is a point on the unit, circle. In fact, every interval of the form, x0 −, , 1, 1, < x ≤ x0 +, , x0 fixed,, 2k0, 2k0, , maps one-to-one onto the unit circle. The line segment, −, , 1, 1, <x≤, , y = y0 ,, 2k0, 2k0, , maps onto the circle |w| = e−2k0 πy0 . The line, x = x0 , −, , 1, 1, < x0 ≤, ,, 2k0, 2k0, , maps onto the ray Arg w = 2k0 πx0 . Hence the infinite strip, , 1, 1, z: −, < Re z ≤, 2k0, 2k0, maps onto the plane, excluding the origin. Finally, the upper half-plane is, mapped onto the interior of the punctured unit disk, and lower half-plane, onto its exterior (see Figure 4.13)., We have previously examined the close relation between the exponential, and trigonometric functions. It is not surprising that their inverses also have, much in common. We will show that the inverse trigonometric functions may, be defined in terms of logarithm., Given a complex number z, we wish to find all the complex numbers w, such that z = sin w. If w0 is one such solution, then w0 + 2kπ must also be a, solution. If, z = sin w = (eiw − e−iw )/2,, then
Page 128 :
118, , 4 Elementary Functions, , Figure 4.13. Illustration for mapping properties of 1z, , e2iw − 2izeiw − 1 = 0,, a quadratic in eiw . Solving, we obtain, eiw = iz + (1 − z 2 )1/2 ; i.e., iw = log[iz + (1 − z 2 )1/2 ],, from which we define the multiple-valued function, w = sin−1 z = −i log[iz + (1 − z 2 )1/2 ]., Remark 4.19. The expression (1 − z 2 )1/2 is itself multiple-valued. To compute sin−1 z, we must first find each of the values of (1 − z 2 )1/2 . For each of, •, these values, we must then determine all the logarithms., Example 4.20. To find all possible determinations of sin−1 0, we write, sin−1 0 = −i log 11/2 = −i, , 2kπi, = kπ,, 2, , k ∈ Z., , For the principal branch, k = 0 and sin−1 0 = 0., , •, , Example 4.21. To find all possible determinations of sin−1 i, we write, sin−1 i = −i log(−1 + 21/2 ), , √, √ , = −i ln | − 1 ± 2| + i arg(−1 ± 2), √, √, = arg(−1 ± 2) − i ln | − 1 ± 2|., Given any determination for the square root and logarithm, the real part of, sin−1 i is an integral multiple of π. The imaginary part depend only on the, determination of the square root. Hence
Page 129 :
4.4 Complex Exponents, , sin−1 i = kπ − i ln | − 1 ±, , √, , 2|,, , 119, , k ∈ Z., , A choice of the positive square root and the principal, value for the logarithm, √, gives the specific determination sin−1 i = −i ln( 2 − 1)., We may similarly find the inverses of the remaining trigonometric functions. For z = cos w = (eiw + e−iw )/2, we have, w = cos−1 z = −i log[z + (z 2 − 1)1/2 ]., , •, , Questions 4.22., 1., 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., 12., 13., 14., 15., , Does log z α = α log z if α = 0?, Does 11/2 + 11/2 = 2(11/2 )?, When will a complex number to a complex power be real?, If z α assumes m distinct values and z β assumes n distinct values, what, can we say about z α z β ?, For the complex number α and β, how does z αβ compare with (z α )β ?, For the multiple-valued function w = z 1/2 , why does the origin play, such an important role?, For the function w = z 1/2 , could we have chosen any rays other than, Arg z = π for our branch cut?, How do the functions w = z 1/n and w = z n compare?, )1/n = (z 1/n )m when m and n are integers?, Does (z m√, When is z 2 = z? Is (z 2 )1/2 = z?, In what regions are the inverse trigonometric functions one-to-one?, How can mapping properties of the inverse trigonometric functions be, determined from those of the trigonometric functions?, If | cos z| ≤ 1, then what can you say about z?, If cos z = α, where −1 ≤ α ≤ 1, then what can you say about z?, If log 4 is real, what must be the value of log(4i)? What must be the, value of log(−4i)?, , Exercises 4.23., 1. Find all values for the following expressions., (b) (πi)e, (c) (2i )i, (d) log(1+i)πi ., (a) 5i, 2. For z = 0, α and β complex numbers, show that every value of z αβ is a, value of (z α )β . When is the converse true?, 3. For z = 0 and α irrational, show that θ0 < Arg (z α ) < θ0 + for infinitely, many values of z α , where −π < θ0 < π and > 0., 4. Separate into real and imaginary parts., (a) xx (x real, x = 0), (b) (iy)iy (y real, y = 0), (c) z z (z = 0)., z, 5. For any nonzero complex number a, show that a is either constant or an, unbounded function, depending on the branch chosen for its logarithm.
Page 130 :
120, , 4 Elementary Functions, , 6. Discuss the image of the circle |z| = r for the following multiple-valued, functions., (b) w = z 2/3 ., (a) w = z 1/n , n a positive integer, 7. Prove the following identities:, , , 1 + zi, 1, log, (a) tan−1 z =, 2i, 1 − zi, z+i, 1, log, (b) cot−1 z =, 2i, z−i, , , 1, + (1 − z 2 )1/2, 1, (c) sec−1 z = log, i, z, , , 2, i, +, (z, 1, − 1)1/2, −1, (d) csc z = log, ., i, z, 8. Find all values of, √, 2, −1 1, −1, (b) cos, (c) tan−1 (1 + i) (d) sec−1 i., (a) sin, 2, 2, 9. Determine all values of the following:, √, , (−1), , 2, , ,, , (cos i)i ,, , √, , 21−i ,, , (1 + i), , (1 + i)1+i ,, , (ii )i ,, , 3, , ,, , arg(1 − i),, isin i ,, , (−1)1/3, √, ( 3 + i)i/2 ., , 10. Evaluate the limits, 1/z, , (i) lim (1+z), z→0, , √, (ii) lim, , z→2, , z−2, z−2, , (iii) lim zArg (z)., z→i, , 11. Find all the points of discontinuity of, (i) f (z) = Log (z 2 − 1), (iii) f (z) = Log (z 3 − 1), , (v) f (z) = z 2 + 1, , (ii) f (z) = Arg (z 2 ), (iv) f (z) = Arg (z 3 ), , (vi) f (z) = z 2 − 1.
Page 131 :
5, Analytic Functions, , In this chapter, we will define differentiation for single-valued functions of a, complex variable, and we will see how the derivative of a complex variable, sometimes behaves like the derivative of a real function of one real variable,, and other times it is comparable to the partial derivatives of a real function, of two real variables. We also learn to appreciate the importance of neighborhoods. If we do not require differentiability in a neighborhood, the smoothness, of a function along one path may obscure potential difficulties along some, other route., , 5.1 Cauchy–Riemann Equation, As we have seen before, a function of a complex variable may be separated into, its real and imaginary parts. Writing f (z) = u(x, y) + iv(x, y) it is interesting, to compare properties of f (z) with those of its real-valued components u(x, y), and v(x, y). In the case of continuity, the comparison is quite straight forward., Let g be a function of the two real variables x and y. We say that, lim, , (x,y)→(x0 ,y0 ), , g(x, y) = L, , if for every, , > 0, there exists a δ > 0 such that, , |g(x, y) − L| < whenever 0 < (x − x0 )2 + (y − y0 )2 < δ., , If L = g(x0 , y0 ), then g(x, y) is said to be continuous at (x0 , y0 )., Theorem 5.1. The function f (z) = u(x, y) + iv(x, y) is continuous at a point, z0 = x0 + iy0 if and only if u(x, y) and v(x, y) are both continuous at the point, (x0 , y0 )., Proof. We first suppose f (z) to be continuous at z = z0 . Then for any, , > 0,
Page 132 :
122, , 5 Analytic Functions, , |u(x, y) − u(x0 , y0 )| ≤ |f (z) − f (z0 )| < ,, |v(x, y) − v(x0 , y0 )| ≤ |f (z) − f (z0 )| < ,, whenever, |z − z0 | =, , , , (x − x0 )2 + (y − y0 )2 < δ., , Thus both u(x, y) and v(x, y) are continuous at (x0 , y0 ). Conversely, if u(x, y), and v(x, y) are continuous at (x0 , y0 ), the continuity of f (z) follows from the, inequality (recall that |z| ≤ |x| + |y| for z ∈ C), |f (z) − f (z0 )| ≤ |u(x, y) − u(x0 , y0 )| + |v(x, y) − v(x0 , y0 )|., In view of Theorem 5.1, it is worthwhile to examine more carefully the, notion of the limit of a function of real two variables. Recall that for a function, of one real variable, when there are only two directions to travel, a limit exists, and only if the right- and left-hand limits coincide. There is no analog for a, function of two variables, since infinitely many modes of approach are possible., Example 5.2. Let f (x, y) = xy/(x2 + y 2 ), (x, y) = (0, 0). Since f (x, y) ≡ 0, as (x, y) → (0, 0) along either of the coordinate axes, we have, lim f (0, y) = lim f (x, 0) = 0., , y→0, , x→0, , However, choosing the straight-line path y = mx, we obtain, m, mx2, =, ., x→0 x2 + mx2, 1 + m2, , lim f (x, mx) = lim, , x→0, , Because f (x, y) approaches different values along different straight lines, the, limit at the origin does not exist., •, Example 5.3. Let f (x, y) = x2 y 2 /(x + y 2 )3 , (x, y) = (0, 0). Here,, m2 x4, m2 x, =, lim, = 0,, x→0 (x + m2 x2 )3, x→0 (1 + m2 x)3, , lim f (x, mx) = lim, , x→0, , and f (x, y) approaches 0as (x, y) → (0, 0) along any straight line. But along, the parabola x = my 2 (m = 0),, m2, m2 y 4 y 2, =, ., y→0 (my 2 + y 2 )3, (m2 + 1)3, , lim f (my 2 , y) = lim, , y→0, , Hence, lim(x,y)→(0,0) f (x, y) does not exist., , •, , Example 5.4. Let f (x, y) = (x3 − 2y 3 )/(x2 + y 2 ), (x, y) = (0, 0). We wish, to show that this function does have a limit at origin. Obviously, we can, not try all modes of approach; so it will be necessary to obtain appropriate, inequalities. We have
Page 133 :
5.1 Cauchy–Riemann Equation, , 123, , |x3 − 2y 3 | ≤ |x|3 + 2|y|3 = |x|x2 + 2|y|y 2, , ≤ x2 + y 2 (x2 + 2y 2 ) ≤ 2(x2 + y 2 )3/2 ., , Thus, |f (x, y)| ≤ 2 x2 + y 2 , so that, , |f (x, y)| < whenever x2 + y 2 < /2 = δ., Therefore,, lim, , (x,y)→(0,0), , f (x, y) = 0., , •, , Our purpose in the above examples was not to develop sophisticated methods for evaluating limits, but to emphasize the importance of the path. A, function may be very well-behaved along one route, while impossible to deal, with along another. This phenomenon will produce some surprising results, within the theory of differentiable functions., A function f is said to be differentiable at a point z if, f (z + h) − f (z), h→0, h, lim, , exists. We then write, f (z) = lim, , h→0, , f (z + h) − f (z), ., h, , Note that h approaches 0 through points in the plane, not just along the, real axis or along the line y = mx. For example, the function f (z) = z 2 is, everywhere differentiable because, f (z + h) − f (z), (z + h)2 − z 2, 2zh + h2, = lim, = lim, = 2z., h→0, h→0, h→0, h, h, h, lim, , More generally, if f (z) = z n , n an integer, then f (z) = nz n−1 . From the, identity, , , f (z + h) − f (z), h (h = 0),, f (z + h) − f (z) =, h, we let h → 0 to obtain the familiar real-variable theorem that a differentiable, function is continuous. That is,, , , f (z + h) − f (z), h = f (z) · 0 = 0., lim (f (z + h) − f (z)) = lim, h→0, h→0, h, Therefore, limh→0 f (z + h) = f (z) and so we have, Theorem 5.5. If f is differentiable at a point z ∈ C, then f is continuous, at z.
Page 134 :
124, , 5 Analytic Functions, , , On the other hand, f (z) = |z| = x2 + y 2 is continuous on C but not, differentiable at the origin. Clearly, limz→0 f (z) = f (0) = 0 and, ⎧, 1 for positive real values of h, ⎪, ⎪, ⎪, ⎪, −1, for negative real values of h, ⎪, ⎪, |h| ⎨ −i for values of h on the positive, f (h) − f (0), =, =, imaginary axis, ⎪, h−0, h, ⎪, ⎪, ⎪, i, for, values, of h on the negative, ⎪, ⎪, ⎩, imaginary axis., Hence, f (0) does not exist. This example shows that continuity at a point is, not sufficient for a function f to be differentiable at that point. Also it follows, that each of the functions z, Re z and Im z is continuous on C but nowhere, differentiable., Note the similarity thus far between the derivative of the complex-valued, function f (z) and the real-valued function f (x). In fact, all the formal differentiation rules are the same. We collect the following properties as, Theorem 5.6. For f (z) and g(z) differentiable,, (i) (f (z) + g(z)) = f (z) + g (z),, , , , (ii) , (f (z)g(z)), = f (z)g(z) +f (z)g (z),, f (z), f (z)g(z) − g (z)f (z), (iii), =, when g(z) = 0., g(z), (g(z))2, Suppose g is differentiable at z, and f is differentiable at g(z). If F (z) =, f (g(z)), then, (iv) F (z) = f (g(z))g (z)., Proof. These properties are proved, word for word, as they are for a function, of a real variable., The function f (z) = |z|2 is differentiable only at the origin. In contrast,, g(x) = |x|2 (x ∈ R) is differentiable everywhere in R. Similarly, the function, φ(x) = x is differentiable on R, but the function ψ(z) = x = Re z (z ∈ C) is, nowhere differentiable., Lest the reader be fooled by the apparent similarity between real and, complex derivatives, we now discuss some of the far-reaching consequences, of requiring (f (z + h) − f (z))/h to approach the same value regardless of, the path of h. To this end, properties of the real and imaginary parts of the, differentiable function f (z) = u(x, y)+iv(x, y) will be deduced by specializing, the mode of approach., Suppose first that h approaches 0 along the real axis. Separating into real, and imaginary parts,, u(x + h, y) + iv(x + h, y) − u(x, y) − iv(x, y), f (z + h) − f (z), =, h, h, u(x + h, y) − u(x, y), v(x + h, y) − v(x, y), =, +i, ., h, h
Page 135 :
5.1 Cauchy–Riemann Equation, , 125, , If f is differentiable at z = x + iy, the limits of both expressions on the right, must exist, and we recognize them to be the partial derivatives of u(x, y) and, v(x, y) with respect to x. Hence, f (z) = lim, , h→0, , ∂u, ∂v, ∂f, f (z + h) − f (z), =, +i, :=, ., h, ∂x, ∂x, ∂x, , (5.1), , Next let h approach 0 along the imaginary axis. Then for h = ik, k real, we, have, u(x, y + k) − u(x, y), v(x, y + k) − v(x, y), f (z + ik) − f (z), =, +i, ., ik, ik, ik, Thus, f (z) = lim, , k→0, , 1 ∂u ∂v, ∂v, ∂u, ∂f, f (z + ik) − f (z), =, +, =, −i, = −i . (5.2), ik, i ∂y, ∂y, ∂y, ∂y, ∂y, , But the expressions in (5.1) and (5.2) must be equal. So, f (z) = ux + ivx = vy − iuy ., , (5.3), , Equating real and imaginary parts in (5.3), we obtain, ux = vy ,, , vx = −uy ; or fx = −ify ., , (5.4), , The two equations in (5.4), known as the Cauchy–Riemann equations, furnish, us with a necessary condition for differentiability at a point., Theorem 5.7. If f = u + iv is differentiable at z, then fx , fy exist at z and, satisfy the Cauchy–Riemann equations at z:, fy (z) = ifx (z);, or equivalently, ux = vy and uy = −vx ., For the function f (z) = z = x − iy, we have ux ≡ 1 and vy ≡ −1. Since, the Cauchy–Riemann equations are never satisfied, the function is nowhere, differentiable., To see that the Cauchy–Riemann equations are not a sufficient condition, for differentiability at a point, consider, ⎧ 3, 3, , x + y3, ⎨ x − y3, +, i, if (x, y) = (0, 0),, f (x + iy) = u + iv = x2 + y 2, x2 + y 2, ⎩, 0 if x = y = 0., For this function the corresponding u and v are continuous at the origin and, the partial derivatives of u and v all exist at the origin, because, u(h, 0) − u(0, 0), h3 /h2 − 0, = lim, = 1,, h→0, h→0, h, h, , ux (0, 0) = lim
Page 136 :
126, , 5 Analytic Functions, , u(0, h) − u(0, 0), (−h3 /h2 ), = lim, = −1, h→0, h→0, h, h, and similarly vx (0, 0) = 1 = vy (0, 0). Thus we see that the Cauchy–Riemann, equations are certainly satisfied at the origin. But f is not differentiable at, the origin, because for h = h1 + ih1 , h1 ∈ R,, uy (0, 0) = lim, , ih1, 1+i, i, f (0 + h) − f (0), =, =, =, h, h1 + ih1, 1+i, 2, and for h = h1 + i0,, h1 + ih1, f (0 + h) − f (0), =, = 1 + i., h, h1, Therefore, , f (0 + h) − f (0), h, does not exist. This shows that the partial derivatives exist and satisfy the, Cauchy–Riemann equations at the origin even though the function is not, differentiable there., A similar conclusion continues to hold for the following function:, % xy, if z = 0,, f (z) = x2 + y 2, 0 if z = 0., lim, , h→0, , This function vanishes on both coordinate axes. Hence at z = 0,, ux = uy = vx = vy = 0,, and the Cauchy–Riemann equations are satisfied. But on the line y = mx, (m = 0), we have, h · mh/(h2 + m2 h2 ), m, f (h + imh) − f (0), =, =, ,, 2, h + imh, h + imh, (1 + m )(1 + im)h, which approaches ∞ as h approaches 0. Therefore, f (0) does not exist. In fact,, despite the existence of partial derivatives the function is not even continuous, at z = 0., However, even if a function is continuous and has partial derivatives that, satisfy the Cauchy–Riemann equations at a point, we still are not guaranteed, differentiability. Consider the function, ⎧, ⎨ xy 2, if z = 0,, f (z) = x2 + y 2, ⎩, 0 if z = 0., Since |f (z)| ≤ |x|, the function is continuous at the origin. Moreover, as f = 0, on both the axes, fx = fy = 0 at the origin, so that the Cauchy–Riemann, equations are satisfied at the origin. But on the line y = mx (m = 0),
Page 137 :
5.1 Cauchy–Riemann Equation, , 127, , m2 h3 /(1 + m2 )h2, m2, f (h + imh) − f (0), =, =, ,, h + imh, h + imh, (1 + im)(1 + m2 ), which does not tend to a unique limit independent of m. Hence, once again,, f (0) does not exist, meaning that f is not differentiable at the origin. It, follows that the converse of Theorem 5.7 is not true., Thus far, the negative character of the Cauchy–Riemann equations has, been emphasized. They have been utilized primarily to prove the nonexistence of a derivative. In the next section, we will use these equations in conjunction with an additional criterion to formulate a sufficient condition for, differentiability., As indicated earlier, (5.3) can be expressed more concisely as, f (z) = fx = −ify ., , (5.5), , Equation (5.5) provides a method for calculating the derivative if the derivative is known to exist. Note that z = x + iy gives that, x=, , z+z, ,, 2, , y=, , z−z, ., 2i, , In view of this note, we may treat f (x + iy) as a function of z and z, and so, , , , , z+z z−z, z+z z−z, ,, ,, + iv, ., f (z) = u, 2, 2i, 2, 2i, Thus, fz =, , ∂f ∂x ∂f ∂y, ∂f, =, +, ∂z, ∂x ∂z, ∂y ∂z, , , 1 ∂f, ∂f, =, +i, 2 ∂x, ∂y, 1, = [(ux − vy ) + i(vx + uy )], 2, , and the equation fz = 0 is equivalent to the system, fx = −ify , or ux = vy and uy = −vx ., Thus, we have, Theorem 5.8. A necessary condition for a function f to be differentiable at, a point a is that it satisfies the equation fz = 0 at a., The differential equation fz = 0 is known as the homogeneous Cauchy–, Riemann equation or simply the complex form of Cauchy–Riemann equations., For instance if f (z) = z then fz = 1 = 0, and so it is nowhere differentiable., Similarly, using this we conclude that
Page 138 :
128, , 5 Analytic Functions, , f1 (z) = Re z =, , z+z, ,, 2, , f2 (z) = Im z =, , z−z, and f3 (z) = ez, 2i, , are all nowhere differentiable functions. We now prove an analog of a wellknown real-variable theorem., Theorem 5.9. If f (z) ≡ 0 in a domain D, then f (z) is constant in D., Proof. In view of (5.5), fx = fy = 0 for all points in D. Now if the derivative, of a function of one real variable vanishes in an interval, the function must, be constant in that interval. Hence, ux = uy = 0 in D implies that u(x, y), is constant along every horizontal and vertical line segment in D. Similarly,, vx = vy = 0 in D implies that v(x, y) is constant along every horizontal and, vertical line segment in D. Thus, f (z) = u(z) + iv(z) is constant along every, polygonal line in D whose sides are parallel to the coordinate axes. According, to Remark 2.7, any two points in D can be joined by such a line. Therefore,, f (z1 ) = f (z2 ) for any pair of points z1 , z2 ∈ D, so that f (z) must be constant, in D., Theorem 5.9 is not true if D in this statement is an open set which is not, connected. For example, if D = {z : |z| > 2} ∪ {z : |z| < 1} and, , 2 if |z| < 1, f (z) =, 3 if |z| > 3,, then f (z) = 0 on D but f is not constant on D., Theorem 5.10. If f (z) is real-valued and differentiable on a domain D, then, f is constant on D., Proof. Suppose that f is differentiable at z0 . Then, f (z0 ) = lim, , h→0, , f (z0 + h) − f (z0 ), ., h, , Allowing h → 0 along both the real and imaginary axes, we see that f (z0 ) is, both real and pure imaginary. Consequently, f (z0 ) = 0. By Theorem 5.9, f, is a constant., An immediate consequence of Theorem 5.9 is the following result., Corollary 5.11. If f = u + iv is differentiable in a domain D with either, u(x, y), v(x, y), or arg f (z) constant in D, then f (z) is constant in D., Proof. Suppose that u(x, y) = c on D. Then ux = uy = 0 on D, and by the, Cauchy–Riemann equations, vy = vx = 0. Therefore, v is constant on D. The, other cases can be handled by a similar argument., Corollary 5.12. If f and g are two differentiable functions in a domain D,, and Re f (z) = Re g(z) on D, then f (z) = g(z) + constant.
Page 139 :
5.1 Cauchy–Riemann Equation, , 129, , This corollary says that “the real part Re f (z), completely determines the, differentiable function f (z) in a domain except for an additive constant”. Similarly, every differentiable function f in a domain D is completely determined, by its imaginary part except for an additive real constant., Questions 5.13., 1. What is the geometric interpretation of a continuous function of two, variables?, 2. For a function of two variables, why is it usually easier to show that a, limit does not exist?, 3. Why did the behavior of functions along the coordinate axes play a, central role?, 4. Is there a function f having every directional derivative at a point without f being continuous at that point?, 5. Can the Cauchy–Riemann equations be written in polar coordinate, form?, 6. Can we give a geometric interpretation to the existence of a derivative, of a complex function?, 7. If a derivative exists, is the derivative a continuous function?, 8. When will it be easier to evaluate limits by use of polar coordinates?, 9. In this section, we used the word “path” without defining it. How would, you define “path”?, 10. If a function f = u + iv is differentiable, can we say anything about, uxx ?, 11. If f (z) assumes only real values in a domain, what can we say about, the existence of f (z)?, 12. Where is |z| differentiable? Where is |z|2 differentiable?, 13. Where is z|z| differentiable? Where is z|z| continuous?, 14. Where is Re z differentiable? Where is (Re z)2 differentiable?, 15. Where is zRe z differentiable? Where is zRe z continuous?, Exercises 5.14., 1. If f (0, 0) = 0, which of the following functions are continuous at the, origin?, x2 y 2, x2 y 2, (b), f, (x,, y), =, (a) f (x, y) = 4, x + y4, (x2 + y 2 )2, 2, x3 y 2, x + ye−x, (c) f (x, y) = 2, (d), f, (x,, y), =, (x + y 2 )2, 1 + y2, , (x + y 2 )2, x |xy|, (e) f (x, y) = 2, (f) f (x, y) = 2, ., x + y2, x + y2, 2. Determine where the following functions satisfy the Cauchy–Riemann, equations, and where the functions are differentiable.
Page 140 :
130, , 5 Analytic Functions, , (a) f (z) = z 2, (b) f (z) = x2 − y 2, (c) f (z) = 2xyi, (d) f (z) = x2 − y 2 + 2xyi, (e) f (z) = zRe z, (f) f (z) = z|z|, (h) f (z) = |Re z Im z|1/3, (g) f (z) = |Re z Im z|1/2, (i) f (z) = zIm z, (j) f (z) = 2z + 4z + 5., 3. Show that at z = 0 the function f defined by, ⎧, 3, 3, ⎨ (1 + i)x − (1 − i)y for x + iy = 0, 2, 2, x +y, f (x + iy) =, ⎩, 0 for x = y = 0, satisfies the Cauchy–Riemann equations but it is not differentiable., 4. If f (z) is continuous at a point z0 , show that f (z) is also continuous at, z0 . Is the same true for differentiability at z0 ? What does the function, f (z) = |z|2 show? How about f (z) = z?, 5. If f (z) is continuous at a point z0 , then show that f (z) is also continuous, at z0 . Is the same is true for the differentiability at z0 ?, 6. Is f : C → C given by f (z) = z 2 + z|z|2 differentiable at z = 0? If so,, what is f (0)? Does f (n) (0) exist for n ≥ 2? Give your reasons., 7. a) Give an integer n and nonzero real number m, construct a function, f (z) such that limz→0 f (z) = 0 along each curve of the form y =, mxk (k = 0, 1, 2, . . . , n − 1), but for which limz→0 f (z) = 0 along, the curve y = mxn ., b) Construct a function f (z) for which limz→0 f (z) = 0 along each, curve of the form y = mxn (n = 1, 2, 3, . . . ), but for which, limz→0 f (z) does not exist., 8. If f (z) is differentiable at z, show that, 2 2 2 2, ∂u, ∂v, ∂u, ∂v, +, =, +, |f (z)| =, ∂x, ∂x, ∂y, ∂y, ∂(u, v), ∂u ∂v, ∂u ∂v, u u, −, = x y =, ., =, vx vy, ∂x ∂y, ∂y ∂x, ∂(x, y), , , 2, , , , The last expression is the Jacobian of u and v with respect to the variables x and y., 9. If fx and fy exist at a point (x0 , y0 ), show that, lim f (x, y0 ) = lim f (x0 , y) = f (x0 , y0 )., , x→x0, , y→y0, , Must f (x, y) be continuous at (x0 , y0 )?, , 5.2 Analyticity, Try as we did, we were unable to extract differentiability from the Cauchy–, Riemann equations. This “smoothness” requirement along the coordinate axes
Page 141 :
5.2 Analyticity, , 131, , could not sufficiently control the behavior of a function along different paths., If we focus on neighborhoods rather than on isolated points, many of our, difficulties will be eliminated., A function is said to be analytic at a point if it is differentiable everywhere, in some neighborhood of the point. A function is analytic in a domain if it is, analytic at every point in the domain. A function analytic at every point in, the complex plane is called an entire function., The function, f (z) = |z|2 = x2 + y 2, is differentiable only at the origin, and hence is not analytic anywhere. The, function f (z) = x2 y 2 is differentiable at all points on each of the coordinate, axes, but is still nowhere analytic. On the other hand, all the polynomials are, entire functions, and f (z) = z/(1 − z) is analytic everywhere except at z = 1., Remark 5.15. If f (z) is analytic at z0 , then there exist an > 0 such that, f (z) is differentiable at each point in N (z0 ; ). But for any point z1 ∈ N (z0 ; ),, we can find a δ > 0 such that N (z1 ; δ) ⊂ N (z0 ; ). Hence f (z) is also analytic, at z1 . Consequently, f (z) is analytic at a point if and only if f (z) is analytic, in some neighborhood of the point. Thus, the set of all values for which a, given function is analytic must form an open set. In particular, if a function is, analytic in a closed set, then there is always an open set containing the closed, •, set in which the function is analytic., Returning to functions of two variables, we prove the following mean-value, theorem:, Theorem 5.16. Let f (x, y) be defined in a domain D, with fx and fy continuous at all points in D. Given a point, √ (x, y) ∈ D, choose δ so that, (x + h, y + k) ∈ D for all points satisfying h2 + k 2 < δ. Then, f (x + h, y + k) − f (x, y) = fx (x, y)h + fy (x, y)k +, where, , 1, , → 0 and, , 2, , 1h, , +, , 2 k,, , → 0 as h → 0 and k → 0., , Proof. We write, f (x + h, y + k) − f (x, y), (5.6), = {f (x + h, y + k) − f (x, y + k)} + {f (x, y + k) − f (x, y)}., Now f (ξ, y + k) may be viewed as a differentiable function of the one real, variable ξ, where ξ assumes values between x and (x + h). Applying the, mean-value theorem for one real variable, we have, f (x + h, y + k) − f (x, y + k) = fx (x + θ1 h, y + k)h, Similarly,, , (|θ1 | < 1). (5.7)
Page 142 :
132, , 5 Analytic Functions, , f (x, y + k) − f (x, y) = fy (x, y + θ2 k)k, , (|θ2 | < 1)., , (5.8), , By the continuity of fx and fy ,, fx (x + θ1 h, y + k) = fx (x, y) +, , 1, , and fy (x, y + θ2 k) = fy (x, y) +, , 2,, , where 1 → 0 and 2 → 0 as h → 0 and k → 0. In view of the last equation,, we may rewrite (5.7) and (5.8) as, f (x + h, y + k) − f (x, y + k) = fx (x, y)h +, , 1h, , f (x, y + k) − f (x, y) = fy (x, y)k +, , 2 k., , The result now follows from (5.6)., This mean-value theorem enables us to utilize the Cauchy–Riemann equations to establish sufficient conditions for analyticity., Theorem 5.17. Let f (z) = u(x, y) + iv(x, y) be defined in a domain D,, and let u(x, y) and v(x, y) have continuous partials that satisfy the Cauchy–, Riemann equations, ux = vy , vx = −uy, for all points in D. Then f (z) is analytic in D., Proof. Set Δz = h + ik, where h and k are arbitrary real numbers. Given a, point z ∈ D, we must show that, f (z + Δz) − f (z), Δz→0, Δz, lim, , exists. For h and k sufficiently small, z +Δz is in D. Since u and v are assumed, to have continuous partials, an application of Theorem 5.1 shows that fx and, fy must also be continuous. Thus the conditions of Theorem 5.16 are satisfied, and we have, fx (x, y)h + fy (x, y)k +, f (z + (h + ik)) − f (z), =, h + ik, h + ik, , 1h, , +, , 2k, , ,, , (5.9), , where 1 → 0 and 2 → 0 as h → 0 and k → 0. As we saw in (5.5), the, Cauchy–Riemann equations may be expressed as fx (x, y) = −ify (x, y), or, ifx (x, y) = fy (x, y)., , (5.10), , Substituting (5.10) into (5.9), we obtain, fx (x, y)h + ify (x, y)k + 1 h + 2 k, f (z + (h + ik)) − f (z), =, h + ik, h + ik, h, k, = fx (x, y) + 1, + 2, ., h + ik, h + ik, , (5.11)
Page 143 :
5.2 Analyticity, , But, , h, ≤ 1,, h + ik, , 133, , k, ≤ 1,, h + ik, , so that the last two expressions in (5.11) approach zero as h and k approach, zero. Therefore,, lim, , h→0, k→0, , f (z + (h + ik)) − f (z), = fx (x, y)., h + ik, , Because no assumptions were made about the manner in which h and k approached zero, the derivative f (z) exists, with f (z) = fx (x, y). Since z was, arbitrary, the function is differentiable everywhere in D, and hence is analytic, in D., Observe that we could similarly have shown that f (z) = −ify (x, y) for all, points in D., It pays, at this point, to extract the important steps of Theorem 5.17 in, order to understand more fully the relationship between the Cauchy–Riemann, equations and differentiability. Requiring continuity of the partials in a neighborhood allowed us to apply Theorem 5.16 to obtain (5.9). A substitution, of the Cauchy–Riemann equations into (5.9) led to (5.11). Analyticity then, followed from (5.11)., Had differentiability at a point been our main objective, we would have, proved a theorem analogous to Theorem 5.17 that required continuity of the, partials at only one point. For complex functions, however, analyticity (rather, than differentiability at a point) is the important concept., Example 5.18. Consider f (z) = u+iv, where u = x2 and v = y 2 . Each of the, partial derivatives is continuous in C, whereas u and v satisfy the Cauchy–, Riemann equations only when y = x. Thus f is differentiable at (1 + i)x,, •, x ∈ R, and nowhere else., Example 5.19. If u(x, y) = x2 + y 2 and v(x, y) = xy then all the partial, derivatives exist and are continuous in C. However f = u + iv is differentiable, only at z = 0 because the Cauchy–Riemann equations are satisfied only at the, origin; so f is nowhere analytic since f is differentiable at z = 0 and nowhere, •, else., In Chapter 8, we will show that if f (z) is analytic at a point, then f (z), has derivatives of all orders at the point. In particular, the existence of f (z), tells us that, ∂f, ∂f, = −i, f (z) =, ∂x, ∂y, is continuous. In view of Theorem 5.1, the partial derivatives of its real and, imaginary components are also continuous., Therefore, the converse of Theorem 5.17 is also true. That is, a function, is analytic in a domain D if and only if the function has continuous partials, that satisfy the Cauchy–Riemann equations for all points in D.
Page 144 :
134, , 5 Analytic Functions, , Remark 5.20. In real analysis, if f (x) is differentiable on (a, b), then the, derivative f (x) need not be differentiable on (a, b) (need not even be continuous as shown by an example below). Clearly, the function, f (x) =, , 3 4/3, x, 4, , is differentiable on (−1, 1), whereas f (x) = x1/3 is not differentiable at the, •, origin., Note that differentiability in a neighborhood does not assume the same, importance for real as for complex functions. The real-valued function, f (x) = x|x|, is differentiable for all real values, but does not have a second derivative at, the origin. The everywhere-differentiable function in R, %, 1, if x = 0, x2 sin, f (x) =, x, 0 if x = 0, does not even have a continuous derivative at the origin., It is now quite simple to establish analyticity for the elementary functions., Examples 5.21. Let f (z) = ez = ex (cos y + i sin y). We have, ux = vy = ex cos y and vx = −uy = ex sin y., Theorem 5.17 may be applied to deduce that f (z) = ez is an entire function, and that, ∂f, = ex cos y + iex sin y = ez ., f (z) =, ∂x, Similarly, if f (z) = sin z = (eiz − e−iz )/2i, then by Theorem 5.6 and Example, 5.21,, eiz + e−iz, ieiz + ie−iz, =, = cos z., f (z) =, 2i, 2, Also, if f (z) = cos z = (eiz + eiz )/2, then, f (z) =, , −eiz + e−iz, ieiz − ie−iz, =, = − sin z., 2, 2i, , •, , The above examples illustrate a recurrent pattern in the theory of complex, variables. The real-valued exponential furnishes us with information about, the complex exponential, which then enables us to derive results about the, complex trigonometric functions. We leave it for the reader to verify that, tan z, cot z, sec z and csc z have the “expected” derivatives., For certain functions, it is easier to express real and imaginary parts in, terms of r and θ of z = reiθ ,
Page 145 :
5.2 Analyticity, , 135, , f (z) = u(r, θ) + iv(r, θ)., For example, if f (z) = z 5 , then expressing this in terms of x and y is uninviting. On the other hand, De Moivre’s theorem quickly gives, f (z) = (reiθ )5 = r5 cos 5θ + ir5 sin 5θ., A sometimes useful expression for the Cauchy–Riemann equations in polar, coordinates will now be proved., Theorem 5.22. Let f = u + iv be differentiable with continuous partials at a, point z = reiθ , r = 0. Then, 1 ∂v, ∂u, =, ,, ∂r, r ∂θ, , ∂v, 1 ∂u, =−, ., ∂r, r ∂θ, , In most textbooks, the polar form of the two Cauchy–Riemann equations, is either left as an exercise or worked out by using the chain rule for functions, of two variables to write ux , uy , vx and vy in terms of r and θ and then, solving/comparing the equations which involve these partial derivatives. Let, us provide a simpler proof of Theorem 5.22., Proof. Recall that f (z) = fx = −ify . We have for z = reiθ = 0,, fr =, , ∂f ∂z, = f (reiθ )eiθ and fθ = f (reiθ )ireiθ ,, ∂z ∂r, , (5.12), , fr = eiθ f (reiθ ) and fθ = ireiθ f (reiθ ), , (5.13), , or equivalently,, , which shows that, i, i, fr = − fθ , i.e. ur + ivr = − (uθ + ivθ ), r, r, , (r = 0)., , Taking the real and imaginary parts of this equation yields the desired, Cauchy–Riemann equations in polar form., To demonstrate that the polar form is useful, we consider f : C \ {0} → C, given by, f (z) = Log z := ln |z| + iArg z., If z = reiθ = 0, then this equation becomes, f (z) = ln r + iθ := u(r, θ) + iv(r, θ), , (−π < θ ≤ π), , and it is now easier to check the polar form of the Cauchy–Riemann equations., As (see (5.13))
Page 146 :
136, , 5 Analytic Functions, , i, f (z) = e−iθ fr = − e−iθ fθ ,, r, , (5.14), , it follows that, e−iθ, 1, d, ( Log z) =, = ,, dz, r, z, , z ∈ C \ {x + iy : x ≤ 0, y = 0}., , Note that v is not continuous at points on the negative real axis, and the, partial derivatives are continuous at all points except those points on the, negative real axis., For another illustration, consider, , , 2xy, y 2 − x2, f (z) = u(x, y) + iv(x, y) := 2, +, i, ., (x + y 2 )2, (x2 + y 2 )2, It is not easy to use the Cartesian form to verify the analyticity of this function., In terms of polar coordinates, we write x = r cos θ and y = r sin θ so that, f (z) = −, , sin 2θ, e−2iθ, cos 2θ, +, i, =, −, r2, r2, r2, , (r = 0)., , Thus,, 2 −2iθ, 2i, e, and fθ = 2 e−2iθ, r3, r, which gives, according to the formula (5.14),, fr =, , f (z) = e−iθ fr =, , 2, r3 e3iθ, , =, , 2, ,, z3, , z = 0., , Example 5.23. We wish to obtain a branch of (z 2 − 1)1/2 that is analytic for, |z| > 1., To do this we need to identify a solution w = f (z) that is analytic for, |z| > 1 and satisfies the condition, w2 = z 2 − 1., , (5.15), , If we use the principal branch of (z 2 − 1)1/2 , then, w = e(1/2) Log (z, , 2, , −1), , ., , Does this function do the job? Note that this function is analytic on the open, set Ω = {z : z 2 − 1 ∈ C \(−∞, 0]}. Note that a point z fails to belong to Ω iff, z 2 − 1 ∈ (−∞, 0]., To exhibit Ω we just need to remove from C those complex numbers z for, which, Re (z 2 − 1) = x2 − y 2 − 1 ≤ 0 and 2xy = 0.
Page 147 :
5.2 Analyticity, , 137, , This gives the set, {x + iy : x = 0} ∪ {x + iy : y = 0, |x| ≤ 1}., Thus, the branch cut for the principal branch of (z 2 − 1)1/2 is the whole y-axis, as well as the real interval [−1, 1] of R. So we need to look for an alternate, solution w satisfying (5.15). In view of this, we rewrite (5.15) as, , , 1, 2, 2, w =z 1− 2, z, and consider, w = g1 (z) = z(1 − 1/z 2 )1/2 = ze(1/2) Log (1−1/z, or, , 2, , ), , 2, , w = g2 (z) = −ze(1/2) Log (1−1/z ) ,, , each of which is a solution of (5.15). As Log (1−z) is analytic for z ∈ C \[1, ∞),, Log (1 − 1/z) is, in particular, analytic for |z| > 1. Consequently, g1 and g2, are analytic for |z| > 1. What are their derivatives for |z| > 1?, Finally, to obtain the branch of (z 2 − 1)−1/2 which are analytic for |z| < 1,, we rewrite (5.15) as, w2 = −(1 − z 2 ) = i2 (1 − z 2 )., This allows us to consider two solutions of (5.15) which are analytic for |z| < 1,, namely,, w = φ1 (z) = ie(1/2) Log (1−z, , 2, , ), , or, , 2, , w = φ2 (z) = −ie(1/2) Log (1−z ) ., , •, , Example 5.24. We wish to determine the largest open set Ω in which the, function Log (1 − z n ) (n ∈ N) is analytic., To do this, we first recall that g(z) = Log z is analytic in C \ (−∞, 0] but, not in any larger open set. Also, g (z) = 1/z. Set h(z) = 1 − z n . Then g is, analytic in C minus those points z for which, 1 − z n ∈ (−∞, 0], i.e., z n ∈ [1, ∞)., For instance, if n = 2, then, 1 − z 2 ∈ (−∞, 0] ⇐⇒ z ∈ (−∞, −1] ∪ [1, ∞), and, , 1 − z 3 ∈ (−∞, 0] ⇐⇒ z ∈ R0 ∪ R1 ∪ R2 ,, , where Rk = {rei2kπ/3 : r ≥ 1} (k = 0, 1, 2) is the ray starting from ei2kπ/3 (a, cube root of unity) to ∞. Thus, the largest open set in which Log (1 − z n ) is, analytic is therefore the set, Ω = C \ ∪n−1, k=0 Sk, where Sk = {rei2kπ/n : r ≥ 1}, k = 0, 1, 2, . . . , n − 1., , •
Page 148 :
138, , 5 Analytic Functions, , Example 5.25. Let f be an entire function. Suppose that f = u + iv has the, property that, uy − vx = −2 for all z ∈ C., What can we say about the function f ? Can it be a constant function? Clearly, not! Can this be a polynomial of degree > 1? The given condition shows that, this is not the case (how?). Let us try to find this function. By the Cauchy–, Riemann equation uy = −vx , the given condition is the same as, vx = 1, which, by the fact that f (z) = ux + ivx , is equivalent to, Im f (z) = 1., This observation implies that f (z) is a constant, say a, so that f has the form, f (z) = az + b, with Im a = 1 (Alternatively, as Im f (z) = 1, h(z) is defined by, h(z) = eif, , , , (z), , is entire and |h(z)| = e−1 which is a constant. Thus, h and hence, f (z) is a, •, constant)., We close this section with a theorem requiring only differentiability at a, point, although a more general theorem for analytic functions will be proved, later., Theorem 5.26. Let f (z) and g(z) be differentiable at z0 , with f (z0 ) =, g(z0 ) = 0. If g (z0 ) = 0, then limz→z0 (f (z)/g(z)) = f (z0 )/g (z0 )., Proof. The result is a consequence of the definition of a derivative, for, f (z0 ), =, g (z0 ), , limz→z0, limz→z0, , Examples 5.27., , f (z) − f (z0 ), f (z) − f (z0 ), f (z), z − z0, z − z0, ., = lim, = lim, z→z0 g(z) − g(z0 ), z→z0 g(z), g(z) − g(z0 ), z − z0, z − z0, (i) Let f (z) = |z|2 and g(z) = z. Then, f (0), 0, f (z), = , = = 0., z→0 g(z), g (0), 1, lim, , Note that for z = 0, f (z)/g(z) equals z., (ii) For f (z) = sin z, we have, lim, , z→0, , cos 0, f (az), sin az, = lim, =a, = a,, z→0 sin z, f (z), cos 0, , where a is any complex number.
Page 149 :
5.2 Analyticity, , 139, , (iii) Let f (z) = 1 − cos z and g(z) = sin2 z. Here g (2nπ) = 0 for each n ∈ Z,, but Theorem 5.26 may be avoided by solving directly. Because, sin2 z = 1 − cos2 z = (1 − cos z)(1 + cos z),, we have for each n ∈ Z, 1, 1 − cos z, 1 − cos z, 1, = lim, = lim, = ., lim, z→2nπ sin2 z, z→2nπ 1 − cos2 z, z→2nπ 1 + cos z, 2, , •, , Questions 5.28., 1. If we do not require continuity of the partials in Theorem 5.16, what, kind of mean-value theorem can we obtain?, 2. What important differences are there between differentiable functions, and analytic functions?, 3. Let f be defined on a domain D ⊆ C and a, b ∈ D such that [a, b] ⊆ D., Does there exists a point c on this line segment [a, b] such that, f (b) − f (a) = f (c)(b − a)?, 4. What alternate definitions could we have given for an entire function?, 5. What relationships are there between continuity of a function and the, continuity of its partials?, 6. If f (z) and f (z) are both analytic in a domain D, what can be said, about f throughout D?, 7. Let f (z) be analytic in the unit disk Δ = {z ∈ C : |z| < 1}. Is g(z) =, f (z) analytic in Δ?, 8. Is f (z) = (z)3 − 3z differentiable at ±1? Is f (z) analytic?, 9. Does there exist a function f that is analytic only for Imz ≥ 2004 and, nowhere else?, 10. How might we define a real analytic function?, 11. If f (z0 ) = g(z0 ) = 0, f (z0 ) and g (z0 ) exist with g (z0 ) = 0, does, lim, , z→z0, , f (z), f (z), = lim ?, g(z) z→z0 g (z), , 12. If a function is analytic in a bounded region, is the function bounded?, 13. If f (z) satisfies the Cauchy–Riemann equations for all points in the, plane, is f (z) an entire function?, 14. Why isn’t the polar form of the Cauchy–Riemann equations valid at the, origin?, 15. Are the branch cuts of (z 2 − 4)1/2 the whole of imaginary axis and the, interval [−2, 2] of the real axis? If so, what is the region of analyticity, of √, the chosen √, branch?, √, 16. Is z 2 − 4 = z + 2 z − 2? If so, when?, 17. What is the branch cut for principal branch of (1 − z 2 )1/2 ?, 18. Does iz + (1 − z 2 )1/2 take the value 0 at some point z ∈ C, regardless, of the choice of the square root?
Page 150 :
140, , 5 Analytic Functions, , √, 19. What is the derivative of ( z)3 at z = 1 − i?, 20. Where is Arg z continuous? Where is (Arg z)2 continuous? Where is, (Arg z)3 continuous?, 21. Does there exist a function f that is analytic for Re (z) ≥ 2006 and is, not analytic anywhere else?, Exercises 5.29., 1. Suppose that f (x) = x2 − y 3 + i(x3 + y 2 ). Does it satisfy the Cauchy–, Riemann equations for points on the line y = x? If so, can we say that, f is differentiable at these points? If so, can we say that f is analytic at, these points?, 2. Place restrictions on the constants a, b, c so that the following functions, are entire:, (a) f (z) = x + ay − i(bx + cy), (b) f (z) = ax2 − by 2 + icxy, (c) f (z) = ex cos ay + iex sin(y + b) + c, (d) f (z) = a(x2 + y 2 ) + ibxy + c., 3. Let, , 4, e−1/z if z = 0, f (z) =, 0 if z = 0., Show that, (a) f (z) satisfies the Cauchy–Riemann equations everywhere, (b) f (z) is analytic everywhere except at z = 0, (c) f (z) is not continuous at z = 0., 4. Let f (z) and g(z) be entire functions. Show that the following functions, are also entire., (a) f (z) + g(z), , (b) f (z)g(z), , (c) f (g(z))., , 5. Let f (z) and g(z) be analytic at z0 . Show that f (z)/g(z) is analytic at, z0 if and only if g(z0 ) = 0., 6. Let f (z) = a0 + a1 z + · · · + an z n . Prove that ak = (f (k) (0))/k! (k =, 0, 1, . . . , n)., 7. If f (z) is analytic at z0 , show that g(z) = Re f (z) is analytic at z0 if, and only if g(z) is constant in some neighborhood of z0 ., 8. Using the Cauchy–Riemann equations, find the most general entire functions such that Re f (z) = 0., 9. If f is entire such that f (z1 + z2 ) = f (z1 ) + f (z2 ) for all z1 , z2 ∈ C,, then show that f (z) = f (1)z. Give an example of a continuous nowhere, differentiable function f satisfying the given condition for all z1 , z2 ∈ C., Find also all continuous functions satisfying this condition., 10. Find the most general entire functions of the form f (z) = u(x) + iv(y),, where u and v depend only one real variable., 11. If f = u + iv is differentiable at a point z, show that the first-order, partials of u and v exist at z.
Page 151 :
5.3 Harmonic Functions, , 141, , 12. If f , is analytic ina domain D, prove that, ∂2, ∂2, + 2 |f (z)|n = n2 |f (z)|n−2 |f (z)|2, (a), 2, ∂y , ∂x2, ∂, ∂2, (b), +, |Re f (z)|n = n(n − 1)|Re f (z)|n−2 |f (z)|2 ., ∂x2, ∂y 2, 13. (a) Let f (z) be analytic with continuous partials in a domain D that, excludes the origin. Use Theorem 5.22 to show that, f (z) = e−iθ, , e−iθ, 1 ∂f, ∂f, = e−iθ (ur + ivr ) =, =, (vθ − iuθ ), ∂r, iz ∂θ, r, , at all points in D., (b) Conversely, if f (z) has continuous partials in a domain D that excludes the origin, and, e−iθ, , 1 ∂f, ∂f, =, ∂r, iz ∂θ, , at all points in D, show that f (z) is analytic in D., 14. Use the above exercise to determine whether the following functions are, analytic in the domain of definition. What is its derivative in terms of, z?, (a) r + iθ, (b) θ + ir, (c) ln r + i(θ + 2π) (d) r10 cos(10θ), 10r2 − sin(2θ), (e) r10 sin(10θ), (f), ., r2, where 0 < r and −π < θ < π., 15. Use the result of the previous exercise to show that the derivative of z n ,, n an integer, is nz n−1 ., 16. Evaluate the following limits, if they exist., ez − 1, z2, (b) lim, z→0, z→0 |z|, 3z, 1, 2 sin z, (c) lim z, (d) lim z sin ., z→0 e − 1, z→0, z, 17. Construct a branch f (z) of log z such that f (z) is analytic at z = −1, and takes on the value 5πi there., (a) lim, , 5.3 Harmonic Functions, If f (z) is analytic in a domain, then its derivative can be expressed in several, forms. For instance, according to (5.5) we may write, f (z) =, , ∂f, ∂f, , or f (z) = −i ., ∂x, ∂y, , (5.16)
Page 152 :
142, , 5 Analytic Functions, , In Chapter 8 it will be shown that an analytic function has derivatives of, all orders. From the first equation in (5.16), we see that the second derivative, can be expressed as, f (z) =, , ∂ , ∂2f, f (z) =, ., ∂x, ∂x2, , In view of the second equation in (5.16), we also have, , , ∂, ∂, ∂f, ∂2f, −i, = − 2., f (z) = −i f (z) = −i, ∂y, ∂y, ∂y, ∂y, , (5.17), , (5.18), , Equating (5.17) and (5.18), we get the identity, ∂2f, ∂2f, + 2 = 0,, 2, ∂x, ∂y, , (5.19), , which is valid for any analytic function f (z). Thus if f (z) = u(x, y) + iv(x, y), is analytic in a domain, equation (5.19) shows that its real and imaginary, components must satisfy the partial differential equations, ∂2v, ∂2v, ∂2u ∂2u, +, =, 0, and, +, = 0., ∂x2, ∂y 2, ∂x2, ∂y 2, , (5.20), , A continuous real-valued function U (x, y), defined in a domain D, is said, to be harmonic in D if it has continuous first- and second-order partials that, satisfy the equation, Uxx + Uyy = 0,, , (5.21), , known as Laplace’s equation, throughout D. Thus, in the case of the functions of two variables, the above discussion provides the intimate connection, between analytic functions and harmonic functions in the following form., Theorem 5.30. If f = u + iv is analytic on a domain D, and the functions, u and v have continuous second order partial derivatives on D, then u and v, are harmonic on D., Using the (as yet unproved) result that a function analytic in a domain has, derivatives of all orders at each point in the domain (hence a continuous second, derivative), we see from (5.20) the following result which is a reformulation of, Theorem 5.30., Theorem 5.31. Both the real and imaginary parts of an analytic function, are harmonic., Let us now obtain the polar form of Laplace’s equation. From (5.13), we, have
Page 153 :
5.3 Harmonic Functions, , 143, , ∂ iθ iθ, (e f (re )) = eiθ [eiθ f (reiθ )] = e2iθ f (reiθ ),, ∂r, ∂, (ireiθ f (reiθ )), =, ∂θ, ∂, = ireiθ (f (reiθ )) + i2 reiθ f (reiθ ), ∂θ, = ireiθ [ireiθ f (reiθ )] − reiθ f (reiθ ), , frr =, fθθ, , = −r2 e2iθ f (reiθ ) − rfr, = −r2 frr − rfr ., We thus obtain, ∂, 1 ∂2f, 1 ∂f, ∂2f, = 0, i.e. r, +, +, 2, 2, 2, ∂r, r ∂θ, r ∂r, ∂r, , , r, , ∂f, ∂r, , , +, , ∂2f, =0, ∂θ2, , which is the polar form for Laplace’s equation., If u is harmonic on D such that f (z) = u(x, y) + iv(x, y) is analytic, then, v is called a harmonic conjugate of u., Remark 5.32. We have the antisymmetric property that v is a harmonic, conjugate of u if and only if u is harmonic conjugate of −v. This follows upon, observing that the function if = i(u + iv) = −v + iu is analytic whenever f, is analytic., •, Although f (z) = x+iy is analytic so that v(x, y) = y is harmonic conjugate, of u(x, y) = x, g(z) = v+iu = i(x−iy) = iz is nowhere analytic. This example, is to illustrate the following: “If v is a harmonic conjugate of u in some domain, D, then u is not a harmonic conjugate of v unless u + iv is a constant”., Laplace’s equation furnishes us with a necessary condition for a function, to be the real (or imaginary) part of an analytic function., Example 5.33. For the function u(x, y) = x2 + y, we have, uxx + uyy ≡ 2,, so that u satisfies Laplace’s equation nowhere. Hence, u(x, y) = x2 + y cannot, •, be the real part of any analytic function., We will now show how the Cauchy–Riemann equations may be applied to, find a harmonic conjugate. For instance, the function u(x, y) = x + e−x cos y, can easily be shown to be harmonic everywhere. If there exists a function, v(x, y) for which f (z) = u(x, y) + iv(x, y) is analytic in C, then, ux = 1 − e−x cos y = vy ., , (5.22), , Antidifferentiation of (5.22) with respect to y yields, v = y − e−x sin y + φ(x),, , (5.23)
Page 154 :
144, , 5 Analytic Functions, , where φ(x) is a differentiable function of x. But in view of (5.23), an application of the other Cauchy–Riemann equation leads to, uy = −e−x sin y = −vx = −e−x sin y − φ (x),, which can be valid only if φ (x) ≡ 0. Hence,, v(x, y) = y − e−x sin y + c,, where c is a real constant. Therefore, v(x, y) is a harmonic conjugate of u(x, y), and, according to Theorem 5.17, f (z) is an entire function. In fact,, f (z) = x + e−x cos y + i(y − e−x sin y + c), = x + iy + e−x (cos y − i sin y) + ic, = z + e−z + ic., Example 5.34. Suppose we wish to find all analytic functions f (z) whose, real part is u(x, y) = ex (x cos y − y sin y). We may simply rewrite u as, u(x, y) = ex Re ((x + iy)(cos y + i sin y)) = Re [(x + iy)ex eiy ] = Re [zez ],, and so the desired analytic functions are of the form f (z) = zez + ic for some, c ∈ R., Similarly, we may rewrite, e−x (x sin y − y cos y) = e−x Re [(x + iy)(sin y + i cos y)], = Re [e−x (x + iy)i(cos y − i sin y)], = Re [i(x + iy)e−(x+iy) ], = Re [ize−z ]., Thus, every analytic function, whose real part is, u(x, y) = e−x (x sin y − y cos y),, must be of the form f (z) = ize−z + ic for some real constant c., In this way one can write, ex [(x2 − y 2 ) cos y − 2xy sin y] = ex Re [(x2 − y 2 + 2ixy)(cos y + i sin y)], = Re [z 2 ex (cos y + i sin y)], = Re [z 2 ez ],, and conclude that every analytic function, whose real part is given by, u(x, y) = ex [(x2 − y 2 ) cos y − 2xy sin y],, must be of the form z 2 ez + ic for some real c., , •
Page 155 :
5.3 Harmonic Functions, , 145, , Example 5.35. Suppose that f (z) = u + iv is an analytic function in a, domain D such that u − v is given as, u − v = ex (cos y − sin y)., We wish to find f (z) in terms of z. The easy procedure is as follows. As, f = u + iv, we have if = −v + iu so that, Re [(1 + i)f ] = u − v, and therefore, we may write the given expression for u − v as, Re ((1 + i)f (z)) = ex (cos y − sin y), = ex Re [(1 + i)(cos y + i sin y)], = Re [(1 + i)ex eiy ], = Re [(1 + i)ez + ic], , (c ∈ R), , which shows that, f (z) = ez + i(c/(1 + i)), for some real constant c. The same procedure may be adopted for problems, similar to this., •, Example 5.36. Let us determine the entire function f = u + iv for which, f (0) = i and u(x, y) = x4 + y 4 − 6x2 y 2 − 4xy. To do this, it suffices to compute, ux = 4x3 − 12xy 2 − 4y and uy = 4y 3 − 12x2 y − 4x., Clearly, u is harmonic in C. Now, we have, f (z) = ux − iuy = 4x3 − 12xy 2 − 4y + i(−4y 3 + 12x2 y + 4x) = 4z 3 + 4iz, which gives, , f (z) = z 4 + 2iz 2 + c., , Setting c = i, we get the desired function., , •, , Two questions now arise: To what extent is Laplace’s equation sufficient, to guarantee the existence of a harmonic conjugate, and how do we, in general, determine all such conjugate functions? Both of these questions will be, answered in Chapter 10 when we construct a harmonic conjugate for any, function harmonic in some neighborhood of a point. Thus, a function will be, shown to be harmonic in a neighborhood of a point if and only if it is the real, part of some analytic function., As might be expected, any two harmonic conjugates of a given harmonic, function differ by a real constant. For, if v(x, y) and v ∗ (x, y) are harmonic, conjugates of u(x, y), then both u + iv and u + iv ∗ are analytic so that the
Page 156 :
146, , 5 Analytic Functions, , difference i(v − v ∗ ) is also analytic. Consequently, v − v ∗ is a real-valued, analytic function. Thus,, v(x, y) = v ∗ (x, y) + C,, where C is a constant., Many properties of analytic functions are inherited from their real or imaginary parts. For example, if u = Re f (z) is constant in a region where f (z) is, analytic, then f (z) must be constant. This follows on applying the Cauchy–, Riemann equations to obtain ux = uy = vx = vy = 0., While the real and imaginary parts of an analytic function are harmonic,, its modulus need not be harmonic. However, properties of the analytic function may still be deduced by studying the behavior of its modulus., Theorem 5.37. Let |f (z)| be constant in a domain D where f (z) is analytic., Then f (z) is constant in D., Proof. If |f (z)| = |u + iv| = C, then u2 + v 2 = C 2 . Differentiating, we have, uux + vvx = 0,, , uuy + vvy = 0., , (5.24), , An application of Cauchy–Riemann equations to (5.24) yields, uux − vuy = 0,, , uuy + vux = 0., , (5.25), , Eliminating uy from (5.25), we get, (u2 + v 2 )ux = 0,, so that ux = 0. In a similar manner, we can show that uy = vx = vy = 0., Thus, we observe that, 0 = f (z) = ux + iuy, which gives that u and v are constants., Theorem 5.37 is actually guided through the start of the proof of Theorem, 5.9. Here is an alternate proof of Theorem 5.37: Suppose that |f (z)|2 = c for, all z ∈ D and for some constant c ∈ R. If c = 0, then |f (z)|2 = 0 implies that, f vanishes on D. If c = 0, then f (z) = 0 on D and 1/f (z) is analytic on D., But then, c = |f (z)|2 = f (z)f (z), shows that f (z) = c/f (z) is analytic on D. Consequently, the real-valued, function (see Theorem 5.10), Re f (z) =, , f (z) + f (z), 2, , is analytic on D and so it is constant. Similarly,
Page 157 :
5.3 Harmonic Functions, , Im f (z) =, , 147, , f (z) − f (z), 2i, , is a real-valued analytic function on D and so it is a constant. Therefore, , , , , f (z) − f (z), f (z) + f (z), +i, f (z) =, 2, 2i, is constant. Thus, we complete the proof of Theorem 5.37., The word “domain” in Theorem 5.37 cannot be replaced by “circle”. To, see this, observe that the nonconstant entire function f (z) = z n , n a positive, integer, has constant modulus on any circle centered at the origin., Thus far, our discussion of analyticity has been confined to single-valued, functions. It makes no sense for us to talk about derivatives of multiple-valued, functions; because in considering the expression, lim, , z→z0, , f (z) − f (z0 ), ,, z − z0, , we have no consistent rule to determine which value for f (z) to take as z, varies towards z0 . It does, however, make sense to discuss the analyticity for, a fixed branch of a multiple-valued function. Recall that, f (z) = Log z = ln |z| + iArg z, , (z = 0, −π < Arg z < π), , is a single-valued function, continuous when −π < Arg z < π. Since f (z) is, not continuous on the negative real axis, it certainly is not differentiable there., We will now show that f (z) is analytic at all points of continuity. Switching, to the polar representation, we get, f (z) = Log z = ln r + iθ, , (z = reiθ , −π < θ < π)., , (5.26), , Note that, , 1, ∂f, ∂f, =, and, = i., ∂r, r, ∂θ, In view of Exercise 5.29(13), it follows that f (z) exists, with, f (z) = e−iθ, , 1 ∂f, 1, ∂f, =, =, ∂r, iz ∂θ, z, , (−π < Arg z < π)., , Alternatively (since w = log z ⇐⇒ z = ew = e Log z+2kπi ), we simply use the, chain rule to differentiate, z = e Log z ,, and obtain, 1 = e Log z, , z ∈ D = C \{x + i0 : x ≤ 0}, d, d, ( Log z) = z ( Log z)., dz, dz
Page 158 :
148, , 5 Analytic Functions, , Thus,, 1, d, ( Log z) = , z ∈ D., dz, z, Any other branch of log z, with branch cut along the negative real axis, differs, from Log z in the cut plane by a multiple of 2πi, and hence also has the same, derivative 1/z in the cut plane., While log z is discontinuous for negative real values when the branch cut, is along the negative real axis, we can find a branch of log z for which log z is, analytic on the negative real axis. To illustrate, we can easily show that the, function, f (z) = log z (0 < Arg z < 2π), is analytic at all points in the plane cut along the positive real axis, with, f (z) = 1/z at all such points. Note that this function cannot be defined to be, continuous for positive real values. Thus, given any nonzero complex number, z0 , there exists a branch for which log z is analytic at z0 with derivative 1/z0 ., As we have seen in the previous chapter, determining properties of the logarithm enables us to determine properties of several related classes of functions., To illustrate, we can choose a definite branch of the logarithm, and write, f (z) = z 1/2 = e(1/2) log z, , (z = 0)., , Then by the chain rule,, f (z) =, , 1, 1 1 log z, 1, 1, 1, e2, = log z e 2 log z = 1 log z = 1/2, 2z, 2e, 2z, 2e 2, , where we have used the same branch of z 1/2 on both sides of the identities., More generally, if f (z) = z α = eα log z for some complex number α and some, determination of log z for z on the cut plane (which depends on our choice),, then, α, f (z) = eα log z = αz α−1 ., z, Example 5.38. We √wish to determine the largest domain D in which the, principal branch of ez + 1 is analytic, and compute its derivative, for z in, √, that domain. To do this, we recall that the principal branch of ez + 1 is, f (z) = e(1/2) Log (e, , z, , +1), , ., , Note that Log z is analytic in C \ (−∞, 0], but not in any larger domain., Consequently, the largest domain of analyticity is C minus those points in the, complex plane for which ez + 1 is real and ≤ 0. To find these points, we note, that, •, •, •, , ez = ex eiy is real iff y = nπ for some n ∈ Z, ez > 0 if n is even, ez < 0 if n is odd.
Page 159 :
5.3 Harmonic Functions, , 149, , For even n, and z = nπ, we have ez = ex > 0 for all x and so, ez + 1 = ex + 1 > 0, , for all x ∈ R., , On the other hand, if n is odd and z = nπ, then ez = −ex so that, ez + 1 = −ex + 1 ≤ 0 iff x ≥ 0., Thus, the domain of analyticity is C \ {x + iy : x ≥ 0, y = (2k + 1)π, k ∈ Z}, and, z, ez, ez−(1/2) Log (e +1), =, ., •, f (z) = √ z, 2, 2 e +1, Remark 5.39. The functions z α and αz should not be confused. The former, is a multiple-valued function of z when α is not an integer; each branch is, analytic in the cut plane and has derivative equal to αz α−1 . The latter, αz ,, which can be expressed as ez log α , is a single-valued entire function once a, branch is chosen for log α (α = 0); the derivative of αz is then given by, •, αz log α., We can also use the chain rule and the logarithm to find the derivatives, of the inverse trigonometric functions. Recall that, sin−1 z = −i log[iz + (1 − z 2 )1/2 ]., Since, , d, (1 − z 2 )1/2 = −z(1 − z 2 )−1/2 ,, dz, where the same branch is used on both sides of the equation, it follows that, d, −i[i − z(1 − z 2 )−1/2 ], 1, (sin−1 z) =, =, ., 2, 1/2, dz, iz + (1 − z ), (1 − z 2 )1/2, More generally, if f is analytic and nonzero at a point z, then a branch may, be chosen for which log f is also analytic in a neighborhood of z, with, f (z), d, [log f (z)] =, ., dz, f (z), , Furthermore, f (z) = |f (z)|ei arg f (z) so that, log f (z) = log |f (z)| + i arg f (z)., Hence log |f (z)|, since it is the real part of an analytic function, is harmonic, at all points where f (z) is analytic, and nonzero. Setting f = u + iv, the har√, monicity of log |f (z)| = log u2 + v 2 may also be proved directly by laborious, computation.
Page 160 :
150, , 5 Analytic Functions, , Questions 5.40., 1. Under what conditions may we deduce properties of analytic functions, from those of harmonic functions?, 2. What properties do analytic and harmonic functions not have in common?, 3. When will arg f (z) be a harmonic function?, 4. How do the properties of |f (z)| and log |f (z)| compare?, 5. If a function u(x, y) is harmonic everywhere in a domain D, does there, exist a function v(x, y) for which u(x, y)+iv(x, y) is analytic everywhere, in D?, 6. Is f = u + iv analytic on a domain whenever u and v are harmonic on, D?, 7. Does f (z) = ux − iuy represent an analytic function on a domain D, whenever u = u(x, y) is harmonic on D?, 8. Let u and v be harmonic on a domain D, and U = uy − vx and V =, ux + vy on D. Is F (z) = U + iV analytic on D?, 9. What is the largest domain on which f (z) = z z is analytic?, 3, 10. What is the largest domain on which f (z) = 3z is analytic?, 11. For what values of z does u = x3 − y 3 satisfy the Laplace equation?, Why is this function not harmonic?, 12. Does there exist an analytic function with real or imaginary part as, y 2 − 2xy?, 13. Suppose that f (z) is analytic for |z| < 1 such that f (1/2) = 3 and |f (z)|, is constant for |z| < 1. Does f (z) = 3 for |z| < 1?, Exercises 5.41., 1. Let, , ⎧, ⎨, f (z) =, , ⎩, , xy, , x2 − y 2, if x2 + y 2 =, 0,, x2 + y 2, 2, 2, 0 if x + y = 0., , Show that fxy (0, 0) = (fx )y (0, 0) = fyx (0, 0) = (fy )x (0, 0)., 2. Show that the following functions are harmonic, and then determine, their harmonic conjugates., (a) u = ax + by, a and b real constants, y, , x2 + y 2 = 0, (b) u = 2, x + y2, (c) u = x3 − 3xy 2, (d) u = Arg z, −π < Arg z < π, 2, 2, (e) u = ex −y cos 2xy, (f) u = 2xy + 3x2 y − y 3 ., 3. Using the Cauchy–Riemann equation or the idea of Example 5.34 find, all the analytic functions f = u + iv, where u is given as below:, (a) u(x, y) = ex (x sin y + y cos y), (x, y) ∈ R2
Page 161 :
5.3 Harmonic Functions, , 151, , x, , (x, y) ∈ R2 \(0, 0), x2 + y 2, (c) u(x, y) = sin x cosh y, (x, y) ∈ R2, (d) u(x, y) = x3 − 3xy 2 − 2x, (x, y) ∈ R2, (e) u(x, y) = x2 − y 2 − 2xy + 2x − 3y, (x, y) ∈ R2 ., 4. Let f = u + iv be an entire function such that, (b) u(x, y) =, , u + v = 2e−x (cos y − sin y)., Construct f in terms of z., 5. For what nonnegative integer values of n is the real-valued function, u(x, y) = xn −y n harmonic? Find the corresponding harmonic conjugate, function in each possible values of n., 6. Show that u(x, y) = xy is harmonic in R2 . Find the conjugate harmonic, function v(x, y) in R2 . Write u + iv in terms of z., 7. Choose the constant a so that the function u = ax2 y − y 3 + xy is, harmonic, and find all harmonic conjugates., 8. Choose the constants a, b, c so that u = ax3 + bx2 y + cxy 2 + dy 3 gives, the more general harmonic polynomial, and find all the harmonic conjugates., 9. Choose the constant a so that u = ax3 + xy 2 + x is harmonic in C and, find all analytic functions f whose real part is the given u., 10. Show that neither xy(x − y) nor xy(x − 2y) can be a real part of an, analytic function., 11. Use the method of this section to attempt to find a function v(x, y) for, which x2 + iv(x, y) is analytic, and explain where the method breaks, down., 12. If u1 and u2 are harmonic on D, show that au1 + bu2 , a and b real, constants, is also harmonic at D., 13. Suppose u and v are conjugate harmonic functions. Show that uv is a, harmonic function. What are the most general conditions for which the, product of two harmonic functions is harmonic?, 14. If f = u + iv is analytic on a domain D, then show that uv is harmonic, on D., 15. Find all harmonic functions u(x, y) in the unit disk x2 + y 2 < 1 such, that uy = 0 for x2 + y 2 < 1. What can be said about u(x, y)?, 16. Suppose f (z) has a derivative of order n. Show that, f (n) (z) =, , ∂ (n) f, ∂ (n) f, = (−i)n (n) ., (n), ∂x, ∂y, , 17. Show that if the real and imaginary parts of both f (z) and zf (z) are, harmonic in a domain D, then f (z) is analytic in D., 18. Define a branch of (1 − z 2 )1/2 so that f (z) = iz + (1 − z 2 )1/2 is analytic, in the domain Ω = C \{z : Im z = 0 and |Re z| ≥ 1}., 19. Find the derivative of the following functions:, (a) cos−1 z, , (b) tan−1 z, , (c) sec−1 z.
Page 162 :
6, Power Series, , From sequences of numbers, we turn to sequences of functions. Then our, concern is with both the form of convergence and the behavior of the limit, function. Convergence, determined at each point in a set, need not require, the limit function to retain any of the properties common to each function in, the sequence. But if a certain “rapport” exists between the sequence of functions and the set, then the limit function will be forced to confirm to definite, standards established by the sequence. This stronger type of convergence, in, which the set takes precedence over its points, is called uniform convergence., The most important sequences of functions are those expressible as power, series. The limit functions for this class are always analytic inside their regions of convergence. In many instances, a power series behaves like a “big”, polynomial., , 6.1 Sequences Revisited, Given a (real or complex) sequence {an }n≥1 , we associate a new sequence, {sn }n≥1 of partial sums sn related by, sn =, , n, , , ak ., , k=1, , ∞, The symbol k=1 ak is called a series. The series is said to converge or to diverge according as the sequence {sn }n≥1 is convergent or divergent. If {sn }n≥1, converges to s, the sum (or value) of the series is said to be s, and we write, s = lim sn =, n→∞, , ∞, , , ak ., , k=1, , Hence for convergent series the same symbol is used to denote both the series, and its sum. We call sn the nth partial sum of the series and an the nth term, in the series.
Page 163 :
154, , 6 Power Series, , By definition, every theorem about series may be formulated as a theorem, about sequences (of partial sums). The converse is also true., Theorem 6.1., Given any sequence {sn }n≥1 , there exists a sequence {an }n≥1, n, such that sn = k=1 ak for every n., Proof. We choose a1 = s1 and an = sn − sn−1 for n > 1. Then, n, , , ak = a1 +, , k=1, , n, , , (sk − sk−1 ) = s1 + (sn − s1 ) = sn ., , k=2, , Thus, the definition of series does not furnish us with a “new” concept. It, merely provides an additional way of stating new theorems and restating old, ones. Because taking limits of sequences is a linear operation, it follows that, summing series is also a linear operation. For instance, if sn → s and tn → t,, then, sn ± tn → s ± t and csn → cs,, where c is a complex constant. If we apply these properties to partial sums of, series we conclude the following:, ∞, ∞, Proposition 6.2. If k=1 ak = α and k=1 bk = β, then, ∞, , , (ak ± bk ) = α ± β and, , k=1, , ∞, , , cak = cα., , k=1, , The Cauchy criterion, for sequences , (Theorem 2.20) may be reworded as, n, ∞, follows: Let sn = k=1 ak . The series k=1 ak converges if and only if for, every > 0, there exists an integer N such that m, n > N implies, |sm − sn | =, , m, , , ak < ., , (6.1), , k=n+1, , By letting m = n + p, (6.1) may be written as, |sn+p − sn | =, , n+p, , , ak <, , (p = 1, 2, . . . ),, , for n > N., , (6.2), , k=n+1, , The Cauchy criterion is the most general test for convergence of a series., Some of the methods frequently used in elementary calculus, like the ratio, and integral tests, require very restrictive hypotheses, and even then do not, supply necessary as well as sufficient conditions for convergence., Many of the familiar properties of series are immediate consequences of, (6.2). For example,, ∞, Proposition 6.3. A necessary condition for the series k=1 ak to be convergent is that ak → 0 as k → ∞.
Page 164 :
6.1 Sequences Revisited, , 155, , ∞, Proof. This result follows on setting p = 1 in (6.2). Indeed, if k=1 ak converges, {sn } converges. Therefore, sn − sn−1 = an → 0 as n → ∞., ∞, It is often useful to consider, given series k=1 ak the corre∞ along with the, ∞, sponding series, moduli k=1 |ak |. A series k=1 ak is said to be absolutely, of, ∞, convergent if k=1 |ak | converges. Applying, n+p, , , ak ≤, , k=n+1, , n+p, , , |ak |, , k=n+1, , to (6.2), we see that the absolute convergence of a series guarantees its convergence. More precisely, we have, , , Proposition 6.4. If a series ak converges absolutely, then ak converges., Similarly, if |an | ≤ K|bn | for every n and for some K> 0, the convergence, , ∞, ∞, of n=1 |an | may be deduced from the convergence of n=1 |bn | by applying, (6.2). Proposition 6.4 can be proved directly with the help of a little trick., Express, Re ak = (Re ak + |ak |) − |ak |., Since |Re ak | ≤ |ak |, we have, 0 ≤ Re ak + |ak | ≤ 2|ak |., , , Re, Hence, (Re ak + |ak |) converges and, ak , being a difference between, two, convergent, series,, converges., Similarly,, Im ak converges. Consequently,, , ak converges., Suppose all the terms of the sequence {an } are real and positive. Then, sn − sn−1 = an > 0,, and the sequence of partial sums {sn } is a monotonically increasing sequence., Since a monotone sequence of real numbers, converges if and only if it is, ∞, bounded (Theorem 2.15), the series k=1 ak of positive real numbers converges if and only if the sequence {sn } is bounded. That the positivity of, dropped from the hypotheses is seen by letting an = (−1)n ., {an } cannot be , ∞, Here, the series k=1 ak does not converge despite the fact that, sn =, , n, , , (−1)k, , k=1, , is bounded in absolute value by 1., Our next theorem shows an interesting relationship between a sequence, and its sequence of partial sums., ∞, Theorem 6.5., Suppose that an > 0 for every n ∈ N and that n=1 an din, verges. If sn = k=1 ak , then
Page 165 :
156, , 6 Power Series, , (i), , ∞, , (ii), , ∞, , n=1, n=1, , an, diverges;, sn, an, converges., s2n, , Proof. According to the Cauchy criterion, the series in (i) will diverge if, for, any integer n, an integer p can be found such that, an+2, an+p, 1, an+1, +, + ··· +, > ., sn+1, sn+2, sn+p, 2, Since {sn } is an increasing sequence,, n+p, , k=n+1, , ak, ≥, sk, , n+p, k=n+1, , ak, , sn+p, , =, , sn+p − sn, sn, =1−, ., sn+p, sn+p, , (6.3), , But sn+p → ∞ as p → ∞. Thus, p may be chosen so large that sn+p > 2sn ., For such a choice of p, it follows from (6.3) that, n+p, , k=n+1, , Hence,, , ∞, , n=1 (an /sn ), , 1, ak, > ., sk, 2, , diverges. To prove (ii), observe that, , an, sn − sn−1, 1, 1, an, ≤, =, =, − ., s2n, sn sn−1, sn sn−1, sn−1, sn, Applying the Cauchy criterion, we have, n+p, , k=n+1, , , n+p, 1, 1, 1, 1, 1, ak, ≤, −, −, <, ., =, s2k, sk−1, sk, sn, sn+p, sn, k=n+1, , For any preassigned, > 0, we may choose n large enough so that 1/sn < ., ∞, Therefore, n=1 (an /s2n ) converges., ∞, ∞, Corollary 6.6. The series n=1 (1/n) diverges and n=1 (1/n2 ) converges., Proof. Apply Theorem 6.5, with an ≡ 1., Moreover, a standard argument from real variable theory gives that, ∞, , 1, nα, n=1, , converges for α > 1., ∞, Corollary 6.7. If an > 0 and n=1 an , diverges, then there exists a positive, ∞, sequence {bn } such that bn /an → 0 and n=1 bn diverges.
Page 166 :
6.1 Sequences Revisited, , 157, , Proof. Choose bn = an /sn , and apply Theorem 6.5., This corollary shows that there is no “slowest” diverging series. The reader, will be asked, in Exercise 6.19(4) to show that there is no slowest converging, series., We shall return to series after establishing additional properties for sequences. We prove the obvious, but useful, Theorem 6.8. (Mousetrap Principle) Let 0 ≤ an ≤ bn for every n, and, assume that limn→∞ bn = 0. Then limn→∞ an = 0., Proof. Given > 0, there exists an integer N such that |bn | < for n > N ., But then we also have |an | ≤ |bn | < for n > N . Hence, an → 0., Theorem 6.9. For α > 0, β > 0, and x real, we have, (ln x)α, = 0;, x→∞, xβ, α, x, (ii) lim βx = 0., x→∞ e, (i) lim, , Proof. Setting f (x) = ln x and g(x) = xβ/α , we apply l’Hôpital’s rule for, functions of a real variable to obtain, f (x), f (x), 1, = lim , = lim, = 0., x→∞ g(x), x→∞ g (x), x→∞ (β/α)xβ/α, lim, , Therefore,, , , , ln x, xβ/α, , α, =, , (ln x)α, → 0 as x → ∞,, xβ, , and (i) is proved., Letting y = ln x in (i), and observing that y → ∞ as x → ∞, we obtain (ii)., Corollary 6.10. For α > 0 and β > 0, we have, (ln n)α, = 0;, n→∞, nβ, nα, (ii) lim βn = 0;, n→∞ e, (iii) lim n1/n = 1;, (i) lim, , n→∞, , (iv) lim rn = 0, n→∞, , (|r| < 1)., , Proof. Letting x = n in Theorem 6.9, we obtain (i) and (ii). Next, n1/n → 1, if (ln n)/n → 0. But this is a special case of (i), and (iii) is proved. Finally,, setting β = − ln |r| in (ii), we have, nα, e−n ln |r|, , = nα |r|n ≥ |r|n ., , Finally, (iv) follows from (ii) and the mousetrap principle.
Page 167 :
158, , 6 Power Series, , Our next example is useful, for establishing the convergence of many series., ∞, Consider the geometric series k=0 z k . Then we may express the partial sum, n−1, in explicit form. If z = 1, then sn = n; if z = 1, then, sn = 1 + z + · · · + z, we have, n, , 1, zn, 1 − zn, sn =, =, −, ., z k−1 =, 1−z, 1−z, 1−z, k=1, , When z = 1, {sn } is unbounded and so has no limit. Thus, the geometric, series diverges for z = 1. For z = 1, we may use the convergence of {z n } to, zero for |z| < 1 to show that, ∞, , , z k−1 =, , k=1, , 1, 1−z, , (|z| < 1)., , On the other hand, if |z| ≥ 1, then the kth term z k−1 does not converge to 0, so that the series does not converge. To summarize, %, 1, ∞, , for |z| < 1, k, z =, lim sn =, 1−z, n→∞, diverges for |z| ≥ 1, k=0, ∞ k, and, k=0 z is absolute when |z| < 1. Hence, a series, ∞ the convergence of, a, converges, absolutely, if there exists a constant r, 0 ≤ r < 1, and a, n=1 n, real number M such that |an | ≤ M rn for n > N . For example, consider the, series, ∞, , 1, 1, an , an = n + i n ,, 3, 4, n=1, which converges. But then, how do we sum the series? In view of Proposition, 6.2, we have, ∞, , , an =, , n=1, , ∞, , , −n, , 3, , n=1, , +i, , ∞, , , 4−n =, , n=1, , 1/4, 1, i, 1/3, +i, = + ., 1 − 1/3, 1 − 1/4, 2 3, , The convergence properties of complex series may be deduced from those, of real series. If {ak } is a sequence of complex numbers, we write ak = αk +iβk ,, of real numbers. By Theorem, where {αk } and {βk } are sequences, ∞ 2.12, we, ∞, a, converges, if, and, only, if, have, that, the, complex, series, k, k=1, k=1 αk and, ∞, β, both, converge., That, is,, k, k=1, ∞, , , (αk + iβk ) = α + iβ ⇐⇒, , k=1, , For instance, the series, , ∞, , , αk = α and, , k=1, , ∞, , 1 + i cos(1/k), k=1, , 3k, , ∞, , k=1, , βk = β.
Page 168 :
6.1 Sequences Revisited, , 159, , converges., , If, in the series, ak , the summation range is (−∞, ∞) instead of (0, ∞), (or (1, ∞) or (p, ∞), where pis a fixed integer), then we have a series of, ∞, complex numbers ak , namely, k=−∞ ak . The most efficient way of handling, this is to discuss separately the two series, ∞, , , −1, , , ak and, , k=0, , ak =, , k=−∞, , ∞, , , a−k, , k=1, , so that the convergence of this series depends on the convergence of both, series., followingdefinition. “A series, the, ∞ This approach helps us to formulate, ∞, ∞, a, converges, if, and, only, if, both, k=−∞ k, k=0 ak and, k=1 a−k converge.”, In other words, we write, ∞, , , ak = lim, , k=−∞, , n→∞, , n, , k=0, , ak + lim, , m→∞, , m, , , a−k, , k=1, , provided both the limits on the right exist. Note that m and ntend inde∞, pendently to ∞.According to the definition, it isclear that if k=−∞ |ak |, ∞, ∞, converges, then , k=−∞ ak converges., ∞ Moreover, k=−∞ |ak | converges pre∞, cisely when both k=0 |ak | and k=1 |a−k | converge., Suppose {an } is a bounded sequence of complex numbers, and that A is, the set of subsequential limits of {an }. Some properties of A have already been, discussed. For instance, the set A is nonempty (Theorem 2.17) and consists of, one point if and only if {an } converges (Theorem 2.14). To insure a pleasant, treatment of Taylor and Laurent series, it is necessary to introduce the socalled “limit superior” of a sequence {an } of nonnegative real numbers. If, the sequence {an } is real and bounded, the set A has a least upper bound, (Dedekind property)., Let {an } be a real bounded sequence, and let A be the set of subsequential, limits of {an }. Setting a∗ = lub A, we call a∗ the limit superior of {an }, and, write, lim sup an = a∗ or limn→∞ an = a∗ ., n→∞, , This definition may be extended to unbounded real sequences. If {an } is unbounded above, we say that, lim sup an = ∞;, n→∞, , whereas, if all but a finite number of an are less than any preassigned real, number, we say that, lim sup an = −∞., n→∞, , A useful counterpart to the limit superior is the limit inferior. For a, bounded real sequence {an } we set a∗ =glb { subsequential limits }, and write
Page 169 :
160, , 6 Power Series, , lim inf an = a∗, , or limn→∞ an = a∗ ., , n→∞, , If {an } is unbounded below, we say that, lim inf an = −∞;, n→∞, , whereas, if all but a finite number of an are greater than any preassigned real, number, we say that, lim inf an = ∞., n→∞, , Formulating theorems in terms of limit superior or limit inferior, rather, than in terms of limit, has one distinct advantage. In the extended real number, system (±∞ included), the limit superior and limit inferior of a real sequence, always exist. This enables us to prove theorems about sequences without worrying about the existence of limits., Examples 6.11., (i) If an = n1 , then lim supn→∞ an = 0., n, (ii) If an = (−1) , then lim supn→∞ an = 1 and lim inf n→∞ an = −1., (iii) If an = 3n , then {an } is not bounded above and lim supn→∞ an = ∞., (iv) If an = −n+(−1)n n, then the sequence −2, 0, −2(3), 0, −2(5), . . . is not, bounded below. Thus, lim supn→∞ an = 0 and lim inf n→∞ an = −∞., (v) If an = 1 − (1/2)n for n ∈ N, then lim supn→∞ an = 1., (vi) If an = (1 + c)n with c > 0 and n ∈ N, then lim supn→∞ an = ∞., (vii) Let {an } = n sin2 (nπ/4). Since 0 ≤ an < ∞ for every n, no subsequence, of {an } can approach a value less than 0. In order to show that, lim sup an = ∞ and lim inf an = 0,, n→∞, , n→∞, , it suffices to find one (out of many) subsequences that approach the, desired value. We have, lim sup an = lim a4n+2 = lim (4n + 2) = ∞, n→∞, , n→∞, , n→∞, , and, lim inf an = lim a4n = lim 4n.0 = 0., n→∞, , n→∞, , n→∞, , (viii) Let an = (1 + 1/n) cos nπ. Then, lim sup an = lim a2n = 1 and lim inf an = lim a2n+1 = −1., n→∞, , n→∞, , n→∞, , n→∞, , (ix) Let an = sin(nπ/2) + sin(nπ/4). Then, √, lim sup an = lim a4n+1 = 1 +, n→∞, , n→∞, , 2, 2, , √, , and, lim inf an = lim a4n−1 = −1 −, n→∞, , n→∞, , 2, ., 2
Page 170 :
6.1 Sequences Revisited, , 161, , (x) Let an = n cos nπ. Then, lim sup an = lim a2n = ∞ and lim inf an = lim a2n−1 = −∞., n→∞, , n→∞, , n→∞, , n→∞, , (xi) Let an = 5n cos nπ − n2 . Then an ≤ 5n − n2 for every n, so that, , •, , lim sup an = lim inf an = −∞., n→∞, , n→∞, , It is clear that the limit superior and limit inferior are both unique, and, that a real sequence {an } converges to L, L finite, if and only if, lim sup an = lim inf an = L., n→∞, , n→∞, , But some of the standard limit theorems are false for both the limit superior, and the limit inferior. For instance, if an = (−1)n and bn = (−1)n+1 , then, lim sup(an + bn ) = 0 = lim sup an + lim sup bn = 2, n→∞, , n→∞, , n→∞, , and, lim inf (an + bn ) = 0 = lim inf an + lim inf bn = −2., n→∞, , n→∞, , n→∞, , However, we do have the following inequalities:, Theorem 6.12. Let {an } and {bn } be real, bounded sequences. Then, we have, (i) lim sup(an + bn ) ≤ lim sup an + lim sup bn, n→∞, , n→∞, , n→∞, , (ii) lim inf (an + bn ) ≥ lim inf an + lim inf bn ., n→∞, , n→∞, , n→∞, , Proof. Let lim supn→∞ an = a and lim supn→∞ bn = b. Assume inequality (i), is false. Then for some > 0 there is a subsequence {ank + bnk } of {an + bn }, such that ank + bnk > a + b + for all nk . But then either, ank > a +, , 2, , or, , bnk > b +, , 2, , infinitely often. This implies that either, lim sup an ≥ a +, n→∞, , 2, , or lim sup bn ≥ b +, n→∞, , 2, , (Why?). This contradicts our assumption, and (i) is proved. The proof of (ii), is similar, and will be omitted., It turns out that absolute convergence for (power) series plays a central, role in complex analysis as it is much easier to test for absolute convergence, than for convergence by other means. The nth root test is one such useful, result in determining convergence properties of power series.
Page 171 :
162, , 6 Power Series, , Theorem 6.13. (Root test) Let {an }n≥1 be a complex sequence, and suppose, that, lim sup |an |1/n = L., Then, , ∞, n=1, , n→∞, , an converges absolutely if L < 1, and diverges if L > 1., , Proof. If L < 1, choose r such that L < r < 1. Then, we have |an |1/n < r for, large values of n; that is,, |an | < rn, , for n > N ., , ∞, ∞, The convergence of n=1 |an | now follows from the convergence of n=1 rn, (|r| < 1)., of n. But then |an | > 1, If L > 1, then |an |1/n > 1 for infinitely many, values, ∞, infinitely often. Hence, an 0 and the series n=1 an cannot converge., Remark, 6.14. When L =, 1, the root test gives no information. The series, ∞, ∞, 2, (1/n), diverges, and, n=1, n=1 (1/n ) converges. However,, lim sup, n→∞, , 1, n, , 1/n, , = lim sup, n→∞, , 1, n2, , 1/n, , = 1., , •, , Theorem 6.15. (Ratio test) Let {an }n≥1 be a complex sequence, and suppose that, an+1, an+1, = L and lim inf, = l., lim sup, n→∞, an, an, n→∞, ∞, Then n=1 an converges absolutely if L < 1, and diverges if l > 1. The test, offers no conclusion concerning the convergence of the series if l ≤ 1 ≤ L., We leave the proof as an exercise. To show that the test fails in the last, case, we simply consider the series in Remark 6.14 and observe that L = l = 1,, in both cases., As usual we easily have the following corollary., Corollary 6.16. Let {an }n≥1 be a complex sequence, and suppose that, lim, , n→∞, , an+1, = L., an, , (6.4), , ∞, Then n=1 an converges absolutely if L < 1, and diverges if L > 1. The test, offers no conclusion concerning the convergence of the series if L = 1., Clearly, from Theorem 6.13, the conclusion of the corollary continues to, hold if (6.4) is replaced by limn→∞ |an |1/n = L., ∞, Example 6.17. Consider the series n=1 (1/n!). Then the ratio test is easier, to apply and it is not obvious that lim supn→∞ |(1/n!)|1/n < 1. Similarly, the, ratio, be easier to examine the convergence property of the series, ∞ test may, n, (−1), (1, −, i)n /n!., •, n=1
Page 172 :
6.1 Sequences Revisited, , 163, , Questions 6.18., 1. What types of theorems for real sequences remain valid for complex, sequences?, ∞, 2. Does there exist a convergent series n=1 an for which limn→∞ an = 0?, ∞, 3. Does there exist, series n=1 an for which limn→∞ an = 0?, ∞a divergent , ∞, 4. Do the series n=N an and n=1 an converge or diverge together?, 5. What does it mean to say that there is no slowest converging series?, 6. Why is the Mousetrap Principle so named?, 7. For which convergent series can you determine the sum?, 8. What alternate definitions for limit superior and limit inferior might we, have given?, 9. Can limit inferior be defined in terms of limit superior?, 10. What are some advantages and disadvantages of allowing the limit superior of a sequence to assume the values ±∞?, 11. How does the limit superior of a product compare with the product of, limit superiors?, 12. Can Theorem 6.12 be modified to include, ∞ unbounded sequences?, 13. If |an |1/n < 1 for every n ≥ 1, does, n n=1 an necessarily converge?, 1, w, ,, does, 14. If zn → z0 and wn →, 0, ! n k=1 zk wk → z0 w0 ?, n, 15. If zn → z0 , does 21n k=1 nk zk → z0 ?, Exercises 6.19., ∞, 1. Let an = αn + iβ, that n=1 |an | converges if, n , αn and βn real. Show , ∞, ∞, and only if, n=1 |αn | converges and, n=1 |βn | converges., both, n, 2. Let sn = k=1 ak . If ak = 1/k, show that s2n+1 − s2n > 12 for every n., If ak = (−1)k+1 /k, show that |sn+p − sn | < 2/n, ∞for every n and p., 3. Suppose an > 0 for every n. Show that, n=1 an diverges if and, only, if, for, any, integers, M, and, N, ,, there, exists, an integer p such that, N +p, a, >, M, ., n=N n, ∞, ∞, 4. Let an > 0, and suppose n=1 an converges. If rn = k=n ak , show, that, (a), , ∞, , an, diverges, r, n=1 n, , (b), , ∞, , an, converges., √, rn, n=1, , 5. Let A be the set of subsequential limits of a complex sequence. Show, that A is closed set., 6. Find the infimum and supremum of the following, (i) 5 + sin(nπ/3), , (ii) 1/n + sin(nπ/3), , (iii) 1/n + cos(nπ/3)., , 7. Suppose that {an } is a real sequence and limn→∞ an = a, a = 0. For, any sequence {bn }, show that, (a) lim sup(an + bn ) = lim an + lim sup bn, n→∞, , n→∞, , n→∞
Page 173 :
164, , 6 Power Series, , (b) lim inf (an + bn ) = lim an + lim inf bn, n→∞, , n→∞, , n→∞, , (c) lim sup an bn = lim an lim sup bn, n→∞, , n→∞, , 8., , 9., , 10., , 11., , n→∞, , (d) lim inf an bn = lim an lim inf bn ., n→∞, n→∞, n→∞, Show that lim supn→∞ an = L, L finite, if and only if the following, conditions hold: For any > 0,, a) an < L + for all but a finite number of n;, b) an > L − infinitely often., Show that lim inf n→∞ an = L, L finite, if and only if the following, conditions hold: For any > 0,, a) an < L + infinitely often;, b) an > L − for all but a finite number of n., Let {an } be a complex sequence., ∞, a) If lim supn→∞ |an+1 /an | = L < 1, show that, n=1 an converges, absolutely., ∞, b) lim inf n→∞ |an+1 /an | = L > 1, show that n=1 an diverges., Show by an example that this result provides no information about the, convergence or divergence of the series when L = 1., Suppose an > 0 for every n. Show that, lim inf, n→∞, , an+1, an+1, ≤ lim inf a1/n, ≤ lim sup a1/n, ≤ lim sup, ., n, n, n→∞, an, an, n→∞, n→∞, , 6.2 Uniform Convergence, A sequence of functions {fn } converges pointwise to a function f on a set, E (fn → f ) if to each z0 ∈ E and > 0, there corresponds an integer, N = N ( , z0 ) for which, |fn (z) − f (z0 )| <, , whenever n > N ( , z0 )., , To say that a sequence of functions {fn } converges pointwise on a set E is, equivalent to saying that the sequence of numbers {fn (z0 )} converges for each, z0 ∈ E. The limit function f is then defined by, lim fn (z0 ) = f (z0 ), , n→∞, , (z0 ∈ E)., , The integer N in the definition of pointwise convergence may, in general,, vary with the points in the set. If, however, one integer can be found that, works for all such points, the convergence is said to be uniform. That is, a, sequence of functions {fn } converges uniformly to f on a set E (fn ⇒ f ) if to, each > 0, there corresponds an integer N = N ( ) such that, for all z ∈ E,, |fn (z) − f (z)| <, , whenever n > N ( )., , Thus, when n is large, fn (z) is required to be uniformly “close” to f (z) on E.
Page 174 :
6.2 Uniform Convergence, , 165, , We emphasize that uniform convergence on a set implies (pointwise) convergence. A formulation of the negation of uniform convergence will be helpful, when producing examples that show the converse to be false. The convergence, of {fn } to f on E is not uniform if there exists an > 0 such that to each, integer N there corresponds an integer n(> N ) and a point zn ∈ E for which, |fn (zn ) − f (zn )| ≥ ., Recall the distinction between continuity and uniform continuity. A continuous function is uniformly continuous on a set if a single δ = δ( ) can be found, that works for all points in the set. In Chapter 2 the function f (z) = 1/z was, shown to be continuous, but not uniformly continuous, on the set 0 < |z| < 1., The following example is an analog for convergence., Example 6.20. Let fn (z) = 1/(nz). Then we see that the sequence {fn (z)}, converges pointwise, but not uniformly, to the function f (z) ≡ 0 on the set, 0 < |z| < 1., For a given > 0, in order that, |fn (z) − 0| = |1/(nz)| < ,, it is necessary that n > 1/( |z|). So the corresponding N = N (z; ) is an, integer greater than 1/( |z|). Note that, as |z| decreases, the corresponding N, increases without bound. Thus, we say that the sequence converge pointwise, but not uniformly on {z : 0 < |z| < 1}., Alternatively, we argue in the following manner. If this convergence were, uniform, there would exist an integer N for which the inequality |1/N z| <, < 1 would be valid for all z, 0 < |z| < 1. But the inequality does not hold, •, for z = 1/N ., We have shown that the convergence in the above example is not uniform, because, , , 1, 1, fn, = 1 for all n., −f, n, n, Example 6.21. Let fn (z) = 1/(1+nz). Then the sequence {fn (z)} converges, uniformly in the region |z| ≥ 2, but does not converge uniformly in the region, |z| ≤ 2. Indeed, the sequence {fn } converges pointwise everywhere to the, function, , 0 if z = 0, f (z) =, 1 if z = 0., If |z| ≥ 2, then, |fn (z)| =, , 1, 1, 1, 1, ≤, ≤, ≤ ., 1 + nz, |nz| − 1, 2n − 1, n, , Therefore, |fn (z)| < whenever n > 1/ , which proves uniform convergence, in the region |z| ≥ 2.
Page 175 :
166, , 6 Power Series, , If the convergence were uniform for |z| ≤ 2, there would exist an integer, N for which the inequality |fN (z) − f (z)| < 12 would be valid for all z in the, region. But, , , 1, 1, 1, 1, fn, =, −0 =, for every n., −f, •, n, n, 1 + n · (1/n), 2, Remark 6.22. Theorem 2.44 states that a function continuous on a compact set is uniformly continuous. The above example shows that pointwise, convergence, even on a compact set, need not imply uniform convergence. •, Example 6.23. Set fn (z) = z n . Then the sequence {fn (z)} converges pointwise on the set |z| < 1 and uniformly on the set |z| ≤ r < 1., The pointwise convergence for |z| < 1 follows from Corollary 6.10(iv). Note, that for r < 1, given > 0,, rn <, , ⇐⇒ n >, , ln, ., ln r, , Since rn → 0 (r < 1), an integer N = N ( ) may be found for which r n <, ln, (n > N > ln, r ). But then,, |z n | ≤ rn <, , (|z| ≤ r, n > N ( ))., , Hence, {fn (z)} converges uniformly to zero in the disk |z| ≤ r. Remember, ln, > 0 and, that the choice of N with N > ln, r is possible for an arbitrary, 0 < r < 1. However, for each fixed > 0, as r → 1− , N must be increased, without bound. It follows that the convergence is not uniform for |z| < 1., Alternatively, if the convergence were uniform on the set |z| < 1, then for, sufficiently large n the inequality |z n | < would be valid for all z, |z| < 1., Choosing z = 1−1/n, we have z n = (1−1/n)n , and, from elementary calculus, , n, 1, 1, 1−, as n → ∞., →, n, e, Hence, |z n | >, |z| < 1., , 1, 3, , (z = 1 − 1/n, n > N ), and the convergence is not uniform on, , •, , Example 6.24. Let fn (z) = z/n on C. Then fn (z) → f (z) ≡ 0 on C but not, uniformly. Note that for a given > 0, |fn (z) − 0| =, , |z|, z, < ⇐⇒ n >, n, , showing that, as |z| increases, the corresponding N increases without bound., It follows that the sequence converges pointwise but not uniformly on C., On the other hand if we restrict the domain to a bounded subset of C, say, •, Ω = {z ∈ C : |z| < 2006}, then the convergence is uniform.
Page 176 :
6.2 Uniform Convergence, , 167, , Example 6.25. We wish to show the sequence fn (z) = z/n2 converges uniformly to f (z) ≡ 0 for |z| ≤ R, but does not converge uniformly in the plane., To do this, we recall that for |z| ≤ R, the inequality, |fn (z)| ≤, , R, <, n2, , (n >, , , R/ ), , shows the convergence to be uniform on any disk |z| ≤ R. But if the convergence were uniform in C, then we would have |z/N 2 | < 1 for some integer N, •, and for all z. Choosing z = N 2 elicits the appropriate contradiction., The uniform convergence of a sequence {fn } on E is often deduced from, the convergence of {fn } at some point z = z0 by showing that the inequality, |fn (z) − f (z)| ≤ |f (z0 ) − f (z)|, is valid for all z ∈ E., If a sequence does not converge uniformly, there is usually some “bad, point” to be exploited. In Examples 6.20, 6.21, 6.23, and 6.25 the bad points, were, respectively, z = 0, z = 0, z = 1, and z = ∞. For each such point z0 ,, the expression limz→z0 (limn→∞ fn (z)) could not be replaced by, lim ( lim fn (z))., , n→∞ z→z0, , In Example 6.21, for instance,, , , 1, lim lim, = 0 while, z→0 n→∞ 1 + nz, , , lim, , n→∞, , 1, lim, z→0 1 + nz, , , = 1., , The importance of uniform convergence is that it does allow for the interchange of many limit operations. This, in turn, compels the limit function to, retain many properties of the sequence., Theorem 6.26. Suppose {fn } converges uniformly to a function f on E. If, each fn is continuous at a point z0 ∈ E, then the limit function f is also, continuous at z0 . That is,, , , , , lim lim fn (z) = lim, lim fn (z) ., z→z0, , n→∞, , n→∞, , z→z0, , Proof. We must show that for any > 0, there exists a δ > 0 such that, |f (z) − f (z0 )| < for all z in E satisfying |z − z0 | < δ. The inequality, |f (z) − f (z0 )| ≤ |f (z) − fn (z)| + |fn (z) − fn (z0 )| + |fn (z0 ) − f (z0 )|, , (6.5), , is valid for every n. The uniform convergence of {fn } allows us to choose N, independent of z so that, |f (z) − fN (z)| <, , 3, , (z ∈ E).
Page 177 :
168, , 6 Power Series, , Letting n = N in (6.5), we have, |f (z) − f (z0 )| <, , 3, , + |fN (z) − fN (z0 )| + ., 3, , (6.6), , By the continuity of fN at z = z0 ,, |fN (z) − fN (z0 )| <, , (6.7), , 3, , for z sufficiently close to z0 . Combining (6.6) and (6.7), the desired result, follows., Remark 6.27. Theorem 6.26 furnishes us with a necessary, but not sufficient,, condition for uniform convergence. In Example 6.23, it was shown that the, sequence of continuous functions {z n } converges nonuniformly in the region, |z| < 1 to the continuous function f (z) ≡ 0. However, in Example 6.21, the, discontinuity of the limit function at z = 0 rules out uniform convergence for, the sequence {1/(1 + nz)} in any region containing the origin., •, Our definition and discussion of uniform convergence remains valid for realvalued functions of a real variable. In fact, the same conclusions may be drawn, from the preceding four examples when z is replaced by x and the regions are, replaced by their corresponding intervals. Moreover, there is an interesting, geometric interpretation to uniform convergence of real-valued functions. If, {fn (x)} converges uniformly to f (x) on a set E, then for sufficiently large n, we have, f (x) − < fn (x) < f (x) + for all x in E., This means that there is some curvilinear strip of vertical width 2 that contains the graph of all functions y = fn (x), with n > N , and that each such, curve is never a distance more than away from the curve y = f (x)., Example 6.28. Let fn (x) = x2 + sin nx/n. We have, |fn (x) − x2 | =, , 1, sin nx, ≤, n, n, , (x real),, , and |fn (x) − x2 | < for n > 1/ . Hence, the sequence {fn (x)} converges, •, uniformly to f (x) = x2 on the set of real numbers (see Fig. 6.1)., Certainly no discussion of convergence is complete without a Cauchy criterion. Rewording Theorem 2.20 for functions, we have that the sequence, {fn (z)} convergence pointwise on E if and only if {fn (z0 )} is a Cauchy sequence for each z0 ∈ E. That is, to each z0 ∈ E and > 0, there corresponds an integer N = N ( , z0 ) for which |fn (z0 ) − fm (z0 )| < whenever, n, m > N ( , z0 )., A sequence {fn } is said to be uniformly Cauchy on E if to each > 0, there corresponds an integer N = N ( ) such that, for all z ∈ E,, |fn (z) − fm (z)| <, , whenever n, m > N ( ).
Page 178 :
6.2 Uniform Convergence, , 169, , Figure 6.1. Illustration for uniform convergence, , Theorem 6.29. A sequence of functions converges uniformly on a set E if, and only if the sequence is uniformly Cauchy on E., Proof. Suppose {fn } converges uniformly to f on E. Then, given, |fn (z) − f (z)| <, , 2, , > 0,, , for n > N ( ) and for all z ∈ E., , But then, for n, m > N ( ), we have, |fn (z) − fm (z)| ≤ |fn (z) − f (z)| + |fm (z) − f (z)| <, , 2, , +, , 2, , = ., , Hence, the sequence {fn } is uniformly Cauchy on E., Conversely, suppose {fn } is uniformly Cauchy on E. In particular, {fn (z0 )}, is a Cauchy sequence for each z0 ∈ E, and thus {fn } converges pointwise to, a function f . We wish to show that this convergence is uniform. Given > 0,, there exists an integer N = N ( ) such that n, m > N implies, |fn (z) − fm (z)| <, , 2, , for all z ∈ E., , (6.8), , Fixing n(> N ) and letting m vary, (6.8) leads to, |fn (z) − f (z)| = lim |fn (z) − fm (z)| ≤, m→∞, , 2, , < ., , (6.9), , Since (6.9) is valid for all z ∈ E and n > N ( ), the convergence of {fn } to f, is uniform on E., In the previous section, we saw that properties for sequences of complex, numbers could be reworded as properties for series of complex numbers. The, remainder of this section will be devoted to the conversion from sequences of
Page 179 :
170, , 6 Power Series, , complex functions to series of complex functions. As expected, our work will, parallel that of the previous section., Given a sequence of functions {fn (z)} defined on a set E, we associate a, new sequence of functions {Sn (z)} defined by, Sn (z) =, , n, , , fk (z)., , (6.10), , k=1, , For all values, ∞of z for which limn→∞ Sn (z) exists, we say that the series,, denoted by k=1 fk (z), converges, and write, f (z) = lim Sn (z) =, n→∞, , ∞, , , fk (z)., , k=1, , ∞, If {Sn (z)} converges uniformly on a set E, then the series, ∞ k=1 fk (z) is said, to be uniformly convergent, on E. Further, the series, k=1 fk (z) is abso∞, |f, (z)|, converges., Moreover,, a, necessary, condition, lutely, convergent, if, k=1 k, ∞, for k=1 fk (z) to converge uniformly on E is that fk (z) → 0 on E. This fact, is evident if we write, fn = Sn − Sn−1, and allow n → ∞. Also, rewording the Cauchy criterion, we have, ∞, Theorem 6.30. The series, n=1 fn (z) converges uniformly on a set E if, and only if, to each > 0, there corresponds an integer N = N ( ) such that, for all z ∈ E, we have, n+p, , , fk (z) <, , whenever n > N ( ), , (p = 1, 2, 3, . . . )., , k=n+1, , Proof. Define Sn by (6.10), and apply Theorem 6.29., Theorem 6.30 may be used to establish a sufficient condition for the uniform convergence of a series, called the Weierstrass M-test or dominated convergence test., Theorem 6.31. Let {Mn }n≥1 be a sequence of real, ∞numbers, and suppose, (z)|, ≤, M, for, all, z, ∈, E, and, each, n, ∈, N., If, that, |f, n, n=1 Mn converges, then, ∞ n, f, (z), converges, uniformly, (and, absolutely), on, the, set E., n=1 n, ∞, Proof. That n=1 |fn (z)| converges on E follows immediately from, the com∞, parison test for real series. To verify the uniform, convergence, of, n=1 fn (z), ∞, on E, we invoke the Cauchy criterion for n=1 Mn . Thus, given > 0, there, exists an integer N such that, for n > N , we have, n+p, , k=n+1, , Mk <, , (p = 1, 2, 3, . . . ).
Page 180 :
6.2 Uniform Convergence, , But, , n+p, , k=n+1, , fk (z) ≤, , n+p, , , n+p, , , |fk (z)| ≤, , k=n+1, , 171, , Mk < ., , k=n+1, , The uniform convergence now follows from Theorem 6.30., ∞ n, Example 6.32. The geometric series, converges absolutely for, n=0 z, |z| < 1 and , uniformly for |z| ≤ r < 1. Here we are dealing with a series, ∞, of functions n=1 fn (z) in the unit disk |z| < 1, where fn (z) = z n . We have, ∞, ∞, , , 1, 1, =, zn ≤, |z|n =, 1−z, 1, −, |z|, n=0, n=0, , (|z| < 1),, , and this proves absolutely convergence. Uniform convergence for |z| ≤ r then, follows from the , M -test (Theorem 6.31). Indeed, if we fix r < 1 and define, ∞, Mk = rk , then, Mk converges, and |z k | ≤ Mk for |z| ≤ r. By the, k=0, ∞, Weierstrass M -test, k=0 z k converges uniformly for |z| ≤ r, for each r < 1., We can also show that the series does not converge uniformly in the unit disk, |z| < 1. Setting, n−1, , 1 − zn, Sn (z) =, ,, zk =, 1−z, k=0, , the sequence {Sn (z)} converges pointwise to f (z) = 1/(1 − z) for |z| < 1., Choosing z = 1 − 1/n, we have, , n, 1, zn, |Sn (z) − f (z)| =, =n 1−, → ∞ as n → ∞, 1−z, n, (because (1 − 1/n)n → 1/e) showing that the partial sums do not converge, uniformly for |z| < 1. Hence, the uniform convergence of the series cannot be, •, extended to the disk |z| < 1., ∞, Example 6.33. The series n=1 (cos nz)/n2 converges uniformly and absolutely on the real line. Indeed, since | cos nz| ≤ 1 for all z real, Theorem 6.31, may be applied with Mn = 1/n2 to obtain the desired result. By writing, cos nz = (einz + e−inz )/2,, the reader may verify , that cos nz/n2 does not approach 0 as n tends to ∞, ∞, unless z is real. Thus, n=1 (cos nz)/n2 converges if and only if z is real. •, ∞, Example 6.34. The series n=1 2z 2 /(n2 + |z|) converges absolutely in the, plane and uniformly for |z| ≤ R, for each R > 0., To see this, it suffices to observe that for any point z0 in the plane,, 2|z02 |, 2|z02 |, ≤, ., n2 + |z0 |, n2
Page 181 :
172, , 6 Power Series, , Now, the absolute convergence in the plane is a consequence of the convergence, of, ∞, , 1, 2|z0 |2, ,, 2, n, n=1, and the uniform convergence on the disk |z| ≤ R follows from the M -test,, with Mn = 2R2 /n2 ., •, Example 6.35. We show that the Riemann zeta function, ∞, , 1, z, n, n=1, , converges absolutely for Re z >1 and uniformly for Re z ≥ 1 + , > 0. The, ∞, given series is concerned with n=1 fn (z), where fn (z) = n−z . We have, n−z = e−z Log n = e−(x+iy) Log n and n−z = e−x ln n = n−x, ∞, so that n=1 |n−z | converges for x = Re z > 1. The uniform convergence for, •, Re z ≥ 1 + follows from the M -test, with Mn = 1/n1+ ., Questions 6.36., 1., 2., 3., 4., 5., 6., 7., 8., , 9., 10., , What kinds of sequences of functions converge uniformly in the plane?, Can a sequence of unbounded functions converge uniformly?, How would you define: {fn } converges uniformly to infinity?, Can a sequence of functions converge uniformly on every compact subset, of a region and not converge uniformly in the region?, Can a sequence of functions converge pointwise on every compact subset, of a region and not converge pointwise in the region?, Can a sequence of discontinuous functions converge uniformly to a continuous function?, Can a sequence of functions converge uniformly, but not absolutely, in, a region? Absolutely, but not uniformly?, Suppose that for every > 0, there exists an integer N such that, |fN (z) − f (z)| < for all z in E. Does {fn } converge uniformly to, f in E? Does {fn } converge pointwise in E?, How would Theorem 6.31 be stated as a theorem for sequences?, If a sequence of differentiable functions converges uniformly, must the, limit function be differentiable?, , Exercises 6.37., 1. Show that fn = un + ivn converges uniformly to f = u + iv if and only, if {un } converges uniformly to u and {vn } converges uniformly to v., 2. Suppose {fn } converges uniformly to f and {gn } converges uniformly, to g on E.
Page 182 :
6.3 Maclaurin and Taylor Series, , 3., , 4., , 5., 6., , 7., , 8., 9., 10., , 173, , (a) Show that {fn + gn } converges uniformly to f + g on E., (b) If, in addition, |fn | ≤ M and |gn | ≤ M for all z ∈ E and all n, show, that {fn gn } converges uniformly to f g on E., n, Show that fn (z) = zn converges uniformly for |z| < 1. Show also that, fn (z) does not converge uniformly for |z| < 1 but it does converge, uniformly for |z| ≤ r for r < 1., Suppose that f (z) is unbounded on a set E. Let fn (z) ≡ f (z) for all n,, and let gn (z) = 1/n., a) Show that fn (z) and gn (z) both converge uniformly on the set E., b) Show that fn (z)gn (z) converges pointwise, but not uniformly, on E., Show that {fn } converges uniformly on a finite set if and only if {fn }, converges pointwise., Prove the following generalization of Theorem 6.26: Suppose {fn } converges uniformly to a function f on E, and fn is continuous at z0 ∈ E, for infinitely many n. Then the limit function f is also continuous at z0 ., Suppose {fn } converges uniformly to f on a compact set E, and each, fn is uniformly continuous on E. Prove that f is uniformly continuous, on, E. May compactness be omitted from the hypothesis?, ∞, If n=1 fn (z) converges uniformly on E, show that {fn (z)} converges, uniformly to zero on E. Is the converse true?, Let 0, < r < 1 and E = {z : |z| ≤ r} ∪ {z : r ≤ z ≤ 1, z ∈ R}. Show, ∞, that n=1 (−1)n z n /n converges uniformly, but not absolutely, on E., Find where the following sequences converge pointwise and where uniformly., (a), , z2, , z, + n2, , (b) ze−nz, , (c), , enz, n, , (d), , 1, ., 1 + zn, , 11. In what regions are the following series uniformly convergent? Absolutely convergent?, (a), (c), , ∞, , n=1, ∞, , , (1 − z)z n, , 2z, 2 − z2, n, n=1, , (b), (d), , ∞, , , z2, (1 + z 2 )n, n=1, ∞, , 1, n=1, , 1 + zn, , ∞, 12. Show that the series n=0 1/[(z + n)(z + n + 1)] converges to 1/z for, z ∈ C \{0, −1, −2, . . . }., , 6.3 Maclaurin and Taylor Series, ∞, Unfortunately, knowing that an arbitrary series, n=1 fn (z) converges (or, diverges) at some point z = z0 gives no information about the series at other, points. However, we can specialize the sequence {fn (z)} to obtain a class, of functions for which the behavior at a point always determines properties
Page 183 :
174, , 6 Power Series, , in a region. This class will play an important role in the theory of analytic, functions., Let fn (z) = an (z − b)n , where an and b are complex constants. The resultant expression,, a0 +, , ∞, , , an (z − b)n =, , n=1, , ∞, , , an (z − b)n ,, , (6.11), , n=0, , is called a power series in z − b. When b = 0, (6.11) reduces to, ∞, , , an z n ,, , (6.12), , n=0, , a power series in z. Our efforts will be focused on properties of power series, defined by (6.12). Upon substituting z − b for z, it becomes a simple matter, to translate these properties to those of series defined by (6.12). We state and, prove a result which provides a “qualitative” behavior of a convergent power, series., ∞, n, Theorem 6.38.Suppose the power series, converges , at a point, n=0 an z, ∞, ∞, n, n, |a, |, |z|, converges, for, |z|, <, |z, |, (that, is,, z = z0 . Then, n, 0, n=0, n=0 an z, converges absolutely for |z| < |z0 |)., ∞, Proof. Suppose that n=0 an z0n converges. Then an z0n → 0 as n → ∞. Hence, there exists a constant M such that |an z0n | ≤ M for all n, and, n, n, z, z, ≤M, |an | |z|n = an z0n, ., (6.13), z0, z0, ∞, For |z| < |z0 |, the geometric series n=0 |z/z0 |n converges. Thus, by (6.13),, ∞, , n=0, , |an | |z|n ≤, , ∞, , n=0, , M, , z, z0, , n, , =, , M, 1 − |z/z0 |, , (|z| < |z0 |)., , ∞, ∞, Corollary 6.39. If n=0 an z n diverges at a point z = z0 , then n=0 an z n, diverges for |z| > |z0 |., ∞, n, Proof. Suppose, on the contrary, that the series n=0, for some, ∞an z converges, point z1 with |z1 | > |z0 |. Then, by Theorem 6.38, n=0 an z n converges absolutely, ∞for |z| < |z1 |. In particular, this would imply the absolute convergence, of n=0 an z0n , contradicting our assumption., ∞, n, Corollary 6.40. If, converges for all real values of z, then the, n=0 an z, series also converges for all complex values., ∞, Proof.Suppose n=0 an z0n diverges for some complex value z0 . By Corollary, ∞, 6.39, n=0 an Rn diverges for R > |z0 |, contradicting our assumption.
Page 184 :
6.3 Maclaurin and Taylor Series, , 175, , Theorem 6.38 can be used to determine precise bounds for the region in, which a power series converges., ∞, Theorem 6.41. To every power series n=0 an z n , there corresponds an R,, 0 ≤ R ≤ ∞, for which the series, (i) converges absolutely in |z| < R if 0 < R ≤ ∞, (ii) converges uniformly in |z| ≤ r < R if 0 < R ≤ ∞, (iii) diverges for |z| > R if 0 ≤ R < ∞., Proof. For z = 0, the series becomes a0 and hence, the power series converges, at the origin. If the series diverges for all nonzero values of z, then clearly, R = 0. If the series converges for some nonzero value, we let, ), %, ∞, , n, |an | |z| converges for |z| < r ,, S= r:, n=0, , and define, , , R=, , lub S if S is bounded,, ∞ if S is unbounded., , We wish to show that R, so chosen, satisfies conditions (i), (ii), and (iii)., For any point z0 , |z0 | < R, we can find a real number ρ such that, |z0 | < ρ < R., , (6.14), , The number ρ must be in S; for otherwise, R could not be , its least upper, ∞, bound. Hence, by (6.14) and the definition of set S, the series n=0 |an | |z0 |n, converges. Since z0 was arbitrary, (i) is proved., Next, if R > 0, choose r so, that 0 < r < R. Then there exists z0 such that, ∞, r < |z0 | < R and the series n=1 |an z0n | is convergent (if R = ∞, this works, for any r). In particular, the nth term |an z0n | is bounded, say by a number K., Now for |z| ≤ r,, n, n, , , r, r, ≤K, = KMn ., |an z n | ≤ |an |rn = |an z0n |, |z0 |, |z0 |, , Since, Mn is, a convergent geometric series, the Weierstrass M -test applies,, and the series, an z n converges uniformly for |z| ≤ r. This proves (ii)., Finally, the convergence of the series at some point z1 , |z1 | > R, would,, according to Theorem 6.38, imply absolute convergence for |z| < |z1 |. But, then |z1 | would be an element of S; this would contradict the assumption, that R is an upper bound for the set S. This proves (iii)., The number R, defined by Theorem 6.41, is called the radius of convergence, and the circle |z| = R is often referred to as the circle of convergence. If R, is the radius of convergence of a series, then the disk |z| < R is called the
Page 185 :
176, , 6 Power Series, , disk/domain of convergence of the corresponding power series. The radius of, convergence depends only on the tail of the series. If we alter a finite number, of coefficients an of the series, the radius of convergence remains unchanged., Further, a power series always converges inside and diverges outside the circle, |z| = R. But there is no general principle regarding its behavior on the circle., Therefore, if it is required to find the convergence of a power series on its, circle of convergence, then this has to be investigated separately because it, may converge at all, none, or some of the points. We illustrate the last fact, by the following examples:, ∞, Examples 6.42., (i) The geometric series n=0 z n converges for |z| < 1, and diverges everywhere on |z| = 1, its circle of convergence. Note that, the series is, not uniformly convergent on the open disk |z| < 1., ∞, n, (ii) The series, n=1 z /n converges at z = −1 and diverges at z = 1., Therefore, , its radius of convergence must be R = 1 (Why?)., ∞, (iii) The series n=1 (z n /n2 ) converges absolutely (and uniformly) for |z| ≤, 1. This follows from the Weierstrass M -test with majorants Mn = 1/n2 ., If z = 1 + > 1, then (1 + )n /n2 → ∞ as n → ∞. Hence, the series, does, for |z| > 1, and the radius of convergence of the series, ∞ notnconverge, 2, (z, /n, ), is, R, = 1., n=1, ∞, (iv) The series n=1 nn z n cannot converge for any nonzero complex values, n, n n, because |nz|, ∞= |z|n n n→ ∞ (z = 0). Therefore, R = 0., (v) The series n=1 (z /n ) converges everywhere. To see this, choose z =, z0 . Then for N > |z0 |,, ∞, ∞, , , z0, z0n, <, nn, N, , n=N, , n, , =, , n=N, , |z0 /N |N, ,, 1 − |z0 /N |, , ∞, n, n, and the absolute convergence of, n=1 (z0 /n ) follows. Since z0 was, arbitrary, R = ∞. By Theorem 6.41, the series converges uniformly on, •, all compact subsets of the plane., Example 6.43. Let us discuss the convergence of the series, ∞, , , 3−|n| z 2n ., , n=−∞, , According to the discussion on geometric series, we may rewrite the given, series as, ∞, , , 3−|n| z 2n =, , n=−∞, , =, , ∞, , , 3−n z 2n +, , n=0, ∞ 2 n, , n=0, , z, 3, , +, , −1, , n=−∞, ∞, , , 3n z 2n, , (3z 2 )−n ., , n=1
Page 186 :
6.3 Maclaurin and Taylor Series, , 177, , Note that, |z|2 < 3, i.e., when, √, √ the first series on the right converges when, |z| < 3 and that it diverges for all z with |z| ≥ 3. Similarly,√the second, series on the right converges, when |3z 2 | > 1, i.e., when |z| > 1/ 3 and that, √, it diverges for |z| ≤ 1/ 3. It follows that, ∞, , , 3−|n| z 2n =, , n=−∞, , 1/(3z 2 ), 8z 2, 1, +, =, 2, 2, 2, 1 − z /3 1 − (1/3z ), (3 − z )(3z 2 − 1), , √, √, •, for 1/ 3 < |z| < 3, whereas the series diverges for all remaining z., ∞, More generally, Theorem 6.41 shows that a power series n=0 an (z − b)n, either converges absolutely in C or only at the origin or else there exists an, R > 0, for which the series converges absolutely in |z − b| < R and diverges, for |z − b| > R. If 0 ≤ r < R (if r > 0 for R = ∞), then the series converges, uniformly in |z − b| ≤ r., Thus far the radii of convergence for different power series have been determined only by the sometimes cumbersome method of testing distinct points, for convergence and divergence and applying Theorem 6.38. But a power series, is defined by its coefficients, and it is these coefficients alone that determine, its radius of convergence., ∞, Theorem 6.44. (Cauchy–Hadamard) The power series n=0 an z n has radius of convergence R, where 1/R = lim supn→∞ |an |1/n (Here we observe the, conventions 1/0 = ∞ and 1/∞ = 0)., Proof. For any point z0 = 0, we have, lim sup |an z0n |1/n = lim sup |an |1/n |z0 | =, n→∞, , n→∞, , 1, |z0 |., R, , ∞, According to Theorem 6.13, the series n=0 an z0n converges absolutely when, |z0 | < R and diverges when |z0 | > R. In view of Theorem 6.41, the radius of, convergence is R., When R = ∞, the series converges everywhere; and when R = 0, the series, converges only at z = 0., Most often the following result suffices to examine the convergence properties of the power series., ∞, Corollary 6.45. The radius of convergence R of the power series n=0 an z n, is determined by, (a), , 1, = lim |an |1/n, R n→∞, , provided these limits exist., , (b), , 1, an+1, = lim, R n→∞ an
Page 187 :
178, , 6 Power Series, , ∞, Examples 6.46., (i) Consider n=1 an z n , 4an = n/(4n2 + 1). Then the, ratio test given by Corollary 6.16 is applicable, as an = 0 for each, n ∈ N. We have, n+1, an+1, 4n2 + 1, → 1 as n → ∞., =, an, 4(n + 1)2 + 1, n, So, the series, for |z| < 1 and diverges for |z| > 1., ∞ converges, 1, 3n, (ii) Consider n=0 (3+i), . Note that the ratio test is not directly applinz, cable. However, we may think of this as a series in a new variable z 3, rather than in z:, ∞, , 1, wn , w = z 3 ., (3, +, i)n, n=0, Now, we can apply the ratio test to this new series. It follows that, √, 1, an+1, (3 + i)n, 1, √, = lim, = lim, =, 10, ,, i.e.,, R, =, n→∞ (3 + i)n+1, R n→∞ an, 10, √, and, √ so the new series converges for |w| < 10 and diverges for |w| >, 10. This is equivalent to saying that the original series converges for, |z| < 101/6 and diverges for |z| > 101/6 ., ∞, n3, (iii) Consider n=1 n+1, n! z . Again the ratio test is not applicable. But we, may fix z = 0 and let, n + 1 n3, an =, z ., n!, Then,, 3, , n!, (n + 2)z (n+1), (n + 2) (n+1)3 −n3, an+1, ×, =, =, z, an, (n + 1)!, (n + 1)2, (n + 1)z n3, and so, , (n + 2) 3n(n+1)+1, an+1, = lim, |z|, n→∞, n→∞ (n + 1)2, an, which is less than 1 provided |z| < 1, showing that the given series, converges for |z| < 1 , and diverges |z| > 1., ∞, (iv) Finally, we consider n=0 an z n , where, n, i2 for n even, an =, −3n for n odd ., lim, , , , Then, 1/n, , |an |, , =, , 2 for n even, 3 for n odd,, , showing that |an |1/n oscillates finitely. So, by Theorem 6.44 we see that, 1, 1/n, = lim sup |an |, = 3, i.e., R = 1/3,, R, n→∞, and the series converges for |z| < 1/3 and diverges for |z| > 1/3., , •
Page 188 :
6.3 Maclaurin and Taylor Series, , 179, , ∞, n, Example 6.47. Suppose that the power, n=0 an z has a positive ra∞ series, an n, dius of convergence. Then the series n=1 nn z is entire. , ∞, n, ∞To see nthis, let R > 0 be the radius of convergence of nn=1 an z . Then, n=0 an r converges, where 0 < r < R. In particular, {an r } is a bounded, sequence. Thus, there exist an M such that |an rn | ≤ M for all n ≥ 0. Now, M 1/n /r, → 0 as n → ∞, n, ∞, which, according to the root test, shows that n=1 an n−n z n is entire., ∞, Example 6.48. It is easier to show that the radius of convergence of n=0 (5+, (−1)n )z n is 1. Indeed, we note that, , 4 if n is odd, n, an = 5 + (−1) =, 6 if n is even., an, nn, , 1/n, , ≤, , Clearly, the ratio test is not applicable. On the other hand, as |z|1/n → 1 for, z = 0, it follows that, 1, = lim sup |an |1/n = 1, R, n→∞, which gives R = 1. How can we find the sum f (z) of the given series? Rewrite, the given series as, ∞, ∞, , , zn +, (−1)n z n ., f (z) = 5, n=0, , n=0, , Note that both the series on the right are known to converge for |z| < 1, and, diverge for |z| ≥ 1. The sum is then given by, f (z) =, , 1, 2(3 + 2z), 5, +, =, ., 1−z, 1+z, 1 − z2, , •, , Example 6.49. Let us find the radius of convergence of the series, ∞, , , sin(nπ/4)z n, , n=0, , and also its sum f (z). To do this, we set an = sin(nπ/4). As |an | ≤ 1 for all, n ≥ 0, we have, 1, = lim sup |an |1/n ≤ 1, i.e., R ≥ 1., R, n→∞, On the other hand an = ±1 for infinitely many n which shows that 1/R ≥ 1,, i.e. R ≤ 1. Hence, R = 1. Note that, a2(4k+1) = sin(2kπ + π/2) = 1 and a2(4k+3) = sin(2kπ + 3π/2) = −1., To find the sum, we easily compute that
Page 189 :
180, , 6 Power Series, , ⎧, ⎪, ⎪, ⎨, , √0, (−1)k / 2, an =, k, (−1), ⎪, ⎪, √, ⎩, k, (−1) / 2, , if, if, if, if, , n = 4k, n = 4k + 1, , k ∈ N0 ,, n = 4k + 2, n = 4k + 3, , and therefore, we may rewrite the given series in the form, z3, z7, z5, z, f (z) = √ + z 2 + √ + 0 − √ − z 6 − √ − 0 + · · ·, 2, 2, 2, 2, , , 3, z, z, (1 − z 4 + z 8 − · · · ), = √ + z2 + √, 2, 2, √, z(1 + 2z + z 2 ), √, •, ., =, 2(1 + z 4 ), ∞, Suppose the power series n=0 an z n has radius of convergence R. We wish, to characterize as fully as possible the behavior of the function f (z), defined, by the power series, at points interior to its circle of convergence. Implicit, in our work is the continuity of f (z) for |z| < R. To show continuity at an, arbitrary , point z0 , |z0 | < R, we note (by Theorem 6.41) that the sequence, n, Sn (z) = k=0 ak z k converges uniformly to f (z) in the disk |z| ≤ r = |z0 |., Since Sn (z) is continuous at z = z0 for every n, Theorem 6.26 may be applied, to establish continuity of the limit function f (z) at z = z0 ., The differentiability of f (z) is not so straightforward. We might expect the, derivatives of a sequence of uniformly convergent differentiable functions to, converge to a differentiable function. However, consider the sequence {fn (z)},, where, √, fn (z) = (sin nz)/ n., , Although {fn (z)}, √ converges uniformly on the real line, the sequence {fn (z)},, , where fn (z) = n cos nz, converges for no real values., Fortunately, no such pathological behavior can occur for the sequence of, partial sums of a power series. In a sense, a power series may be thought of as, a polynomial of infinite degree; indeed, a polynomial can be defined as a power, series in which all but a finite, n number of coefficients are zero. The derivative, of a polynomial Pn (z) = k=0 ak z k is, , Pn (z) =, , n, , , kak z k−1 ., , k=1, , We will prove a similar result for power series. But first we need the following:, Lemma 6.50. The two power series, ∞, , , an z n and, , n=0, , have the same radius of convergence., , ∞, , n=1, , nan z n
Page 190 :
6.3 Maclaurin and Taylor Series, , 181, , Proof. Using properties of the limit superior (Exercise 6.19(7)), we have, lim sup |nan |1/n = lim n1/n lim sup |an |1/n = lim sup |an |1/n ., n→∞, , n→∞, , n→∞, , n→∞, , The result now follows from Theorem 6.44., Theorem, 6.51. If a function f (z) is the pointwise limit of a power series, ∞, n, a, z, in |z| < R, then f (z) is analytic for |z| < R, with, n=0 n, f (z) =, , ∞, , , nan z n−1 ., , n=1, , Proof. Given z0 , |z0 | < R, we will show that, ∞, , f (z) − f (z0 ) , −, nan z0n−1 <, z − z0, n=1, whenever |z − z0 | < δ = δ( , z0 ) (see Fig. 6.2)., , Figure 6.2., , For z sufficiently close to z0 , there is a real number ρ satisfying the inequalities, |z0 | ≤ ρ,, , |z| ≤ ρ (ρ < R)., , For any integer N , we write, PN (z) =, , N, , n=0, , an z n, , (6.15)
Page 191 :
182, , 6 Power Series, , so that PN (z) =, , N, n=1, , nan z0n−1 . Therefore, we have, , ∞, , f (z) − f (z0 ) , −, nan z0n−1, z − z0, n=1, , (6.16), , n, , ∞, , PN (z) − PN (z0 ), z − z0n, − PN (z0 ) +, an, − nz0n−1, z − z0, z − z0, n=N +1, n, , ∞, , PN (z) − PN (z0 ), z − z0n, ≤, − PN (z0 ) +, an, − nz0n−1 ., z − z0, z − z0, , =, , n=N +1, , Denote the last two expressions by A1 and A2 , respectively. We shall first, choose N large enough so that A2 < /2, and then choose δ small enough so, that A1 < /2. From (6.15),, z n − z0n, − nz0n−1 ≤ |z n−1 + z n−2 z0 + · · · + z0n−1 | + n|z0 |n−1, z − z0, , (6.17), , ≤ |z|n−1 + |z|n−2 |z0 | + · · · + |z0 |n−1 + n|z0 |n−1, ≤ nρn−1 + nρn−1 = 2nρn−1 ., ∞, According to the lemma, n=1 n|an |ρn−1 converges and, by the Cauchy criterion, an integer N may be found such that, ∞, , , n|an |ρn−1 <, , n=N +1, , 4, , ,, , i.e., A2 < /2., , (6.18), , The inequality, A1 <, , 2, , for |z − z0 | < δ, , (6.19), , is a consequence of the differentiability of the polynomial PN (z) at z = z0 . If, we combine (6.18) and (6.19), the result follows from (6.16)., Remark 6.52. A power series may always be written as a polynomial plus a, “tail”. The essence of this proof consisted of showing the tail to be inconsequential., •, Remark 6.53. Theorem 6.51 says that every function defined by its power, series is analytic inside its radius of convergence. In Chapter 8, the converse, will also be proved. That is, every function analytic in a disk may be expressed, •, as a power series., An examination of Theorem 6.51 reveals that much, more has been proved, ∞, than was originally intended. The power series f (z) = n=0 an z n was shown, to have derivative
Page 192 :
6.3 Maclaurin and Taylor Series, , , f (z) =, , ∞, , , 183, , nan z n−1, , n=1, , which, itself, is a power series having the same radius of convergence as f (z)., Thus, Theorem 6.51 may be applied repeatedly to obtain, f (k) (z) =, , ∞, , , n(n − 1) · · · (n − (k − 1))an z n−k, , n=k, , = k!ak +, , (k + 1)!, (k + 2)!, ak+1 z +, ak+2 z 2 + · · · ,, 1!, 2!, , (6.20), , which is valid inside the circle of convergence of f (z). Setting z = 0 in (6.20), we see that the coefficients ak are related to the sum function f (z) of the, power series through the expression, f (k) (0) = k!ak ,, , i.e., ak =, , f (k) (0), k!, , (6.21), , for k = 0, 1, 2, . . . . Here we have used the conventions f (0) (z) = f (z) and, 0! = 1. We sum up these results as, Theorem 6.54. If a function f (z) is the pointwise limit of a power series, in some neighborhood of the origin, then f (z) has derivatives of all orders, at each point interior to the circle of convergence of f (z). Furthermore, the, coefficients of the power series are uniquely determined and are related to the, derivatives of f (z) at the origin by (6.21)., The representation, f (z) =, , ∞, , f (n) (0) n, z, n!, n=0, , is called the Maclaurin series expansion of f (z)., ∞, Example 6.55. Let us sum the series n=1 n(n + 3)z n for |z| < 1. To do, this, for z = 0, the geometric series shows that, ∞, , z3, =, z n+3, 1 − z n=0, , from which one obtains that, 3 , ∞, z, 1, =, (n + 3)z n ., z2 1 − z, n=0, Differentiating the left-hand side and then multiplying the resulting expression, by z would yield the desired sum., •
Page 193 :
184, , 6 Power Series, , All of our results on power series may easily be reworded to accommodate, expansionsin powers of z − b, where b is any complex number. For instance,, ∞, the series n=0 an (z − b)n converges absolutely to an analytic function f (z), inside the circle |z − b| = R, where R−1 = lim supn→∞ |an |1/n . Moreover, the, Taylor series expansion, f (z) =, , ∞, , f (n) (b), (z − b)n, n!, n=0, , is valid for |z − b| < R., ∞, Theorem 6.56. (see also Theorem 8.44) Let f (z) = n=0 an z n be a power, series with positive radius of convergence and {zk }k≥1 a sequence which converges to zero such that zk = 0 for all k ∈ N. Further assume that f (zk ) = 0, for k ∈ N. Then an = 0 for all n ∈ N0 ., Proof. As f is analytic at z = 0, f is continuous at z = 0. We have, zk → 0 ⇒ f (zk ) → f (0) ⇒ f (0) = 0., ∞, Next consider the function f1 (z) = n=1 an z n−1 which has the same radius, of convergence as f , and, 0 = f (zk ) = f1 (zk )zk ,, , k ∈ N., , Since zk = 0, f1 (zk ) = 0 for k ∈ N. Hence a1 = f1 (0) = 0. Continuing this, process we obtain the desired result., Questions 6.57., 1. Suppose a power series converges at z = z0 and diverges at z = z1 ., What is the relationship between z0 and z1 ?, 2. Suppose a power series converges at all the positive integers. What kind, of function does it represent?, ∞, 3. Can a power series n=0 an (z − 5i)n converge at z = 0 and diverge at, z = 1 + 7i?, 4. Is the set S, defined in Theorem 6.41, a closed set?, 5. When will the regions of absolute and uniform convergence of a power, series coincide?, ∞, ∞, 6. How do the convergence properties of n=0 an z n and n=0 nan z n compare?, ∞, 7. If n=0 an converges, what can be said about the radius of convergence, ∞, of n=0 an z n ?, 8. In what ways do power series having radius of convergence R = 0 or, R = ∞ differ from other power series?, 9. Suppose {fn } is a sequence of differentiable functions, and {fn } converges uniformly on a set E to a differentiable function. Must {fn }, converge on E?
Page 194 :
6.3 Maclaurin and Taylor Series, , 185, , 10. What can be said about the sum and product of power series?, 11. What analytic functions can be shown to have power series representations?, 12. Are the theorems in this section valid for real power series?, Exercises 6.58., ∞, n, has radius of convergence R. Show that the se1. Suppose, n=0 an z, n, quence{an z0 } is unbounded if |z, 0 | > R., ∞, ∞, z1n converges and n=0 an z2n diverges, with |z1 | = |z2 |,, 2. (a) If n=0 an, ∞, n, show, ∞ of convergence R = |z1 |., ∞that n=0 an z has radius, (b) , If n=0 an converges and n=0 |an | diverges, show that the series, ∞, n, n=0 an z has radius of convergence R = 1., 3. Suppose , |an | is a decreasing sequence. Show that the radius of conver∞, gence of n=0 an z n is at least 1., ∞, n, converges at an unbounded sequence of points., 4. Suppose, n=0 an z, Show that the power series converges everywhere. , ∞, 5. Suppose {an } is a sequence of integers. Prove that n=0 an z n is either, an entire function or has radius of convergence at most one., 6. Show that a power series converges uniformly on all compact subsets, interior to its circle of convergence., ∞, 7. Suppose limn→∞ |an /an+1 | = R. Show that n=0 an z n has radius of, convergence R., ∞, 8. Show that the radius of convergence of any power series n=0 an z n is, given by R = lim inf n→∞ |an |−1/n, ∞., 9. Show, for any integer k, that n=1 (nk /nln n )z n has radius of convergence R = 1., 10. Find the radius of convergence for, ∞, ∞, ∞ n, , , , nk n, i −1 n, k, n n, z, z, (b), (n, +, a, )z, (c), n, a, n, n=1, n=1, n=1, n2, ∞ , ∞, ∞, , , , 1, n2 + 5n + 3in n, (n!)2 n, z (f), z, (d), z n (e), 1+, n, 2n + 1, (2n)!, n=1, n=0, n=1, ∞, ∞, ∞, , , , 3, n, (g), n1/n (z + i)n, (h), (ln n)n z n, (i), n3 (z + 1)3, , (a), , (j), , n=1, ∞, , , n!z n, , (k), , n=1, , n=1, ∞, , , n!z n!, , (l), , n=0, , n=1, ∞, , , n! n, z ., n, n, n=1, , 11. Find the radius of convergence for, ∞, , z 2n, (a), 4n nk, n=1, , 3n, z 2n, (b), 2 + 4n, n, n=1, , ∞, , zn, (c), 2n 2, n=0, , (d), , (e), , (f), , (g), , 2, ∞, , zn, 2n, n=0, , ∞, , n=0, , ∞, , , cos(nπ/6)z n (h), , ∞, , 2n + 3n n, z, 4n + 5n, n=0, ∞, , n=0, , sin(nπ/16)z n ., , ∞, , , n=0, , 2n z n, , 2
Page 195 :
186, , 6 Power Series, , n, 12. Suppose {an } is a complex sequence whose partial sums i=1 ai are, bounded. If {b, n } is a real sequence that is monotonically decreasing to, , ∞, 0, show that , n=1 an bn converges., ∞, 13. (a) Show that n=1 (z n /n) converges everywhere on the circle |z| = 1, except z = 1., ∞, (b) Show, for |z1 | = 1, that n=1 (1/n)(z/z1 )n converges everywhere, on the unit circle except z = z1 ., (c) Suppose |z1 | = |z2 | = · · · = |zp | = 1. Show that, , , ∞, , 1, 1 1, +, ·, ·, ·, +, zn, n, n, n, z, z, p, 1, n=1, converges everywhere on the unit circle except z1 , z2 , . . . , zp ., 14. Write Taylor expansions for the polynomial P (z) = z 3 + 3z 2 − 2z + 1 in, powers of, (a) z − 1 (b) z + 2 (c) z − i., 15. Show that the series, , ∞, , (−1)n z 2n, 8n, n=0, , √, converges for |z| < 2 2. Find its sum., 16. Suppose that an = 0 for all n ∈ N. If R is the radius of convergence of, both the series, ∞, ∞, , , zn, an z n and, ,, a, n=1, n=1 n, then show that , R = 1., ∞, 17. Sum the series n=0 cos(nπ/3)z n ., , 6.4 Operations on Power Series, Our study of power series has revolved around the following three questions:, ∞, (i) For what values of z does n=0 an (z − b)n converge?, ∞, (ii) What properties may be attributed to f (z) = n=0 an (z −b)n at points, where the series converges?, (iii) Under what conditions may a function f (z) be represented by a power, series in some neighborhood of a point?, The first two questions are almost completely solved. The series, ∞, , , an (z − b)n, , n=0, , either converges everywhere, only at z = b, or there exists a circle for which, the series converges absolutely inside and diverges outside. The function
Page 196 :
6.4 Operations on Power Series, , f (z) =, , ∞, , , 187, , an (z − b)n, , n=0, , is analytic with derivatives of all orders inside the circle of convergence. Only, its behavior on the circle remains a mystery., The third question has not yet been properly dealt with. We know that, f (z) cannot be represented by a power series in a neighborhood of a point unless it has derivatives of all orders at each point in the neighborhood. Furthermore, if f (z) does have a power series representation in some neighborhood, of z = b, then that representation is unique, and its coefficients are related to, the derivatives at z = b by, f (z) =, , ∞, , f (n) (b), (z − b)n ., n!, n=0, , But we still have no criteria that will guarantee a power series development., Consider, for example, the function, f (z) =, , ∞, , z2, z3, zn, =1+z+, +, + ··· ., n!, 2!, 3!, n=0, , By the ratio test, the radius of convergence is ∞, and so f (z) is entire. Alternatively, in view of Theorem 6.44, it suffices to show that (n!)1/n → ∞., Beginning with the inequality, , n n n/2, ≥, n! ≥ n(n − 1) · · · n −, ,, 2, 2, we take nth roots of both sides to obtain, n 1/2, (n!)1/n ≥, → ∞., 2, Thus, f (z) is analytic everywhere. Moreover, f (0) = 1 and, by Theorem 6.51,, f (z) = f (z) for all z., We would like very much, at this point, to say that f (z) = ez . In fact, if, z, e does have a power series representation, then, f (z) = ez =, , ∞, ∞, , f (n) (0) n z n, z =, ., n!, n!, n=0, n=0, , (6.22), , To give us even more faith in the truth of (6.22), note that the identity, ex =, , ∞, , xn, n!, n=0, , is valid for all real x. This is proved in elementary calculus by use of Taylor’s, formula with remainder [T]. That is,
Page 197 :
188, , 6 Power Series, , f (x) = ex =, , n−1, , k=0, , f (k) (0) k, x + Rn (x),, k!, , where Rn (x), defined in terms of a real integral, approaches zero as n approaches infinity., The proof of (6.22) for complex z will be postponed until the theory of, complex integration has been developed, but even the most impatient reader, will find it worth the wait. Not only will we prove that all the familiar real, power series identities like, ex =, , ∞, , x2, xn, =1+x+, + ··· ,, n!, 2!, n=0, , sin x =, , ∞, , x3, x5, (−1)n+1 x2n−1, =x−, +, − ··· ,, (2n − 1)!, 3!, 5!, n=1, , cos x =, , ∞, , x2, x4, (−1)n x2n, =1−, +, − ···, (2n)!, 2!, 4!, n=0, , remain valid in the complex plane, but also that there is a converse to Theorem, 6.51; namely, that every analytic function admits a power-series expansion. In, particular, if f (z) is an entire function, then its Taylor series representation, f (z) =, , ∞, , f (n) (b), (z − b)n, n!, n=0, , is valid for all complex b and z., This becomes even more striking in view of the absence of a real variable, analog., Example 6.59. We remark that the function f (x) = x|x| is differentiable for, all real x, but cannot be expanded in a Maclaurin (real) series because f (0), •, does not exist., Example 6.60. The function f (x) = 1/(1 + x2 ) has derivatives of all orders, for all real x, although the Maclaurin expansion, f (x) =, , ∞, , f (n) (0) n, x = 1 − x2 + x4 − x6 + · · ·, n!, n=0, , is valid only in the real interval (−1, 1). There appears to be nothing in the, nature of the function to account for this restriction. But replacing x by the, complex variable z, we see that the function f (z) = 1/(1 + z 2 ) is not analytic, at z = ±i. This prevents a Maclaurin series from converging outside the circle, |z| = 1. In particular, for real values of z the series cannot converge outside, •, the real interval [−1, 1].
Page 198 :
6.4 Operations on Power Series, , 189, , Example 6.61. The function, , f (x) =, , 2, , e−1/x if x = 0,, 0 if x = 0, , has derivatives of allorders for all real values. Since f (n) (0) = 0 for every, ∞, (n), (0)/n!)xn ≡ 0. Hence, the Maclaurin series, integer n, we have, n=0 (f, •, represents the function only at the origin., Returning, once again, to functions of a complex variable, the sum of two, polynomials of degree n is a polynomial of degree at most n and is formed by, adding coefficients termwise. That is,, n, , , ak z k +, , k=0, , n, , , bk z k =, , k=0, , n, , , (ak + bk )z k ., , k=0, , The product of two polynomials of degree n is a polynomial of degree 2n, but, the relationship between coefficients is not as simple. We have, (a0 + a1 z + a2 z 2 + · · · + an z n )(b0 + b1 z + b2 z 2 + · · · + bn z n ), = a0 b0 + (a0 b1 + a1 b0 )z + (a0 b2 + a1 b1 + a2 b0 )z 2, + · · · + an bn z 2n ., More concisely,, n, , , , ak z, , k, , k=0, , , , n, , , bk z, , k=0, , k, , =, , 2n, , , ck z k ,, , ck =, , k, , , am bk−m ., , m=0, , k=0, , If two functions are known to have power series representations, then information about their sum and product can be obtained., ∞, ∞, n, n, and g(z) =, have, Theorem 6.62. Suppose f (z) =, n=0 an z, n=0 bn z, radii of convergence R1 and R2 , respectively. Then f (z) + g(z) and f (z)g(z), have power series representations whose radius of convergence is at least R =, min{R1 , R2 }., n, n, Proof. Set Sn (z) = k=0 ak z k and Tn (z) = k=0 bk z k . Then, Sn (z) + Tn (z) =, , n, , , (ak + bk )z k, , and Sn (z)Tn (z) =, , k=0, , where ck =, , k, m=0, , k=0, , am bk−m . For any point z0 , |z0 | < R, we have, , lim Sn (z0 ) = f (z0 ) and, , n→∞, , Hence, , 2n, , , lim Tn (z0 ) = g(z0 )., , n→∞, , ck z k ,
Page 199 :
190, , 6 Power Series, , lim (Sn (z0 ) + Tn (z0 )) = f (z0 ) + g(z0 ) =, , n→∞, , ∞, , , (an + bn )z0n, , n=0, , and, lim Sn (z0 )Tn (z0 ) = f (z0 )g(z0 ) =, , n→∞, , ∞, , , cn z0n ., , n=0, , Since z0 was arbitrary, both sides converge for |z| < R., ∞, ∞, Remark 6.63. The radii of convergence for n=0 (an +bn )z n and n=0 cn z n, may actually be greater than min{R1 , R2 }. If an ≡ 1, bn ≡ −1, then R1 =, R2 = 1; but, ∞, ∞, , , (an + bn )z n =, (1 − 1)z n ≡ 0,, n=0, , n=0, , and the series converges for all values of z. If, , , 2 if n = 0, −1 if n = 0, and bn =, an =, 2n if n ≥ 1,, 1 if n ≥ 1,, then R1 =, , 1, 2, , and R2 = 1. However, c0 = a0 b0 = −2 and for n ≥ 1,, cn =, , n, , , ak bn−k = a0 bn + an b0 +, , k=0, , n−1, , , ak bn−k, , k=1, , = 2 − 2n +, , n−1, , , 2k, , k=1, , = 2 − 2n +, Therefore,, , ∞, , n=0 cn z, , n, , 2 − 2n, = 0., 1−2, , = c0 = −2, and the series converges for all z. Note that, , ∞, , , 2(1 − z), 2z, =, 2 z =2+, f (z) = 2 +, 1 − 2z, 1 − 2z, n=1, n n, , , , 1, |z| <, 2, , and, g(z) = −1 +, , ∞, , n=1, , z n = −1 +, , 1 − 2z, z, =−, 1−z, 1−z, , (|z| < 1)., , •, , The only function, essentially, whose Maclaurin expansion we know as yet, in “closed form” is the geometric series, ∞, , 1, =, zn, f (z) =, 1 − z n=0, , (|z| < 1)., , For any nonzero complex number a, this leads to the identity
Page 200 :
6.4 Operations on Power Series, ∞, ∞, 1, 1 z n 1 n, 1, =, =, =, z, a−z, a(1 − z/a), a n=0 a, an+1, n=0, , 191, , (|z| < |a|)., , Also, for any two distinct complex numbers a and b, we have, , , 1, 1, 1, 1, =, −, (z − a)(z − b), a−b z−a z−b, &∞ , ', , 1, 1, 1, =, − n+1 z n, a − b n=0 bn+1, a, valid in the region, |z| < R = min{|a|, |b|}., , (6.23), , A Maclaurin expansion for 1/(1 − z)2 can be found by two different methods., Method 1: Setting an ≡ bn ≡ 1 in the proof of Theorem 6.54, we have, 1, 1, 1, =, ·, 2, (1 − z), 1−z 1−z, ∞, , ∞, , , zn, zn, =, n=0, , =, , ∞, , , n=0, , (n + 1)z n, , (|z| < 1)., , n=0, , Method 2: By Theorem 6.51 and the geometric series defined above,, f (z) =, , ∞, , 1, =, nz n−1, (1 − z)2, n=1, , (|z| < 1)., , (6.24), , It is usually much easier to decide whether or not a given series converges than, it is to find the value of a known convergent series., For instance, any student, ∞, of elementary calculus can show that the series n=0 (1/n3 ) converges, but, the finest mathematicians in the world have not yet developed methods to, find its sum. However, all is not lost; the closed form of the geometric series, does enable us to find the , value of many series., ∞, Let us find the sum of n=1 n/2n . To do this, by (6.24), we have, zf (z) =, Letting z =, , 1, 2, , ∞, , z, =, nz n, (1 − z)2, n=1, , (|z| < 1)., , in (6.25), we obtain, ∞, 1, , n, 2, =, 1 2 = 2., n, 2, (1, −, 2), n=1, , (6.25)
Page 201 :
192, , 6 Power Series, , ∞, ∞, It is interesting to note that n=0 2−n = n=0 n2−n . The apparent paradox, may be resolved by noting that the first term on the right vanishes., ∞, Example 6.64. Let us find the sum of n=1 (n2 /3n ). Differentiating (6.25), and multiplying by z, we have, ∞, , z + z2, =, n2 z n, (1 − z)3, n=1, , Setting z =, , 1, 3, , (|z| < 1)., , (6.26), , in (6.26) leads to, ∞, , ( 13 ) + ( 13 )2, 3, n2, =, = ., 1, n, 3, 3, 2, (1 − 3 ), n=1, , •, , The method employed in these examples may be used to evaluate any, series of the form, ∞, , nk, zn, n=1 0, , (k a positive integer, |z0 | > 1)., , It is sometimes possible to obtain, in closed form, a power series whose coefficients are defined recursively., Example 6.65. The Fibonacci sequence is defined by, an+2 = an+1 + an for all n ≥ 0,, ∞, with a0 = 0 and a1 = 1. Suppose f (z) = z + n=2 an z n . Then, f (z) = z +, =z+, , ∞, , n=0, ∞, , , an+2 z n+2, (an+1 + an )z n+2, , n=0, ∞, , , = z+z, , an+1 z n+1 + z 2, , n=0, , ∞, , , an z n, , n=0, , = z + zf (z) + z 2 f (z)., Solving, we obtain, , z, ., 1 − z − z2, The above manipulations are valid only at points√where the series converges., The roots of the denominator of f (z) are, √ z = (1± 5)/2. By (6.23), the radius, of convergence of f (z) is seen to be ( 5 − 1)/2., The geometric series may also be manipulated to obtain Taylor series, expansions. For example, we know that for any complex number b, b = 1,, the identity, f (z) =
Page 202 :
6.4 Operations on Power Series, , 193, , ∞, , 1, 1, 1, =, =, (z − b)n, 1−z, (1 − b)[1 − (z − b)/(1 − b)] n=0 (1 − b)n+1, , is valid whenever |z − b| < |1 − b|. Hence, f (z) = 1/(1 − z) has a Taylor, expansion about every point except z = 1. The reader may wish to check that, 1, f (n) (b), =, for every n., n!, (1 − b)n+1, , •, , Questions 6.66., 1. What are the differences between real and complex power series?, 2. Does the quotient of two power series have a power series representation?, 3. Suppose f (z) and g(z) have power series representations. What can be, said about f (g(z))?, 4. Suppose f (z) has a Maclaurin expansion with radius of convergence R., Can f (z) be analytic at a point z0 , |z0 | > R? Can f (z) be analytic, everywhere on the circle |z| = R?, 5. Suppose a function is known to have a power series representation. What, operations may then be used to evaluate specific infiniteseries?, ∞, 6. , If an = 0, how do the radii of convergence of the series n=0 an z n and, ∞, n, n=0 (1/an )z compare?, 7. If a function has a Taylor expansion about two distinct points, how will, the radii of convergence of the two power series compare?, 8. What can be said about power series representations for rational functions?, 9. Can a power series converge in an open disk |z| < R without being, absolutely convergent there? What about a closed disk |z| ≤ R?, Exercises 6.67., ∞, n, 1. Suppose, n=0 an z has radius of convergence R1 , 0 < R1 < ∞, and, ∞, n, bn z has radius of convergence R2 , 0 < R2 < ∞. Show that, n=0, ∞, (a) n=0 an bn z n has radius of convergence at least R1 R2 ., ∞, n, (b), n=0 (an /bn )z (bn = 0) has radius of convergence at most R1 /R2 ., Give examples, ∞ to show that inequality may hold in (a) and (b)., 2. Suppose n=0 an z n has radius of convergence R, 0 < R < ∞. Find the, radii of convergence for, (a), , ∞, , n=0, , an nk z n, , (b), , ∞, , an n, z, nk, n=0, , (c), , ∞, , an n, z, n!, n=0, , (d), , ∞, , , an n!z n ., , n=0, , Which of these answers are different if R = 0 or R = ∞?, 3. Derive the power series of (1 − z)−4 (about a = 1) from the geometric, series (about a).
Page 203 :
194, , 6 Power Series, , ∞, 4. Suppose, an z n has radius of convergence R. Give examples in, ∞, ∞ n=0 kn, 2, which n=0 an z (k a positive integer) and n=0 an z n have radius, of convergence R, and radius of convergence greater than R., 5. (a) Suppose f (z) can be expanded in a Taylor series about the point, z = a. Show that f (z) is entire if |f (n) (a)| ≤ M for some constant, M and for every integer n., (b) Show that f (z) is entire if |f (n) (a)| ≤ nk for some integer k and all, n., 6. Find the sum of the following series., (a), , ∞, , n2 + 2n − 1, 3n, n=1, , (b), , ∞, , n(3n − 2n ), 6n, n=2, , (c), , ∞, , , 5in, ., (1, + i)n, n=1, , !, ∞, 7. Find the radius of convergence of the series n=1 n12 + (−3)n z n ., 8. Suppose that an + Aan−1 + Ban−2 = 0 (n = 2, 3, 4, . . . ). Show that, ∞, , n=0, , an z n =, , a0 + (a1 + a0 A)z, 1 + Az + Bz 2, , at all points where the power series converges. What is the radius of, convergence?, = an+1 + an , with a0 = 0, 9. For the Fibonacci sequence defined, √ by a!n+2, n, and a1 = 1, show that an ≤ 2/( 5 − 1) for every n., 10. Suppose a, b, and c are distinct nonzero complex numbers. Find a Taylor, series expansion for f (z) = 1/(z − a)(z − b) about the point z = c, and, determine its radius of convergence., 11. Assume that f is analytic for |z| < r for some r > 0 and f satisfies the, functional equation f (2z) = (f (z))2 for all z sufficiently close to zero, (which is to make sure that both z and 2z lie in some disk about 0 that, is contained in Δr ). Show that f can be extended to an entire function., Determine all such entire functions explicitly.
Page 204 :
7, Complex Integration and Cauchy’s Theorem, , One of the most important theorems in calculus is properly named the fundamental theorem of integral calculus. On the one hand it relates integration, to differentiation, and on the other hand it gives a method for evaluating, integrals. In this chapter, we mainly look for a complex analog to develop, a machinery of integration along arcs and contours in the complex plane., The problem, of course, is that between any two points there are an infinite, variety of paths along which to integrate. The antiderivative of a complexvalued function f (z) of a complex variable z is completely analogous to that, for a real function; it is indeed a complex function F whose derivative is f ., Cauchy’s theorem, the fundamental theorem of complex integration says that, for analytic functions, one path over special domains is as good as another., , 7.1 Curves, We begin by recalling some properties of the Riemann integral. Suppose f (t), and g(t) are real-valued functions continuous on the interval a ≤ t ≤ b. Then, *b, *b, the Riemann integrals a f (t) dt and a g(t) dt exist. Further, for any real, constants c1 and c2 , we have the linearity property, +, , +, , b, , a, , +, , b, , (c1 f (t) + c2 g(t)) dt = c1, , f (t) dt + c2, a, , b, , g(t) dt, , (7.1), , a, , and the integral inequality, +, , +, , b, , f (t) dt ≤, a, , b, , |f (t)| dt., , (7.2), , a, , The integration of a complex-valued function of complex variable along a, contour leads to results of great importance both in pure and applied sciences., Consider now the complex-valued function
Page 205 :
196, , 7 Complex Integration and Cauchy’s Theorem, , F (t) = F1 (t) + iF2 (t),, where F1 (t) and F2 (t) are real-valued functions continuous on the interval, [a, b]. For example, eit and (1 + 2 cos t) − 2i sin t are complex-valued functions, defined on every interval in R. Obviously, as F1 and F2 are integrable over, the interval [a, b], the definite integral of F (t) is defined by, +, , +, , b, , +, , b, , F (t) dt =, a, , F1 (t) dt + i, a, , and, , +, , +, , b, , F1 (t) dt, , +, , a, , +, , b, , F (t) dt =, , b, , Im F (t) dt =, , a, , (7.3), , b, , Re F (t) dt =, a, , b, , Im, , F2 (t) dt., a, , First, we observe that, + b, +, Re, F (t) dt =, a, , b, , a, , F2 (t) dt., a, , Many familiar rules of integration for real-valued functions can be carried over, to the complex case. For instance, the linearity property expressed in (7.1) is, true for complex-valued functions and complex constants. The proof consists, of separating into real and imaginary parts. To prove (7.2) for continuous, complex-valued functions F (t) on [a, b], suppose, +, , b, , (R > 0, − π < α ≤ π)., , F (t) dt = Reiα, a, , Then, +, , b, , e−iα F (t) dt = e−iα, , +, , a, , +, , b, , b, , F (t) dt = R =, a, , F (t) dt ., , (7.4), , a, , In view of (7.4) and properties of the real integral,, +, R = Re, , b, , e−iα F (t) dt =, , a, , +, , b, , !, Re e−iα F (t) dt, , a, , +, ≤, , b, , −iα, , |e, a, , +, , b, , |F (t)| dt., , F (t)| dt =, a, , *b, The inequality is obvious when a F (t) dt = 0. Thus, the magnitude of an, integral does not exceed the integral of the absolute value of the integrand., Later in Theorem 7.19, we show that a similar inequality holds for integration, along contours., Suppose that f (z) is a complex-valued function of a complex variable z, defined on a subset Ω ⊆ C. Suppose that z1 and z2 are two points in Ω. At, first, we are concerned with the following:
Page 206 :
7.1 Curves, , 197, , Problem 7.1. How do we define the integral of a complex-valued function f, of complex variable z from z1 to z2 ?, Our discussion thus far provides no help in dealing with Problem 7.1. Overcoming this problem will necessitate integrating along more general curves, than real intervals. First, we must define the notion of a curve in C., A continuous curve (an arc) C in the complex plane is defined parametrically by, C : z(t) = x(t) + iy(t), , (t ∈ [a, b], a < b),, , (7.5), , where x(t) and y(t) are real-valued, continuous functions of the real variable, t. We will henceforth assume that all curves are continuous curves, so that, the terms “curve” and “arc” may be used interchangeably. So, by a curve C, in C, we mean a continuous function from [a, b] into C., A curve may have more than one parameterization. For instance,, z1 (t) = t (t ∈ [0, 1]) or z2 (t) = t2 (t ∈ [0, 1]), represents the interval [0, 1]. A natural ambiguity arises when dealing with, curves. Though a curve is defined to be a function, and its properties are, those of functions, we shall also refer to the point set representing the graph, of the function as “the curve”. Thus, a curve is a continuous function as well, as a compact, connected set of points. When the topological properties of a, curve are being discussed, the curve will sometimes be denoted by C, without, regard to the parameterization from which the curve arose., For a parameterized curve C defined by (7.5), the point z(a) is called, the initial point of C and z(b) the terminal point of C. If the initial and, terminal points coincide, i.e., z(a) = z(b), then C is said to be a closed curve., If z(t1 ) = z(t2 ) when t1 = t2 , so that C does not intersect itself, the curve, is said to be simple. A closed curve C : z(t), t ∈ [a, b], that is simple in the, interval (a, b) with the possible exception that z(a) = z(b) is said to be a, simple closed curve or Jordan curve., Every simple closed curve cuts the plane into two separate domains. In, other words, we say that every simple closed curve has an interior (inside), and an outside (exterior). We warn the reader that Jordan curve can be more, complicated than Figure 7.1. More formally, we have, Jordan Curve Theorem. If C is a simple closed (Jordan) curve, then the, complement of C consists of two disjoint domains, one bounded domain, and, the other an unbounded domain each of which has C as its boundary., This geometrically intuitive theorem is remarkably difficult to prove. The, reader unwilling to accept the theorem on faith can find a proof in Newman, [Ne]. However, given a drawing of some particular simple closed curve, it is, usually easy to distinguish the inside from the outside., A domain D is simply connected if each simple closed curve contained in, D contains only points of D inside.
Page 207 :
198, , 7 Complex Integration and Cauchy’s Theorem, , Figure 7.1. Illustration for Jordan curve theorem, , For instance, consider the punctured unit disk D = {z : 0 < |z| < 1}., Then D is a domain but is not simply connected. Some curves, such as C2 ,, C3 and C4 in Figure 7.2 contain only points on D, but C1 contains z = 0 and, z = 0 does not belong to D. We have the following heuristic interpretations., , Figure 7.2. Illustration for multiply connected domains, , Topologically, a simply connected domain can be continuously shrunk to a, point. Note that the punctured unit disk D = {z : 0 < |z| < 1} can be, shrunk to an arbitrarily small domain, but not to a point in D. Geometrically,, a “simply connected domain” has “no holes” inside, for if a simple closed, curve should surround a hole, then the curve could not be shrunk beyond, the hole. Here again, removal of a single point from a domain is akin to, punching a hole in it. A domain that is not simply connected is said to be, multiply connected.Open disks, open rectangles and star shaped domains are, simple examples of simply connected domains. Punctured disks, punctured, rectangles, and the punctured plane all have one “hole” and hence are not, simply connected. The domain in Figure 7.3 has three holes and hence is not, simply connected., Remark 7.2. In discussing simply connected domains, we will confine ourselves to the finite complex plane. Consequently, the exterior of a circle is, not simply connected, since the domain is prevented from being shrunk to a
Page 208 :
7.1 Curves, , 199, , Figure 7.3. Multiply connected domain with three holes, , point by the circle from one end and by the point at ∞ from the other. In the, extended plane, the exterior of a circle is simply connected because it can be, •, “shrunk” to the point at ∞., Remark 7.3. While we have required analytic functions to be single-valued,, it is of interest to discuss a slightly more general concept than analytic, which, incorporates multiple-valued functions. We are tempted to say that log z is, analytic in the punctured plane because for each value z0 = 0 a branch of, log z may be found in which log z is analytic at z0 . As we shall see in Chapter 13, this property of the logarithm will enable us to call the multiple-valued, function log z regular in the punctured plane. Note that a branch cut for log z, transforms the multiple-valued function into a single-valued function, and also, transforms a multiply connected domain (the punctured plane) into a simply, connected domain. After the term “regular” is carefully defined in Chapter 13,, we shall prove that a function regular in a simply connected domain must also, be single-valued (hence analytic) there. This is known as the Monodromy The•, orem., The boundary C of a domain is said to have positive orientation, or to be, traversed in the positive sense if a person walking on C always has the domain, to his left. The boundary |z − a| = R of the disk |z − a| < R has positive orientation if traversed in a counterclockwise direction and negative orientation, if traversed in a clockwise direction. A word of caution: Don’t equate positive, with counterclockwise. For the annulus in Figure 7.4, the positive direction, along the outer circle is counterclockwise, while along the inner circle it is, clockwise., , Figure 7.4. An annulus region
Page 209 :
200, , 7 Complex Integration and Cauchy’s Theorem, , However, if a simple closed curve is given without reference to a region, it, will be assumed that the domain is inside so that the positive orientation will, be counterclockwise., Remark 7.4. To unquestioningly accept the idea of counterclockwise is to, be deluded by the term “simple” closed curve. There are examples of simple, closed curves that are not “simple” in the intuitive sense of the word, which, occupy almost an entire square. For such a curve, the reader could spend a, lifetime tracking down the counterclockwise direction. Although our definition, of orientation is more intuitive than rigorous, it will be adequate for all curves, •, encountered in this text., Suppose that, C1, C2, C3, C4, , :, :, :, :, , z1 (t) = eit = cos t + i sin t, z2 (t) = e−it = cos t − i sin t, z3 (t) = −eit = − cos t − i sin t, z4 (t) = −e−it = − cos t + i sin t, , (0 ≤ t ≤ 2π),, (0 ≤ t ≤ 2π),, (0 ≤ t ≤ 2π),, (0 ≤ t ≤ 2π)., , All four of these simple closed curves traverse the unit circle. They differ from, one another either in initial point or in orientation. The curves C1 and C2, have initial point (1, 0), whereas C3 and C4 have initial point (−1, 0). The, curves C1 and C3 have positive orientation, and the curves C2 and C4 have, negative orientation (see Figure 7.5)., , Figure 7.5. Oriented curves, , A curve z(t) = x(t) + iy(t), t ∈ [a, b], having a continuous derivative (i.e.,, z(t) and z (t) are continuous on [a, b]) is said to be smooth or continuously, differentiable on [a, b]. Of course, by the derivatives at the end points a, b, we, mean the appropriate one sided derivatives z (a+) and z (b−). For example,, z (a+) = lim+, t→a, , z(t) − z(a), ., t−a, , A curve γ that is not smooth consists of a finite sequence of smooth curves,, γ1 , γ2 , . . . , γn joined together end-to-end. In other words, by a curve we mean, a continuous piecewise smooth curve defined on a closed interval.
Page 210 :
7.1 Curves, , 201, , Suppose that f (x) = u(x) + iv(x) is a complex-valued continuous function, defined on [a, b]. As in the construction of Riemann integral of a real-valued, function over [a, b], we consider a partition, P : a = x0 < x1 < x2 < · · · < xn = b, and form the corresponding Riemann sum:, n, , , f (x∗k )(xk − xk−1 ) =, , k=1, , n, , , u(x∗k )(xk − xk−1 ) + i, , k=1, , n, , , v(x∗k )(xk − xk−1 ), , k=1, , where x∗k is a point in [xk−1 , xk ]. As u and v are real-valued continuous on, [a, b], the Riemann sum on the right converges to, +, , +, , b, , b, , u(x) dx + i, a, , v(x) dx, a, , which leads us to define the integration of a complex-valued continuous function of a real variable:, + b, + b, + b, f (x) dx =, u(x) dx + i, v(x) dx., a, , a, , a, , If f (x) is piecewise continuous on [a, b], then apply the above result to each, subintervals (ak−1 , ak ) (1 ≤ k ≤ m) on which f (x) is continuous, and define, +, , b, , f (x) dx =, a, , m +, , k=1, , ak, , u(x) dx + i, , ak−1, , m +, , k=1, , ak, , v(x) dx., , ak−1, , Thus, (7.1) continues to hold if c1 , c2 are complex constants and f, g are, piecewise continuous complex-valued function defined on [a, b]., More generally, if f (z) is a complex-valued continuous function defined on, a smooth curve, C : z(t) = x(t) + iy(t), t ∈ [a, b],, then it follows that t → f (z(t)) is a continuous function from [a, b] into C., Consequently, t → f (z(t)) is continuous for a ≤ t ≤ b. We wish to prove that, the integral of f (z) on C is given by, +, , +, f (z) dz =, C, , b, , f (z(t))z (t) dt., , (7.6), , a, , We call the right-hand side of the last equation as a “pullback” of the lefthand side of the equation to the interval [a, b]. Let us now first define this, integral as a limit of sums, analogous to the definition of the Riemann integral., An advantage of (7.6) is that it enables us to use familiar properties of the, Riemann integral.
Page 211 :
202, , 7 Complex Integration and Cauchy’s Theorem, , Let P be a partition of [a, b] and zk∗ = z(t∗k ) denotes the point on the, subarc, with end points z(tk−1 ) and z(tk ). The Riemann sum approximating, *, f, (z), dz corresponding to the partition P is given by, C, Sn =, , n, , , f (z(t∗k ))(z(tk ) − z(tk−1 ))., , k=1, , We have, Sn =, , n, , , [u(x(t∗k ), y(t∗k )) + iv(x(t∗k ), y(t∗k ))], , k=1, , [x(tk ) + iy(tk ) − (x(tk−1 ) + iy(tk−1 ))], =, , n, , , u(x(t∗k ), y(t∗k ))(x(tk ) − x(tk−1 )) − v(x(t∗k ), y(t∗k ))(y(tk ) − y(tk−1 )), , k=1, , +i[u(x(t∗k ), y(t∗k ))(y(tk ) − y(tk−1 )) + v(x(t∗k ), y(t∗k ))(x(tk ) − x(tk−1 ))]., Interpreting each sum on the right as a Riemann sum over the interval [a, b],, we have the complex line integral (or contour integral) of f along C as follows:, +, f (z) dz = lim Sn, |P |→0, , C, , +, , b, , +, , , , b, , u(z)x (t) dt −, , =, a, , +, +i, , +, =, , v(z)y (t) dt, , a, b, , u(z)y (t) dt + i, , a, , +, , b, , v(z)x (t) dt, , a, , f (z(t))γ (t) dt,, , C, , where |P | denotes the maximum of the length of the subintervals. Thus, we, have actually proved (7.6). We may conveniently write the last expression as, +, , +, f (z) dz =, , b, , [ux − vy ] dt + i, , a, , C, , +, , b, , [uy + vx ] dt., , (7.7), , a, , Thus, just as a complex function may be expressed in terms of real-valued, functions, so may a complex integral clearly be expressed in terms of realvalued integrals. We formulate the above discussion as, Theorem 7.5. Suppose that f (z) = u(x, y) + iv(x, y) is continuous on a parameterized smooth curve C : z(t) = x(t) + iy(t), t ∈ [a, b]. Then, +, +, +, f (z) dz =, u dx − v dy + i, u dy + v dx., (7.8), C, , C, , C
Page 212 :
7.1 Curves, , 203, , Observe that the right side of (7.8) would be obtained by the formal substitution, f = u + iv, dz = dx + i dy, into the left side of (7.8). Either equation (7.7) or (7.8) could have been taken, as the definition of the complex integral, instead of (7.6)., Also, we observe that the integrand on the right side of (7.6) would be, obtained by the formal substitution, dz = z (t) dt, , z = z(t),, , into the left side of (7.6). Moreover, in the special case that z(t) = t, the curve, is a real interval and (7.6) reduces to an integral of the form (7.3)., For a general piecewise smooth curve C, the derivative z (t) need not be, continuous but is piecewise continuous so that, t → f (z(t))z (t), is piecewise continuous. In this case we evaluate the integral as a finite sums, of integrals of continuous functions. The above discussion leads to, Definition 7.6. Let C be a piecewise smooth curve on [a, b] and f a continuous function on the graph/trace of C. The contour integral of f along C is, defined to be, +, + b, f (z) dz =, f (z(t))z (t) dt., C, , Sometimes the notation, smooth curve., , a, , *, γ, , f (z) dγ or, , *, γ, , f dγ is used when γ is a piecewise, , *, An expression of the form C P (x, y) dx + Q(x, y) dy is called a real line, integral. From (7.7), we see that the complex (line) integral may be expressed, in terms of two real line integrals. We give here an example to illustrate, different methods for computing a complex integral., *, Example 7.7. Consider the problem of evaluating I = γ z 2 dz, where, (i) γ is an arc of a circle centered at the origin, (ii) γ is the union of the horizontal segment from 0 to 1 and the vertical, segment from 1 to 1 + 2i, (iii) γ is the line segment from 0 to 1 + 2i, (iv) γ is the contour parameterized by γ : z(t) = t2 + it (0 ≤ t ≤ 1)., Let f (z) = z 2 . In the first case we may write γ in the form, γ(t) = reit , a ≤ t ≤ b., Then γ (t) = ireit and f (γ(t))γ (t) = ir3 e3it so that
Page 213 :
204, , 7 Complex Integration and Cauchy’s Theorem, , +, , b, , I=, , f (γ(t))γ (t) dt = r3, , a, , e3ib − e3ia, ., 3, , In particular if γ is a closed circle, then I = 0, since b = a + 2kπ for some, integer k. In the second case, we may write γ as, %, t if 0 ≤ t ≤ 1,, γ(t) =, 1 + (t − 1)i if 1 ≤ t ≤ 3., Therefore, we have, + 1, +, , I=, f (γ(t))γ (t) dt +, 0, , 1, , 3, , , , +, , f (γ(t))γ (t) dt =, , 1, , 2, , +, , t dt +, 0, , 1, , 3, , [1 + (t − 1)i]2 i dt, , and, it is a simple exercise to see that I = −(11 + 2i)/3., In the third case, the path γ is given by γ(t) = t(1 + 2i), 0 ≤ t ≤ 1., Therefore the integral is, + 1, 1, 11 + 2i, t3, ., [t2 (1 + 2i)2 ](1 + 2i) dt = (1 + 2i)3, =−, 3 0, 3, 0, In the final case, according to (7.6), we can easily see that, +, 2 2, z 2 dz = − + i., 3, 3, γ, , •, , Remark 7.8. At first glance, it appears that (7.8) serves no purpose other, than to introduce a cumbersome method for evaluating the complex integral., We will rarely use (7.8) to compute integral directly. However, it will enable us, to formulate theorems about the complex integral from theorems about real, line integrals. This will lead to a method for evaluating the complex integral, that is far simpler than (7.6)., •, Example 7.9. Consider the curve γ given by, , t(1 + it sin(1/t)) if t = 0, z(t) =, 0 if t = 0., Then, z (t) =, , , , 1 + i(2t sin(1/t) − cos(1/t)) if t = 0, 1 if t = 0., , Note that z (t) is discontinuous at 0 and neither the left nor the right limit of, z (t) exists at 0. So, z (t) is not piecewise continuous, for example on [−π, π]., •, Consequently, the restriction of the curve γ to [−π, π] is not smooth., *b, Remark 7.10. The value of the real integral a f (x) dx depends on the function f (x)* and the end points of the interval [a, b]. The value of the complex, integral C f (z) dz may depend on the function f (z) and all the points on the, •, curve C, not just the end points of C.
Page 214 :
7.1 Curves, , 205, , y, 1+i, C1, C2, x, , Figure 7.6. Graph of curves C1 and C2, , Example 7.11. We wish to find, , *, C, , |z|2 dz along the curves, , (a) C = C1 : z1 (t) = t + it (0 ≤ t ≤ 1),, (b) C = C2 : z2 (t) = t2 + it (0 ≤ t ≤ 1)., Clearly, C1 and C2 are smooth and f (z) = |z|2 is continuous on C. According, to (7.6),, +, + 1, + 1, 2 2, |z|2 dz =, |t + it|2 (1 + i) dt = (1 + i), 2t2 dt = + i, 3 3, C1, 0, 0, and similarly,, +, , |z|2 dz =, , +, , 1, , 0, , C2, , |t2 + it|2 (2t + i) dt =, , 8, 5, + i., 6 15, , Despite the fact that the straight line C1 and the parabola C2 have the, same initial and terminal points (see Figure 7.6), we have, +, +, 2, |z| dz =, |z|2 dz., C1, , C2, , Note, however, that, +, +, + 1, 1, z 2 dz =, (t + it)2 (1 + i) dt = 2i(1 + i) =, z 2 dz., 3, 0, C1, C1, It is no coincidence that, , +, , 2, , +, , z 2 dz., , z dz =, C1, , C2, , *, It will later be shown that the value of C z 2 dz depends only on the initial, and terminal points of the smooth curve C. Our goal in this chapter is to, characterize the class of functions for which the integral is independent of, path, i.e., functions for which, +, +, f (z) dz =, f (z) dz, C1, , C2, , along any smooth curves C1 and C2 having the same initial and terminal, •, points.
Page 215 :
206, , 7 Complex Integration and Cauchy’s Theorem, , Questions 7.12., 1. Is a single “point” a curve?, 2. Is a simple closed curve a simple curve?, 3. In our definition of curve, would it have made any difference had the, parameter t been restricted to the interval [0,1]?, 4. What is the relationship between simply connected and connected?, 5. Can a domain and its complement both be simply connected?, 6. Can a domain have finitely many holes?, 7. Can a domain have infinitely many holes?, 8. When a point is removed from a simply connected domain, is the new, domain simply connected?, 9. Why was it important for the derivative of a curve to be continuous?, 10. What is the relationship between an integral being independent of path, and an integral around a closed curve being zero?, 11. Why is every curve compact and connected?, 12. Let D be the complement in C of the real axis, i.e., D = C \R. Is D, simply connected?, 13. Let D be the complement in C of the nonnegative real axis, i.e., D =, C \{x ∈ R : x ≥ 0}. Is D simply connected?, Exercises 7.13., 1. Describe the curve z(t) = a cos t + ib sin t (−π ≤ t ≤ π), where a and b, are positive real numbers., 2. Describe the curve z(t) = t3 + it2 (−1 ≤ t ≤ 1). Is it a “smooth curve”?, 3. Describe the curve, z(t) =, , 2t, 1 − t2, +i, (−R ≤ t ≤ R)., 2, 1+t, 1 + t2, , What happens as R → ∞?, 4. Plot the given, ⎧ curves, t if −3 ≤ t ≤ −1, ⎨, (i) z(t) = eiπ(1−t)/2 if −1 ≤ t ≤ 1, ⎩, t if 1 ≤ t ≤ 3., ⎧, t(1 + i) if 0 ≤ t ≤ 1, ⎨, 3 + i − 2t if 1 ≤ t ≤ 2, (ii) z(t) =, ⎩, (−1 + i)(3 − t) if 2 ≤ t ≤ 3., 5. Find a parameterized curve tracing out the following loci:, (a) The line segment from z = i to z = 1 − i, (b) The line segment from z = 1 to z = 2 + 3i, (c) The square whose vertices are ±1±i, traversed in the positive sense,, with initial point −1 − i, (d) The part of the circle |z − 1| = 2 in the right half-plane., 6. Find a parameterized curve for the parabola y = 2x2 − 3 that has initial, point z = −1 − i and terminal point z = 2 + 5i.
Page 216 :
7.2 Parameterizations, , 207, , 7. Parameterize the following simple closed curves in polar coordinates., (a) x2 + y 2 = 4 (b) 4x2 + y 2 = 1 (c) x2 + (y + 1)2 = 9., *, 8. Find C z dz along the following curves., (−π ≤ t ≤ π), (a) z(t) = eit, (−π ≤ t ≤ π), (b) z(t) = e2it, (−π ≤ t ≤ π), (c) z(t) = eit − 1, (d) z(t) = t + it, (0 ≤ t ≤ 2), (0 ≤ t ≤ π), (e) z(t) = 5eit + 3, (f) z(t) = 1 − t + it, (0 ≤ t ≤ 1), (g) z(t) = 1 + it, (0 ≤ t ≤ 1), (h) z(t) = 1 + i − t, (0 ≤ t ≤ 1)., *, 9. Along the curve C : z(t) = eit (−π ≤ t ≤ π), evaluate C f (z) dz for, 1, (b) f (z) =, (a) f (z) = z 2, z, , , 1, 1, ., (d) f (z) = 2z − i z +, (c) f (z) = 2, z, z, , 7.2 Parameterizations, Suppose C : z(t) is a smooth curve defined on the interval [a, b]. Breaking, the interval into two subintervals [a, c] and [c, b], we obtain two curves C1 and, C2 from z(t) by restricting the parameter t to the intervals [a, c] and [c, b],, respectively. For any function f (z) continuous on C,, +, , +, , b, , f (z) dz =, C, , a, , +, , c, , =, +a, =, , f (z(t))z (t) dt, + b, f (z(t))z (t) dt +, f (z(t))z (t) dt, c, +, f (z) dz +, f (z) dz., , C1, , C2, , Similarly, the curve C can be expressed as the “sum” of n curves with, +, +, f (z) dz =, f (z) dz, (7.9), C, C1 + ··· +Cn, +, +, +, =, f (z) dz +, f (z) dz + · · · +, f (z) dz., C1, , C2, , Cn, , Remark 7.14. By the sum of two curves, we mean the curve formed by, joining the initial point of one curve to the terminal point of the other; this, is not to be confused with termwise addition of functions defined on the same, •, set.
Page 217 :
208, , 7 Complex Integration and Cauchy’s Theorem, , A function is sectionally continuous on an interval if it has at most a finite, number of discontinuities, with right- and left-hand limits at each point in the, interval. More precisely, f is sectionally continuous on [a, b] if, (i) f is continuous at all but finitely many points on (a, b)., (ii) At any point c in (a, b) where f fails to be continuous, both left limit, limt→c− f (t) and the right limit limt→c+ f (t) exist and are finite., (iii) At the end points, the right limit limt→a+ f (t) and the left limit, limt→b− f (t) exist and are finite., A curve having a sectionally continuous derivative is called a contour. In, other words, piecewise smooth curve is called a contour. Recall that a path, C : z(t), t ∈ [a, b] is said to be piecewise smooth if there exists a partition, P : a = t0 < t1 < · · · < tn = b of [a, b] such that the restriction of C to each, of the subintervals [tk−1 , tk ], k = 1, 2, . . . , n, is a smooth curve., Since every contour C may be expressed as the sum of a finite number of, smooth curves, C1 + C2 + · · · + Cn , the integral of a continuous function along, a contour is defined by (7.9)., Example 7.15. The curve, C : z(t) = t + i|t|,, , t ∈ [−1, 1],, , is a piecewise smooth but not a smooth curve. It is easy to see that the, derivative at the origin fails to exist. The restriction of C to [−1, 0] and to [0, 1], is clearly seen to be smooth. Hence, C is referred to as a contour. Note also that, C is simple but not closed. How about the curve described by z(t) = |t| + it,, t ∈ [−1, 1]? How about the curve described by z(t) = |t3 | + it3 on the interval, •, t ∈ [−1, 1]?, Define C = C1 + C2 + C3 + C4 , where Cj ’s are the line segments given by, C1 = [0, 1], C2 = [1, 1 + i], C3 = [1 + i, i], C4 = [i, 1]; see Figure 7.8. Then, C describes the boundary of a square. Note that the curve C is piecewise, smooth, simple and closed., *, Example 7.16. We find the value of the integral C z dz along the contour, , Figure 7.7. The piecewise smooth curve C : z(t) = t + i|t|, t ∈ [−1, 1]
Page 218 :
7.2 Parameterizations, , 209, , Figure 7.8. The curve C = C1 + C2 + C3 + C4, , , , 2t if 0 ≤ t ≤ 1,, 2 + i(t − 1) if 1 ≤ t ≤ 2., , C : z(t) =, , Defining curves C1 and C2 by restricting the parameter t of C to the intervals, [0, 1] and [1, 2], respectively (see Figure 7.9), we have, +, +, +, z dz =, z dz +, z dz, C, , C1, 1, , C2, , +, =, , +, , 2, , 2t(2 dt) +, 0, , +, , +, , 1, , =, 0, , = 2−, , 1, , 4t dt −, , {2 + i(t − 1)}(i dt), , 2, 1, , +, , (t − 1) dt +, , 2, , 2i dt, 1, , 1, 3, + 2i = + 2i., 2, 2, , •, , Recall, from elementary calculus, that the (arc) length L of a smooth curve, in the plane defined parametrically by the equations, x = φ(t),, , y = ψ(t), , is given by, , Figure 7.9., , (a ≤ t ≤ b)
Page 219 :
210, , 7 Complex Integration and Cauchy’s Theorem, , +, , b, , L=, , , , +, , b, , , , (φ (t))2 + (ψ (t))2 dt =, , a, , a, , dx, dt, , 2, , , +, , dy, dt, , 2, dt., , Note that the integrand on the right integral is recognized as ds/dt, where s, is the arc measured from the point z(a) of C. Using the parameterization, (a ≤ t ≤ b), , C : z(t) = x(t) + iy(t), , for C, a smooth curve (or contour) in the plane, the length of C is given by, +, , b, , L=, , |z (t)| dt =, , a, , +, a, , b, , dy, dx, dt., +i, dt, dt, , (7.10), , In the special case that the curve is a line segment from z0 to z1 , parameterized, by, z(t) = tz0 + (1 − t)z1 (0 ≤ t ≤ 1),, we have z (t) = z1 − z0 . Hence, as expected,, +, L=, 0, , 1, , +, , , , |z (t)| dt =, , 1, , |z1 − z0 | dt = |z1 − z0 |., , 0, , When z is on C, the symbol |dz| = |z (t)| dt so that (7.10) may also be, written as, +, +, L=, |dz| =, ds,, (7.11), C, , C, , it being understood, that C is parameterized by z(t). This observation partly, *, explains why C f (z) |dz| defines an integral of f along C with respect to arc, length:, +, + b, f (z) |dz| =, f (z(t))|z (t)| dt., C, , a, , The arc length integrals of this type play significant roles in certain areas of, mathematics and physics., *, Example 7.17. Let us now evaluate C z −n |dz| where C = z(t) = reit (r >, 0, 0 ≤ t ≤ 2π and n ∈ Z). Set f (z) = 1/z n . Note that C is smooth and, z (t) = ireit ,, , f (z(t)) =, , e−int, ., rn, , Therefore,, , +, + 2π, + 2π, 1, 1, r dt, 0 if n = 0, −int, |dz| =, = n−1, e, dt =, n, n eint, 2πr, if n = 0., z, r, r, C, 0, 0, , •
Page 220 :
7.2 Parameterizations, , 211, , Remark 7.18. It is meaningful to talk about the length of an arbitrary curve., However, if the curve is not a contour, then the length may not be finite., Consider an arbitrary curve z(t), with a ≤ t ≤ b. Define, V (P ) =, , n, , , |z(tk ) − z(tk−1 )|,, , (7.12), , k=1, , where a = t0 < t1 < · · · < tn = b is a partition P . By the triangle inequality,, V (P ) increases monotonically as the subintervals are further subdivided into, smaller subintervals. The length of this curve can be defined as the least upper, bound of all sums of the form (7.12), that is,, sup, P, , n, , , |z(tk ) − z(tk−1 )|., , k=1, , If the length is finite, the curve is said to be rectifiable. The reader should, verify that every contour is rectifiable and that, in the case of a contour, this, definition agrees with (7.11). In Exercise 7.29(1), an example of a nonrectifiable curve is given., •, What follows is the complex analog to a well-known real variable theorem., Theorem 7.19. (M-L Inequality) Suppose f (z) is continuous on a contour, C having length L, with |f (z)| ≤ M on C. Then, +, +, +, f (z) dz ≤, |f (z)| |dz| ≤ M, |dz| = M L., C, , C, , C, , Proof. Since C is parameterized by z(t) on the interval [a, b], we have, +, , +, , b, , f (z(t))z (t) dt, , f (z) dz =, C, , +, , a, , ≤, a, , b, , |f (z(t))| |z (t)| dt =, , +, , ≤M, , +, |f (z)| |dz|, C, , b, , |z (t)| dt = M L, , (since |f (z)| ≤ M on C), , a, , and the conclusion follows., * For example, we can use the M-L Inequality to find an upper bound for, | C (z 2 + 10)−1 dz|, where C is the circle C : z(t) = 2eit (−π ≤ t ≤ π). In, fact, for z ∈ C, |z 2 + 10| ≥ 10 − |z|2 = 10 − |z(t)|2 ≥ 10 − 4 = 6, and so, we, have, +, +, +, 1, 2π, dz, |dz|, ≤, ≤, ., |dz| =, 2 + 10, 2 + 10|, z, |z, 6, 3, C, C, C
Page 221 :
212, , 7 Complex Integration and Cauchy’s Theorem, , Similarly, we easily see that, +, , ez, dz ≤ 4πe2 ., z+1, , C, , Strictly speaking, a curve C is associated with a definite parametric form, C : z = z(t), a ≤ t ≤ b and so, the length of a curve is defined in terms of its, parameter; but it is geometrically evident that, for simple curves, the length, is independent of the parameterization. For instance, the curves, (0 ≤ t ≤ π),, , C1 : z1 (t) = eit, , and C2 : z2 (t) = e2it, , (0 ≤ t ≤ π/2), , both traverse the upper half of the unit circle. Moreover, by (7.11), we have, +, + π, + π, |dz1 | =, |z1 (t)| dt =, dt = π,, 0, , C1, , and, , +, , +, |dz2 | =, C2, , π/2, , 0, , 0, , |z2 (t)| dt =, , +, , π/2, , 2 dt = π., 0, , The contours C1 and C2 , although different in formal sense because they arise, from different parameterizations, have the same length., Remark 7.20. The curves eit (0 ≤ t ≤ 2π) and e2it (0 ≤ t ≤ 2π) both, represent the set of points on the unit circle. The length of the first curve, is 2π and that of second is 4π. Note, however, that the second curve is not, •, simple because it traverses the unit circle twice., The next theorem gives general criteria for changing parameters without, affecting arc length., Theorem 7.21. Let C : z(t) = x(t) + iy(t), a ≤ t ≤ b, be a contour. Suppose, t = φ(s) with a = φ(c), b = φ(d), and φ (s) > 0, so that t increases with s., If φ (s) is sectionally continuous on the interval [c, d], then the length of C is, given by, + d, L=, |z (φ(s))|φ (s) ds., c, , Proof. By (7.10) and the chain rule,, +, , b, , L=, a, , +, , |z (t)| dt =, , +, , b, , |x (t) + iy (t)| dt, , a, b, , =, , |x (φ(s)) + iy (φ(s))| dφ(s), , a, , +, =, , a, , b, , |z (φ(s))|φ (s) ds.
Page 222 :
7.2 Parameterizations, , 213, , Let C be an arc z(t), −2 ≤ t ≤ 1 and let C1 be the arc z(s) = γ(3s − 5),, 1 ≤ s ≤ 2. Clearly, C and C1 have the same initial and the same end point,, and the same trajectory. It is easy to see that, +, +, f (z) dz =, f (z) dz, (7.13), C, , C1, , holds for every continuous function in D which contains C1 and C2 . Indeed,, if we set t = φ(s) = 3s − 5, then φ(1) = −2, φ(2) = 1 and φ (s) = 3. Hence, +, + 2, f (z) dz =, f (γ(3s − 5))3γ (3s − 5) ds., 1, , C1, , *2, By the substitution t = φ(s), the change of variable leads to 1 f (z(t))z (t) dt, which is nothing but the right-hand side of (7.13). This continues to hold for, a broader class of functions as we see next., Suppose that f (z) is continuous on a contour C : z(t), a ≤ t ≤ b, and that, t = φ(s) satisfies the conditions of Theorem 7.21. Then the chain rule may be, applied to obtain, +, + b, f (z) dz =, f (z(t))z (t) dt, (7.14), C, , a, , +, , d, , =, , f (z(φ(s)))z (φ(s))φ (s) ds., , c, , , , , Since z (φ(s))φ (s) is the derivative with respect to s of z(φ(s)), the curve, C could have been parameterized by z(φ(s)), c ≤ s ≤ d, without affecting the, value of the integral., The contour, −C : z(−t) = x(−t) + iy(−t), , (−b ≤ t ≤ −a), , represents the same curve, traversed in the opposite direction, as, C : z(t) = x(t) + iy(t), We have, , +, , +, , −a, , f (z) dz =, −C, , (a ≤ t ≤ b)., , f (z(−t))z (−t)(−1) dt, , −b, , and, upon making the substitution s = −t,, +, + a, f (z) dz =, f (z(s))z (s) ds, −C, , b, , +, , b, , =−, , f (z(s))z (s) ds, , +, , a, , =−, , f (z) dz., C, , (7.15)
Page 223 :
214, , 7 Complex Integration and Cauchy’s Theorem, , Remark 7.22. Loosely speaking, equations (7.14) and (7.15) say that the, value of the integral along a simple contour C, viewed as a point set in the, plane, depends on the parameterization of the contour only with regard to, orientation., •, Remark 7.23. Since integrating, around a circle is such a common occurrence,, *, we introduce the notation |z−z0 |=r f (z) dz, which will be interpreted as the, integral of f (z) around the contour consisting of the circle |z −z0 | = r oriented, •, in the positive sense., *, Examples 7.24. Let us evaluate C |z|n dz (n ∈ N0 := N ∪ {0}) along the, straight line C joining the origin to the point 1 + i. We parameterize the line, by, C : z(t) = t + it (0 ≤ t ≤ 1)., Then z (t) = 1 + i, and, +, + 1, +, n, n, n/2, |z| dz =, |t + it| (1 + i) dt = 2 (1 + i), 0, , C, , (7.16), , 1, , tn dt =, , 0, , 2n/2 (1 + i), ., n+1, , The parameterization (7.16) was chosen because it was the most natural. We, could have parameterized C by, C : z(t) = x(t) + ix(t), , (a ≤ t ≤ b),, , where x(a) = 0, x(b) = 1, and x (t) > 0. Then z (t) = (1 + i)x (t), and, +, + b, n, |z| dz =, |x(t) + ix(t)|n (1 + i)x (t) dt, C, , a, , +, , b, , = 2n/2 (1 + i), , (x(t))n x (t) dt, , a, , , = 2n/2 (1 + i), Next, to evaluate, , (x(b))n+1 − (x(a))n+1, n+1, , , =, , 2n/2 (1 + i), ., n+1, , +, |z|=r, , |z|n dz, , (n ∈ Z),, , we parameterize the specified circle by z(t) = reit , 0 ≤ t ≤ 2π, so that, + 2π, + 2π, +, n, it n, it, n+1, |z| dz =, |re | ire dt = ir, eit dt = 0., C, , 0, , 0, , •, , Example 7.25. Let, * us now consider one more similar integral over a closed, contour. Consider C |z| dz along the rectangle C having corners −1, 1, 1 +, i, −1 + i. This contour is the sum of four smooth curves (straight lines). We, have
Page 224 :
7.2 Parameterizations, , +, , +, , +, , |z| dz =, C, , |z| dz +, C1, , +, |z| dz +, , C2, , where, C1, C2, C3, C4, , :, :, :, :, , z1 (t) = t, z2 (t) = 1 + it, z3 (t) = −t + i, z4 (t) = −1 − it, , Solving, we obtain, +, + 1, +, |z| dz =, |t| dt + i, −1, 1, , C, , +, =, , −1, , +, , 0, , 1, , |z| dz +, C3, , |z| dz,, C4, , (−1 ≤ t ≤ 1),, (0 ≤ t ≤ 1),, (−1 ≤ t ≤ 1),, (−1 ≤ t ≤ 0)., , +, , 2, 1 + t dt −, , , (|t| − t2 + 1) dt, , 215, , +, , 1, , −1, , +, , 2, t + 1 dt − i, , 0, , , 1 + t2 dt, , −1, , , (t − t2 + 1) dt, 0, √, √, = 1 − 2 − ln( 2 + 1)., 1, , =2, , •, , Remark 7.26. The contour C is not the sum of the four curves C1 , C2 , C3 , C4, as defined in (7.9), because these curves are not parameterized on four distinct subintervals of the interval on which C is parameterized; but according, to (7.14), the parameterization is not critical, so we will adopt a more liberal definition of “sum” that does not require a specific parameterization. In, fact, we may even integrate without expressing x and y in terms of common, parameter t. Therefore, in the last example, we could just as well have written, +, + 1, + 1, + −1, + 0, |z| dz =, |x| dx +, |1 + iy|i dy +, |x + i| dx +, | − 1 + iy|i dy., C, , −1, , 0, , 1, , 1, , Here, we are actually using x for the parameter on C1 and C3 , and y for the, •, parameter on C2 and C4 ., Examples 7.27. For each integer n, we have, , +, + 2π, + 2π, z n dz =, (reit )n ireit dt = irn+1, ei(n+1)t dt =, |z|=r, , 0, , 0, , 0 if n = −1,, 2πi if n = −1., , Note that the value of this integral is independent of the radius of the given, circle., *, Our final example of this section is to compute I = |z|=r x dz. Note that, |z|2 = zz = r2 and x = (z + z)/2. Thus, the linearity property gives, +, +, +, 1, 1, r2, dz, I=, = iπr2 ., (z + r2 /z) dz =, z dz +, 2 |z|=r, 2 |z|=r, 2 |z|=r z, Note that the value of the integral in this case depends on the radius of the, given circle. Why is this so? Again, as z = r2 /z, we have
Page 225 :
216, , 7 Complex Integration and Cauchy’s Theorem, , +, , (z)n dz = r2n, , |z|=r, , +, |z|=r, , 1, dz =, zn, , , , 2πir2 if n = 1, 0 if n ∈ Z \ {1}., , •, , Questions 7.28., 1. Can (7.9) be extended to the case of infinitely many curves?, 2. How may the following expressions be interpreted?, +, +, +, |f (z)| dz, (b), f (z) |dz|, (c), |f (z)| |dz|., (a), C, , C, , C, , *, , 3. Can C f (z) dz be defined without requiring f (z) to be continuous at, all points of C?, 4. If f (z) is continuous on a contour C, does |f (z)| necessarily assume a, maximum on C?, 5. If f (z) is continuous for |z| < r for some r > 0 such that f (0) = 0, is, * 2π, *, limδ→0 0 f (δeiθ ) dθ = 0? Is limδ→0 |z|=δ f (z), z dz = 0?, 6. Does the orientation affect the length of a curve?, 7. Why is it usually easier to integrate along a circle than along a square?, *, 8. If |f (z)| ≤ 2 on the circle |z| = 3, is |z|=3 f (z) dz ≤ 3?, Exercises 7.29., 1. Show that the curve C parameterized by, !, !!, , if 0 < t ≤ 1, t cos 1t + i sin 1t, z(t) =, 0, if t = 0,, is nonrectifiable., 2. Prove that every (continuous) curve is bounded., 3. Show that, +, +, 3π, dz, 2z + 1, (a), ≤, π, (b), ., dz ≤, 2, 2, 2, |z|=1 3 + 5z, |z|=1 5 + z, 4. Find the length of the following contours., (−π ≤ t ≤ π), (a) z(t) = 3e2it + 2, (b) z(t) =* et cos t + *iet sin t * (−π ≤ t ≤ π)., y dz,, z dz along the following contours:, 5. Evaluate C x dz,, C, C, (a) The line segment from the origin to 1 + i, (b) The line segment from the origin to 1 − i, (c) The circle |z| = 1, (d) The curve C consisting of the line segment from 0 to 1 followed by, the line segment from 1 to 1 + i, (e) The curve C consisting of the line segment from 0 to i followed by, the line segment from i to 1 + i.
Page 226 :
7.3 Line Integrals, , 217, , *, *, *, *, *, 6. Evaluate C z dz,, |z| dz,, z|z| dz,, z |dz|,, |z| |dz| along, C, C, C, C, the same contours as above. Do the same for the closed contour C, consisting of the upper semicircle |z| = 1 from 1 to −1, and the line, segment [−1,, *, *, * 1]., Re z |dz| and, Im z |dz|, where C : z(t) =, 7. Evaluate C |z|2 dz,, C, C, 0, ≤, t, ≤, 1., t2 /3 + it for, *, 8. Evaluate C (az + bz) dz where a, b are some nonzero fixed constants, and C is the contour given by C = [0, eiπ/6 ] ∪ {eiθ : π/6 ≤ θ ≤ π/2} ∪, [eiπ/3 , 0]. Do the same by replacing π/6 by α and π/3 by β, 0 ≤ α <, 2, β ≤ 2π. Do, * the same for C : z(t) = −t + i(t + 2), 0 ≤ t ≤ 2., 9. Evaluate C (1/z) dz along the square having corners ±1 ± i., 10. Evaluate, the following integrals:, *, (a) *C ez dz along the line segment from the origin to 2 + 2i, (b) *C (ez + z + 1) dz along the line segment from −1 + i to 1 − i, (c) *C cos z dz along the line segment from the origin to the point 1 + i, (d) *C |z|2 dz along the square with vertices 0, 1, 1 + i, i, (e) C (x2 + iy 2 ) dz along the line segment from 0 to 1 + i followed by, the line, * segment from 1 + i to 1 + 2i., 11. Evaluate C (z/z) dz along the simple closed contour C shown in Figure 7.10., , Figure 7.10., , *, 12. Evaluate C z|z| dz along the upper semicircle |z| = R from R to −R,, and the line segment [−R, R]., , 7.3 Line Integrals, In order to draw a useful analogue with single-variable calculus, we begin, by reviewing the (first) fundamental theorem of calculus. Suppose f (x) is, continuous on the interval [a, b]. The first fundamental theorem of calculus, asserts the existence of an antiderivative F (x) for f (x) (i.e., a function F, such that F (x) = f (x) on [a, b]) with, + b, f (x) dx = F (b) − F (a)., a
Page 227 :
218, , 7 Complex Integration and Cauchy’s Theorem, , This theorem relates the behavior of a function on the boundary of a set (two, points) to the behavior of an associated function, its derivative, on the whole, set (a closed interval). From our earlier discussion, it is clear that the familiar, properties of Riemann integral is carried over to the case of complex integrals., For instance, if f, g : [a, b] → C are continuous and if c is a complex constant,, then, +, +, +, b, , b, , (f (x) + cg(x)) dx =, a, , b, , f (x) dx + c, a, , g(x) dx., a, , The fundamental theorem of calculus is also valid in this setting. More precisely, we have, Theorem 7.30. If f : [a, b] → C is continuous and if there exists a function, F (x) such that F (x) = f (x) on [a, b], then, +, a, , b, , b, , f (x) dx = F (x)|a = F (b) − F (a)., , For instance, if f1 (t) = 3t2 − 2it and f2 (t) = e2πit , then the corresponding, antiderivatives are F1 (t) = t3 − it2 and F2 (t) = e2πit /(2πi) so that, +, 0, , 1, , (3t2 − 2it) dt = t3 − it2, , 1, 0, , +, =1−i, , and, 0, , 1, , e2πit dt =, , e2πit, 2πi, , 1, , = 0., 0, , The second fundamental theorem of calculus asserts that “if f : [a, b] → R, is continuous, then the indefinite integral, + t, f (x) dx, a ≤ t ≤ b,, F (t) =, a, , is an antiderivative for f (t). Moreover, each antiderivative for f (t) differs from, F (t) by a constant.” Our next theorem is a two-dimensional analogue of the, first fundamental theorem of calculus., Theorem 7.31. (Green’s Theorem) Let P (x, y) and Q(x, y) be continuous, with continuous partials in a simply connected closed region R whose boundary, is the contour C. Then, , +, + + , ∂Q ∂P, −, P dx + Q dy =, dx dy,, (7.17), ∂x, ∂y, C, R, where C is traversed in the positive sense., Proof. We prove the theorem in the special case that R is a rectangle (and, its interior) whose sides are parallel to the coordinate axes (see Figure 7.11)., Let C = C1 + C2 + C3 + C4 in Figure 7.11. Observing that dy ≡ 0 on C1 and, C3 while dx ≡ 0 on C2 and C4 , we have
Page 229 :
220, , 7 Complex Integration and Cauchy’s Theorem, , +, , xy dx + (x2 + y 2 ) dy =, , +, , x · 0 dx +, , 0, , C, , +, , 1, , +, , +, , 1, , +, , x · 1 dx +, +, , 1, , =, 0, , (1 + y 2 ) dy, , 0, , 0, , +, , 1, , dy −, , 0, , (0 + y 2 ) dy, , 1, , 1, , x dx =, 0, , 1, ., 2, , Alternately, by Green’s theorem,, +, + 1+ 1, 1, 2, 2, xy dx + (x + y ) dy =, (2x − x) dx dy = ., 2, C, 0, 0, , •, , Enough playing around. We now return to complex variables to show the, reason for introducing Green’s theorem., Theorem 7.33. (Cauchy’s “Weak” Theorem) If f (z) is analytic (with a, continuous derivative) in a simply, * connected domain D, and C is closed contour lying in D, then we have C f (z) dz = 0., Proof. Set f (z) = u(x, y) + iv(x, y). By the Cauchy–Riemann equations for, analytic functions,, u x = vy ,, , uy = −vx, , for (x, y) ∈ D., , (7.20), , Since f (z) is presumed continuous, the four partials must also be continuous. Suppose, for the moment, that C is a simple closed contour. Then an, application of Green’s theorem to (7.8) yields, +, +, +, f (z) dz =, u dx − v dy + i, v dx + u dy, C, C, C, , , + + , + + , ∂u, ∂u ∂v, ∂v, −, −, =, dx dy + i, dx dy,, −, ∂x ∂y, ∂x ∂y, R, R, where R is the region enclosed by C. In view of (7.20), both integrands on, the right are identically zero in R. This proves the theorem when C is simple, closed contour., For a general closed contour, the proof follows in like manner from a more, general statement of Green’s theorem. See Apostol [Ap]., Corollary 7.34. Under the conditions of Theorem 7.33, let C1 and C2 be any, contours in the domain with the same initial and terminal points. Then, +, +, f (z) dz =, f (z) dz., C1, , C2, , Proof. Suppose C1 and C2 both have initial and terminal points z0 and z1, respectively (see Figure 7.12). Let C = C1 − C2 . Then
Page 230 :
7.3 Line Integrals, , 221, , Figure 7.12., , +, , +, f (z) dz =, , C, , contour,, *, *Since C is a closed, f, (z), dz, =, f, (z), dz., C1, C2, , f (z) dz, +, +, =, f (z) dz +, f (z) dz, C1, −C2, +, +, =, f (z) dz −, f (z) dz., C1 −C2, , C1, , *, C, , C2, , f (z) dz = 0, from which we conclude that, , Remark 7.35. Corollary 7.34 (as well as Theorem 7.33) says that the integral is independent of path in the domain. That is, the value of the integral, just depends on the initial and terminal points, provided only that the contour stays inside the domain where the function is continuously differentiable., Hence, under, * z the conditions of the theorem, we can give meaning to the expression z01 f (z) dz. Its value may be found by computing the complex line, *, integral C f (z) dz along any contour C in the domain that has initial point, z0 and terminal point z1 . In particular, if the contour C is closed (z1 = z0 ),, then, +, + z1, f (z) dz =, f (z) dz = 0., •, C, , z0, , Cauchy’s theorem, in its present form, is weak because the analytic function was required to have a continuous derivative (so that Green’s theorem, could be applied). While this may seem like a minor restriction, it does not, allow us to apply Cauchy’s theorem to the class of all analytic functions. However, in the next section, this restrictive hypothesis will be eliminated. Then,, in Chapter 8, it will be shown that every analytic function does, in fact, have, a continuous derivative., We will now examine the extent to which Cauchy’s theorem is valid for, multiply connected regions. Recall that, +, + 2π iθ, 1, ie, dz =, dθ = 2πi., z, eiθ, |z|=1, 0
Page 231 :
222, , 7 Complex Integration and Cauchy’s Theorem, , Thus even though the function f (z) = 1/z is analytic everywhere on the unit, circle, the above integral is not zero. Note that f has derivatives of all orders, in C \ {0}. Cauchy’s theorem is not applicable because the punctured disk is, not simply connected. To trace what went wrong with this function, observe, that 1/z is the derivative of many different branches of log z. Suppose we start, with any point w on the unit circle and integrate counterclockwise around the, circle through one revolution. While the terminal point we2πi has the same, location in the plane as the initial point w, its argument has increased by 2π., Choosing a specific branch for the logarithm, with a branch cut on the ray, arg z = −α, w = eiα , we may now write, +, |z|=1, , 1, dz =, z, , +, |z|=1, , d, log z = log z, dz, , we2πi, , = log we2πi − log w., w, , This last expression simplifies to, ln |we2πi | + i arg we2πi − (ln |w| + i arg w) = i(arg we2πi − arg w) = 2πi., Thus the integral is nonzero because the function 1/z has many antiderivatives. The value of the integral is related to the change in the argument of the, multiple-valued function log z. Note also that the value of the integral is independent of the choice of the initial branch. For simply connected domains, this, problem does not arise because analytic functions then have single-valued antiderivatives as we shall see. Moreover, this idea can be extended to any closed, contour C : z(t), a ≤ t ≤ b that does not pass through the origin. Indeed, if, C is a closed contour that avoids the origin, then we have, +, dz, b, b, = log z(t)|a = i arg z(t)|a = i θ|C, C z, where θ is the angle which the line segment [0, z(t)] joining 0 to the variable, point z(t) makes with the horizontal line. Thus, the total variation is 2π times, the number of times z winds around 0 as z traverses C. In other words,, +, 1, dz, 2πi C z, is an integer which is called the winding number of C with respect to the, origin (see for example [A, P1])., Suppose we integrate 1/z along the boundary C of the multiply connected, region consisting of the annulus r0 < |z| < r1 (r0 > 0). If the integration is, performed in the positive sense (where the domain always remains on the left), as shown in Figure 7.13, then, ,, ,, +, 1, 1, 1, dz =, dz +, dz, z, z, z, C, |z|=r1, |z|=r0, + 2π, + −2π, ir1 eiθ, ir0 eiθ, =, dθ, +, dθ, r1 eiθ, r0 eiθ, 0, 0, = 2πi − 2πi = 0.
Page 232 :
7.3 Line Integrals, , 223, , Figure 7.13., , The fact that this integral is zero is a consequence of the following Cauchy, theorem for multiply connected regions., Theorem 7.36. Suppose that f (z) is analytic (with a continuous derivative), in, * a multiply connected domain and on its boundary C. Then we have, f (z) dz = 0, where the integration is performed along C in the positive, C, sense., We indicate a method of proof that involves “transforming” a multiply, connected region into a simply connected region. To illustrate, consider the, multiply connected region in Figure 7.14. Suppose we construct the line segment AB, called a cross-cut, which connects the outer boundary C1 with the, inner boundary C2 . Then the domain bounded by the contour C1 , the line, segment AB, the contour C2 , and the line segment BA (traversed as illustrated in Figure 7.14) is simply connected. This is so because no closed curve, in the new region is allowed to cross the line segment AB. Let C denote the, boundary of this domain. Then by Cauchy’s theorem for simply connected, regions, we have, ,, +, ,, +, +, f (z) dz =, f (z) dz +, f (z) dz +, f (z) dz +, f (z) dz, C, , C1, , AB, , = 0., , Figure 7.14., , C2, , BA
Page 233 :
224, , 7 Complex Integration and Cauchy’s Theorem, , +, , Note that, , +, f (z) dz = −, , AB, , so that, , f (z) dz,, BA, , +, , ,, f (z) dz =, C1 +C2, , ,, f (z) dz +, , C1, , f (z) dz = 0., C2, , This finishes the discussion of Cauchy’s theorem for the domain in Figure 7.14., In Figure 7.15, we illustrate Cauchy’s theorem for domain with (n − 1) holes., , Figure 7.15., , In a manner similar to that used for one hole, we get, +, f (z) dz = 0., , (7.21), , C1 +C2 + ··· +cn, , Equation (7.21) can be written in the form, ,, , ,, ,, f (z) dz = −, f (z) dz + · · · +, f (z) dz, C1, C, , C2, , n, =, f (z) dz + · · · +, f (z) dz., −C2, , −Cn, , In other words, by integrating along each inner contour in the counterclockwise, direction, so that the (n−1) inner contours have negative orientation, it follows, that the value of the integral along the outer contour is equal to the sum of, the values along the inner contours., In a more complicated multiply connected region, it may not be possible to, connect an inner boundary to an outer boundary by a straight line segment;, but a polygonal line can always be found that furnishes us with the necessary, cross-cut for any multiply connected region. In fact, Green’s theorem can also, be generalized from simply to multiply connected regions, thus affording us, with a direct proof of Cauchy’s theorem for multiply connected regions., Finally, we remark that requiring analyticity on the boundary C in, Theorem 7.36 means that the function is actually analytic in a domain containing C.
Page 234 :
7.3 Line Integrals, , 225, , Questions 7.37., 1. What are the differences between real and complex line integrals? Between a line integral and a Riemann integral?, 2. Where was continuity of the partials used in proving Green’s theorem, for rectangles?, 3. Why is Green’s theorem a two-dimensional analog to the fundamental, theorem of calculus?, 4. In Green’s theorem, if C were traversed in the negative direction, what, could we conclude?, 5. How does Green’s theorem give us a way to compute the area of a, region?, 6. Suppose f (z) has a continuous derivative in a simply connected region, *whose boundary is C. May Cauchy’s theorem be applied to conclude, f (z) dz* = 0?, C, 7. Suppose C f (z) dz = 0 for some contour C. Can anything be said about, f (z)?, 8. Does Cauchy’s theorem apply to a function having a continuous derivative in a region exterior to a disk?, 9. Can Cauchy’s theorem be used to evaluate Riemann integrals?, 10. Let u(z) = u(x, y) be a real-valued harmonic* function on the unit disk, Δ, and γ be a simple closed contour in Δ. Is γ u(z) dz = 0? How about, if Δ, * is replaced by a general domain D?, 11. Is *C z dz independent of the path C between 0 and 1 + i?, 12. Is *C (Re z) dz independent of the path C between 0 and 1 + i?, 13. Is C z dz independent of the path C between 0 and 1 + i?, Exercises 7.38., 1. Evaluate, the following line integrals:, +, xy dx+(x2 +y 2 ) dy along the quarter-circle C in the first quadrant, (a), C, , having radius r = 2., +, (b), x2 y dx + (2x + 1)y 2 dy along the square having vertices (1, 0),, C, , (1,, + −1), (2, −1), and (2, 0)., (c), y 2 dx + x2 dy along the curve C parameterized by x = a cos3 t,, C, , y+ = a sin3 t (0 ≤ t ≤ 2π)., xy 2, dy along the circle |z| = r., (d), 2, 2, +C x + y, , (x2 + xy) dy along the parabola y = x2 from (−2, 4) to (2, 4)., , (e), C, , 2. Let C be any simple closed contour bounding a region having area A., Prove that
Page 235 :
226, , 7 Complex Integration and Cauchy’s Theorem, , +, , 1, A=, 2, , +, , +, , i, x dy − y dx = −, y dz = −i, x dz = −, 2, C, C, C, , +, z dz., C, , 3. Modify the proof of Green’s theorem for a rectangle to show that, +, + +, + +, +, ∂P, ∂Q, dx dy,, dx dy., P (x, y) dx = −, Q(x, y) dy =, ∂y, C, R, C, R ∂x, 4. Verify Cauchy’s theorem for the functions 3z − 2 and z 2 + 3z − 1 if C is, the square having, * corners ±1 ± i., 5. By evaluating |z|=1 ez dz, show that, +, , π, , cos θ, , e, , +, , π, , ecos θ sin(θ + sin θ) dθ = 0., , cos(θ + sin θ) dθ =, , −π, , −π, , 6. Evaluate the following integrals along any contour between the points, represented by the limits of integration., + πi, + π+i, + 2+i, (a), ez dz, (b), eiz dz, (c), (z 2 + 3z − 2) dz., −πi, , 0, , 1−i, , 7.4 Cauchy’s Theorem, The central theme of this section is to investigate conditions to cover general, situations so that the integral of an analytic function along a closed contour vanishes. We will actually prove several forms of Cauchy’s theorem (also, called the Cauchy–Goursat theorem), each involving different geometric and, topological considerations. Goursat showed that Theorem 7.33 can be proved, without assuming the continuity of f (z). In its simplest form, the theorem is, proved for a rectangle. The proof involves a construction similar to that used, in the proof of the Heine–Borel theorem (Theorem 2.26). The ultimate aim is, to understand precisely the local structure of analytic functions., Theorem 7.39. (Cauchy’s theorem for a rectangle) Let *f (z) be analytic in, a domain containing a rectangle C and its interior. Then C f (z) dz = 0., Proof. Divide C into four congruent rectangles, C (1) , C (2) , C (3) and C (4) as, *, j, indicated in Figure 7.16, and let I1 = C (j) f (z) dz for 1 ≤ j ≤ 4. The integrals over the common sides have opposite orientation, and hence cancel, one another. Therefore, from the known properties in the complex integral, it, follows that, +, 4, , I :=, f (z) dz =, I1j ., C, , By the triangle inequality,, , j=1
Page 236 :
7.4 Cauchy’s Theorem, , 227, , Figure 7.16., , |I| ≤, , 4, , , |I1j |., , (7.22), , j=1, , If every term in this sum were less than |I|/4, then we would get a contradiction. Thus, for at least, * one of the terms on the right side of (7.22), denoted, conveniently by I1 = C1 f (z) dz, we have, |I1 | ≥ |I|/4., Next divide the rectangle C1 into four congruent rectangles, and, as above,, observe that for at least one, denoted by C2, +, |I|, |I1 |, ≥ 2., |I2 | =, f (z) dz ≥, 4, 4, C2, Continuing the process, we obtain a nested sequence of rectangles {Cn } (see, Figure 7.17), each satisfying the inequality, +, |In−1 |, |I|, |In | =, ≥ ··· ≥ n, f (z) dz ≥, 4, 4, Cn, so that, |I| ≤ 4n |In | for n = 1, 2, 3, . . . ., , Figure 7.17., , (7.23)
Page 238 :
7.4 Cauchy’s Theorem, , 229, , For instance, by Theorem 7.39, we have, +, 2z 2 + 1, dz = 0, 5, 2, 3, C (z + 3) (z + 8)z (z − 1), where C is the positively oriented square with vertices 1 + i, 1 + 3i, 2 + 3i,, 2 + i., Corollary 7.40. Let f be continuous in a domain D containing a rectangle, C and*its interior. Suppose that f is analytic in D \{a} for some point a ∈ D., Then C f (z) dz = 0., Proof. It suffices to prove the theorem when a lies inside C. As before divide C, into n2 congruent rectangles Cjk (see Figure 7.18 for illustration when n = 4)., From the elementary properties of complex line integrals, we have, +, n , n +, , f (z) dz =, f (z) dz., C, , j=1 k=1, , Cjk, , *If a is neither an interior point nor a point of Cjk , then, by Theorem 7.39,, f (z) dz = 0. On the other hand, if a is inside or on the rectangle Cjk ,, Cjk, then the M-L inequality shows that, +, +, M L(C), ,, f (z) dz ≤, |f (z)| |dz| ≤ M L(Cjk ) =, n, Cjk, Cjk, where M = maxz∈C |f (z)|, L(Cjk ) and L(C) represent the perimeter of Cjk, and C, respectively. Note that |f (z)| is a continuous function on the compact, set C, and the point a at the worst can belong to one of the four rectangles, Cjk . It follows that, +, f (z) dz =, C, , +, a∈Cjk, , Cjk, , f (z) dz ≤, , +, a∈Cjk, , Figure 7.18., , Cjk, , f (z) dz ≤, , 4M L(C), ., n
Page 239 :
230, , 7 Complex Integration and Cauchy’s Theorem, , Since n was arbitrary,, , *, C, , f (z) dz = 0 and the proof is complete., , Let us pause for a moment to summarize what we have shown and where, we are headed. In the previous section, it was shown that for a function having, a continuous derivative in a domain, the integral around any closed contour, in the domain is zero, or, equivalently, the integral along any contour in the, domain depends only on the end points of the contour. The previous theorem, eliminates the requirement of continuity for the derivative when the contour, is a rectangle., Our goal is to show that the rectangle in Theorem 7.39 may be replaced, by an arbitrary closed contour in the domain. This will be accomplished by, first showing that every continuous function having an antiderivative in a, domain also has the property that the integral is independent of path. Next, we will show that a function analytic in a disk has an antiderivative, and then, that a function analytic in a simply connected domain has an antiderivative., Finally, Cauchy’s theorem will be extended to multiply connected domains, by “transforming” them into simply connected domains, as was done in the, previous section. We start with the following theorem which is an analogue of, the first fundamental theorem of calculus., Theorem 7.41. (Fundamental Theorem of Integration) Let f (z) be continuous in a domain D, and suppose there is a differentiable function F (z), such that F (z) = f (z) in D. Then for any contour C in D parameterized by, z(t), a ≤ t ≤ b, we have, +, f (z) dz = F (z(b)) − F (z(a))., C, , In particular, if C is closed then, , *, C, , f (z) dz = 0., , Proof. Since F (z) has a continuous derivative in D, we get, +, +, f (z) dz =, F (z) dz, C, , C, b, , +, =, , F (z(t))z (t) dt, , a, , +, =, , a, , b, , d, (F (z(t)) dt = F (z(b)) − F (z(a)),, dt, , the last equality following from the fundamental theorem of integral calculus., If we use a more familiar notation z(a) = z0 and z(b) = z1 , then the conclusion, may be expressed as, + z1, f (z) dz = F (z1 ) − F (z0 ), z0, , along any contour C in the domain having initial point z0 and terminal point, z1 . If the contour is closed, then z(a) = z(b) = z0 so that
Page 240 :
7.4 Cauchy’s Theorem, , +, , +, , z0, , f (z) dz =, C, , 231, , f (z) dz = F (z0 ) − F (z0 ) = 0., , z0, , Moreover, Theorem 7.41 is a consequence of the corresponding formula for, line integrals. Indeed, as F (z) = Fx = −iFy , it follows that, + z1, dF, F (z1 ) − F (z0 ) =, z0, + z1, =, Fx dx + Fy dy, z, + 0z1, =, F (z)(dx + i dy), z0, + z1, + z1, , =, F (z) dz =, f (z) dz., z0, , z0, , Example 7.42. The function f (z) = z n (n ∈ N) is continuous everywhere, and has an antiderivative F (z) = z n+1 /(n + 1). Hence for any contour C in, the plane from z0 to z1 ,, + z1, +, z n+1, z n+1, n, − 0 ., z dz =, z n dz = 1, n+1 n+1, C, z0, In particular,, if C is a closed curve (z0 = z1 ) then, n ∈ N, we have, *, n for each, k, that C f (z) dz = 0. More generally, if p(z) =, k=0 ak z is a polynomial,, then, n, , ak k+1, P (z) =, z, +c, k+1, k=0, , , , is primitive of p(z), P (z) = p(z), and so, + z1, p(z) dz = P (z1 ) − P (z0 )., In particular,, , *, C, , z0, , p(z) dz = 0 if C is closed curve in C., , •, , Examples 7.43. By Theorem 7.41, we obtain the following:, *, (i) Clearly, |z|=1 csc2 z dz = 0. Indeed if f (z) = csc2 z, then F (z) = − cot z, has the property that F (z) = csc2 z and F (z) is analytic in C \ {nπ :, n ∈ Z}. In particular, f and F are analytic for 0 < |z| < π. Similarly,, we obtain that, +, sec2 z dz = 0., |z|=1, , *, (ii) Suppose we wish to evaluate C (z + a)ebz dz (b = 0), where C is the, parabolic arc x2 = y from (0, 0) to (1, 1). First we note that if f (z) =, (z + a)ebz , then F (z) for which F (z) = f (z) is given by
Page 241 :
232, , 7 Complex Integration and Cauchy’s Theorem, , F (z) = (z + a), , ebz, ebz, ebz, − 2 = 2 (b(z + a) − 1), b, b, b, , (which may be obtained by integrating (z + a)ebz by parts). Thus, by, Theorem 7.41, we have, +, (z + a)ebz dz = F (1 + i) − F (0)., C, , Similarly,, + 1, eb, 1, eb, a, (z + a)ebz dz = F (1) − F (0) = (1 + a) − 2 − + 2 ., b, b, b, b, 0, (iii) If C is the quarter circle |z| = 2 in the first quadrant joining 2 to 2i,, then, according to Theorem 7.41, we have, +, 2i, z n+1, 2n+1 n+1, (i, z n dz =, =, − 1) (n ∈ Z \ {−1})., •, n+1 2, n+1, C, Theorem 7.41 looks deceptively similar to the fundamental theorem of, integral calculus. There is an important difference. The fundamental theorem, says that a continuous function f (x) defined on [a, b] has an antiderivative, F (x) satisfying, + x1, f (t) dt = F (x1 ) − F (x0 ) (a ≤ x0 < x1 ≤ b)., x0, , Theorem 7.41 merely asserts that if the continuous function f (z) has an antiderivative, then the conclusion follows. That continuity is not a sufficient, condition for the existence of an antiderivative can be seen by the following, example., Example 7.44. If the everywhere continuous function f (z) = z had an antiderivative, then the conclusion of Theorem 7.41 would follow. But, + π, +, z dz =, e−it ieit dt = 2πi = 0., |z|=1, , −π, , This shows that f (z) = z does not have an antiderivative., , •, , However, as seen in Example 7.42, Theorem 7.41 provides, a powerful tool, *z, for evaluating definite integrals. So, in order to evaluate z01 f (z) dz, it suffices, to find a analytic function F (z) such that F (z) = f (z). But finding such an, F (z) is not always easy. For instance, if f (z) = sin(1/z) or cos(1/z), how do, we know whether F (z) exists or what precisely is F (z)?, We now examine the relationship between antiderivatives and analytic, functions. The following theorem, at least on the local level provides a condition which guarantees existence of the antiderivatives of a function (see also, Theorem 7.39).
Page 242 :
7.4 Cauchy’s Theorem, , 233, , Theorem 7.45. (Cauchy’s theorem for a disk) Let* f (z) be analytic in a, domain containing the closed disk |z − z0 | ≤ r. Then |z−z0 |=r f (z) dz = 0., Proof. In view of Theorem 7.41, it suffices to find a function F (z) such that, F (z) = f (z) for |z − z0 | ≤ r. Choose any point z = x + iy in the disk and let, C1 be the contour consisting of the horizontal line segment from z0 = x0 + iy0, to x + iy0 followed by the vertical line segment from x + iy0 to x + iy. Also, let, C2 be the contour consisting of the vertical line segment from z0 = x0 + iy0, to x0 + iy followed by the horizontal line segment from x0 + iy to x + iy (see, Figure 7.19)., , Figure 7.19., , By Theorem 7.39 and basic properties of integrals,, +, +, +, f (z) dz =, f (z) dz −, f (z) dz = 0., C1 −C2, , C1, , (7.27), , C2, , Define, +, , +, C1, , +, , x, , f (z) dz =, , F (z) =, , y, , f (t + iy0 ) dt +, x0, , f (x + it)i dt., , In view of (7.27), F (z) may also be expressed as, +, + y, +, F (z) =, f (z) dz =, f (x0 + it)i dt +, C2, , y0, , (7.28), , y0, , x, , f (t + iy) dt., , (7.29), , x0, , Taking the partial derivative of F (z) with respect to y in (7.28), we obtain, (since the first term in right side of (7.28) is independent of y), , + y, ∂, ∂F, =i, f (x + it) dt = if (x + iy) = if (z);, (7.30), ∂y, ∂y, y0, (here the fundamental theorem of calculus is applied to
Page 243 :
234, , 7 Complex Integration and Cauchy’s Theorem, , +, , +, , y, , y0, , +, , y, , g(x, t) dt =, , y, , u(x, t) dt + i, y0, , v(x, t) dt,, , g = u + iv)., , y0, , Similarly, taking the partial derivative of F (z) with respect to x in (7.29), we, have (since the first term in (7.29) is independent of x), , + x, ∂, ∂F, =, f (t + iy) dt = f (x + iy) = f (z);, (7.31), ∂x, ∂x, x0, (again this is a consequence of the fundamental theorem of calculus). In view, of (7.30) and (7.31),, Fx (z) = −iFy (z) = f (z)., , (7.32), , But (7.32) is just the Cauchy–Riemann equations for F (z). Furthermore, the, continuity of partials Fx and Fy on the disk follows from the continuity of, f (z). Hence Theorem 5.17 may be applied to establish the analyticity of F, at z. Since z was arbitrary, F (z) is analytic in the disk |z − z0 | ≤ r. Finally,, from (5.5), we conclude that F (z) = Fx (z) = f (z), i.e., F is a primitive of f, in the disk |z − z0 | ≤ r., Corollary 7.46. Let f be analytic for |z − z0 | < r except, at some point a, *, inside the disk and continuous for |z − z0 | ≤ r. Then |z−z0 |=r f (z) dz = 0., Proof. Follows if we combine Corollary 7.40 and Theorem 7.45., A circle has the property that any point inside can be joined to the center by two distinct broken line segments, which, when taken together, form, the perimeter of a rectangle whose sides are parallel to the coordinate axes., Furthermore, this is the only property that was used in going from Cauchy’s, theorem for a rectangle to Cauchy’s theorem for a circle. In a more general, domain, no such construction is possible; however according to Remark 2.1,, every pair of points in a domain D can be joined by a polygonal line lying in, D (with sides parallel to the coordinate axes). In Ahlfors [A], it is shown that, if the domain is simply connected, then two such polygonal lines C1 and C2, can be constructed so that their difference C1 − C2 consists of a finite number, of boundaries of rectangles traversed alternately in the positive and negative, directions, as illustrated in Figure 7.20., This fact will be used in proving our main theorem., Theorem 7.47. (Cauchy’s Theorem) If f (z) is analytic in* a simply connected domain D and C is a closed contour lying in D, then C f (z) dz = 0., Proof. According to Theorem 7.41, it suffices to find a function F (z) such, that F (z) = f (z) in the simply connected domain D. Fix a point z0 in D, and choose an arbitrary z in D. Then with C1 and C2 constructed as in, Figure 7.20, we define F (z) by
Page 244 :
7.4 Cauchy’s Theorem, , 235, , +, F (z) =, , f (z) dz., C1, , According to Theorem 7.31,, +, +, f (z) dz =, C1 −C2, , +, f (z) dz −, C1, , f (z) dz = 0,, C2, , because the integral around each rectangle is zero. Hence, we also have, +, F (z) =, f (z) dz., C2, , Suppose z1 = x1 +iy1 is the last point of intersection of C1 and C2 between, z0 and z = x + iy. Also suppose that, in this last rectangle, C1 consists of, the horizontal followed by the vertical line, whereas C2 consists of the vertical, followed by the horizontal, as shown in Figure 7.20., , Figure 7.20., , In view of Theorem 7.39, the value for the integral of f (z) from z0 to z1, is the same along both the contours C1 and C2 , and we denote their common, value by K. The remainder of the proof is similar to that of Theorem 7.45,, for we have, + x, + y, +, f (z) dz = K +, f (t + iy1 ) dt +, f (x + it)i dt (7.33), F (z) =, C1, , and, , x1, , +, F (z) =, , +, , From (7.33), we get, , +, , y, , f (z) dz = K +, C2, , y1, , x, , f (x1 + it)i dt +, y1, , f (t + iy) dt. (7.34), x1
Page 245 :
236, , 7 Complex Integration and Cauchy’s Theorem, , ∂, ∂F, =, ∂y, ∂y, , +, , , , y, , f (x1 + it) dt, , = if (x + iy) = if (z), , y1, , and, from (7.34),, ∂, ∂F, =, ∂x, ∂x, , +, , , , x, , f (t + iy) dt, , = f (x + iy) = f (z)., , x1, , Hence, Fx (z) = −iFy (z) = f (z)., , (7.35), , Equation (7.35) represents the Cauchy–Riemann equations for F (z). The partials of F (z) are continuous because f (z) is continuous. Therefore, F (z) is, analytic in D, with F (z) = Fx (z) = f (z) in D., Remark 7.48. The topological notions utilized in the proof of Theorem 7.47, allowed us to deal with finitely many rectangles inside the simply connected, domain, from whence Theorem 7.39 was applicable. But the essence of Theorem 7.39 consisted of “shrinking” a rectangle to* a point. Consequently,, z, •, Cauchy’s theorem ultimately relies on the fact that z00 f (z) dz = 0., Theorem 7.36, which generalized Cauchy’s “weak” theorem (Theorem, 7.33) from simply to multiply connected domains, was purely topological in, nature, and nowhere used the continuity of the partials. Hence, Cauchy’s theorem is also valid for a multiply connected region, the proof consisting of, “transforming” a multiply connected region into a simply connected region,, as in the proof of Theorem 7.36. We remark that if the contour encloses, singularities of the function, we cannot use Cauchy’s theorem. For example,, consider, +, 1, dz, I=, (z, −, 1)2, C, along a simple closed contour having the point 1 as an interior point. Note, that F (z) = −1/(1−z) is an antiderivative of f (z) = (z −1)−2 for z ∈ C\{1}., According to Theorem 7.41, I = 0. But Theorem 7.47 is not applicable. On, the other hand, for example, if, f (z) =, , ez, sin z, or, ,, (z − 1)2, (z − 1)2, , then it is not clear whether these functions have antiderivatives. To cover, a situation like this, we need to develop another theorem called Cauchy’s, integral formula which we shall do in Chapter 8. Its extension in the form of, the Residue theorem will be discussed in Chapter 9. On the other hand, for, certain situations the following theorem is helpful to simplify the problem by, replacing the given contour by another, or others.
Page 246 :
7.4 Cauchy’s Theorem, , 237, , Theorem 7.49. (Cauchy’s Theorem for multiply connected domains) Let, D be a multiply connected domain bounded externally by a simple closed, contour C and internally by n simple closed nonintersecting contours C1 ,, C2 , . . . , Cn . Let f be analytic on D ∪ C ∪ C1 ∪ C2 ∪ · · · ∪ Cn . Then, +, f (z) dz =, C, , n +, , k=1, , f (z) dz, , Ck, , where C is taken counterclockwise around the external boundary C and clockwise around the internal boundaries C1 , C2 , . . . , Cn ., This generalization of Cauchy’s theorem aids us in evaluating integrals, along a contour enclosing, a region in which the function is not analytic., *, First we evaluate C (z − z0 )−1 dz along a simple closed contour C having, z0 is an interior point., For some > 0, the circle C : |z − z0 | = is interior to the contour C., Also, the function f (z) = 1/(z − z0 ) is analytic in the multiply connected, region between C and |z − z0 | = (see Figure 7.21). Hence, ,, ,, +, 1, 1, 1, dz =, dz +, dz., 0=, z, −, z, z, −, z, z, −, z0, 0, 0, C+C1, C, C1, Therefore,, ,, C, , 1, dz = −, z − z0, , ,, C1, , 1, dz =, z − z0, , ,, −C1, , 1, dz., z − z0, , Note that the positive orientation of C1 is clockwise so that the positive, orientation of −C1 is counterclockwise. Parameterizing −C1 by z(t) = eit ,, 0 ≤ t ≤ 2π, we have, + 2π, ,, + 2π , ,, 1, 1, z (t), i eit, dt =, dz =, dz =, dt = 2πi., z(t), eit, C z − z0, −C1 z − z0, 0, 0, , Figure 7.21.
Page 247 :
238, , 7 Complex Integration and Cauchy’s Theorem, , Thus the validity of Cauchy’s theorem for multiple connected regions takes, the worry out of parameterizing, ugly contours. The method used in the above, example shows that C (z −z0 )−1 dz = 2πi or 0 according to whether the point, z0 is inside or outside the simple closed contour C. No additional information, is required in order to evaluate the integral. On the other hand, for n ∈ N\{1}, the function f (z) = (z − z0 )−n is not analytic at z0 and for the same contour, C, we can easily see that, ,, 1, dz = 0., (z, −, z0 )n, C, Observe that this result does not contradict Theorem 7.47. Note also that in, all these cases, the value of the integral does not depend on the radius , as, long as C1 lies inside C., We now illustrate Theorem 7.49 by evaluating the integral, +, 1, ., I=, f (z) dz, f (z) = 2, z, +4, |z|=4, Letting C1 = {z : |z − 2i| = 1} and C2 = {z : |z + 2i| = 1}, we have, by, Theorem 7.49,, +, +, I=, f (z) dz +, f (z) dz, C1, C2, , , + , + , 1, 1, 1, 1, 1, 1, =, −, −, dz +, dz, 4i C1 z − 2i z + 2i, 4i C2 z − 2i z + 2i, 1, 1, = (2πi + 0) + (0 − 2πi) = 0., 4i, 4i, Similarly, one can easily show that, +, +, dz, dz, =0=, 2 + r2, 2 − r2, z, z, C, C, , (r > 0), , where C is the positively oriented circle |z| = r + 1/2., Example 7.50. We wish to evaluate the integral, +, dz, ., 3, |z|=2 z − 3, This integral may be evaluated without using Cauchy’s residue theorem which, will be discussed in Chapter 9. Define D = {z : 2 < |z| < R}. Then, f (z) =, 1/(z 3 − 3) is analytic in D for each R > 2. By Cauchy’s theorem for multiply, connected domains,, +, +, dz, dz, =, 3−3, 3−3, z, z, |z|=2, |z|=R
Page 248 :
7.4 Cauchy’s Theorem, , 239, , and the value of the integral must be independent of R. By the M-L Inequality,, +, 2πR, dz, ≤ 3, → 0 as R → ∞, 3, R −3, |z|=R z − 3, which shows that the value of the given integral is 0. The same argument may, be used for similar integrals. For example, we see that, +, dz, = 0., 1, +, z, +, z2 + z3, |z|=2, Note that (1 + z + z 2 + z 3 )−1 = (1 − z)/(1 − z 4 )., Cauchy’s theorem for multiply connected regions cannot be proved directly by the same method as was used for simply connected regions, because, analytic functions need not have (single-valued) analytic antiderivatives in, multiply connected domains. In the above example, log(z − z0 ) is the analytic, antiderivative of 1/(z − z0 ) only when confined to a branch. This concept of, analytic logarithm in simply connected domains is made more explicit in the, following theorem., Theorem 7.51. If f (z) is analytic and nonzero in a simply connected domain, D, then there exists a function g(z), analytic in D, such that eg(z) = f (z)., Proof. Since f (z) never vanishes in D, the function f (z)/f (z) is analytic in, D. Furthermore, the integral of f (z)/f (z) between any two points in D is, independent of the path in the simply connected domain D. We define g(z), by the formula, + z , f (ζ), dζ + Log f (z0 ),, (7.36), g(z) =, z0 f (ζ), where z0 is fixed point in D, z is an arbitrary point in D, and the path of, integration is any path that lies in D. Set h(z) = f (z)e−g(z) , and observe that, h (z) = f (z)e−g(z) − f (z)g (z)e−g(z), f (z) −g(z), e, = f (z)e−g(z) − f (z), f (z), = 0., Thus h(z) is a constant in D. To determine the constant, we set z = z0 to, obtain, h(z0 ) = f (z0 )e−g(z0 ) = f (z0 )e− Log f (z0 ) = 1., Therefore, f (z)e−g(z) ≡ 1 throughout D, and the theorem is proved., To see that the hypothesis that the domain be simply connected is essential, observe that 1/z never vanishes in the punctured plane and cannot, be expressed as eg(z) for an analytic function g(z). (Recall that no branch of, − log z is analytic in the punctured plane.)
Page 249 :
240, , 7 Complex Integration and Cauchy’s Theorem, , Corollary 7.52. If f (z) is analytic and nonzero in a simply connected domain, D, then an analytic branch of (f (z))1/n (n a positive integer) can be defined, in D., Proof. Set f (z) = eg(z) , where g(z) is analytic in D, and the existence, of g provided in Theorem 7.51. Define the nth root function (f (z))1/n by, (f (z))1/n = e(1/n)g(z) ., Remark 7.53. More generally, each of the n functions e(g(z)+2kπi)/n (k =, •, 0, 1, 2, . . . , n − 1) is an analytic branch of (f (z))1/n ., We end this section with an example., Example 7.54. Consider, +, f (z) dz,, , f (z) = 1/z 1/2 ., , |z|=1, , Then z = 0 is a branch point of f (z). If we choose principal branch, then, z 1/2 = e(1/2) Log z = e(1/2)(ln |z|+iArg z), so that for z = eiθ , we have z 1/2 = ei(1/2)Arg z = e(1/2)iθ and, + π, + π, ieiθ, dθ, =, i, e−iθ/2 dθ = 4i., iθ/2, −π e, −π, If C is the line segment [1, 1 + i] connecting 1 and 1 + i, and if we choose the, principal branch for z 1/2 , then we have F (z) = z 1/2 = e(1/2) Log z , and so, F (z) = e(1/2) Log z, , 1 e(1/2) Log z, 1, 1, 1, 1, =, =, = f (z), (1/2), Log, z, z, Log, 2z, 2 e, 2e, 2, , where F is analytic on C \(−∞, 0] with F (z) = f (z). Using this we compute, that, +, +, dz, =, 2, F (z) dz = 2 [F (1 + i) − F (1)] = 2 [21/4 eiπ/8 − 1]., •, 1/2, z, C, C, Questions 7.55., 1. If f (z), * and C is a closed contour in the domain,, * is analytic in a domain, does C |f (z)| dz = 0? Does C f (z) |dz| = 0?, 2. Where in the proof of Theorem 7.47 did we use the fact that the domain, was simply connected?, *, 3. Suppose f (z) is analytic on a contour C., * Does C f (z)*dz = 0?, 4. If f is continuous on the contour C, is C f (z) dz = − −C f (z) dz?
Page 250 :
7.4 Cauchy’s Theorem, , 241, , 5. For what type of simple closed contours, is, +, (1 + z + z 2 + · · · + z n )−1 dz = 0?, γ, , 6. What is the relationship between Theorem 7.41 and Green’s theorem?, 7. Where, in the proof of Theorem 7.45, was the hypothesis of analyticity, needed?, 8. What are the differences between Cauchy’s theorem and Cauchy’s weak, theorem?, *, 9. What can be said about C (1/z) dz if the contour C passes through the, origin?, *, 10. What values may be assumed by C (1/z) dz if C is a closed curve that, is not simple?, 11. Is the function g(z) in Theorem 7.51 unique?, 2, 2, 12. What is an antiderivative of cos(z, * )? of2 sin(z )? Is it possible to use, Theorem 7.41 to conclude that C sin(z ) dz = 0 for any simple closed, contour C?, 13. What is a domain D of analyticity of f (z) = z(z − 1)1/2 ? Find an, antiderivative of f in D?, *, 14. When can a contour integral C f (z) dz be independent of the path?, 15. What is a complex version of the fundamental, theorem of *calculus?, *, 16. Let C* be a closed contour. Does, z, dz, =, 0?, Does, * C, * Re C z dz = 0?, Does C (Re z) dz = 0? Does Im C z dz = 0? Does C (Im z) dz = 0?, Exercises 7.56., *i, 1. Evaluate −i |z| dz along different contours. Does |z| have an antiderivative?, *, 2. Evaluate γ f (z) dz, where, (i) f (z) = z 3 and γ(t) = t2 + it for t ∈ [0, π], (ii) f (z) = sin z and γ(t) = t + it2 for t ∈ [0, π/2], (iii) f (z) = 1/z and γ(t) = cos t2 − i sin t for t ∈ [0, π/2], (iv) f (z) = 1/z and γ(t) = − cos t − ie sin t for* t ∈ [0, π/2]., 3. Give an example of a function f (z) for which |z|=r f (z) dz = 0 for each, r > 0 even though f (z) is not analytic everywhere., 4. Let f = u+iv be analytic inside and on a simple closed contour γ. Show, by an example that Cauchy’s theorem does not hold separately for the, real and, * imaginary parts of f ., 5. Find C (1 + z 2 )−1 dz, where C is the circle, (a) |z − i| = 1, , (b) |z + i| = 1, , (c) |z| = 2, , (d) |z − 1| = 1., , 6. Separate the integrand into real, * and imaginary parts and evaluate,, where possible, the expression C (ez /z) dz, where C is, (a) |z| = 1, , (b) |z−2| = 1, , (c) the square having vertices ±1 ± i.
Page 251 :
242, , 7 Complex Integration and Cauchy’s Theorem, , *z, 7. Suppose Re z0 > 0 and Re z1 > 0. Evaluate z01 (1/z) dz along contours, in the right half-plane., 8. Let a, b ∈ C and r > 0. Then, by decomposing the integrand into partial, fractions, show that, ⎧, ⎪, ⎪, ⎪, ⎪, ⎪, ⎨, , +, |z|=r, , 0, 0, dz, 2πi, =, (z − a)(z − b) ⎪, a−b, ⎪, ⎪, ⎪, ⎪, ⎩ 2πi, b−a, , if |a| > r and |b| > r (a, b = 0), if |a| < r and |b| < r (a, b ∈ C), if |a| < r < |b|, if |b| < r < |a|., , 9. Suppose that f is analytic for |z| < 2 and α is a complex constant., Evaluate, +, f (z), dz., I=, (Re z + α), z, |z|=1, *z, 10. Evaluate z12 az dz (a = 0 is given)., *, 11. Find |z|=1 f (z) dz, when, (a) f (z) = (z sin z)/(z + 2) + z, , (b) f (z) = z 4 + iz + 2Im z.
Page 252 :
8, Applications of Cauchy’s Theorem, , Most of the powerful and beautiful theorems proved in this chapter have, no analog in real variables. While Cauchy’s theorem is indeed elegant, its, importance lies in applications. In this chapter, we prove several theorems that, were alluded to in previous chapters. We prove the Cauchy integral formula, which gives the value of an analytic function in a disk in terms of the values, on the boundary. Also, we show that an analytic function has derivatives of all, orders and may be represented by a power series. The fundamental theorem, of algebra is proved in several different ways. In fact, there is such a nice, relationship between the different theorems in this chapter that it seems any, theorem worth proving is worth proving twice., , 8.1 Cauchy’s Integral Formula, *If f (z) is analytic in a simply connected domain D, then we know already that, f (z) dz = 0 along every closed contour C contained in D. An interesting, C, variation occurs when a function is analytic at all but a finite number of, points. As we have seen in the previous chapter,, +, 1, dz = 2πi, (8.1), C z − z0, along every positively oriented simple closed contour C containing z0 . We now, develop the Cauchy integral formula which is indeed a generalization of (8.1)., Moreover, Cauchy’s integral formula leads to three important properties of, analytic functions that are unparalleled in real variable methods:, •, •, •, , every analytic function is infinitely differentiable, see Theorem 8.3;, every analytic function can be expressed locally as a Taylor series in the, vicinity of a point of analyticity, see Theorem 8.8;, every analytic function can be expressed as a Laurent series in the vicinity of an isolated singularity, see Section 9.2.
Page 253 :
244, , 8 Applications of Cauchy’s Theorem, , These facts hinge on the following result., Theorem 8.1. (Cauchy’s First Integral Formula) Let f (z) be analytic in, a simply connected domain containing the simple closed contour C. If z0 is, inside C, then, +, 1, f (z), dz., f (z0 ) =, 2πi C z − z0, Proof. Given > 0, construct a circle Cr : |z − z0 | = r inside C and small, enough so that |f (z) − f (z0 )| < for all z on Cr . According to Cauchy’s, theorem for multiply connected regions,, +, +, f (z), f (z), dz =, dz., C z − z0, Cr z − z0, Thus, writing f (z) = f (z0 ) + (f (z) − f (z0 )) for the integral on the right, one, has, +, +, +, f (z), f (z0 ), f (z) − f (z0 ), dz =, dz +, dz, z, −, z, z, −, z, z − z0, 0, 0, C, Cr, Cr, +, *, f (z) − f (z0 ), 1, dz, since Cr z−z, dz = 2πi., = 2πif (z0 ) +, 0, z, −, z, 0, Cr, Since the integral in the left side has a fixed value, as does 2πif (z0 ), it follows, that for each Cr , the value of the contour integral, +, f (z) − f (z0 ), dz, z − z0, Cr, is constant. We now show that this value must be zero. We have, +, +, f (z) − f (z0 ), |f (z) − f (z0 )|, |dz| < (2πr) = 2π ., dz ≤, z − z0, |z − z0 |, r, Cr, Cr, Since, , is arbitrary, the integral is zero. This concludes the proof., , Remark 8.2. When f (z) ≡ 1, the conclusion of the theorem reduces to (8.1)., Moreover, Theorem 8.1 expresses the value of f (z) at any point inside C in, terms of its values on C. In other words, if f (z) is known to be analytic inside, and on the boundary of a simply connected domain, then the values of f (z), on the boundary completely determine the values of f (z) inside. There is no, analog to this theorem for functions of a real variable. More precisely, when, a real-valued function f (x) is differentiable on the closed interval [a, b], its, value at x = a and x = b in no way can dictate the value of f (x) on the open, interval (a, b). For instance, for each n ∈ N, the functions, fn (x) = xn , 0 ≤ x ≤ 1,, all have the same boundary values (f (0) = 0, f (1) = 1) but differ from one, another at all interior points. Similarly, we see that when a function f (x, y) of, two real variables x, y real differentiable inside and on a simple closed contour, •, C, its value on C do not determine the values of f (x, y) inside C.
Page 254 :
8.1 Cauchy’s Integral Formula, , 245, , We next express the derivative of an analytic function in terms of an, integral. Using the notation of Theorem 8.1, choose h small enough in absolute, value so that z0 + h is inside C. Then,, +, +, 1, 1, f (z), f (z), dz., dz and f (z0 + h) =, f (z0 ) =, 2πi C z − z0, 2πi C z − (z0 + h), Hence, for h = 0,, , + , 1, 1, 1, f (z0 + h) − f (z0 ), =, −, f (z) dz, h, 2πih C z − z0 − h z − z0, +, f (z), 1, =, dz., 2πi C (z − z0 − h)(z − z0 ), , (8.2), , When h → 0, the integrand approaches f (z)/(z − z0 )2 . It appears probable, that the limit is, +, f (z), 1, dz,, 2πi C (z − z0 )2, although in general the limit of the integrand is not necessarily the same as, the integrand of the limit. To prove that we may take the limit inside the, integral, we must show that the difference, +, +, 1, 1, f (z), f (z), dz −, dz, (8.3), 2πi C (z − z0 − h)(z − z0 ), 2πi C (z − z0 )2, +, f (z), h, =, dz, 2πi C (z − z0 − h)(z − z0 )2, can be made arbitrarily small. Given a circle C1 : |z − z0 | = r contained, in C, choose h small enough so that |h| ≤ r/2 (see Figure 8.1). Note that, z0 ∈ Int (C1 ) and, |z0 + h − z0 | < r/2, , ⇐⇒ |h| < r/2,, , Figure 8.1.
Page 255 :
246, , 8 Applications of Cauchy’s Theorem, , which shows that z0 + h ∈ Int (C1 ). Further, for z ∈ C1 , we have, |z − z0 − h| ≥ |z − z0 | − |h| ≥ r − |h| ≥ r − r/2 > 0, and so, by Cauchy’s theorem for multiply connected domains, we have, +, +, f (z), f (z), dz =, dz., 2, (z, −, z, −, h)(z, −, z, ), (z, −, z, −, h)(z − z0 )2, 0, 0, 0, C, C1, Since f (z) is continuous on C1 , it is bounded (say |f (z)| ≤ M on C1 ). Thus, +, +, |h|, h, f (z), |f (z)|, dz ≤, |dz|, 2πi C1 (z − z0 − h)(z − z0 )2, 2π C1 |z − z0 − h| |z − z0 |2, +, |h|M, |dz|, ≤, 2πr2 C1 |z − z0 | − |h|, +, |h|M, |dz|, ≤, πr3 C1, , , |h|M, 2M, =, (2πr) = |h|, ., πr3, r2, Clearly, the limit of this expression as h → 0 is zero and so, (8.3) tends to, zero with h. In view of (8.2) and (8.3),, +, 1, f (z), , f (z0 ) = lim, dz, (8.4), h→0 2πi C (z − z0 − h)(z − z0 ), +, 1, f (z), dz., =, 2πi C (z − z0 )2, Equation (8.4) expresses the value of f (z) at any point inside C in terms, of the values of f (z) on C. Moreover, (8.4) shows that the operations of, differentiation and contour integration can be interchanged. We knew, by, hypothesis, that f (z) was differentiable at all points inside C. But the above, process can be repeated. From (8.4), we obtain, +, 1, f (z0 + h) − f (z0 ), {2(z − z0 ) − h}f (z), =, dz., (8.5), h, 2πi C (z − z0 )2 (z − z0 − h)2, As before, we can show that the limit may be taken inside the integral so that, (8.5) leads to, +, 2, f (z), dz., (8.6), f (z0 ) =, 2πi C (z − z0 )3, Equation (8.6) gives much more information than we had a right to expect., First, we have the existence of f (z) at all points inside C. Next, the second, derivative may be expressed in terms of the values of f (z) on C. This argument
Page 256 :
8.1 Cauchy’s Integral Formula, , 247, , can be repeated indefinitely. An induction shows that the nth derivative is, given by, +, n!, f (z), (n), f (z0 ) =, dz., (8.7), 2πi C (z − z0 )n+1, To see this, we assume that (8.7) holds for n = k ≥ 1. Then, +, k!, f (z), dz., f (k) (z0 ) =, 2πi C (z − z0 )k+1, We must show that f (k+1) (z0 ) exists and the formula (8.7) holds for n = k +1., To do this, we use the binomial expansion, (z − z0 − h)k+1 = (z − z0 )k+1 − (k + 1)(z − z0 )k h +, , (k + 1)k, (z − z0 )k−1 h2 − · · ·, 2, , to express, 1, 1, −, (z − (z0 + h))k+1, (z − z0 )k+1, (z − z0 )k+1 − (z − z0 − h)k+1, (z − z0 )k+1 (z − z0 − h)k+1, (k + 1)h, (k + 1)k, h2, =, −, + ···, k+1, 2, (z − z0 )(z − z0 − h), 2, (z − z0 ) (z − z0 − h)k+1, =, , where the dots indicate terms with powers of h up to hk+1 . In view of this, expression,, , , +, k+1, k!, f (k) (z0 + h) − f (k) (z0 ), f (z), =, +, O(h), dz., h, 2πi C z − z0 (z − z0 − h)k+1, Letting h → 0, we obtain (8.7) for n = k + 1,, +, (k + 1)!, f (z), dz., f (k+1) (z0 ) =, k+2, 2πi, C (z − z0 ), Hence, every analytic function has derivatives of all orders and derivatives of, all orders at each point may be expressed in terms of the values of the function, on its boundary. We sum up this remarkable result with, Theorem 8.3. (Generalized Cauchy’s Integral Formula) Let f (z) be analytic in a simply connected domain containing the simple closed contour C., Then f (z) has derivatives of all orders at each point z0 inside C, with, +, n!, f (z), (n), dz., f (z0 ) =, 2πi C (z − z0 )n+1
Page 257 :
248, , 8 Applications of Cauchy’s Theorem, , Remark 8.4. While Theorem 8.3 is stated as a global property, a property, of simply connected domains, it is really a local property. Since a function, analytic at a point is also analytic in a (simply connected) neighborhood of, the point, it follows that a function analytic at a point must have derivatives, of all orders at that point., •, Note how this result is radically different from the theory of functions of, a real variable. In basic calculus, we have learned that the existence of the, derivative of f (x) does not guarantee the continuity of the derivative f (x),, much less the differentiability of f (x), see the example on p. 134. Now, we, consider the function, f (x) = x7/5, which has a first derivative for all x ∈ R and, f (x) = (7/5)x2/5 ., Observe that f (x) does not have a first derivative at x = 0 and therefore,, f (x) does not have a second derivative at the origin. Similarly, f (x) = x11/5, has a first and second derivative on R but has no third derivative at x = 0. It, follows that the existence of n derivatives of a real-valued function f (x) does, not guarantee the (n + 1)th derivative of f (x). Thus, Theorem 8.3 does not, hold in the case of functions of a real variable., This example demonstrated one essential difference between functions of, a complex variable and functions of a real variable., Remark 8.5. Cauchy’s integral formula is also valid for multiply connected, regions. We prove it for the multiply connected region in Figure 8.2. In the, •, exercises, the reader is asked to supply the general proof., , Figure 8.2., , Suppose f (z) is analytic in the multiply connected region R whose boundary consists of the contour C = C1 ∪ C2 . Construct a circle Γ contained in R, and having center at z0 . Then by Cauchy’s theorem for multiply connected, regions,
Page 258 :
8.1 Cauchy’s Integral Formula, , 1, 2πi, , ,, C1, , 1, f (z), dz +, z − z0, 2πi, , ,, C2, , 1, f (z), dz +, z − z0, 2πi, , ,, Γ, , 249, , f (z), dz = 0., z − z0, , Thus, in view of Theorem 8.1,, ,, 1, f (z), dz, f (z0 ) =, 2πi Γ z − z0, ,, ,, 1, 1, f (z), f (z), dz −, dz., =, 2πi C1 z − z0, 2πi C2 z − z0, That is,, 1, f (z0 ) =, 2πi, , +, C, , f (z), dz,, z − z0, , and this proves Theorem 8.1 for the multiply connected region R. Similarly,, we can show that the conclusion of Theorem 8.3 remains valid for the multiply connected region R (see Figure 8.2). Next we have an immediate and, important corollary to Theorem 8.3., Corollary 8.6. If f = u + iv is analytic in a domain D, then all partial, derivatives of u and v exist and are continuous in D., Proof. Let f = u + iv be analytic in D. Then, by (5.3),, f (z) = ux + ivx = vy − iuy ., By the analyticity of f (z), it follows that each of the first partial derivatives, of u and v exist and are continuous in D because f (z) is continuous in D., Because f (z) is analytic in D, from the above equation, we again have, f (z) = uxx + ivxx, = (vx )y − i(ux )y, = (vy )x − i(uy )x ., This process may be continued to conclude that u and v have continuous, partial derivatives of all orders at each point where the function f = u + iv is, analytic., Example 8.7. Setting f (z) = z − 3 cos z, we compute, +, , z − 3 cos z, π, = 8πi., dz, =, 2πif, 2, 2, |z|=2 (z − π/2), , •, , We are now able to prove Taylor’s theorem for complex functions., Theorem 8.8. (Taylor’s Theorem) Let f (z) be analytic in a domain D, whose boundary is C. If z0 is a point in D, then f (z) may be expressed as
Page 260 :
8.1 Cauchy’s Integral Formula, , 251, , 0, 1, , 0, Figure 8.3., , (8.11) leads to, M, Rn ≤, 2π(ρ − r), , n +, n, r, Mρ r, |dζ| =, ., ρ, ρ−r ρ, C1, , Since r < ρ, (r/ρ)n → 0 as n → ∞. Thus,, f (z) =, , ∞, , f (n) (z0 ), (z − z0 )n ,, n!, n=0, , and the proof is complete., Remark 8.9. By the M -test (Theorem 6.31), the Taylor series for f (z) also, •, converges uniformly on compact subsets of |z − z0 | < δ., In Section 6.3, we saw that a power series represents an analytic function, inside its circle of convergence. Theorem 8.8 is essentially the, Thus,, converse., ∞, a function f (z) is analytic at a point z0 if and only if f (z) = n=0 an (z −z0 )n, in some disk |z − z0 | ≤ r, where, 1, f (n) (z0 ), =, an =, n!, 2πi, , +, |z−z0 |=r, , f (z), dz., (z − z0 )n+1, , Theorem 8.8 justifies the Maclaurin expansion, ez =, , ∞, ∞, , , zn, (−1)n 2n−1, , sin z =, z, ,, n!, (2n − 1)!, n=0, n=0, , cos z =, , ∞, , (−1)n 2n, z, (2n)!, n=0, , that were stated, without proof, in Chapter 6 (see also the Examples below)., Examples 8.10., (i) Consider f (z) = sin z. Then f is entire. Also, for each, n ∈ N, we have f (z) = cos z = sin(z + π/2) and
Page 261 :
252, , 8 Applications of Cauchy’s Theorem, , f (z) = cos(z + π/2) = sin(z + π/2 + π/2) = sin(z + π)., Consequently, f (n) (z) = sin(z + nπ/2) and so, , 0 if n = 2k, (n), f (0) = sin(nπ/2) =, ,, (−1)k if n = 2k + 1, , k ∈ N0 ., , It follows that, sin z = z −, , ∞, , z5, (−1)n+1 2n−1, z3, +, − ··· =, z, ,, 3!, 5!, (2n − 1)!, n=1, , z ∈ C., , (8.12), , A similar method gives that, cos z = 1 −, , ∞, , z4, (−1)n 2n, z2, +, − ··· =, z ,, 2!, 4!, (2n)!, n=1, , z ∈ C., , (8.13), , d, (sin z) = cos z, Also, it is much easier to use (8.12), and the relation dz, to achieve (8.13), because of Theorem 6.51., (ii) Suppose we wish to find the Taylor expansion of f (z) = ez about the, point z = 1. We could of course, begin by computing the coefficients, using the formula for an in Theorem 8.8. To avoid this, we may simply, rewrite, , f (z) = ez−1+1 = eez−1 = e, , ∞, , (z − 1)n, ,, n!, n=0, , |z − 1| < ∞,, , because of the known series expansion for ez ., (iii) To find the Taylor expansion for f (z) = zez about a point z = a, we, may simply rewrite, f (z) = (z − a + a)ez−a+a, = ea [(z − a)ez−a + aez−a ], ', &∞, ∞, (z − a)n+1, , (z − a)n, a, +a+a, , |z − a| < ∞,, =e, n!, n!, n=0, n=1, ', &∞, ∞, n, (z − a)n, , (z, −, a), +a+a, = ea, (n − 1)!, n!, n=1, n=1, &, ', ∞, , n+a, (z − a)n , for all z ∈ C., = ea a +, n!, n=1, For instance if a = 1, it follows that, &, ', ∞, , n+1, z, n, ze = e 1 +, (z − 1), n!, n=1, , for all z ∈ C.
Page 262 :
8.1 Cauchy’s Integral Formula, , 253, , (iv) To find the Taylor expansion for 1/z about a = 0, we simply follow the, above path, namely,, , , ∞, 1, 1, 1, 1 (−1)n, 1, =, =, =, (z − a)n, z, a+z−a, a 1 + (z − a)/a, a n=0 an, which is valid for |z − a| < |a|. For example, for a = i, −i, one has, ∞, ∞, , 1, 1 (−1)n, n, •, =, (z, −, i), :=, in−1 (z − i)n , |z − i| < 1,, z, i n=0 in, n=0, ∞, ∞, , 1 (−1)n, (z + i)n, 1, n, •, =, (z, +, i), =, −, , |z + i| < 1., n, z, −i n=0 (−i), in+1, n=0, (v) A similar technique may be adopted to find the Taylor expansion about, z = a = 0, for f (z) = 1/z 2 . To do this, we first recall that for |z| < 1, ∞, , 1, =, zn, 1 − z n=0, , and, because of Theorem 6.51, differentiating with respect to z gives, ∞, ∞, , , 1, n−1, =, nz, =, (n + 1)z n ,, (1 − z)2, n=1, n=0, , |z| < 1., , Using this, we may write, 1, 1, 1, 1, =, = 2, 2, 2, z, (a + z − a), a [1 + (z − a)/a]2, , n, ∞, z−a, 1 , = 2, (n + 1)(−1)n, , |z − a| < |a|., a n=0, a, In particular, the Taylor expansion of 1/z 2 about a = 1 follows:, ∞, , 1, =, (−1)n (n + 1)(z − 1)n ,, z2, n=0, , |z − 1| < 1., , Moreover, it is clear that for |z − 1| < 1,, ∞, z−1 , =, (−1)n−1 n(z − 1)n, z2, n=1, , which is the Taylor series for (z − 1)/z 2 about z = 1. We can often build, •, on results such as this., We now examine some relationships between uniform convergence and, integration.
Page 263 :
254, , 8 Applications of Cauchy’s Theorem, , Theorem 8.11. Let {fn (z)} be a sequence of functions continuous on a contour C, and suppose that {fn (z)} converges uniformly to f (z) on C. Then, +, +, +, fn (z) dz =, lim fn (z) dz =, f (z) dz., lim, n→∞, , C n→∞, , C, , C, , Proof., The statement of *the theorem requires us to show that the sequence, *, f, (z), dz converges to C f (z) dz., n, C, *, Note that, by Theorem 6.26, f (z) is continuous on C so that C f (z) dz, exists. Given > 0, there is an integer N = N ( ) such that, |fn (z) − f (z)| <, , for n > N and all z on C., , Denoting the length of C by L, it follows, for n > N , that, +, +, +, fn (z) dz −, f (z) dz =, (fn (z) − f (z)) dz, C, C, C, +, ≤, |fn (z) − f (z)| |dz| < L., C, , Since, , is arbitrary, the proof is complete., , Corollary, ∞ 8.12. Suppose {fn (z)} is a sequence of continuous functions and, that n=0 fn (z) converges uniformly on a contour C. Then, , + , ∞ +, ∞, , fn (z) dz =, fn (z) dz., C, , n=0, , Proof. Set Sn (z) =, , k=0, , C, , fn (z) dz, , C, , = lim, , n→∞, , But by Theorem 8.11,, +, +, Sn (z) dz =, lim, n→∞, , C, , n=0, , fk (z). Then,, , , , ∞ +, , n=0, , n, , n +, , k=0, , +, fk (z) dz = lim, , lim Sn (z) =, , C n→∞, , n→∞, , C, , + , ∞, C, , Sn (z) dz., C, , , fn (z) dz., , n=0, , Remark 8.13. Our proof of Theorem 8.8 mimicked the proof in the real case., In view of this corollary, we can now give a simpler proof that does not involve, a remainder term. Instead of (8.9), we can write, ∞, , 1, (z − z0 )n, =, ,, ζ − z n=0 (ζ − z0 )n+1, , the convergence being uniform on C1 . Hence we may multiply by f (ζ)/2πi, and integrate term-by-term. This leads directly to
Page 264 :
8.1 Cauchy’s Integral Formula, , 1, 2πi, , 255, , +, , f (ζ), dζ, C1 ζ − z, , + , ∞, 1, (z − z0 )n, =, f (ζ) dζ, 2πi C1 n=0 (ζ − z0 )n+1, , , +, ∞, , 1, f (ζ), =, (z − z0 )n, dζ, 2πi C1 (ζ − z0 )n+1, n=0, , f (z) =, , =, , ∞, , f (n) (z0 ), (z − z0 )n ., n!, n=0, , •, , We might expect a function f (z) to be analytic at a point z0 if, +, f (z) dz = 0, C, , along every simple closed contour in which z0 is an interior point. Unfortunately, this converse of Cauchy’s theorem is not true. For example,, +, 1, dz = 0, 2, z, C, along every simple closed contour C having the origin as an interior point., This is because f (z) is analytic in the region between C and some circle |z| =, contained in C. Thus by Cauchy’s theorem for multiply connected regions,, +, +, +, + 2π, i 2π −iθ, 1, 1, i eiθ, dz, =, dz, =, dθ, =, e, dθ = 0., 2, 2, 2 eiθ, C z, |z|= z, 0, 0, But we do have a partial converse to Cauchy’s theorem even when the domain, D is not simply connected., Theorem, 8.14. (Morera’s Theorem) Let f (z) be continuous in a domain, *, D. If C f (z) dz = 0 along every simple closed contour C contained in D,, then f (z) is analytic in D., Proof. Fixing z0 in D, the value of the function, + z, f (ζ) dζ, F (z) =, z0, , is independent of the path of integration from z0 to z inside D. Choose h, small enough so that the line segment from z to (z + h) lies in D, and consider, the difference quotient, &+, ', +, + z, z+h, 1, 1 z+h, F (z + h) − F (z), =, f (ζ) dζ −, f (ζ) dζ =, f (ζ) dζ., h, h z0, h z, z0
Page 265 :
256, , 8 Applications of Cauchy’s Theorem, , Then, 1, F (z + h) − F (z), − f (z) =, h, h, , +, , z+h, , (f (ζ) − f (z)) dζ,, , (8.14), , z, , where the path from z to (z + h) is taken to be the straight line segment., By the continuity of f (z), we have |f (ζ) − f (z)| < for |h| sufficiently small., Since the line segment from z to (z + h) has length |h|, it follows from (8.14), that, F (z + h) − F (z), 1, − f (z) <, |h| = ., h, |h|, Therefore, F (z) is analytic in D with F (z) = f (z). Moreover, F (z) must have, derivatives of all orders. In particular, F (z) = f (z) at all points in D, thus, proving the analyticity of f (z)., Corollary 8.15. Let f (z) be *analytic in a simply connected domain D. If z0, z, is a point in D, then F (z) = z0 f (ζ) dζ is analytic in D., Proof., According to Cauchy’s theorem for simply connected domains, we have, *, f, (z), dz = 0 along every closed contour C contained in D. Hence f (z), C, satisfies the hypotheses (as well as the conclusion) of Morera’s theorem. The, result is thus implicit in the proof of Morera’s theorem., Recall, that even if f (z) is analytic in a domain D, we are not guaranteed, *, that C f (z) dz = 0 along every simple closed contour C contained in D. The, function f (z) = 1/z, * is analytic in the annulus bounded by the circles |z| = 1/2, and |z| = 2. But |z|=1 (1/z) dz = 2πi even though the circle is contained in, the annulus. Note, however, that the interior of |z| = 1 is not contained in the, annulus, so that Cauchy’s theorem is not applicable. For a simply connected, domain, it is true that the integral around every simple closed contour in the, domain is zero., In view of Morera’s theorem, we can say that a necessary and sufficient, condition for a continuous function to be analytic in a simply connected domain is that the integral be independent of the path of integration. At first, glance, it appears that Morera’s theorem is useless for proving a function to, be analytic, in as much as it is not possible to test all simple closed contours., However, the proof of the next theorem should dispel any doubts as to the, utility of Morera’s theorem., Theorem 8.16. Let {fn (z)} be a sequence of analytic functions converging, uniformly to a function f (z) on all compact subsets of a domain D. Then, f (z) is analytic in D., Proof. It suffices to show that f (z) is analytic at an arbitrary point z0 in D., Construct a neighborhood D of z0 contained in D. By Theorem 6.26, f (z), must be continuous at all points in D . According to Theorem 8.11,
Page 266 :
8.1 Cauchy’s Integral Formula, , +, lim, , n→∞, , 257, , +, fn (z) dz =, , C, , f (z) dz, , (8.15), , C, , along every simple closed contour C contained in D . Since fn (z) is analytic, in D ,, +, fn (z) dz = 0, C, , for each n and each simple closed contour C. In view of (8.15),, +, f (z) dz = 0., C, , By Morera’s theorem, f (z) must be analytic in D . In particular, f (z) is, analytic at z0 . This completes the proof., Remark 8.17. Requiring uniform convergence only on compact subsets,, rather than on the whole domain, will give us the needed flexibility to deal, •, with certain questions in later chapters., Rewriting Theorem 8.16 in terms of series,, ∞we have the following: If {fn (z)}, is a sequence of analytic functions, and k=0 fk (z) converges uniformly to, f (z) on compact subsets of D, then f (z) is analytic in D., This follows on noting that, for each simple closed contour C contained, in D,, , + , +, ∞ +, ∞, , fk (z) dz =, fk (z) dz =, f (z) dz = 0., k=0, , C, , C, , C, , k=0, , (z)} is a sequence of functions analytic in a, Corollary 8.18. Suppose {f, n∞, domain D, and that f (z) = n=0 fn (z), the series being uniformly convergent, on all compact subsets of D. Then for all z in D,, f (z) =, , ∞, , , fn (z)., , n=0, , Proof. According to Theorem 8.16, f (z) is analytic in D. Hence, by Theorem, 8.3,, +, 1, f (ζ), dζ,, f (z) =, 2πi C (ζ − z)2, where C is any simple closed contour in D , a neighborhood of z contained in, D. Also note that, for each n,, +, 1, fn (ζ), dζ., fn (z) =, 2πi C (ζ − z)2, ∞, Since n=0 [fn (ζ)/(ζ − z)2 ] converges uniformly to f (ζ)/(ζ − z)2 for ζ on C,, we have
Page 267 :
258, , 8 Applications of Cauchy’s Theorem, , 1, 2πi, , +, , f (ζ), dζ, 2, C (ζ − z), , , +, ∞, , 1, fn (ζ), =, dζ, 2πi C n=0 (ζ − z)2, , +, ∞ , , 1, fn (ζ), =, dζ, 2πi C (ζ − z)2, n=0, , f (z) =, , =, , ∞, , , fn (z)., , n=0, , More generally, for each integer k,, +, +, k!, k!, f (ζ), fn (ζ), (k), (k), dζ and fn (z) =, dζ., f (z) =, 2πi C (ζ − z)k+1, 2πi C (ζ − z)k+1, ∞, Since n=0 [fn (ζ)/(ζ − z)k+1 ] converges uniformly to f (ζ)/(ζ − z)k+1 for ζ, on C, we can conclude, as above, that, f (k) (z) =, , ∞, , , fn(k) (z) for all z in D., , n=0, , ∞, Example 8.19. For example, n=1 3−n sin(nz) represents an analytic function in the strip |Im z| < ln 3. Indeed, as, |3−n sin(nz)| = 3−n, , einz − e−inz, 2, , , 2en|Im z|, , 3−n, 2, = e−n(ln 3−|Im z|) ,, ∞, the Weierstrass M -test shows that n=1 3−n sin(nz) converges uniformly on, each compact subset of D = {z : |Im z| < ln 3}. By Corollary 8.18, the given, •, series of functions represents an analytic function for |Im z| < ln 3., ≤, , Remark 8.20. Note a difference between real and complex series. The real, series, ∞, , sin nx, f (x) =, n2, n=1, is uniformly convergent on the real line. But a term-by-term differentiation, leads to, ∞, , cos nx, f (x) =, ,, n, n=1, which does not converge at x = 0. In the complex case, a series of analytic, functions uniformly convergent on compact subsets of a domain may be differentiated term-by-term to obtain the derivative of the sum. However, we
Page 268 :
8.1 Cauchy’s Integral Formula, , 259, , cannot extend this to the boundary of the domain. For example, even though, the function, ∞, , zn, f (z) =, n2, n=1, is uniformly convergent in the disk |z| ≤ 1, term-by-term differentiation at, z = 1 would yield, ∞, , 1, ,, f (1) =, n, n=1, , •, , which does not converge., , Suppose f (z) is an entire function. According to Theorem 8.8, f (z) has a, power series representation, f (z) =, , ∞, , , an z n =, , n=0, , ∞, , f (n) (0) n, z, n!, n=0, , valid for all z. By Theorem 6.51 or by Corollary 8.18, the derivative of the, sum is the sum of the derivatives. That is,, f (z) =, , ∞, , , nan z n−1, , n=1, , for all z. Now by Corollary 8.12, we may also integrate term-by-term. In other, words,, , + z , + z, ∞, n, dζ, f (ζ) dζ =, an ζ, F (z) =, 0, , 0, , n=0, , ∞, , an n+1, =, z, ,, n, +1, n=0, , where the integral is taken along any contour joining the origin to z., These results may be combined to obtain useful power series relationships., Example 8.21. Let us now expand f (z) = sin2 z in a Maclaurin series. We, could, of course, take derivatives to obtain the Maclaurin expansion directly., But let us consider other methods., Method 1. We may use (8.12) and obtain, , , , z5, z3, z5, z3, 2, +, + ···, z−, +, + ··· ,, sin z = z −, 3!, 5!, 3!, 5!, , z ∈ C., , Collecting terms and arranging in ascending order, we obtain, , , 1, 2, 1 4, 2, 2, +, z 6 + · · · , z ∈ C., sin z = z − z +, 3!, 5! (3!)2
Page 269 :
260, , 8 Applications of Cauchy’s Theorem, , In this form, it is difficult to find the general term., Method 2. We have, f (z) = 2 sin z cos z = sin 2z =, , ∞, , (−1)n+1, (2z)2n−1 ,, (2n, −, 1)!, n=1, , z ∈ C., , But, as f is analytic in C and f (0) = 0,, + z, f (z) − f (0) = sin2 z =, f (ζ) dζ, 0, , =, =, , ∞ +, , n=1, ∞, , , 0, , z, , (−1)n+1, (2z)2n−1, (2n − 1)!, , , , (−1)n+1 22n−1 2n, z ., (2n)!, n=1, , Method 3. We use the trigonometric identity sin2 z = (1 − cos 2z)/2, and so, by (8.13), we obtain, , , ∞, ∞, , 1, 1 (−1)n+1 22n 2n, (−1)n, 2, 2n, 1−, (2z), z ., •, =, sin z =, 2, (2n)!, 2 n=1, (2n)!, n=0, Example 8.22. To expand f (z) = Log (1 + z) in a Maclaurin series valid for, |z| < 1, we rely upon the geometric series, f (z) =, , ∞, , 1, = 1 − z + z2 − z3 + · · · =, (−1)n z n ., 1+z, n=0, , Hence, as f is analytic for |z| < 1, with f (0) = Log 1 = 0,, + z, ∞ + z, , , f (z) − f (0) = Log (1 + z) =, f (ζ) dζ =, (−1)n ζ n dζ, 0, , n=0, , 0, , ∞, , (−1)n z n+1, =, n+1, n=0, , z2, z3, z4, +, −, + · · · , |z| < 1., 2, 3, 4, Note that we have chosen the principal branch so that Log 1 = 0. More, generally, when log 1 = 2kπi, we have, + z, ∞, , (−1)n z n+1, f (z) − f (0) = log(1 + z) − 2kπi =, ,, f (ζ) dζ =, n+1, 0, n=0, =z−, , and so, log(1 + z) = 2kπi +, , ∞, , (−1)n−1 n, z ., n, n=1, , •
Page 270 :
8.1 Cauchy’s Integral Formula, , 261, , Questions 8.23., *, 1. If f (z0 ) = (1/2πi) C (f (z)/(z − z0 )) dz for all points z0 inside C, is f (z), analytic inside C?, 2. Suppose that f is analytic inside and on* a simple closed contour C, and, z0 lies outside C. What is the value of C (f (z)/(z − z0 )) dz?, 3. *Suppose that f is analytic, * inside and on a2simple closed contour C. Does, , (f, (z)/(z, −, z, )), dz, =, (f (z)/(z − z0 ) ) dz for all z0 not on C?, 0, C, C, 4. If the derivatives of all orders for two different functions agree at one, point, how *do the two functions, compare?, *, 5. If limn→∞ C fn (z) dz = C (limn→∞ fn (z)) dz, does {fn (z)} converge, uniformly on C?, 6. If a sequence of functions converges uniformly on all compact subsets of, a domain, must the convergence be uniform throughout the domain?, 7. If {fn (z)} is a sequence of functions analytic in a domain D, and {fn (z)}, converges to f (z) in D, is f (z) analytic in D?, 8. Suppose {fn (z)} converges uniformly to an analytic function. What can, we say* about the functions {fn (z)}?, *, 9. Does |z|=1 z −3 eiz dz = 0? Does |z|=1 z −3 sin z dz = 0?, 10. *If f (z) is continuous inside and on a simple closed contour C, and, f (z) dz = 0, is f (z) analytic inside C?, C, ∞, (k), 11. If f (k) (z) , = n=0 fn (z), what can we say about f (k−1), ∞(z)?, ∞, n, 12. If f (z)* = n=0 an z , can f (z0 ) exist and not equal n=1 nan z0n−1 ?, 13. Does C (z − a)−1 (z − b)−1 dz = 0 for every simple closed contour C not, passing through a, b?, Exercises 8.24., 1. Prove Cauchy’s integral formula for multiply connected regions., 2. (a) If P (z) is a polynomial of degree n, prove that, +, P (z), dz = 0., (z, −, 1)n+2, |z|=2, (b) If n and m are positive integers, show that, +, +, (n − 1)!ez, (m − 1)!ez, dz, =, dz, n, m, C (z − z0 ), C (z − z0 ), along any contour containing z0 ., 3. Evaluate the following integrals, where C is the circle |z| = 3., +, +, +, 2, 2, ez, ez, ez, (a), dz, (b), dz, (c), dz, 2, C z−2, C z−2, C (z − 2), + −z, +, +, ez sin z, e cos z, 3z 4 + 2z − 6, dz, (e), dz, (f), dz., (d), 2, 3, (z − 2)3, C (z − 2), C (z − 2), C
Page 271 :
262, , 8 Applications of Cauchy’s Theorem, , 4. Use partial fractions to evaluate the following integrals., +, +, 1, 1, dz (b), dz, (a), 2−1, 2+1, z, z, |z|=2, |z|=2, +, +, 1, z 3 + 3z − 1, dz, (d), dz., (c), 4, |z|=2 z − 1, |z|=3 (z − 1)(z + 2), 5. If C = {z : |z| = r} (r = 1), then show that, , +, dz, −2 arctan r if 0 < r < 1, =, 2, π, −, 2 arctan r if 1 < r., C 1+z, 6. Use the Cauchy integral formula to evaluate the following integrals., +, +, +, (Re z)2 dz, (b), (z)2 dz, (c), (Im z)2 dz., (a), |z|=1, , |z−1|=1, , 7. Evaluate the integral, (a) |z| = 2, , *, C, , |z−1|=1, , z/((16 − z 2 )(z + i)) dz, where C is the circle, , (b) |z − 4| = 2 (c) |z + 4| = 2, , (d) |z| = 12 (e) |z| = 5., 8. Let γ : [0, 4π] → C be given by, , 3teit if 0 ≤ t ≤ 2π, γ : γ(t) =, 10π − 2t if 2π ≤ t ≤ 4π., *, Evaluate the integral γ (z 2 + π 2 )−1 dz., 9. Find the first five coefficients in the Maclaurin expansion for, 2, 1, (c) ez+z (d) ez/(1−z) ., cos z, 10. Suppose f (z) and g(z) are analytic at z0 with, , (a) ez sin z, , (b), , f (z0 ) = f (z0 ) = · · · = f (n−1) (z0 ) = 0,, g(z0 ) = g (z0 ) = · · · = g (n−1) (z0 ) = 0., If g (n) (z0 ) = 0, show that, lim, , z→z0, , f (n) (z0 ), f (z), ., = (n), g(z), g (z0 ), , This is a generalization of Theorem 5.26., 11. Evaluate the following limits, using either the previous exercise or Theorem 8.8., ez − 1 − z, sin z, (a) lim, (b) lim, z→0, z→0 z − z 3, z2, sin z, sin z − z, (c) lim z, (d) lim, ., z→0 e − 1, z→0 cos z − 1
Page 272 :
8.2 Cauchy’s Inequality and Applications, , 263, , 12. If f (z) is analytic at z = z0 , and |f (n) (z0 )| ≤ nk for each n (k fixed),, show that f (z) is actually an entire function., 13. Show that the following series represent analytic functions in the given, domain, and find their derivatives., (a), , ∞, , 1, nz, n=1, , (Re z > 1), , (b), , ∞, , n=1, , z2, , 1, − n2, , (z = ±1, ±2, . . . )., , 14. Suppose f (z) and g(z) are analytic in a simply connected domain D., Prove that, + z1, + z1, , f (z)g (z) dz = f (z1 )g(z1 ) − f (z0 )g(z0 ) −, g(z)f (z) dz,, z0, , z0, , where the path of integration is any contour from z0 to z1 that lies in, D., 15. Suppose f (z) is continuous, but not necessarily analytic, on a contour, C. Show that the function, +, f (ζ), dζ, F (z) =, ζ, −z, C, is analytic at each z not on C, with, +, f (ζ), F (z) =, dζ., (ζ, − z)2, C, 16. Choose a specific determination, find Maclaurin expansions for the following functions, and state the region for which the expansion is valid., (a) (1 + z)α (0 < α < 1), , (b) tan−1 (z), , (c) sin−1 z., , 8.2 Cauchy’s Inequality and Applications, In elementary calculus, we often deduce information about a function based, on the behavior of its derivative. For example, if the derivative is positive,, negative, or zero on an interval the function is, respectively, increasing, decreasing or constant on that interval. In complex analysis, the opposite is also, true in the following sense. The behavior of an analytic function is used to estimate the behavior of its derivatives. More precisely, we see that if an analytic, function f is bounded in a neighborhood of a point z0 , then the derivatives, of f cannot be arbitrarily large at z0 ., Theorem 8.25. (Cauchy’s Inequality) Suppose f (z) is analytic inside and, on the circle C having center at z0 and radius r. If |f (z)| ≤ M on C, then, |f (n) (z0 )| ≤ M n!/rn ,, , n = 1, 2, . . . .
Page 273 :
264, , 8 Applications of Cauchy’s Theorem, , Proof. By the generalized Cauchy integral formula,, +, n!, f (z), (n), f (z0 ) =, dz., 2πi C (z − z0 )n+1, Hence, by the hypothesis,, |f, , (n), , +, n!, |f (z)|, (z0 )| ≤, |dz|, 2π C |z − z0 |n+1, +, n!M, |dz|, ≤, 2πrn+1 C, n!M, n!M, =, (2πr) = n ., 2πrn+1, r, , As a comparison with real analysis, we consider f (x) = sin(1/x) for x > 0., Then f is differentiable for x > 0 and |f (x)| ≤ 1 for all x > 0. However,, f (x) = −, , 1, cos(1/x), x2, , which is clearly not bounded because, for example, for each n ∈ N, , 1, , f, = n2 π 2 ., nπ, Here is a surprising application of Cauchy’s inequality., Corollary 8.26. Suppose that f is analytic for |z −z0 | < r. If |f (z)−b| ≤ M ,, then |f (z0 )| ≤ M/r., Proof. Take 0 < δ < r. Then g(z) = f (z) − b is analytic and maps the disk, |z − z0 | ≤ δ into |w| ≤ M , so by Cauchy’s inequality, |f (z0 )| = |g (z0 )| ≤ M/δ., Letting δ → r proves the result., In particular, if f : Δ → Δ is analytic, then |f (0)| ≤ 1 which is a basic, version of the “Schwarz lemma” which will be discussed in the next section., Recall the functions that are analytic in C are called entire functions. We, know that, f : C → C is entire if and only if f (z) has the series expansion, ∞, n, with infinite radius of convergence. An entire function, f (z) =, n=0 an z, that is not a polynomial is said to be a transcendental entire function. The, functions ez , sin z, cos z, etc. are transcendental functions., Cauchy’s inequality enables us to obtain results in complex analysis which, have no real variable counterpart. For example, we have, Theorem 8.27. (Liouville’s Theorem) A bounded entire function must be a, constant.
Page 274 :
8.2 Cauchy’s Inequality and Applications, , 265, , Proof. Suppose f (z) is entire with |f (z)| ≤ M for all z. Given any complex, number z0 , we have by Cauchy’s inequality, |f (z0 )| ≤ M/r for every positive, number r. Letting r → ∞, we deduce that f (z0 ) = 0. Since z0 was arbitrary,, f (z) = 0 for all z, and hence f is constant by Theorem 5.9., Remark 8.28. There is, of course, no real variable analog to Liouville’s theorem. The function f (x) = sin x is a nonconstant, everywhere real differentiable, function on R and |f (x)| ≤ 1 on R. On the other hand, we have already seen, that each of | sin z| and | cos z| approaches to ∞ as y → ±∞ for any fixed, x. Likewise, | sinh z| and | cosh z| are unbounded entire functions. Again, by, Liouville’s theorem, each of these hyperbolic functions must be unbounded,, •, because each is a nonconstant entire function., Corollary 8.29. Every f : C → Δ which is analytic is constant. In particular, there exists no bijective mapping of the unit disk Δ onto C., Example 8.30. The method used in the proof of Liouville’s theorem helps, us to characterize all those entire functions f such that, |f (z)| ≤ |z|4 /ln |z| for |z| > 1., To do this, we choose R with R > 1. Then for |z| = R > 1,, |f (z)| ≤, , R4, |z|4, =, ln |z|, ln R, , and so, by the Cauchy integral formula and the M-L Inequality, +, 1, f (z), 1, |an | =, →0, dz ≤ n−4, n+1, 2π |z|=R z, R, ln R, as R → ∞ and n − 4 ≥ 0. Thus, an = 0 for n ≥ 4 and hence, f (z) is a, •, polynomial of degree at most 3., Liouville’s theorem says that for a nonconstant entire function f (z), there, is a sequence of points {zn } such that f (zn ) → ∞. This result can be sharpened., Theorem 8.31. (Generalized version of Liouville’s Theorem) A nonconstant entire function comes arbitrarily close to every complex number., Proof. Suppose that f (z) is entire and that there exists a complex number a, such that |f (z) − a| ≥ for all z. Then the function, g(z) = 1/[f (z) − a], is entire and, |g(z)| =, , 1, ≤ 1/ ., |f (z) − a|, , By Liouville’s theorem, g(z) is a constant. Hence, f (z) = 1/g(z) + a must also, be a constant.
Page 275 :
266, , 8 Applications of Cauchy’s Theorem, , Corollary 8.32. Suppose f (z) is a nonconstant entire function. Given any, complex number a, there exists a sequence {zn } such that f (zn ) → a., Proof. According to Theorem 8.31, for each n we can find a point zn such, that f (zn ) ∈ N (a; 1/n). Since 1/n → 0, it follows that f (zn ) → a., Remark 8.33. Though a nonconstant entire function come arbitrarily close, to every complex value, it does not necessarily assume every complex value., For example, f (z) = ez is never equal to zero. However, f (−n) = e−n → 0 as, n → ∞. The fact that ez assumes every other complex value can be viewed, •, as a special case of Picard’s Theorem., Theorem 8.34. (Picard’s Theorem) A nonconstant entire function assumes, each complex value, with one possible exception., For a proof of Picard’s theorem, see DePree and Oehring [DO] and Ponnusamy [P1]., Our next theorem gives an estimate on the “rate of growth” of entire, functions., Theorem 8.35. Suppose that f (z) is an entire function and that |f (z)| ≤, M rλ (|z| = r ≥ r0 ) for some nonnegative real number λ. Then f (z) is a, polynomial of degree at most λ., Proof. Let, f (z) =, , ∞, , n=0, , an z n =, , ∞, , f (n) (0) n, z ., n!, n=0, , By Cauchy’s inequality, on the circle |z| = r we have, |an | =, , M rλ, |f (n) (0)|, M, ≤ n = n−λ ., n!, r, r, , Letting r → ∞, we see that an = 0 whenever n > λ. Hence, f (z) is a, polynomial of degree no more than λ., Example 8.36. If f is entire such that |f (z)| ≤ a+b|z| for some a ≥ 0, b > 0,, then f is either constant or a first degree polynomial. Let us use the Cauchy, inequality to provide a proof. We let z0 be an arbitrary point of C. Then for, |z − z0 | ≤ R, one has, |f (z)| ≤ a + b|z| = a + b|z − z0 + z0 |, ≤ a + b(R + |z0 |)., By Cauchy’s inequality, with M = a + b(R + |z0 |), |a2 | =, , a + b(R + |z0 |), f (z0 ), ≤, → 0 as R → ∞, 2!, R2
Page 276 :
8.2 Cauchy’s Inequality and Applications, , 267, , so that f (z0 ) = 0. Since z0 is arbitrary, f (z) = 0 on C, and therefore f (z), is a constant, say c. Thus,, + z, f (ζ) dζ = cz, f (z) − f (0) =, 0, , •, , and hence, f (z) = f (0) + cz., , Remark 8.37. When λ = 0, Theorem 8.35 reduces to Liouville’s theorem., Choosing 0 < λ < 1 shows that we need not assume that f (z) is bounded, (only that it grows at a sufficiently slow rate) in order to deduce that f (z), •, must be constant., Set M (r, f ) = max|z|=r |f (z)|. Theorem 8.35 says that, for a transcendental entire function f (z), the function M (r, f ) grows faster than any power of, r. This does not mean that f (z) → ∞ along every path to ∞. For instance, we, have M (r, ez ) = er , but ez → 0 as z → ∞ along the negative real axis. Polynomials are somewhat different. The growth of a polynomial is determined by, its degree – a fact that has been used in almost every proof of the fundamental, theorem of algebra., Theorem 8.38. (Growth Lemma) Suppose P (z) = a0 + a1 z + · · · + an z n ,, an = 0. Then there exists a sufficiently large r such that, 3|an | n, |an | n, |z| ≤ |P (z)| ≤, |z| ,, 2, 2, , for all z ∈ C with |z| ≥ r., , Proof. For z = 0, we have, , , an−2, an−1, a0 , P (z) = z n an +, + 2 + ··· + n ., z, z, z, , By the triangle inequality,, , a0 , an−1, + · · · + n ≤ |P (z)|, |z|n |an | −, z, z, , a0 , an−1, + ··· + n ., ≤ |z|n |an | +, z, z, For |z| > 1 and n > k, we have |z|n > |z|k and so, an−2, K, |an−1 | + |an−2 | + · · · + |a0 |, a0, an−1, + 2 + ··· + n ≤, :=, ., z, z, z, |z|, |z|, Hence, , , , , , K, K, n, |z| |an | −, ≤ |P (z)| ≤ |z| |an | +, |z|, |z|, n, , The result now follows when K/|z| < |an |/2, i.e., when, |z| ≥ max{1, 2K/|an |}., , (|z| ≥ 1).
Page 277 :
268, , 8 Applications of Cauchy’s Theorem, , Theorems 8.35 and 8.38 provide a comparison between the growth of polynomial and transcendental functions. If P (z) is a polynomial and f (z) is a, transcendental function, then M (r, P )/M (r, f ) → 0 as r → ∞. That is, along, its “best path”, a transcendental function approaches infinity more rapidly, than does a polynomial. However, P (z) → ∞ as z → ∞ along any path. We, shall show in Chapter 9 that no transcendental function has this property., Loosely speaking, a transcendental function grow more rapidly than a polynomial, whereas a polynomial grows more consistently than a transcendental, function., An important property of polynomials is stated in Theorem 8.39., Theorem 8.39. (The Fundamental Theorem of Algebra) Every nonconstant polynomial has at least one zero., Proof. Suppose P (z) = a0 + a1 z + · · · an z n , an = 0. If P (z) never vanishes, then 1/P (z) is entire. By Theorem 8.38, 1/P (z) → 0 as z → ∞. Thus, |1/P (z)| < 1 for |z| ≥ R. But 1/P (z) is continuous (hence bounded) on the, compact set |z| ≤ R. Therefore, 1/P (z) is bounded in the whole plane and,, by Liouville’s theorem, must be a constant. This implies that P (z) is also a, constant, contradicting our assumption., Corollary 8.40. Every polynomial of degree n has exactly n (not necessarily, distinct) zeros., Proof. The fundamental theorem shows the existence of at least one zero r1 ., The expression z − r1 may be factored out, leaving a polynomial of degree, n − 1. Reapplying the theorem to this new polynomial, we obtain another, zero. This process can be repeated n times., Corollary 8.41. Every polynomial of degree n assumes each complex number, exactly n times., Proof. If P (z) is a polynomial of degree n, then, Q(z) = P (z) − a, is also a polynomial of degree n. By Corollary 8.40, Q(z) has n zeros. But the, zeros of Q(z) are the “a” points of P (z)., Remark 8.42. Corollary 8.41 provides a more complete solution, in the case, of polynomials, than does Theorem 8.31. A polynomial of degree n not only, comes arbitrarily close to every complex value, it actually takes on every, value n times. But the existence of n roots tells nothing about their location., In Section 9.4, we shall develop a method for approximating the location of, •, zeros for some analytic functions.
Page 278 :
8.2 Cauchy’s Inequality and Applications, , 269, , Example 8.43. We wish to show that if all the zeros of a polynomial P (z), have positive real parts, then so do the zeros of its derivative P (z)., To do this, let P (z) be a polynomial of degree n and have zeros at zk, (not necessarily distinct) with Re zk > 0 for each k = 1, 2, . . . , n. By the, fundamental theorem of algebra we can express P (z) as, P (z) = c(z − z1 )(z − z2 ) · · · (z − zn ), where c is a nonzero constant. For z = zk , we have, 1, 1, 1, P (z), =, +, + ··· +, ., P (z), z − z1, z − z2, z − zn, We claim that P (z) = 0 whenever Re z ≤ 0. If Re z ≤ 0, as Re zk > 0, we, have Re (z − zk ) < 0. Thus, for each k = 1, 2, . . . , n, we see that, , , 1, <0, Re, z − zk, and therefore,, , , Re, , P (z), P (z), , , <0, , which gives that P (z) = 0 whenever Re z ≤ 0. Hence, the zeros of P (z) must, •, have its real parts positive., We now see how the behavior of an analytic function at a sequence of, points influences its behavior elsewhere (see also Theorem 6.56)., Theorem 8.44. Suppose f (z) is analytic in the disk |z − z0 | < R, and that, {zn }n≥1 is a sequence of distinct points converging to z0 . If f (zn ) = 0 for, each n ∈ N, then f (z) ≡ 0 everywhere in |z − z0 | < R., Proof. We have, f (z) = a0 +, , ∞, , , ak (z − z0 )k, , (|z − z0 | < R)., , (8.16), , k=1, , Since f (z) is continuous at z0 , it follows that f (zn ) → f (z0 ) as zn → z0 ., Therefore,, lim f (zn ) = 0 = f (z0 ) = a0 ., n→∞, , Hence f (z) has no constant term, and we may write, , , ∞, , k−1, f (z) = (z − z0 ) a1 +, ak (z − z0 ), ., k=2, , Setting z = zn and dividing by zn − z0 leads to
Page 279 :
270, , 8 Applications of Cauchy’s Theorem, , f (zn ), = 0 = a1 ., n→∞ zn − z0, lim, , In like manner,, , f (z) = (z − z0 )2, , a2 +, , ∞, , , , ak (z − z0 )k−2, , k=3, , so that, lim, , n→∞, , f (zn ), = 0 = a2 ., (zn − z0 )2, , An induction shows that ak = 0 for each k, and the result follows., Corollary 8.45. (Zeros are isolated) Suppose f (z) is analytic at a point, z = z0 . Then either f (z) ≡ 0 in some neighborhood of z0 , or there exists a, real number r such that f (z) = 0 in the punctured disk 0 < |z − z0 | ≤ r., Proof. Assume that no such r exists. Then in each punctured disk, 0 < |z − z0 | ≤ 1/n, there exists a point zn such that f (zn ) = 0. Since zn → z0 ,, an application of Theorem 8.44 shows that f (z) must be identically zero in, some neighborhood of z0 ., Remark 8.46. There is no real variable analog to this corollary. The function, ⎧, ⎨ x2 sin π if x = 0, x, f (x) =, ⎩, 0 if x = 0, is differentiable for all real x, with f (1/n) = 0 for each n. However, f (x) = 0, in any neighborhood of the origin., •, We now generalize the previous theorem to arbitrary domains., Theorem 8.47. (Uniqueness Theorem) Suppose f (z) is analytic in a domain D, and that {zn } is a sequence of distinct points converging to a point, z0 in D. If f (zn ) = 0 for each n, then f (z) ≡ 0 throughout D., Proof. Consider the following two disjoint sets:, A = {a ∈ D : f (z) ≡ 0 in some neighborhood of a}, B = {a ∈ D : f (z) = 0 for all z in some deleted neighborhood of a}., By Corollary 8.45, every point in D is either in A or in B. It may easily be, verified that both A and B are open sets. Since the domain D = A ∪ B is, connected, either A or B must be the empty set. By Theorem 8.44, the point, z0 is in A. Therefore, B = ∅. This means that A = D, and f (z) ≡ 0 in D., As a consequence of Theorem 8.47, we have the following result which is, also referred to as the uniqueness/identity principle.
Page 280 :
8.2 Cauchy’s Inequality and Applications, , 271, , Theorem 8.48. (Identity Theorem) Suppose {zn } is a sequence of points, having a limit point in a domain D. If f (z) and g(z) are analytic in D, with, f (zn ) = g(zn ) for each n, then f (z) ≡ g(z) throughout D., Proof. Let z0 denote the limit point of {zn }. By Theorem 2.18, there exists a, subsequence {znk } converging to z0 . Setting h(z) = f (z) − g(z), we see that, h(znk ) = 0 for all points of the sequence {znk }. An application of Theorem, 8.47 shows that h(z) ≡ 0 in D and the result follows., Remark 8.49. The requirement that the limit point z0 be in the domain of, analyticity is essential. The nonconstant function f (z) = e1/(1−z) is analytic, in |z| < 1 but not at the point z = 1. For zn = 1 − 1/2nπi, we have, e1/(1−zn ) = e2nπi = 1., Note that zn → 1 as n → ∞, a point at which f (z) is not analytic., , •, , Example 8.50. Suppose that f is entire. We wish to show that f (R) ⊆ R if, and only if f (n) (0) is real for each n ∈ N. It is evident that if each f (n) (0) ∈ R,, (n), ∞, then f (z) = n=0 f n!(0) z n shows that f (R) ⊆ R. To prove the converse, let, f (R) ⊆ R. Then, ∞, , f (n) (0) n, g(z) = f (z) =, z, n!, n=0, is entire and g(z) = f (z) for z ∈ R. By the identity theorem, g(z) = f (z), throughout C. Moreover, the power series representation of an analytic function is unique, and we must have f (n) (0) = f (n) (0) for all n., •, Example 8.51. Does there exist a function f (z) analytic in |z| < 1 and, satisfying, , , , 1, 1, 1, (n = 1, 2, . . . )?, (8.17), =f, =, f, 2n, 2n + 1, 2n, The function f (z) = z is an analytic function and satisfies the condition, f (1/2n) = 1/2n. By the identity theorem, that is the only such analytic, function. Since, , , 1, 1, f, ,, =, 2n + 1, 2n, there does not exist an analytic function that satisfies (8.17). Note, however,, that we can construct a function in |z| < 1 that satisfies, , , 1, 1, =, f, 2n + 1, 2n, for every n. Setting z = 1/(2n + 1), we have 1/2n = z/(1 − z), so that, •, f (z) = z/(1 − z) satisfies the condition.
Page 281 :
272, , 8 Applications of Cauchy’s Theorem, , Remark 8.52. Cauchy’s integral formula says that the behavior of an analytic function on a simple closed contour determines its behavior inside. The, identity theorem tells us even more. It says that the behavior at any sequence, of points, inside or on the simple closed contour, determines the behavior of, the analytic function at all points of the domain., •, Remark 8.53. We now have a simple way to prove the standard trigonometric identities. For example, the function, f (z) = sin2 z + cos2 z, is an entire function that is equal to one on the real axis. Hence f (z) ≡ 1 in, the complex plane; that is, sin2 z + cos2 z ≡ 1 for all z., •, Example 8.54. Let S = {1/n : n = 1, 2, . . . }. Then S ⊂ [0, 1] and S has a, limit point 0. Let f be an entire function and f (z) = p(z) for z = x, x ∈ S,, where p is polynomial p(x) = a0 + a1 x + a2 x2 + · · · + ak xk . By the Uniqueness, Theorem, since 0 ∈ [0, 1] ⊆ C, we have f (z) = a0 + a1 z + · · · + ak z k for all, •, z ∈ C., Questions 8.55., 1. Is there a corresponding “Cauchy inequality” when the circle is replaced, by a simple closed contour?, 2. Does there exist an analytic function in a neighborhood of the origin, such that |f (n) (0)| ≥ (n!)2 for all n ∈ N?, 3. Does there exist a condition which ensures an entire function to be a, polynomial?, 4. Is cos z an entire nonconstant function? Must cos z be unbounded? Must, sin z be unbounded?, 5. Can a nonconstant entire function be bounded in a half-plane?, 6. Can a real part of a nonconstant entire function be bounded?, 7. Why is Theorem 8.31 a generalization of Liouville’s theorem?, 8. If a nonconstant function is analytic everywhere outside a disk, can the, function be bounded?, 9. If f (z) = 1/z, then it is bounded as z → ∞ but is not constant. Does, this contradict Liouville’s theorem?, 10. What can we say about entire functions that omit the value zero?, 11. What is wrong in the following proof?, Since ez = 0, 1/ez is bounded. Therefore, by Liouville’s theorem 1/ez is, constant., 12. Suppose that f is entire such that |f (z)| → ∞ as |z| → ∞. Does f have, at least one zero in C? How do we compare with ez ?, Note: Note that ez is entire and has no zeros in C. We observe that, lim|z|→∞ |ez | = lim|z|→∞ ex does not exist.
Page 282 :
8.2 Cauchy’s Inequality and Applications, , 273, , 13. If two entire functions agree at infinitely many points, must they be, equal?, 14. If two entire functions agree on a segment of the real axis, must they, agree on C?, 15. If f is entire such that either Re f or Im f is bounded, must f be constant?, 16. If f is entire such that either Re f > −2006 or Im f > −2006, must f, be constant?, 17. If f = u + iv is entire such that au + bv ≥ c for some real numbers a, b, and c, must f be constant?, 18. Let f be entire such that f (1/n) = cos(1/n) for all n ∈ Z. Is f (z) = cos z, for all z ∈ C?, 19. Let f be analytic in the punctured complex plane C\{0} such that, f (1/nπ) = sin(nπ) for all n ∈ Z. Is f (z) = sin(1/z) for all z ∈ C\{0}?, Exercises 8.56., ∞, n, is analytic in |z| < R, and f (x) is real when, 1. If f (z) =, n=0 an z, −R < x < R, show that an is real for each n. Also show that f (z) =, f (z)., 2. Let P (z) = a0 + a1 z + · · · + an z n , an = 0. Given > 0, show that there, exists an R such that r > R implies, (1 − )|an |rn ≤ |P (z)| ≤ (1 + )|an |rn, , (|z| = r)., , 3. If all the zeros of a polynomial P (z) have negative real parts, show that, all the zeros of P (z) have negative real parts., 4. Suppose f (z) is an entire function with |f (z)| ≤ |ez | for all z. Prove that, f (z) = Kez , |K| ≤ 1., 5. Find all entire functions f for which there exists a positive constant M, such that |f (z)| ≤ M | cos z| for all z ∈ C. How about if cos z is replaced, by cosh z or sin z or sinh z respectively?, 6. Suppose that f (z) is an entire function such that |f (z)| ≤ |z| for all, z ∈ C. Show that f must be of the form f (z) = az 2 + b where a, b are, complex constants such that |a| ≤ 1/2. What will be the form of f if, f (z) is entire such that |f (k) (z)| ≤ |z| for some fixed k > 2 and for all, z ∈ C?, 7. Suppose that f (z) is entire with a and b positive constants. If, f (z + a) = f (z + bi) = f (z), for all z, show that f (z) is constant., 8. Suppose that f is an entire function such that |f (z)| ≤ 10 on |z − 2| = 3., Find a bound, for |f (3) (2)|., ∞, 9. Let f (z) = n=0 an z n be analytic for |z| ≤ 1, and assume that |f (z)| ≤, 1 for |z| ≤ 1. Use Cauchy’s inequality to prove
Page 283 :
274, , 8 Applications of Cauchy’s Theorem, ∞, , , |z|k, (|z| < 1)., 1 − |z|, n=k, (b) If |z1 | ≤ r, |z2 | ≤ r, 0 < r < 1, and z1 = z2 , then, (a), , an z n ≤, , 1, f (z2 ) − f (z1 ), ≤, ., z2 − z 1, (1 − r)2, 10. If f (z) and g(z) are analytic in a domain D with f (z)g(z) = 0 in D,, prove that either f (z) ≡ 0 or g(z) ≡ 0 in D., 11. Suppose f (z) and g(z) are analytic in domain D and that, f (zn ), g (zn ), =, f (zn ), g(zn ), , 12., 13., 14., , 15., , 16., , 17., , at a sequence of points {zn } converging to a point z0 in D. Show that, f (z) = Kg(z) in D., Give an example of a nonvanishing analytic function f in the unit disc, |z| < 1 having infinitely many zeros., Prove that there is no analytic function f in the unit disk Δ =, {z : |z| < 1} such that f (1/n) = (−1)n /n2 for n = 2, 3, 4, . . . ., Let f and g be analytic in the unit disk Δ., (a) If f (1/n) = g(1/n) for n, √= 2, 3, . . . , show that f = g., (b) Show that f (1/n) = 1/ n for each n = 2, 3, . . . is not possible., Suppose that f is entire and that there exists a bounded sequence of, distinct real numbers {an }n≥1 such that f (an ) is real for each n ≥ 1., Show that f (z) is real on R. In addition, if {an } is decreasing such that, an → 0 as n → ∞, and f (a2n ) = f (a2n+1 ) for all n ≥ 1, then show that, f is a constant., In each case, exhibit a nonconstant f having the desired properties or, explain why no such function exists:, (a) f is entire with f (n) (0) = 3n for n even and f (n) (0) = (n − 1)! for n, odd., (b) f is analytic in |z| < 1 with f (1/n) = n/(2 + n) for n ∈ N., (c) f is analytic in |z| < 1 such that f (1/n) = (1 + (−1)n )/3 for n ∈ N., n ∈ N., (d) f is analytic in |z| < 1 such that f (1/n) = 2n for √, (e) f is analytic in |z| < 1 such that f (1/n) = 1/ n for each n =, 2, 3, . . . . √, √ √, The functions z and sin z/ z are well defined and analytic on the, cut plane C \(−∞, 0]. Show that, √, ∞, , sin z, (−1)k+1 k, z for z ∈ C \(−∞, 0]., f (z) = √, =, (2k + 1)!, z, k=0, , Explain why it is possible to define f (z) on the cut (−∞, 0] so that f is, analytic on C. What values f (x) should be assigned, when x ∈ (−∞, 0]?, √, Can this procedure be applied to g(z) = sin z?
Page 284 :
8.3 Maximum Modulus Theorem, , 275, , 8.3 Maximum Modulus Theorem, Suppose f (x) is a continuous real-valued function defined on the interval [a, b], and that F (x) = f (x) throughout. Then the “average” value of f (x) on the, interval is given by, 1, b−a, , +, , b, , f (x) dx =, a, , F (b) − F (a), ., b−a, , Furthermore, according to the mean-value theorem, this expression is equal, to F (ξ) for some ξ, a < ξ < b., Our next theorem shows that for functions analytic inside and on a circle,, the average of the values on the circumference is equal to the value of the, function at the center of the circle., Theorem 8.57. (Gauss’s Mean-Value Theorem) Suppose f (z) is analytic, in the closed disk |z − z0 | ≤ r. Then, f (z0 ) =, , 1, 2π, , +, 0, , 2π, , f (z0 + reiθ ) dθ., , Proof. By Cauchy’s integral formula,, +, 1, f (z), dz., f (z0 ) =, 2πi |z−z0 |=r z − z0, Write this out in terms of a parameterization z = z0 + reiθ with 0 ≤ θ ≤ 2π,, dz = ireiθ dθ. Then, + 2π, + 2π, 1, 1, f (z0 + reiθ ) iθ, f (z0 ) =, ire, dθ, =, f (z0 + reiθ ) dθ., 2πi 0, reiθ, 2π 0, If f (z) is a constant (say f (z) = C), then Gauss’s theorem gives the, obvious fact that, + 2π, 1, f (z0 + reiθ ) dθ = C., f (z0 ) =, 2π 0, Our next theorem shows that for nonconstant functions there must be, some points on the circle |z| = r for which |f (z)| > |f (z0 )|., Theorem 8.58. (Maximum Modulus Theorem: First Form) If f (z) is analytic in a domain D, then |f (z)| cannot attain a maximum in D unless f (z), is constant., Proof. Suppose |f (z)| attains maximum at a point z0 in D. Choose a disk, |z − z0 | ≤ r contained in D. Gauss’s mean-value theorem gives
Page 285 :
276, , 8 Applications of Cauchy’s Theorem, , f (z0 ) =, , 1, 2π, , |f (z0 )| ≤, , 1, 2π, , so that, , +, , 2π, , f (z0 + reiθ ) dθ, , 0, , +, , (8.18), , 2π, , |f (z0 + reiθ )| dθ., , 0, , By assumption,, |f (z0 + reiθ )| ≤ |f (z0 )|,, , (8.19), , and so, 1, 2π, , +, , 2π, , 1, |f (z0 + re )| dθ ≤, 2π, , +, , iθ, , 0, , 0, , 2π, , |f (z0 )| dθ = |f (z0 )|., , (8.20), , Combining (8.18) and (8.20), we have, 1, |f (z0 )| =, 2π, or, 1, 2π, , +, 0, , +, 0, , 2π, , |f (z0 + reiθ )| dθ, , 2π, , (|f (z0 )| − |f (z0 + reiθ )|) dθ = 0., , By (8.19) the integrand is nonnegative and therefore, |f (z0 + reiθ )| = |f (z0 )| for 0 ≤ θ ≤ 2π., Hence, |f (z)| = |f (z0 )| for each z on |z − z0 | = r. Since r is arbitrary, |f (z)| =, |f (z0 )| for all points inside and on |z −z0 | = r. Recall that an analytic function, with constant modulus in a domain is constant in that domain, hence f (z) is, constant on |z − z0 | < r. From the identity theorem, it follows that f (z) is, constant in the whole domain. Therefore, |f (z)| cannot attain a maximum at, a point of D unless f (z) is constant., Suppose that f is analytic on a domain D and continuous on the boundary, ∂D. By Theorem 8.58, |f (z)| cannot attain its maximum in D unless f (z) is, constant. This raises the following questions:, •, •, , Does |f (z)| attains its maximum on ∂D?, Is |f (z)| ≤ M on D when |f (z)| ≤ M on ∂D?, , In general, the answer to both questions are negative., Theorem 8.59. (Maximum Modulus Theorem: Second Form) If f (z) is, analytic in a bounded domain D and continuous on its closure D, then |f (z)|, attains a maximum on the boundary. Furthermore, |f (z)| does not attain a, maximum at an interior point unless f (z) is constant.
Page 286 :
8.3 Maximum Modulus Theorem, , 277, , Proof. First, observe that D is a compact set because D is bounded. Thus,, by Theorem 2.41, as |f (z)| is a continuous real function on D, |f (z)| attains, a maximum somewhere in D. According to the first form of the theorem, the, maximum cannot occur at an interior point. Hence the maximum must occur, on the boundary., Remark 8.60. The domain need not be simply connected. For instance, the, modulus of any function analytic in the open annulus 1/R < |z| < R and, continuous on the closed annulus 1/R ≤ |z| ≤ R must attain its maximum on, the boundary. The modulus of the function f (z) = z attains its maximum on, the outer boundary, whereas the modulus of f (z) = 1/z attains the maximum, •, on the inner boundary., Theorem 8.59 is not true if D is not bounded. For example, if, f (z) = ez and D = {z ∈ C : Re z > 0},, then ∂D is the imaginary axis and |f (iy)| = |eiy | = 1, i.e., f (∂D) = ∂Δ. Yet, |f (z)| = ex → ∞ as z → ∞ along the positive real axis. Thus, the hypothesis, that D is bounded is essential in Theorem 8.59., 2, 2, 2, Here is another example. Set f (z) = eiz = ei(x −y ) e−2xy so that, |f (z)| = e−2xy ., If D = {z ∈ C : Re z > 0, Im z < 0}, then, for points on the boundary ∂D,, either x = 0 or y = 0 and so |f (z)| = 1 on ∂D. However, for y = −x (x > 0), we have, 2, |f (z)| = e2x → ∞ as x → ∞., Then both examples show that the modulus of an analytic function need not, attain its maximum on the boundary ∂D and the maximum of the modulus, on the boundary may not be the maximum value inside the domain D unless, D is bounded., Example 8.61. Suppose that we wish to find the maximum modulus of, f (z) = 3z − 2i on |z| ≤ 3. To do this, we compute, |f (z)|2 = |3z − 2i|2 = 9|z|2 + 12Re (iz) + 4 = 9|z|2 − 12Im z + 4., By Theorem 8.59, max|z|≤3 |f (z)| occurs on the boundary |z| = 3. Therefore,, on |z| = 3,, , √, |f (z)| = 9(3)2 − 12Im z + 4 = 85 − 12Im z., The last expression attains its maximum when Im z attains its minimum on, |z| = 3, namely, at the point z = −3i. Thus,, , √, max |f (z)| = 85 − 12(−3) = 121 = 11., |z|≤3
Page 287 :
278, , 8 Applications of Cauchy’s Theorem, , Alternatively, as |f (z)| attains its maximum on |z| = 3, we consider z = 3eiθ, and compute, √, |f (z)| = |3(3eiθ ) − 2i| = 85 − 36 sin θ., This expression is maximum when − sin θ is maximum, i.e., when θ = −π/2., Thus, the maximum value of |f (z)| on |z| ≤ 3 is 11. More generally, we have, the following:, If 0 = a ∈ C, 0 = b ∈ C and f (z) = az + b on |z| ≤ R, then, max |az + b| = max |aRz + b| = |a|R + |b|, , |z|≤R, , |z|≤1, , and the maximum value is attained at z0 on the boundary |z| = 1, where, arg z0 = Arg b − Arg a. Clearly, this follows from, , , az + b = |a|eiArg a z + |b|eiArg b = eiArg b |a|ze−i(Arg b−Arg a) + |b| ., Moreover, it is easy to see that max|z|≤1 |az n + b| ≤ |a| + |b|., , •, , Example 8.62. If f (z) = z 2 /(z 3 − 10) for |z| ≤ 2, then f is analytic inside, and on |z| = 2. Then |f (z)| attains its maximum value on |z| = 2. For z = 2eiθ ,, 0 ≤ θ ≤ 2π, we have, |f (z)| =, , |z|2, − 10|, , |(2eiθ )3, , 4, = (, (8 cos 3θ − 10)2 + 82 sin2 3θ, = √, , 4, ., 64 + 100 − 160 cos 3θ, , This expression is maximum when − cos 3θ is minimum. Clearly, the minimum value is when cos 3θ = 1, i.e., when θ = 0, 2π/3, 4π/3, 2π. Thus, the, •, maximum value of |f (z)| is 2., As an application of the maximum modulus theorem, we reprove the fundamental theorem of algebra (see also Theorem 8.39)., Suppose P (z) = a0 + a1 z + · · · + an z n , an = 0, has no zeros. Then, 1/P (z) is a nonconstant entire function with no zeros. In view of Theorem, 8.38, 1/P (z) → 0 as z → ∞. When r is large enough,, 1, 1, 1, <, =, P (z), P (0), a0, , (|z| = r)., , Thus the continuous function 1/P (z) does not attain a maximum on the, boundary of the compact set |z| ≤ r. Hence |1/P (z)| must attain a maximum, at an interior point, contradicting the maximum modulus theorem. Therefore,, P (z) must have a zero in the disk |z| ≤ r, and the proof is complete.
Page 288 :
8.3 Maximum Modulus Theorem, , 279, , It is possible for the modulus of a nonconstant analytic function to attain a, minimum in a domain, and not on its boundary. To see this, consider f (z) = z n, on |z| ≤ r. Then, 0 = |f (0)| = min |f (z)| < min |f (z)| = rn ,, |z|≤r, , |z|=r, , and so the min{|f (z)| : |z| ≤ r} is attained at the interior point 0. The, max{|f (z)| : |z| ≤ r} = rn is attained at the boundary point r, on |z| = r., However, we do have the following counterpart to the maximum modulus, theorem., Theorem 8.63. (Minimum Modulus Theorem) Suppose f (z) is analytic in, a domain D, and that f (z) = 0 in D. Then |f (z)| cannot attain a minimum, in D unless f (z) is constant. If f (z) is also continuous on D, D compact,, then |f (z)| attains a minimum on the boundary., Proof. If f (z) = 0 in D, then 1/f (z) is analytic in D. The function |f (z)|, attains a minimum at a point z0 in D if and only if 1/|f (z)| attains a maximum, at z0 . The result follows now upon applying the maximum modulus theorem, to 1/f (z)., Not surprisingly, the fundamental theorem of algebra can also be proved, from the minimum modulus theorem, for, in view of Theorem 8.38, when r is, large enough the polynomial, P (z) = a0 + a1 z + · · · + an z n, , (an = 0),, , satisfies the inequality, |P (z)| > |P (0)| = |a0 | (|z| = r)., To preserve the validity of the minimum modulus theorem, the hypothesis, P (z) = 0 in |z| < r must be false. That is, P (z) must have a zero in the disk, |z| < r., 2, , Example 8.64. Consider f (z) = ez /z on D = {1 ≤ |z| ≤ 2}. We wish to, find points where |f (z)| has maximum and minimum values. To do this, we, set z = reiθ . Then, on |z| = r,, |f (z)| =, , er, , 2, , cos 2θ, , r, , and the maximum occurs when cos 2θ = 1 and the minimum occurs when, cos 2θ = −1. Thus, 2, , max |f (z)| =, , |z|=r, , er, r, , 2, , and min |f (z)| =, |z|=r, , e−r, ., r
Page 289 :
280, , 8 Applications of Cauchy’s Theorem, , Note that the maximum occurs at θ = 0, π while the minimum occurs at, θ = π/2, 3π/2. In particular,, max |f (z)| =, z∈D, , e4, 2, , and min |f (z)| =, z∈D, , 1, ., 2e4, , •, , If f (z) is a nonconstant analytic function in |z| ≤ R and |f (z)| ≤ M, on |z| = R, then the maximum modulus theorem says that |f (z)| < M for, |z| < R. We next develop methods to improve this bound inside the disk., Lemma 8.65. (Schwarz’s Lemma) Suppose f (z) is analytic for |z| < R with, f (0) = 0. If |f (z)| ≤ M in |z| < R, then, |f (0)| ≤ M/R and |f (z)| ≤ M |z|/R for |z| < R,, with equality only for f (z) = (M/R)eiα z, α real., Proof. Since f (0) = 0, we may write f (z) = a1 z + a2 z 2 + a3 z 3 + · · · . Define, a function g : Δ → C by, , f (z)/z if 0 < |z| < R, g(z) =, f (0) if z = 0., Then g(z) is analytic for |z| < R. Since |f (z)| ≤ M for |z| < R, we have, max |g(z)| ≤, , |z|=r, , M, r, , (8.21), , for all positive real number r < R. By the maximum modulus theorem applied, to g(z), |g(z)| ≤ M/r for all |z| ≤ r, 0 < r < R. Now, since r can come, arbitrarily close to R, we have, |g(z)| ≤ lim, , r→R, , M, M, =, for all |z| < R., r, R, , This proves that, M, |z| for all |z| < R, R, and therefore, |f (0)| = |g(0)| ≤ M/R., In case either |f (0)| = M/R or |f (b)| = (M/R)|b| for some b, 0 < |b| < R,, we get |g(0)| = M/R or |g(b)| = M/R and so, |g(z)| attains a maximum in, |z| < R. Therefore, g(z) is a constant function by the maximum modulus, principle and the result follows., |f (z)| ≤, , Remark 8.66. Schwarz’s lemma can be phrased in terms of the function, M (r, f ). For if f (z) is analytic in |z| < R with f (0) = 0, then M (r, f ) ≤, •, (r/R )M (R , f ) (r < R < R).
Page 290 :
8.3 Maximum Modulus Theorem, , 281, , Corollary 8.67. Suppose f (z) is analytic for |z| < R with f (0) = f (0) =, · · · = f (n−1) (0) = 0. If |f (z)| ≤ M in |z| < R, then, |f (z)| ≤ (|z|/R)n M, , (|z| < R),, , with equality only for f (z) = (M/Rn )eiα z n , α real., Proof. Write f (z) = z n g(z), and apply the maximum modulus theorem to, g(z)., If f (0) = 0, Schwarz’s lemma may be modified to obtain Corollary 8.68., Corollary 8.68. Suppose f (z) is analytic for |z| < R. If |f (z)| ≤ M in, |z| < R, then |f (z) − f (0)| ≤ (2|z|/R)M (|z| < R)., Proof. Set g(z) = f (z) − f (0). Then g(0) = 0 and, |g(z)| ≤ |f (z)| + |f (0)| ≤ 2M., Applying Schwarz’s lemma to g(z), we obtain, |g(z)| = |f (z) − f (0)| ≤, , 2|z|, M, R, , (|z| < R)., , Remark 8.69. Corollary 8.68 supplies another proof of Liouville’s theorem., Suppose f (z) is entire with |f (z)| ≤ M for all z. Given a point z0 , |z0 | = r0 ,, and an > 0, choose R > 2r0 M/ . Then, |f (z0 ) − f (0)| ≤, , 2r0 M, < ., R, , Since was arbitrary, f (z0 ) = f (0). But z0 was also arbitrary, so that f (z) =, •, f (0) for all z. That is, f (z) is a constant., Example 8.70. We wish to characterize those functions f such that f (z) is, analytic for |z| ≤ 1 and |f (z)| = 1 for |z| = 1. To do this, we split the proof, into two parts., Case (i): Let f have no zeros in |z| < 1. As |f (z)| = 1 for |z| = 1, the, maximum and minimum modulus theorem shows that 1 ≤ |f (z)| ≤ 1 for, |z| ≤ 1; i.e., |f (z)| = 1 for |z| ≤ 1. By Theorem 5.37 (see also Corollary 9.57),, f (z) = eiα , α real., Case (ii): Let f have zeros in |z| < 1. Clearly, f cannot have infinitely, many zeros in |z| < 1; otherwise a limit point would be in |z| ≤ 1 in which f, is analytic and so f (z) ≡ 0 by the uniqueness theorem. This is a contradiction, because |f (z)| = 1 for |z| = 1. Define, F (z) = ., n, , f (z), , , k=1, , z−ak, 1−ak z, , ,
Page 291 :
282, , 8 Applications of Cauchy’s Theorem, , where ak s are the finite zeros of f in |z| < 1. Then F is analytic in |z| ≤ 1,, F (z) = 0 in |z| < 1 and |F (z)| = 1 for |z| = 1. By Case (i),, F (z) = eiα or f (z) = eiα, , n, , z − ak, 1 − ak z, , k=1, , •, , for some real α., , Example 8.71. Suppose that f : Δ → Δ is analytic such that f (0) = 0., Then it can be easily seen that, (i) |f (z) + f (−z)| ≤ 2|z|2 in Δ = {z : |z| < 1},, (ii) the inequality in (i) is strict, except at the origin, unless f has the form, f (z) = z 2 with | | = 1., ∞, To see this, we let f (z) = n=1 an z n . Then, F (z) =, , ∞, f (z) + f (−z) , =, a2n z 2n−1, 2z, n=1, , and, by Schwarz’ lemma, |f (z)| ≤ |z|. Therefore, |F (z)| ≤ 1 and (i) follows., If the equality in (i) holds at some point in |z| < 1, then |F (z)| = 1 for all, |z| < 1 and so F (z) = z with | | = 1. This gives, f (z) + f (−z) = 2, , ∞, , , a2n z 2n = 2 z 2 ., , n=1, , Comparing the coefficients of z n on both sides shows that a2 = and a2n = 0, for each n > 1. Thus f has the form, 2, , f (z) = z +, , ∞, , , a2n+1 z 2n+1 = z 2 + h(z),, , n=1, , where h(z) is odd. We need to show that h(z) = 0. Note that,, 1 ≥ |f (z)|2 = | z 2 + h(z)|2 ,, , z ∈ Δ,, , and, since h is odd,, 1 ≥ |f (−z)|2 = | z 2 + h(−z)|2 = | z 2 − h(z)|2 ,, , z ∈ Δ., , Adding the last two inequalities, we see that, 2 ≥ | z 2 + h(z)|2 + | z 2 − h(z)|2 = 2(|z|4 + |h(z)|2 ), so that |h(z)|2 ≤ 1−|z|4 throughout |z| < 1. Note that h is analytic for |z| < 1, and by the maximum modulus theorem, |h(z)|2 ≤ when |z|4 ≤ 1 − . Since, > 0 is arbitrary, we see that h(z) = 0.
Page 292 :
8.3 Maximum Modulus Theorem, , 283, , Alternatively, as |f (z)| < 1 on Δ, for the proof of h(z) = 0, it suffices to, observe that, + 2π, ∞, , 1, iθ 2, lim, |f, (re, )|, d, θ, =, |an |2 ≤ 1., r→1− 2π 0, n=1, Since a2 = , the remaining Taylor’s coefficients of f must be zero, and so, •, f (z) = z 2 ., Questions 8.72., 1. Suppose f is analytic in the annulus 1 ≤ |z| ≤ R, |f (z)| ≤ Rn for, |z| = R and |f (z)| ≤ 1 on |z| = 1. Is |f (z)| ≤ |z|n in the annulus?, 2. Suppose f (z) is analytic inside and on a simple closed contour C. Can, |f (z)| be constant on C without f (z) being constant?, 3. Suppose f (z) is analytic in |z| ≤ r. Do there exist two distinct points, z0 = reiθ0 and z1 = reiθ1 such that |f (z)| ≤ |f (z0 )| and |f (z)| ≤ |f (z1 )|, for all z, |z| < r?, 4. Can the modulus of a nonconstant function, analytic in a region that is, not closed, attain a maximum? What if the region is unbounded?, 5. Suppose f (z) and g(z) are analytic inside and on a simple closed contour, C, and that |f (z) − g(z)| = 0 on C. Does f (z) = g(z) inside C? What, if |f (z)| − |g(z)| = 0 on C?, 6. Define m(r, f ) = min|z|=r |f (z)|. What properties does m(r, f ) have in, common with M (r, f )?, 7. Consider the function f (z) = z 2 + 3z + 1 for |z| ≤ 1. Then the triangle, inequality gives that |f (z)| ≤ 5 for |z| ≤ 1. Can we conclude that, max|z|≤1 |f (z)| = 5? What happens if f (z) = z 2 −3z −1? What happens, if f (z) = z 2 + 3iz + i? What happens if f (z) = z 2 − 3z + 1?, 8. Suppose f is a one-to-one analytic function from the unit disk |z| < 1, onto itself such that f (0) = 0. Does f (z) ≡ eiα z for some real α?, Exercises 8.73., 1. If f (z) is analytic and nonzero in the disk |z − z0 | ≤ r, show that, log |f (z0 )| =, , 1, 2π, , +, 0, , 2π, , log |f (z0 + reiθ )| dθ., , 2. Suppose the nonconstant function f (z) is analytic in a domain D and, continuous on its closure D. If |f (z)| is constant on the boundary of D,, prove that f (z) has a zero in D., 3. Suppose that f is analytic and bounded in the unit disk Δ and f (ζ) → 0, as ζ → 1− along the upper half of the circle C : |ζ − 1/2| = 1/2. Then,, f (x) → 0 as x → 1− along [0, 1)., 4. Set M (r, f ) = max|z|=r |f (z)| and m(r, f ) = min|z|=r |f (z)|. Find, M (r, f ) and m(r, f ) for the following entire function, and indicate all, points on |z| = r where the maximum and minimum occur.
Page 293 :
284, , 8 Applications of Cauchy’s Theorem, , (a) f (z) = ez, , (b) f (z) = z n, , (c) f (z) = z 2 + 1 (d) f (z) = z 2 − z + 1, 5. Find the maximum and minimum values of, (a) |z(1 − z)| on |z| ≤ 1, (b) |z/(z 2 + 9)| on 1 ≤ |z| ≤ 2, (c) |(3 − iz)2 | on |z| ≤ 1, (d) |5 + 2iz 2 | on |z| ≤ 1, z−α, on |z| ≤ 1 (where α with |α| < 1 is fixed)., (e), 1 − αz, 2, 6. Show that max|z|=r |ez −iz | is attained at the point ir when r ≤ 1/4, and at reiθ , θ = sin−1 (1/4r), when r > 1/4., 7. Suppose f is analytic for |z| ≤ 1, f (0) = 0 and |f (z)| ≤ 5 for all |z| = 1., Can |f (0)| > 5?, 8. Let P (z) be a polynomial, and set f (z) = P (z)ez . Show that M (r, f ) →, ∞ and m(r, f ) → 0 as r → ∞., 9. Consider the polynomial P (z) = a0 + a1 z + · · · + an−1 z n−1 + z n . Show, that, for all sufficiently large values of r, there must exist points z0 , z1, on the circle |z| = r such that, n, , eP (z0 ) = e(1/2)r ,, , n, , eP (z1 ) = e−(1/2)r ., , 10. Suppose f (z) is analytic with |f (z)| < 1 for |z| < 1. If f (0) = 0, show, that |f (0)| ≤ 1, with equality only when f (z) = eiα z (α real)., 11. Suppose f (z) is analytic for |z| ≤ 1, and |f (z)| ≥ 1 for |z| ≤ 1. If, f (0) = 1, show that f (z) is a constant., 12. Let f (z) be analytic in |z| < R with f (0) = 0. Prove that, F (z) = f (z) + f (z 2 ) + f (z 3 ) + · · ·, is analytic in |z| < R, and that, |F (z)| ≤, , r, 1−r, , (|z| = r < R)., , 13. Does there exist an analytic function f : Δ → Δ such that f (1/2) = 3/4, and f (1/2) = 2/3, where Δ = {z : |z| < 1}?, 14. Let f be analytic for |z| ≤ 3 such that |f (z)| ≤ 1 for |z| ≤ 3 and, has n roots at wk = e2kπi/3 (k = 0, 1, 2, . . . , n − 1), the nth roots of, unity. What is the maximum value of |f (0)|? Which functions attain a, maximum?
Page 294 :
9, Laurent Series and the Residue Theorem, , In this chapter, we investigate the behavior of a function at points where, the function fails to be analytic. While such functions cannot be expanded, in a Taylor series, we show that a Laurent series expansion is possible. Also,, we introduce the notion of isolated and non-isolated singularities and discuss, different ways of characterizing isolated singularities. The complex integration machinery that was built in Chapter 7 and developed in Chapter 8 is, now ready to be utilized in order to evaluate definite integrals of real-valued, functions., , 9.1 Laurent Series, We know that a function, ∞ f (z), analytic at a point z0 , has a power series, representation f (z) = n=0 an (z − z0 )n that is valid in some neighborhood, of z0 . In this section, we will characterize expressions of the form, ∞, , , an (z − z0 )n ., , n=−∞, , To this end, observe that the series, ∞, , , bn (z − z0 )−n, , n=1, , may be viewed as a power series in the variable 1/(z − z0 ). If R is its radius, of convergence, then the series converges absolutely for, 1/|z − z0 | < R, that is, for |z − z0 | > R1 = 1/R., ∞, −n, Thus the series, represents an analytic function, f1 (z),, n=1 bn (z − z0 ), outside the circle |z − z0 | = R1 . Suppose also that the series
Page 295 :
286, , 9 Laurent Series and the Residue Theorem, ∞, , , an (z − z0 )n, , n=0, , ∞, has radius of convergence R2 . Then, f2 (z) = n=0 an (z − z0 )n is analytic, for |z − z0 | < R2 . If R2 > R1 , then f1 (z) and f2 (z) are both analytic in the, annulus R1 < |z − z0 | < R2 . Hence the function, f (z) = f1 (z) + f2 (z) =, , ∞, , , bn (z − z0 )−n +, , n=1, , ∞, , , an (z − z0 )n, , n=0, , is analytic for all z in the annulus R1 < |z − z0 | < R2 . Setting a−n = bn , we, may rewrite the above expression as, ∞, , , f (z) =, , an (z − z0 )n ., , (9.1), , n=−∞, , An expression of the form (9.1) is called a Laurent series, ∞about the point, z0 . We remark that if R1 > R2 , then the Laurent series n=−∞ an (z − z0 )n, diverges everywhere, and if R1 = R2 , it diverges everywhere except possibly at, points where |z −z0 | = R1 . In the last case, there are three different situations., For example,, •, •, •, , ∞, , n=−∞ n=0, ∞, , n, , zn, converges for |z| = 1, n2, , z diverges for |z| = 1, , n=−∞, ∞, , n=−∞, n=0, , zn, converges everywhere on |z| = 1 except for z = 1., n, , We now show that every function analytic in an annulus has a Laurent series, representation., Theorem 9.1. (Laurent’s Theorem) Suppose f (z) is analytic in the annulus, R1 < |z − z0 | < R2 . Then the representation, f (z) =, , ∞, , , an (z − z0 )n, , n=−∞, , is valid throughout the annulus. Furthermore, the coefficients are given by, +, 1, f (ζ), an =, dζ (n = 0, ±1, ±2, ±3, . . . ), 2πi C (ζ − z0 )n+1, where C is any simple closed contour contained in the annulus that makes a, complete counterclockwise revolution about the point z0 .
Page 298 :
9.1 Laurent Series, , ak =, and, , +, , 1, 2πi, , 1, ak =, 2πi, , C2, , +, C1, , f (ζ), dζ, (ζ − z0 )k+1, , f (ζ), dζ, (ζ − z0 )k+1, , 289, , (k = 0, 1, 2, . . . ), , (k = −1, −2, . . . )., , Now choose any simple closed contour contained in the annulus that makes, a complete counterclockwise revolution about the point z0 . Since f (ζ)/(ζ −z0 ), is analytic in the region bounded by the closed contours C1 and C (C2 and, C), Cauchy’s integral formula implies that the coefficients may be computed, by replacing C1 and C2 by C. Thus,, +, 1, f (ζ), dζ (k = 0, ±1, ±2, ±3, . . . )., (9.4), ak =, 2πi C (ζ − z0 )k+1, Finally, since R1 and R2 may be chosen arbitrarily close to R1 and R2 respectively, the expression (9.3) (with ak defined by (9.4)) is valid for all z in the, annulus R1 < |z − z0 | < R2 ., Remark 9.2. Observe that the series of positive powers of z − z0 converges, everywhere inside the circle |z − z0 | = R2 , whereas the series of negative, powers of z − z0 converges everywhere outside the circle |z − z0 | = R1 . The, series of negative powers of z − z0 is called the principle part of the Laurent, expansion, while the series of positive powers is called the analytic part. •, Remark 9.3. The Laurent expansion, like the power series expansion for analytic functions, is unique. Suppose that, f (z) =, , ∞, , , an (z − z0 )n =, , n=−∞, , ∞, , , bn (z − z0 )n ,, , n=−∞, , which is valid for R1 < |z−z0 | < R2 . Then each series converges uniformly on a, 1, (z −z0 )k, circle C contained in the annulus and enclosing z0 . Multiplying by 2πi, for any integer k, and integrating along C, we obtain, +, +, ∞, ∞, , , an, bn, n+k, (z − z0 ), dz =, (z − z0 )n+k dz., (9.5), 2πi, 2πi, C, C, n=−∞, n=−∞, Since, , 1, 2πi, , , , +, (z − z0 )m dz =, C, , 1 if m = −1, ,, 0 otherwise, , we get from (9.5), a−k−1 = b−k−1, , for each, , k ∈ Z., , Hence, ak = bk for every integer k showing that the Laurent expansion is, •, unique.
Page 299 :
290, , 9 Laurent Series and the Residue Theorem, , Remark 9.4. The coefficients of a Laurent series are usually not found by, the integral representation (9.4). In fact, determining the coefficients ak by, other means will enable us to evaluate the integral given in (9.4)., * Of particular, interest is the coefficient a−1 , for this enables us to determine C f (ζ) dζ. We, •, shall focus our attention on this part in later sections., We now give some examples to show different methods for computing the, coefficients of a Laurent series., Example 9.5. To find the Laurent expansion for, f (z) =, , sin z, z2, , (|z| > 0),, , we expand sin z in a Maclaurin series. This observation leads to, , , z5, 1, z, z3, z3, 1, sin z, f (z) = 2 = 2 z −, +, − ··· = − +, − ···, z, z, 3!, 5!, z, 3!, 5!, , (|z| > 0)., , Similarly, from the identity eu = 1 + u + u2 /2! + u3 /3! + · · · , we get, f (z) = e1/z = 1 +, , 1, 1, 1, +, +, + ···, 2, z, 2!z, 3!z 3, , (|z| > 0)., , •, , Example 9.6. Consider the function, f (z) =, , 1, 1, =, ,, z2 + 1, (z + i)(z − i), , which is analytic in C \{i, −i}. We first expand this function in a Laurent, series valid in a deleted neighborhood of z = i. To do this, we consider, , n, ∞, z−i, 1, 1, 1 , 1, n, =, =, =, (−1), z+i, 2i + (z − i), 2i[1 + (z − i)/2i], 2i n=0, 2i, which is valid for |z − i| < 2. Hence, for 0 < |z − i| < 2, we have, n, ∞ , , 1, 1, 1, f (z) =, =, (z − i)n, −, (z + i)(z − i), 2i(z − i) n=0, 2i, n+2, ∞ , , 1, −, =−, (z − i)n, 2i, n=−1, ∞ n+2, , i, =−, (z − i)n ., 2, n=−1, Similarly, to expand in a Laurent series valid in a deleted neighborhood of, z = −i, we first write
Page 300 :
9.1 Laurent Series, , 291, , n, ∞ , 1, 1, 1 z+i, 1, =, =−, =−, ,, z−i, −2i + (z + i), 2i[1 − (z + i)/2i], 2i n=0, 2i, which is valid for |z + i| < 2. Thus, for 0 < |z + i| < 2,, n+2, n+2, ∞ , ∞ , , , 1, i, f (z) = −, (z + i)n = −, (z + i)n ., −, 2i, 2, n=−1, n=−1, , •, , Example 9.7. The function f (z) = 1/[(z − 1)(z − 2)] is analytic in C \{1, 2}., As in the previous example, we can expand this function in a Laurent series, valid in a deleted neighborhood of z = 1 or z = 2. This observation leads to, f (z) = −, , ∞, , 1, =−, (z − 1)n, (z − 1)[1 − (z − 1)], n=−1, , (0 < |z − 1| < 1), , and, f (z) =, , ∞, , 1, =, (−1)n+1 (z − 2)n, (z − 2)[1 + (z − 2)] n=−1, , (0 < |z − 2| < 1)., , But we can also expand f (z) in Laurent series that are valid in different, regions. For example, f has three Laurent series centered at 0:, (i) |z| < 1, , (ii) 1 < |z| < 2, , (iii) |z| > 2., , (i) Suppose |z| < 1. Then (as |z| < 1 implies |z| < 2), we get, , ∞ , , 1, 1, 1, 1, 1, −, =−, +, =, 1 − n+1 z n ,, f (z) =, z−2 z−1, 2(1 − z/2) 1 − z n=0, 2, which is the Maclaurin series expansion for f (z), i.e., no principal part., (ii) Suppose 1 < |z| < 2. Then, ∞, ∞ n, 1, 1 z n 1 1, 1, −, =−, −, ., f (z) = −, 2(1 − z/2) z(1 − 1/z), 2 n=0 2, z n=0 z, Here we are using the fact that |z/2| < 1 and |1/z| < 1. Hence, f (z) = −, , −1, , , zn −, , n=−∞, , ∞, , , 1 n, z for 1 < |z| < 2., n+1, 2, n=0, , (iii) Suppose |z| > 2. Then (as |z| > 2 implies |2/z| < 1 and |1/z| < 1), 1, 1, −, z(1 − 2/z) z(1 − 1/z), , ∞ n, ∞ n, −1 , , 1, 1 2, 1 1, =, −, =, − 1 zn, n+1, z n=0 z, z n=0 z, 2, n=−∞, , f (z) =, , which has no analytic part., , •
Page 301 :
292, , 9 Laurent Series and the Residue Theorem, , Questions 9.8., 1. What similarities are there between Laurent series and power series?, 2. If the Laurent series for f (z) converges on the boundary of an annulus,, is f (z) analytic on the boundary?, 3. Where does a Laurent series converge uniformly?, 4. Can a function be analytic only in a rectangular strip?, 5. Can f (z) and f (1/z) be analytic at the same set of points?, 6. What can, be said about the sum of two Laurent series? The product?, ∞, 7. If f (z) = n=−∞ an (z − z0 )n , under what circumstances does, f (z) =, , ∞, , , nan (z − z0 )n−1 ?, , n=−∞, , 8. Does the function f (z) = z have Laurent series valid in some domain, 0 < |z| < δ? How about if f (z) = Re z or Im z or |z|2 ?, Exercises 9.9., 1, in Laurent series valid for, (z + 1)(z 2 + 2), √, √, (ii) |z| > 2, (iii) |z| < 1., (i) 1 < |z| < 2, , 1. Expand f (z) =, , 2. Expand f (z) =, , 3z − 1, in Laurent series valid for, z 2 − 2z − 3, , (i) 1 < |z| < 3, , (ii) |z| > 3, , 3. Find the Laurent series for f (z) =, (i) 1 < |z − 1| < 4, , z − 12, valid for, +z−6, , z2, , (ii) |z − 1| > 1, , 2, , (iii) |z| < 1., , (iii) |z − 1| < 4., , 2, , 4. Expand f (z) = ez + e1/z in a Laurent series valid for |z| > 0., 1, in Laurent series valid, 5. For 0 < |a| < |b|, expand f (z) =, (z − a)(z − b), for, (i) |z| < |a|, (ii) |a| < |z| < |b|, (iii) |z| > |b|, , (iv) 0 < |z − a| < |a − b|, , (v) 0 < |z − b| < |a − b|., 6. If 0 < |a| < |b|, expand, f (z) =, , 1, z(z − a)(z − b), , as a Laurent series valid in, (i) 0 < |z| < |a|, , (ii) |a| < |z| < |b|, , (iii) |z| > |b|.
Page 302 :
9.2 Classification of Singularities, , 7. Expand, f (z) =, , 293, , z 2 + 9z + 11, (z + 1)(z + 4), , as a Laurent series about z = 0 valid when, (i) |z| < 1, , (ii) 1 < |z| < 4, , (iii) |z| > 4., , 8. Find the principal part for the following Laurent series., √, z2, (0 < |z − i| < 2), (a) 4, z −1, √, z2, (0 < |z + i| < 2), (b) 4, z −1, ez, (c) 4 (|z| > 0), z, sin z, (d), (|z| > 0), z4, 1, 1, (e), − 2 (0 < |z| < π/2)., z, tan2 z, 9. Expand the following in a Laurent series valid in the region indicated., (a) z n e1/z (|z| > 0), (b) e1/(z−1) (|z| > 1)., 10. Express sin z sin(1/z) in a Laurent series valid for |z| > 0., 11. Use series division to find the principal part in a neighborhood of the, origin for the function ez /(1 − cos z)2 ., , 9.2 Classification of Singularities, A single-valued function is said to have a singularity at a point if the function, is not analytic at the point while every neighborhood of that point contains, at least one point at which the function is analytic. For instance, f (z) = z is, nowhere analytic. Note that we do not say that every point of C is a singularity, for f (z). Basically, there are two types of singularities, (i) isolated singularity, (ii) non-isolated singularity., If the function is analytic in some deleted neighborhood of the point, then the, singularity is said to be isolated. For example, consider, , z + 1 for z = 0, (i) f1 (z) =, 3 for z = 0, (ii) f2 (z) = 1/z for z = 0, (iii) f3 (z) = sin(1/z) for z = 0.
Page 303 :
294, , 9 Laurent Series and the Residue Theorem, , For each of the functions defined above, the point z = 0 is an isolated singularity., Recall from Section 4.3 that f (z) = Log z, the principal logarithm, is, analytic for z ∈ Ω = C \ {x + iy : y = 0, x ≤ 0}. The point z = 0 is a branch, point of Log z since every deleted neighborhood of z = 0 contains points on, the negative real axis. We say that a singularity at z = z0 of f is non-isolated, if every neighborhood of a contains at least one singularity of f other than, a. For example, z = 0 as well as every point on the negative real axis is a, non-isolated singularity of the principal logarithm given by, Log z = ln |z| + iArg z,, , −π < Arg z < π,, , since every neighborhood of z = x (x ≤ 0) contains points on the negative, real axis on which Log z is not analytic. The behavior of a function f (z) near, an isolated singularity z0 can be described by considering the limiting value, of limz→z0 f (z). Then there are three possibilities:, (i) f (z) may be bounded in a deleted neighborhood of z0 . For instance,, in the above example, f1 (z) is bounded in a deleted neighborhood of, the origin. This is an uninteresting example because f1 (z) can be made, analytic by defining f1 (0) = 1. Another example of this type may be, given by, sin z, for z = 0., g1 (z) =, z, (ii) f (z) may approach ∞ as z approaches z0 . For instance,, f2 (z) =, , 1, → ∞ as z → 0., z, , Another example of this type may be given by g1 (z) = z1n (z = 0),, where n ∈ N is fixed., (iii) f (z) may satisfy neither (i) nor (ii). For instance, consider, f (z) = e1/z ,, , z = 0., , For z = x, x = 0 real, note that, lim e1/x = ∞ and, , x→0+, , lim e1/x = 0., , x→0−, , Moreover, for z = iy (y ∈ R \ {0}), we note that, f (iy) = e1/(iy) = e−i/y, which lies always on the unit circle. Consequently, limz→0 e1/z does not, exist. Thus, in a neighborhood of the origin, e1/z is neither bounded nor, does it approach ∞. A similar argument continues to hold for sin(1/z), in the neighborhood of the origin. Indeed, we consider two sequences:
Page 304 :
9.2 Classification of Singularities, , 295, , zn = i/n → 0 and zn = 1/n → 0 as n → ∞., Then, , en − e−n, → ∞,, 2i, whereas {sin(1/zn )} = {sin n} is clearly a bounded sequence. It follows, that z = 0 is an essential singularity of sin(1/z)., sin(1/zn ) = sin(−in) =, , It follows in general that if a function has one non-isolated singularity, then, it will have many singularities, although not necessarily non-isolated. This is, demonstrated by the following example. The function, f (z) =, , 1, sin(1/z), , has a singularity at the origin because limz→0 f (z) does not exist either as, a finite limit or as an infinite limit. Note that limz→0 sin(1/z) does not exist, because, 1, 1, → 0 and zn =, → 0,, zn =, nπ, nπ + π/2, whereas sin(zn ) = 0 and sin(zn ) = cos(nπ) = (−1)n . Furthermore, the zeros, of sin(1/z) are given by zn = 1/(nπ), n ∈ Z \{0}. These points are also, singularities of f (z). Notice that |zn | → 0 as n → ∞ so that each deleted, neighborhood of the origin contains a singularity of f (z). Consequently, the, singularity at z = 0 is a non-isolated singularity of f (z)., Observe that the “clever” way we phrased (iii) makes every isolated singularity satisfy either (i), (ii) or (iii)., Each of the three kinds of isolated singularities have important characterizations which will be obtained in the next three theorems., Theorem 9.10. (Riemann’s Theorem) Suppose that f (z) has an isolated, singularity at z = z0 and is bounded in some deleted neighborhood of z0 . Then, f (z) can be defined at z0 in such a way as to be analytic at z0 ., Proof. Assume the hypothesis. Then, for some R > 0, f (z) is analytic in the, punctured disk, 0 < |z − z0 | ≤ R., Given a point z1 inside this disk, choose r > 0 so that r < |z1 − z0 | < R., The function f (z) is analytic in the annulus bounded by the two circles C :, |z − z0 | = R and C1 : |z − z0 | = r (see Figure 9.2). Hence by Cauchy’s integral, formula,, ,, ,, 1, 1, f (ζ), f (ζ), dζ −, dζ., (9.6), f (z1 ) =, 2πi C ζ − z1, 2πi C1 ζ − z1, Observe that the value of (9.6) is independent of the choice of r. We will prove, that the last integral on the right of (9.6) is zero by showing that its absolute
Page 306 :
9.2 Classification of Singularities, , 297, , We have shown that functions satisfying the conditions of Theorem 9.10, have a singularity simply because they have not been defined “properly” at, the point in question. If f (z) has an isolated singularity at z0 , the singularity, is said to be a removable singularity if limz→z0 f (z) exists. Theorem 9.10 says, that a function which is analytic and bounded in a deleted neighborhood of, a point has at worst a removable singularity at the point. For all practical, purposes, we may consider such a function to be analytic. Thus when we, speak of (sin z)/z as an entire function, it will be understood that f (0) =, limz→0 (sin z)/z = 1. Note that the Maclaurin expansion for this function is, , , sin z, 1, z5, z2, z4, z3, f (z) =, =, +, + ··· = 1 −, +, + · · · , |z| > 0., z−, z, z, 3!, 5!, 3!, 5!, A function f (z), analytic in a deleted neighborhood of z = z0 , has a pole, of order k (k a positive integer) if, lim (z − z0 )k f (z) = A = 0, ∞., , z→z0, , We now characterize singularities of the form (ii) quoted in the beginning., Theorem 9.12. If f (z) has an isolated singularity at z = z0 and f (z) → ∞, as z → z0 , then f (z) has a pole at z = z0 ., Proof. Suppose that f (z) → ∞ as z → z0 . Then for a given R > 0 there exists, a δ > 0 such that f (z) is analytic for 0 < |z − z0 | < δ and, |f (z)| > R whenever 0 < |z − z0 | < δ., In particular, f (z) = 0 for 0 < |z−z0 | < δ and so, g(z) = 1/f (z) is analytic and, bounded by 1/R in this deleted neighborhood of z0 . By Theorem 9.10, g(z) has, a removable singularity at z0 , and we may write g(z) as (using limz→z0 g(z) =, 0), g(z) =, , 1, = a1 (z − z0 ) + a2 (z − z0 )2 + · · · ,, f (z), , 0 < |z − z0 | < δ., , Moreover, g(z) = 0 for 0 < |z − z0 | < δ, and so not all the coefficients of g(z), are zero. This means that there is a k ≥ 1 such that ak is the first nonzero, coefficient of g(z). Then, g(z) =, , 1, = ak (z − z0 )k + ak+1 (z − z0 )k+1 + · · · ,, f (z), , so that, 1, = ak + ak+1 (z − z0 ) + · · · → ak as z → z0 ,, (z − z0 )k f (z), and therefore,, 1, ., ak, Hence, by the definition, f (z) has a pole of order k at z = z0 ., lim (z − z0 )k f (z) =, , z→z0, , (9.9)
Page 307 :
298, , 9 Laurent Series and the Residue Theorem, , Remark 9.13. If f (z) has a pole at z0 , it follows from the definition that, there exists a k ≥ 1 such that (z − z0 )k f (z) → A = 0 as z → z0 . Thus, for, z → z0 ,, (z − z0 )k, 0, 1, =, →, = 0 as z → z0, f (z), (z − z0 )k f (z), A, and hence, f (z) → ∞ as z → z0 . Thus, Theorem 9.12 gives a necessary and, •, sufficient condition for an isolated singularity to be a pole., Corollary 9.14. If f (z) has a pole at z = z0 , then f (z) may be expressed as, f (z) =, , ∞, , , bn (z − z0 )n ,, , n=−k, , where k is the order of the pole., Proof. We use the notation of Theorem 9.12. By (9.9), the function, 1, = ak + ak+1 (z − z0 ) + · · ·, f (z)(z − z0 )k, , (ak = 0), , is analytic at z = z0 . Since ak = 0, a continuity argument shows that there is, a neighborhood of z0 in which, 1, = 0., f (z)(z − z0 )k, Hence, f (z)(z − z0 )k is analytic at z0 and the expansion, , , ∞, , 1, k, m, f (z)(z − z0 ) =, c0 =, cm (z − z0 ), = 0, ak, m=0, is valid in some neighborhood of z0 . That is,, f (z) =, , ∞, , , cm (z − z0 )m−k ., , (9.10), , m=0, , Upon setting n = m − k and bn = cm+k in (9.10), we get the desired form., An isolated singularity that is neither removable nor a pole is said to be an, (isolated) essential singularity. Equivalently, a function f (z) which is analytic, in a deleted neighborhood of z = z0 has an essential singularity at z = z0 if, there exists no nonnegative integer k for which limz→z0 (z − z0 )k f (z) exists, (either as a finite value or as an infinite value). The behavior of a function in, the neighborhood of an isolated essential singularity is most surprising., Theorem 9.15. (Casorati–Weierstrass) If f (z) has an essential singularity, at z = z0 , then f (z) comes arbitrarily close to every complex value in each, deleted neighborhood of z0 .
Page 308 :
9.2 Classification of Singularities, , 299, , Proof. Let f have an essential singularity at z0 . Then f is analytic throughout, a deleted neighborhood of z0 . Suppose, for some complex number a, that, |f (z) − a| ≥ > 0, for all z in a punctured disk 0 < |z − z0 | < δ. Set g(z) = 1/(f (z) − a). Then, |g(z)| =, , 1, 1, ≤, f (z) − a, , for 0 < |z − z0 | < δ., , Thus, by Theorem 9.10, g(z) has a removable singularity at z = z0 and we, may write, g(z) =, , 1, = a0 + a1 (z − z0 ) + a2 (z − z0 )2 + · · · ., f (z) − a, , Observe that limz→z0 1/(f (z) − a) = a0 . There are two cases to consider., Case 1) a0 = 0: Then limz→z0 f (z) = 1/a0 + a, and f (z) has a removable, singularity at z = z0 ., Case 2) a0 = 0: Suppose ak is the first nonzero coefficient. Then, 1, = ak + ak+1 (z − z0 ) + · · · ., (f (z) − a)(z − z0 )k, In this case,, lim (z − z0 )k (f (z) − a) = lim (z − z0 )k f (z) =, , z→z0, , z→z0, , 1, ,, ak, , and f (z) has a pole of order k at z = z0 . Since our hypothesis excludes, both cases, the inequality |f (z) − a| ≥ > 0 cannot be true and the desired, conclusion follows., Corollary 9.16. Suppose f (z) has an essential singularity at z = z0 . Given, any complex number a, there exists a sequence {zn } such that zn → z0 and, f (zn ) → a., Proof. Choose a sequence {δn } for which δn > 0 for each value of n and, limn→∞ δn = 0. By Theorem 9.15, we can find a sequence of points {zn } such, that |f (zn ) − a| < 1/n for 0 < |zn − z0 | < δn . Thus, f (zn ) → a as zn → z0 ., A generalization of Theorem 9.15 is described by Picard in the following, form. For details of this result, we refer to advanced texts, see Hille [Hi]., Theorem 9.17. (Picard’s Great Theorem) In the neighborhood of an isolated essential singularity, a function assumes each complex value, with one, possible exception, infinitely often.
Page 309 :
300, , 9 Laurent Series and the Residue Theorem, , The function e1/z has an isolated essential singularity at z = 0, and never, assumes the value 0. We wish to show that e1/z assumes every other value in, a neighborhood of the origin infinitely often. The behavior near an essential, singularity is strange. Indeed, if c = 0, then the solutions of, e1/z = c, are given by, 1, 1, =, , n ∈ Z., log c, Log c + 2nπi, Observe that infinitely many zn are contained in every neighborhood of the, origin., zn =, , Isolated singularities at z = ∞. We may also refer to the point at ∞ as, being an isolated singularity. A function f (z) has an isolated singularity at, z = ∞ if f (z) is analytic in a deleted neighborhood of ∞, that is there exists, a real number R such that f (z) is analytic for R < |z| < ∞. Note that f (z), is analytic for |z| > R if and only if f (1/z) is analytic for |1/z| < R. Hence,, f (z) has an isolated singularity at z = ∞ if and only if f (1/z) has an isolated, singularity at z = 0. Moreover, we make the definition that the singularity of, f (z) at z = ∞ is removable, a pole, or essential according as the singularity of, f (1/z) at z = 0 is removable, a pole, or essential. For example, the function, 1. f (z) = z 2 +1 has a pole of order 2 at z = ∞ because f (1/z) = (1/z 2 )+1, has a pole of order 2 at z = 0., 2. f (z) = ez has an isolated essential singularity at z = ∞ because, f (1/z) = e1/z has an isolated essential singularity at z = 0., 3. f (z) = 1/[z(z 2 + 4)] has simple poles at z = 0, ±2i. Now, f (1/z) =, , z3, ,, 1 + 4z 2, , showing that z = 0 is a point of analyticity for f (1/z). In fact, it has, zero of order 3 at z = 0., With, definition, we can examine the behavior of entire functions. If, this, k, f (z) = n=0 an z n is a polynomial of degree k, then, , , 1, ak−1, a1, ak, 1, + a0 and lim z k f, = k + k−1 + · · · +, = ak = 0, g(z) = f, z→0, z, z, z, z, z, so that g(z) has a pole of order k at z = 0. Hence, f (z) has a pole of order k at, z = ∞. Theorem 8.38 is now seen to be a special case of Theorem 9.15. That, is, the polynomial f (z) → ∞ as z → ∞ because f (z) has a pole at z = ∞., Note that the entire function f (z) has a pole of order k at z = ∞ if and only, if, , 1, f (z), k, = A = 0, ∞., (9.11), = lim, lim z f, z→∞ z k, z→0, z
Page 310 :
9.2 Classification of Singularities, , 301, , Moreover, if (9.11) is satisfied, then (by Theorem 8.35) f (z) is a polynomial, of degree k. We summarize the discussion as, Theorem 9.18. Let f (z) be a nonconstant entire function. Then f (z) is a, polynomial if and only if f (z) → ∞ as z → ∞., Thus a transcendental entire function cannot have a pole at z = ∞. That, is, transcendental entire functions must have essential singularities at z = ∞., An interesting comparison can now be made between Theorem 8.31 and, Theorem 9.15. Theorem 8.31 merely asserts that an entire function comes, arbitrarily close to every complex value. Theorem 9.15 says that a transcendental entire function comes arbitrarily close to every complex value outside, of every circle |z| = R. Of course the behavior of a non-transcendental entire, function (a polynomial) has already been fully discussed., In the Laurent series expansion (see Theorem 9.1), if R1 = 0, then the, point z0 becomes an isolated singularity of f (z). In view of this, the Laurent, series allows us to classify the type of isolated singularity at z0 . Suppose that, f has an isolated singularity at z0 . Then f (z) is analytic throughout a deleted, neighborhood of z0 , that is f (z) possesses a Laurent series expansion, f (z) =, , −1, , , an (z − z0 )n +, , n=−∞, , ∞, , , an (z − z0 )n, , n=0, , valid for 0 < |z − z0 | < δ for some δ > 0. Then the following situations arise., (i) No principal part: In this case an = 0 for all n < 0. So the above, Laurent series simply reduces to, f (z) = a0 + a1 (z − z0 ) + a2 (z − z0 )2 + · · · ,, , 0 < |z − z0 | < δ., , It follows that limz→z0 f (z) = a0 , i.e., f is bounded in a deleted neighborhood of z0 . In other words, f has a removable singularity at z0 if, and only if an = 0 for all n < 0., (ii) The principal part consists of a finite number of terms: In this, case, an = 0 for all n < −m and a−m = 0 for some m ≥ 1. So the, Laurent series about z0 reduces to, ∞, , f (z) =, , , a−1, a−m, +, ·, ·, ·, +, +, an (z − z0 )n ,, (z − z0 )m, z − z0 n=0, , which is valid for 0 < |z − z0 | < δ so that, lim (z − z0 )m f (z) = a−m = 0., , z→z0, , If we rewrite (9.12) in the form, , (9.12)
Page 311 :
302, , 9 Laurent Series and the Residue Theorem, , , a−m+1, a−m, (z − z0 ) + · · ·, 1+, m, (z − z0 ), a−m, ', ∞, , a−1, an, m−1, n+m, (z − z0 ), +, (z − z0 ), +, a−m, a, n=0 −m, a−m, [1 + g(z)] ,, =, (z − z0 )m, , f (z) =, , we see that g(z) → 0 as z → z0 . Thus, given, δ > 0 such that, , = 1/2, there exists a, , |g(z)| < 1/2 whenever |z − z0 | < δ, and therefore, |1 + g(z)| ≥ 1 − |g(z)| ≥ 1/2 for |z − z0 | < δ. This, observation shows that, |f (z)| ≥, , |a−m |, |a−m |, |a−m |, |1 + g(z)| >, >, |z − z0 |m, 2|z − z0 |m, 2δ m, , which, for δ → 0, implies that f (z) → ∞ as z → z0 . In particular, f (z), behaves like a−m /(z − z0 )m for z near z0 . In other words, f (z) has a, pole of order m at z0 if and only if there exists an m ∈ N such that, a−m = 0 and an = 0 for all n less than −m., (iii) The principal part consists of infinitely many terms: In this case,, an = 0 for infinitely many negative integer values of n. Thus by a process, of elimination, the principal part has infinitely many terms if and only, if the singularity at z0 is an essential singularity., From the above discussion, when expanding in a Laurent series about an, isolated singularity, we are sometimes interested only in the principal part. If, f (z) has a simple pole at z = z0 , then the principal part is particularly easy, to determine. For then limz→z0 (z − z0 )f (z) must exist, and we may set, (z − z0 )f (z) = a−1 + a0 (z − z0 ) + a1 (z − z0 )2 + · · ·, , (a−1 = 0)., , The principal part is then seen to be a−1 /(z − z0 )., Examples 9.19., (i) For instance, consider f (z) = z/(z 2 +4). Suppose that, we wish to find the principal part of the Laurent expansion valid in a, deleted neighborhood of z = 2i. Since, lim (z − 2i)f (z) = lim, , z→2i, , z→2i, , 1, (z − 2i)z, = ,, (z + 2i)(z − 2i), 2, , f (z) has a simple pole at z = 2i. Therefore,, f (z) =, where g(z) is analytic at z = 2i., , 1/2, + g(z),, z − 2i
Page 312 :
9.2 Classification of Singularities, , 303, , (ii) If f (z) = 1/(z 4 + 1), then the solution set of z 4 + 1 = 0 is given by, zk = (−1)1/4 = ei(π+2kπ)/4 ,, , k = 0, 1, 2, 3,, , and these are the simple poles of f (z). For example, as in the previous, example, we get, lim, , z→zk, , 1, zk, zk, z − zk, 1, = lim, = 3 = 4 =− ., 4, 3, z + 1 z→zk 4z, 4zk, 4zk, 4, , Then,, f (z) =, , −zk /4, + gk (z), z − zk, , for k = 0, 1, 2, 3,, , where each gk (z) is analytic at z = zk . In particular, at z0 = eiπ/4, √, −(1 + i)/ 2, + g0 (z),, f (z) =, 4(z − eiπ/4 ), √, where g0 (z) is analytic at z0 = eiπ/4 = (1 + i)/ 2., (iii) Let us now find all singularities for the function f (z) = cot πz, and, determine the principal part of the Laurent expansion about each singularity. Since sin πz has simple zeros at z = n (n ∈ Z), the function, f (z) = cos πz/ sin πz has simple poles at z = n. In a deleted neighborhood of z = n, we have, f (z) =, , a−1, + gn (z),, z−n, , 0 < |z − z0 | < 1,, , where gn (z) is analytic at z = n. It remains to determine a−1 . From the, identity, sin πz = (−1)n sin π(z − n),, we get (as f has simple pole at each z = n), a−1 = lim (z − n)f (z) =, z→n, , 1, 1, π(z − n) cos πz, lim, =, n, π z→n sin π(z − n) (−1), π, , and the principal part of f (z) at each z = n is, 1/π, ., z−n, (iv) If f (z) =, , 2 + ez, , then z = 0 is an isolated singularity and, sin z + z cos z, 3, 2 + ez, = = 0,, z→0 z −1 sin z + cos z, 2, , lim zf (z) = lim, , z→0, , which shows that z = 0 is a simple pole for f (z). Similarly, we see that
Page 313 :
304, , 9 Laurent Series and the Residue Theorem, , g(z) =, , 1 + sin z, cos z − 1 + sin z, , has a simple pole at z = 0, because, lim zg(z) = lim, , z→0, , z→0, , 1 + sin z, = 1., cos z − 1 sin z, +, z, z, , (v) Let us discuss the nature of the singularity of f (z) = (z + 1)−4 sin πz, at z = −1 and write down the principal part of it. To do this, we first, observe that f (z) has a pole of order 3 at z = −1, and f is analytic for, 0 < |z + 1| < ∞. It follows that, f (z) = (z + 1)−4, , ∞, , , an (z + 1)n ,, , n=0, , where an = g (n) (−1)/n!, with g(z) = sin πz. Note that, g (z) = π cos πz = π sin(πz + π/2),, g (z) = π 2 cos(πz + π/2) = π 2 sin(πz + 2(π/2)),, .., .., . =, ., g (n) (z) = π n sin(πz + nπ/2),, and so, g, , (n), , , n, , (−1) = π sin(−π+nπ/2) =, , Thus,, −4, , f (z) = (z + 1), , 0 if n = 2k, −π n (−1)k if n = 2k + 1, k = 0, 1, . . . ., , ∞, , (−1)k+1 π 2k+1, k=0, , (2k + 1)!, , (z + 1)2k+1, , so that f (z) has a pole of order 3 at z = −1. From this expansion, one, •, can easily write down the principal part., Examples 9.20. We wish to characterize all rational functions which have a, removable singularity at ∞., To do this, we let f (z) = p(z)/q(z), where p and q are polynomials. Then, f (z) has a removable singularity at ∞, ⇐⇒ f (1/z) has a removable singularity at z = 0, ⇐⇒ |f (1/z)| ≤ M for 0 < |1/z| < , for some > 0 and M > 0, ⇐⇒ |f (z)| ≤ M for |z| > 1/, ⇐⇒ deg p(z) ≤ deg q(z).
Page 315 :
306, , 9 Laurent Series and the Residue Theorem, , We shall use this method to find the principal part of, f (z) =, , π cot πz, z4, , valid in a deleted neighborhood of the origin. We have, , , 1 1 − (πz)2 /2! + (πz)4 /4! − · · ·, π cos πz, = 5, f (z) = 4, z sin πz, z, 1 − (πz)2 /3! + (πz)4 /5! − · · ·, , , 2, 1, π 2 π4 4, = 5 1+, z −, z + ··· ., z, 3, 45, Thus the principal part of f (z) is, π 2 /3 π 4 /45, 1, ., +, −, z5, z3, z, Questions 9.21., 1. Can a function have infinitely many isolated singularities in the plane?, In a bounded region? In a compact set?, 2. Given a function f (z), does there exist a real number M such that no, pole of f (z) has order greater than M ?, 3. Can a function have poles at a preassigned sequence of points?, 4. Can a function have essential singularities at the preassigned sequence, of points?, 5. Can an entire function omit the value 7 − 2i and assume every other, value infinitely often?, 6. Why are the points 0 and ∞ so often different from all other values?, 7. What kind of function has no singularities in the extended plane?, 8. How do the singularities of f (z) compare with those of 1/f (z)? With, those of f (1/z)? With those of 1/f (1/z)?, 9. Can a pole be a non-isolated singularity?, 10. How does Picard’s great theorem compare with Picard’s theorem stated, in Section 8.2?, 11. If f (z) has a pole of order k at z0 , what are the most and least number, of terms for the principal part of the Laurent expansion?, 12. When can one have an accumulation of singularities?, 13. If a function has an absolute value greater than 1 near an isolated singularity, what kind of singularity can it be?, 14. If f (z) has an isolated singularity at z = 0, what can you say about f, if limz→0 |z|2/3 |f (z)| = 2?, Exercises 9.22., 1. Suppose that f (z) has an isolated singularity at z = z0 , and that, limz→z0 (z − z0 )α f (z) = M = 0, ∞. Prove that α must be an integer.
Page 316 :
9.2 Classification of Singularities, , 307, , 2. If f (z) is analytic in a deleted neighborhood of the origin and, lim |zf (z)| = 0,, , z→0, , show that the origin is a removable singularity of f (z)., 3. Show that tan z does not assume the value ±i. Does this contradict, Picard’s theorem?, 4. Find all singularities for the following functions, and describe their nature., 1, 1, (a) tan z, (b) 1/z, (c) 2 z, z (e − 1), e +1, 1, 1, (e) ez+1/z, (f), (d), sin z − cos z, cos(1/z), (g), , (z − 1)1/2, z+1, , (h), , sin4 z, + cos(3z), z4, , (i), , z, ., ez − 1, , 5. Discuss the singularities of, f (z) =, , z 3 (z 2 − 1)(z − 2)2 1/z2, e, ., sin2 (πz), , Classify which of these are poles, removable singularities and essential, singularity., 6. Describe the singularity at z = ∞ for the following functions., z2, z 2 + 10, 2z 2 + 1, (b), (c), 2, 3z − 10, z+1, ez, z, 1, e, (e) tan z − z, (f) + sin z., (d) 2, z + 10, z, Given arbitrary distinct complex numbers z0 , z1 and z2 , construct a, function f (z) having a removable singularity at z = z0 , a pole of order, k at z = z1 , and an essential singularity at z = z2 ., Show that f (z) has no singularities in the extended plane other than, poles if and only if f (z) is a rational function (quotient of two polynomials)., If f (z) has poles at a sequence of points {zn }, and zn → z0 , show that, f (z) does not have a pole at z = z0 . Illustrate this fact by a concrete, example., Suppose f (z) has a pole of order m at z = z0 , and P (z) is polynomial, of degree n. Show that P (f (z)) has a pole of order mn at z = z0 ., Determine the order of the pole at z = 0 for, (a), , 7., , 8., , 9., , 10., 11., , (i) f (z) =, , z, sin z − z + z 3 /3!, , (ii) f (z) =, , z, ., (sin z − z + z 3 /3!)2, , 12. Use “long division” method (or other method) to find the principal part, in the Laurent series of f (z) = 1/(1 − cos z) about z = 0.
Page 317 :
308, , 9 Laurent Series and the Residue Theorem, , 13. Let f (z) be analytic in the disk |z| < R (R > 1) except for a simple, pole at a point z0 , |z0 | = 1. Consider the expansion f (z) = a0 + a1 z +, a2 z 2 + · · · , and show that limn→∞ (an /an+1 ) = z0 ., 14. Consider, in a neighborhood of the origin, the various determinations of, (1 + z)1/z ., (a) Show that one of them is analytic in |z| < 1, and denote it by f0 (z)., (b) Determine a0 , a1 and a2 in the expansion f0 (z) = a0 + a1 z + a2 z 2 +, ···., (c) Let f (z) be a determination of (1 + z)1/z other than f0 (z). Find, the nature of g(z) = f (z)/f0 (z), and give its Laurent expansion for, |z| > 0., , 9.3 Evaluation of Real Integrals, If f (z) is analytic in a deleted neighborhood of z0 , then by Laurent’s theorem, we may write, f (z) =, , ∞, , , an (z − z0 )n, , (0 < |z − z0 | < δ),, , n=−∞, , +, 1, f (z), an =, dz (n ∈ Z)., 2πi C (z − z0 )n+1, Here C is any simple closed contour enclosing z0 and contained in the neighborhood. In particular,, +, +, 1, a−1 =, f (z) dz, i.e.,, f (z) dz = 2πia−1 ., (9.13), 2πi C, C, where, , Therefore, by hook or crook, we should be able to compute a−1 . The coefficient, a−1 is called the residue of f (z) at z0 and is denoted by, Res [f (z); a]., Equation (9.13) says that evaluating a certain integral of f (z) around C that, encloses no other singularity other than z0 is akin to determining a certain, coefficient in Laurent series., Examples 9.23., (i) As e1/z = 1 + 1/z + 1/(2!z 2 ) + · · · for |z| > 0,, 1/z, Res [e ; 0] = 1 and so, we have, +, e1/z dz = 2πi., |z|=1, , More generally,, +, , , exp, , C, , 1, zk, , , , , dz =, , 0 if k =, 1, , k ∈ Z,, 2πi if k = 1, , where C is a simple closed contour enclosing the origin.
Page 318 :
9.3 Evaluation of Real Integrals, , 309, , (ii) As, , sin, , 1, z2, , , =, , 1, 1, −, 2, z, 3!, , , , 1, z2, , 3, +, , 1, 5!, , , , 1, z2, , 5, − ···, , for |z| > 0,, , we have Res [sin(1/z 2 ); 0] = 0, and so, +, 1, sin 2 dz = 0., z, |z|=1, More generally,, +, , , sin, , C, , 1, zk, , , , , dz =, , 0 if k =, 1, , k ∈ Z,, 2πi if k = 1, , where C is a simple closed contour enclosing the origin., (iii) For z = 0, we have Res [z 2 sin(1/z 2 ); 0] = −1/6. Therefore, , , +, 1, πi, 1, =− ,, z 2 sin dz = 2πi −, z, 6, 3, |z|=1, where C is a simple, + closedz contour enclosing the origin., e −1, (iv) To evaluate I =, dz, we consider, |z|=π 1 − cos z, f (z) =, , ez − 1, ez − 1, =, 1 − cos z, 2 sin2 (z/2), , and note that, , lim z, , z→0, , ez − 1, 2 sin2 (z/2), , , , , = lim 2, z→0, , z/2, sin(z/2), , 2, , ez − 1, = 2., z, , Thus, z = 0 is a simple pole for f (z). Note also that f (z) has no other, singularity inside+the circle |z| = π. Hence, I = 4πi., sin z, (v) To evaluate I =, dz, we may rewrite the integral as, |z|=π 1 − cos z, +, 2 sin(z/2) cos(z/2), I=, dz, 2 sin2 (z/2), |z|=π, +, cos(z/2), dz, =, |z|=π sin(z/2), , , cos(z/2), = 2πiRes, ;0, sin(z/2), cos(z/2), = 2πi lim z, z→0 sin(z/2), = 4πi.
Page 319 :
310, , 9 Laurent Series and the Residue Theorem, , * 2π iθ, (vi) To evaluate I = 0 ee −inθ dθ for n ∈ Z, we first rewrite it in the form, +, ez dz, (z = eiθ , dz = iz dθ), I=, n, |z|=1 z iz, +, ez, 1, =, dz, i |z|=1 z n+1, %, 0 if n = −1, −2, −3, . . . (by Cauchy’s theorem), = 1 2πi, •, if n = 0, 1, 2, . . . (by Cauchy’s integral formula)., i n!, Consider now the following generalization of (9.13)., Theorem 9.24. (Residue Theorem) Suppose f (z) is analytic inside and on, a simple closed contour C except for isolated singularities at z1 , z2 , z3 , . . . , zn, inside C. Then, +, n, , f (z) dz = 2πi, Res [f (z); zk ]., C, , k=1, , Proof. About each singularity zk construct a circle Ck contained in C and, such that Cj ∩ Ck = ∅ when j = k (see Figure 9.3). By Cauchy’s integral, formula for multiply connected domains,, +, +, +, +, f (z) dz =, f (z) dz +, f (z) dz + · · · +, f (z) dz,, C, , C1, , C2, , Cn, , where the integration along each interior contour is counterclockwise. Setting, C = Ck in (9.13), we see that, +, 1, (k), a−1 :=, f (z) dz = Res [f (z); zk ], 2πi Ck, for each k, and the result follows., As a matter of fact, Cauchy’s integral formula is a special case of the, residue theorem. To see this, we suppose that f (z) is analytic inside and on a, , Cn, C3, C2, , C4, C, , C1, Figure 9.3.
Page 320 :
9.3 Evaluation of Real Integrals, , 311, , simple closed contour C containing z0 . Then g(z) = f (z)/(z −z0 ) has a simple, pole at z0 provided that f (z0 ) = 0. The residue of g(z) at z0 is given by, Res [g(z); z0 ] = lim (z − z0 )g(z) = f (z0 ),, z→z0, , and so, , +, , +, g(z) dz =, , C, , C, , f (z), dz = 2πif (z0 )., z − z0, , Thus, the Cauchy integral formula is a special case of the residue theorem., Next, suppose f (z) has a pole of order k at z = z0 . To find the residue, a−1 in terms of f (z), by Laurent’s series, we write, (z − z0 )k f (z) = a−k + a−k+1 (z − z0 ) + · · ·, +a−1 (z − z0 )k−1 + g(z)(z − z0 )k ,, , (9.14), , where g(z) is analytic at z = z0 . Differentiating (9.14) k − 1 times and evaluating at z = z0 , we get the following result., Theorem 9.25. (Residue at a pole of order k) If f (z) has a pole of order, k at z = z0 , then, Res [f (z); z0 ] =, , 1, dk−1, lim, (z − z0 )k f (z)., (k − 1)! z→z0 dz k−1, , (9.15), , In particular, if f has a simple pole at z0 , then, Res [f (z); z0 ] = lim (z − z0 )f (z)., z→z0, , The following special case is particularly useful., Theorem 9.26. (Residue at a simple pole) Let f (z) and g(z) be analytic, at z0 . If g(z) has a simple pole at z0 and f (z0 ) = 0. Then, we have, , , , , f (z), 1, f (z0 ), 1, ; z0 = , and Res, ; z0 = , ., Res, g(z), g (z0 ), g(z), g (z0 ), Proof. By hypothesis f (z)/g(z) has a simple pole at z0 . Consequently (as, g(z0 ) = 0 and g (z0 ) = 0), by Theorem 9.25, , , , , f (z), z − z0, f (z0 ), ; z0 = lim, ., Res, f (z) = , z→z0, g(z), g(z) − g(z0 ), g (z0 ), To illustrate the use of (9.15), we provide a couple of more examples., *, Examples 9.27., (i) To evaluate I = |z|=1 |z − a|−4 |dz| for a > 1, we may, first rewrite
Page 321 :
312, , 9 Laurent Series and the Residue Theorem, , dθ, |dz|, =, (z = eiθ ⇒ |dz| = |iz dθ| = d θ), 4, |z − a|, (z − a)2 (z − a)2, dz, 1, =, (z − a)2 (1/z − a)2 iz, dz, z2, =, 2, 2, (z − a) (1 − az) iz, z/(z − a)2, dz,, = 2, ia (z − 1/a)2, and therefore, since a > 1,, +, z/(z − a)2, 1, I= 2, dz, ia |z|=1 (z − 1/a)2, &, , , 1, z, d, = 2 2πi, ia, dz (z − a)2, 2, , a +1, = 2π, ., (a2 − 1)3, , ', z=1/a, , (ii) If f (z) = (z 2 + a2 )−n for some a > 0 and n ∈ N, then the singularities, of f (z) are given by, z 2 + a2 = 0,, , i.e., z = ±ia., , Clearly, z = ±ia are poles of order n for f (z). If n = 1, then, Res[f (z); ia] =, , 1, ., 2ia, , For n > 1, the residue is given by, dn−1, 1, lim, ((z − ia)n f (z)), (n − 1)! z→ia dz n−1, , , 1, dn−1, 1, =, lim, (n − 1)! z→ia dz n−1 (z + ia)n, , , −n(−n − 1) · · · (−n − (n − 2)), 1, =, lim, (n − 1)! z→ia, (z + ia)n+n−1, , , (−1)n−1 n(n + 1) · · · (2n − 2), 1, =, (n − 1)!, (2ia)n+n−1, , Res [f (z); ia] =, , (2n − 2)! i2n−2−(2n−1), ((n − 1)!)2 (2a)2n−1, (2n − 2)!, =−, ., ((n − 1)!)2 (2a)2n−1, , =, , •
Page 322 :
9.3 Evaluation of Real Integrals, , 313, , Suppose that z = ∞ is an isolated singularity of f (z). Then f (z) is analytic, in a deleted neighborhood of z = ∞ and so, by Laurent’s theorem, we may, write, ∞, , f (z) =, an z n (δ < |z| < ∞),, n=−∞, , for some δ > 0. Choose R > δ and let γ be the circle of radius R centered at, 0, which is traversed in the clockwise direction, so that the point *at infinity is, to the left as in the case of finite isolated singularity. Note that γ z n dz = 0, *, for n = −1 and γ z −1 dz = −2πi. Thus, because of the uniform convergence, on |z| = R, we have, 1, 2πi, , +, f (z) dz =, γ, , +, ∞, 1 , an z n dz = −a−1 ., 2πi n=−∞, γ, , Therefore, we define the residue of f (z) at z = ∞ as, +, +, 1, 1, Res [f (z); ∞] =, f (z) dz = −, f (z) dz = −a−1, 2πi γ, 2πi |z|=R, ∞, where R > δ. Also, as f (z) = n=−∞ an z n is analytic for |z| > R iff f (1/z) =, , ∞, −n, is analytic for 0 < |z| < 1/R, we have, n=−∞ an z, a−1 = coefficient of 1/z in, , = Res, , 1, f, z2, , , 1, ;0, z, , 1, f, z2, , , ∞, , 1, an, ,, =, n+2, z, z, n=−∞, , and hence,, , , Res [f (z); ∞] = −Res, , 1, f, z2, , 0 < |z| < 1/R,, , , 1, ;0 ., z, , For instance, if f (z) = 1 − 1/z for 0 < |z| < ∞, then, g(z) = f (1/z) = 1 − z and (1/z 2 )f (1/z) = z −2 − z −1 ,, showing that g(z) has a removable singularity at the origin. In other words,, f (z) has a removable singularity at the point at infinity, and Res [f (z); ∞] = 1., Note also that z = 0 is the only singularity of f (z) in C and is a simple pole, with Res [f (z); 0] = −1. Thus,, Res [f (z); 0] + Res [f (z); ∞] = 0, which is a demonstration for the following result.
Page 323 :
314, , 9 Laurent Series and the Residue Theorem, , Theorem 9.28. (Residue Theorem for C∞ ) Suppose f (z) is analytic in C∞, except for isolated singularities at z1 , z2 , z3 , . . . , zn , ∞. Then the sum of its, residues (including the point at infinity) is zero. That is,, Res [f (z); ∞] +, , n, , , Res [f (z); zk ] = 0., , k=1, , Proof. Choose R large enough so that all the isolated singularities in C are in, |z| < R. Then, by Theorem 9.24,, 1, 2πi, , +, f (z) dz =, |z|=R, , n, , , Res [f (z); zk ]., , k=1, , But the integral on the left is −Res [f (z); ∞], and the result follows., Example 9.29. Consider the evaluation of the integral, +, 1, 1, (n ∈ N)., f (z) dz, f (z) =, I=, 2πi |z|=2, (z − 3)(z n − 1), n, By Residue theorem, I = k=1 Res [f (z); zk ] where zk ’s are nothing but the, nth roots of unity. However, by the residue theorem for the extended complex, plane, we must have, n, , , Res [f (z); zk ] = −{Res [f (z); 3] + Res [f (z); ∞]}., , k=1, , We note that Res [f (z); 3] = limz→3 (z − 3)f (z) = (3n − 1)−1 and, , , , , z n−1, 1, 1, ;, 0, = 0., ; 0 = −Res, Res [f (z); ∞] = −Res 2 f, z, z, (1 − 3z)(1 − z n ), , •, , Hence, I = −1/(3n − 1)., , Example 9.30. We illustrate Theorem 9.28 by finding residues at all singularities of, z n e1/z, f (z) =, , n ∈ N., 1+z, This function has a simple pole at z = −1 and has an essential singularity at, z = 0. Therefore, Res [f (z); −1] = (−1)n /e. If we let w = z −1 we obtain, f (z) = f (1/w) =, , ew, wn−1 (1, , + w), , , 0 < |w| < 1., , (9.16), , This implies that z = ∞ is a pole of order n − 1 for f (z). Since z = 0 is an, essential singularity of f (z), we must rely on our ability to find the Laurent, series expansion of f (z) around zero. Thus we form
Page 324 :
9.3 Evaluation of Real Integrals, , , f (z) =, , ∞, , , , k n+k, , (−1) z, , k=0, , ∞, , z −m, m!, m=0, , 315, , , , 0 < |z| < 1., , Collecting the terms involving 1/z (use Cauchy product of two convergent, series), we have, ∞, , (−1)k−1, Res [f (z); 0] = a−1 =, ., (n + k)!, k=1, , Next we determine the residue at z = ∞. For this, using (9.16), we write, f (1/w), w2, ew, = n+1, w, (1 + w), ', &∞, '& ∞, , wm, −n−1, k k, , 0 < |w| < 1., =w, (−1) w, m!, m=0, , F (w) =, , k=0, , Again collecting the terms involving 1/w (again use Cauchy product of two, convergent series), we have, (−1)n−k, (−1)n−1, (−1)0, (−1)n, +, + ··· +, =, ., 0!, 1!, n!, k!, n, , Res [F (w); 0] =, , k=0, , Therefore, by the definition of the residue at ∞, we find that, Res [f (z); ∞] = −Res [F (w); 0] = −, , n, , (−1)n−k, k=0, , k!, , ., , We see that Res [f (z); 0] + Res [f (z); −1] + Res [f (z); ∞] = 0., , •, , Armed with several ways to determine residues, we turn now to an important application, that of evaluating a real integral by integrating a complex, function along a simple closed contour. The usual method involves a complex, function that is real on the real axis. Then a real interval is one of the smooth, curves that make up the contour along which we integrate. Recall that the, *∞, *R, improper integral a f (x) dx is defined to be limR→∞ a f (x) dx if this limit, exists., Example 9.31. We wish to use contour integration to show that, + ∞, dx, = π., 2+1, x, −∞, The complex function f (z) = 1/(z 2 + 1) has singularities at z = ±i. Let C be, the contour consisting of the real axis from −R to R (R > 1) followed by the
Page 325 :
316, , 9 Laurent Series and the Residue Theorem, , Figure 9.4., , semicircle in the upper half-plane (see Figure 9.4). Then the only singularity, of f (z) inside C is at z = i, and its residue is, lim (z − i)f (z) =, , z→i, , Hence, , +, C, , 1, ., 2i, , 1, dz, = 2πi = π, z2 + 1, 2i, , and the value of this integral is independent of R, R > 1. Also,, + R, + π, +, dz, dx, iReiθ, =, +, dθ., 2, 2, 2 2iθ + 1, C z +1, −R x + 1, 0 R e, Observe that, , +, 0, , π, , iReiθ, dθ ≤, 2, R e2iθ + 1, , +, 0, , π, , πR, R, dθ = 2, ., −1, R −1, , R2, , (9.17), , (9.18), , (9.19), , In view of (9.19), the second integral on the right of (9.18) approaches 0 as, R → ∞. Thus, +, + ∞, dz, dx, lim, =, ,, 2+1, R→∞ C z 2 + 1, x, −∞, and the result follows from (9.17)., , •, , In evaluating a real integral by contour integration, an appropriate complex function and an appropriate contour must be chosen. In Example 9.31,, the choice of the complex function was easy. That this is not always the case, will be seen shortly. The reader should verify that the desired result in Example 9.31 could also have been obtained using the contour consisting of the, real axis from −R to R followed by the semicircle in the lower half-plane. The, technique of Example 9.31 can be adopted to evaluate integrals of the form, + ∞, p(x), I=, dx,, −∞ q(x), where p(x) and q(x) are polynomials such that
Page 326 :
9.3 Evaluation of Real Integrals, , (i), (ii), (iii), , 317, , q(x) = 0 for x ∈ R, p(x) and q(x) have real coefficients, deg q(x) ≥ deg p(x) + 2., , In view of these assumptions, if we proceed exactly as in Example 9.31, it, follows that the integral over the semicircular contour, ΓR = {z = Reiθ : 0 ≤ θ ≤ π}, in the upper half-plane approaches zero as R → ∞. Consequently, the value, of the integral I is 2πi times the sum of the residues evaluated at those, singularities which lie in the upper half-plane., The same contour may be used to evaluate integrals of the form, + ∞, + ∞, p(x), p(x), cos(ax) dx and, sin(ax) dx (a > 0),, −∞ q(x), −∞ q(x), where p and q are as above. One may weaken the condition (iii) described, above by replacing it by, deg q(x) ≥ deg p(x) + 1,, with a slightly different argument (as we shall see in some examples below)., The integrals of this form are encountered in applications of Fourier analysis,, and so they are often referred to as a special case of Fourier integrals., Clearly, to use the semicircular contour, we cannot start with, f (z) =, , p(z), p(z), cos(az) or, sin(az), q(z), q(z), , (a > 0), , because both cos z and sin z grow faster than polynomials along the imaginary, axis. The trick is to start with, f (z) =, , p(z) iaz, e, q(z), , and recover the cosine and sine integrals at the end by taking real and imaginary parts, respectively. Note that for Im z ≥ 0 and a > 0,, |eiaz | = |eia(x+iy) | = e−ay ≤ e0 = 1,, so that eiaz is bounded by 1 for all z in the upper half-plane {z : Im z ≥ 0}., Note that for a < 0, eiaz is bounded on the lower half-plane {z : Im z ≤ 0}, but not on the upper half-plane. In this situation either one has to choose the, lower half-plane or start with, p(z) −iaz, e, ., q(z), As another example, we next show that
Page 327 :
318, , 9 Laurent Series and the Residue Theorem, , +, , ∞, , −∞, , πe−a, cos(ax), dx =, 2, 2, x +m, m, , for a, m > 0., , Consider f (z) = eiaz /(z 2 + m2 ) and C is same contour as above, namely, C is, the boundary of the semi-disk in the upper half-plane bounded by the interval, [−R, R] on the real axis and the semicircular contour ΓR of radius R (large, enough to enclose im inside C) in the upper half-plane. Note that f (z) has, only one simple pole inside C at z = im with, Res [f (z); im] = lim (z − im), z→im, , z2, , e−a, eiaz, eia(im), =, ., =, 2, +m, 2im, 2im, , As usual,, +, , +, C, , f (x) dx +, −R, , , , +, , R, , f (z) dz =, , f (z) dz = 2πi, ΓR, , e−a, 2im, , , =, , πe−a, ., m, , (9.20), , As |eiaz | = e−ay ≤ e0 = 1 for Im z = y ≥ 0, the M L-estimate yields, +, +, πR, eiaz, f (z) dz =, dz ≤ 2, → 0 as R → ∞., 2, 2, z, +, m, R, − m2, ΓR, ΓR, Consequently, passing to the limit R → ∞ in (9.20) shows that, + ∞, πe−a, eiax, ., dx, =, 2, 2, m, −∞ x + m, Equating real parts, we have, + ∞, + ∞, πe−a, πe−a, cos ax, cos ax, ,, i.e.,, ., dx, =, dx, =, 2, 2, m, x2 + m2, 2m, −∞ x + m, 0, Note also that if we take the imaginary part of the integral, we get, + ∞, sin ax, dx = 0 (a, m > 0)., 2 + m2, x, −∞, For our next example, we need the following result., Lemma 9.32. If 0 < θ ≤ π/2, then sin θ ≥ (2/π)θ., Proof. Geometrically, the result is clear because the graph of sin θ lies above, the line segment connecting (0, 0) and (π/2, 1)., Alternatively, if we set f (θ) = (sin θ)/θ then, as f (π/2) = 2/π, it suffices, to show that f (θ) is a decreasing function in the interval [0, π/2], where we, define f (0) = limθ→0 f (θ) = 1. An application of the mean-value theorem, yields
Page 328 :
9.3 Evaluation of Real Integrals, , 319, , θ cos θ − sin θ, cos θ − (sin θ)/θ, =, 2, θ, θ, cos θ − cos ξ, (0 < ξ < θ)., =, θ, , f (θ) =, , Since the cosine is a decreasing function in the interval [0, π/2], f (θ) < 0 and, the result follows., Example 9.33. We wish to show that, + ∞, π, sin x, dx = ., x, 2, 0, Our first inclination is to integrate (sin z)/z along the same contour as in the, previous example. This does not work for two reasons. First, (sin z)/z has a, singularity at z = 0 and we can not usually integrate along a path that passes, through a singularity point. But the singularity is removable; so this difficulty, can be overcome. Second, and more important as was indicated earlier, the, integral of (sin z)/z along the semicircle does not approach a finite limit as, the radius tends to infinity, because for z = iR one sees that, sin(iR), e−R − eR, = lim, → ∞ as R → ∞., R→∞, R→∞, iR, 2i2 R, lim, , We will consider the function eiz /z, whose imaginary part on the real, axis is (sin x)/x. Our contour C will consist of the real axis from to R, the, semicircle in the upper half-plane from R to −R, the real axis from −R to, − , and the semicircle in the upper half-plane from − to (see Figure 9.5)., The function eiz /z is analytic inside and on C, so that, + iz, e, dz, 0=, C z, + π iReiθ, + 0 i eiθ, + − ix, + R ix, e, e, e, e, iθ, dx +, dx, +, iRe, dθ, +, i eiθ dθ, =, iθ, iθ, x, Re, x, e, 0, −R, π, + π, + π, + R ix, iθ, iθ, e − e−ix, dx + i, eiRe dθ − i, ei e dθ, =, x, 0, 0, , 0, Figure 9.5.
Page 331 :
322, , 9 Laurent Series and the Residue Theorem, , and so, I = πe−am /2., The next integral is found by methods from calculus. It will be used in, conjunction with contour integration to evaluate a different real integral., + ∞, √, 2, e−x dx = π/2., Lemma 9.34., 0, , *R, , 2, , Proof. Set I = 0 e−x dx. Then, +, +, +, R, R, 2, −x2, −y 2, e, dx, e, dy =, I =, 0, , 0, , R+ R, , 0, , e−(x, , 2, , +y 2 ), , dxdy., , 0, , Here we are integrating along a square S in the first quadrant whose sides, have length R. Let C1 and C2 be the quarter, √ circles in the first quadrant, centered at the origin having radii R and R 2, respectively (see Figure 9.7)., Evaluating along the circles in polar coordinates, we have, +, , π/2 + R, , −r 2, , e, 0, , R+ R, , +, r dr dθ <, , 0, , −(x2 +y 2 ), , e, 0, , or, 2, π, (1 − e−R ) <, 4, , +, dxdy <, , 0, , 0, , +, , R, , 2, −x2, , e, , dx, , <, , 0, , ∞, , 2, , e−x dx, , 2, , 0, , and the result follows., , Figure 9.7., , =, , 0, , 2, π, (1 − e−2R )., 4, , Letting R → ∞, we see that, +, , √, π/2 + R 2, , π, ,, 4, , 2, , e−r r dr dθ,
Page 333 :
324, , 9 Laurent Series and the Residue Theorem, , +, , ∞, , +, , 2, , eix dx = eπi/4, , 0, , ∞, , 2, , e−t dt., , 0, , Applying Lemma 9.34, we get, /, /, √, + ∞, + ∞, 1 π, i π, 2, 2, πi/4 π, =, +, ., cos x dx + i, sin x dx = e, 2, 2 2, 2 2, 0, 0, The result now follows upon equating real and imaginary parts., , •, , Our next example involves integrals of trigonometric functions. If z traverses the unit circle |z| = 1, then we may parameterize z by, z = eiθ, , (0 ≤ θ ≤ 2π)., , The identities, sin θ =, , z − 1/z, eiθ + e−iθ, z + 1/z, eiθ − e−iθ, =, , and cos θ =, =, 2i, 2i, 2, 2, , enable us to evaluate certain integrals of the form, + 2π, g(sin θ, cos θ) dθ, 0, , by the residue theorem in the normal way., To illustrate this, we let a and b real, |a| > |b|, and show that, +, I=, 0, , 2π, , 2π, dθ, =√, ., 2, a + b cos θ, a − b2, , (9.25), , The idea is to convert this into a contour integral around the unit circle. First,, we observe that there is nothing to prove if b = 0. For b = 0, we may rewrite, I as, +, +, 1 2π, dθ, dθ, 1 2π, I=, =, b 0 a/b + cos θ, b 0 α + cos θ, where α = a/b ∈ R with |α| > 1. Thus, it suffices to deal with the case b = 1;, i.e., to evaluate, + 2π, dθ, J=, , for α real with |α| > 1., α, +, cos θ, 0, Now, setting z = eiθ , we see that dz = ieiθ dθ = iz dθ. Thus, + 2π, +, +, 2, dθ, 1, dz, =, =, f (z) dz,, α + cos θ, i C, C α + (1/2)(z + 1/z) iz, 0, where C is the unit circle |z| = 1 and, , (9.26)
Page 338 :
9.3 Evaluation of Real Integrals, , 329, , Questions 9.36., 1. If f has a removable singularity at ∞, is Res [f (z), ∞] = 0?, 2. If Res *[f (z), ∞] = 0, does f have a removable singularity at z = ∞?, 3. Since C sin(1/z 2 ) dz = 0 along any simple closed contour containing, the origin, why is sin(1/z 2 ) not analytic?, 4. Is the residue theorem valid for non-isolated singularities?, 5. For what kinds of functions will we be able to evaluate complex integrals, by means of the residue theorem but not Cauchy’s integral formula?, 6. In evaluating real integrals by contour integration, what general criteria, do we have for choosing the proper complex function and the proper, contour?, 7. Why does the residue theorem not hold for multiple-valued functions?, 8. Why is it easier to evaluate integrals of the form, + ∞, 1, dx, 1, +, x2n, −∞, than integrals of the form, +, , ∞, , −∞, , 1, dx?, 1 + x2n+1, , Exercises 9.37., 1. Determine the residue at each singularity for the following functions., z, 1, 1, (a), (b), (c) z n cos, cos z +, (z − 1)2 (z − 2), z, 2. Show that, |z|=R, , |(sin z)/z| |dz| → ∞ as R → ∞., , 3. Let f be analytic on an open set D, and f (a) = 0 for some a ∈ D., Show that, +, 2πi, dz, = , f (a), C f (z) − f (a), where C is a sufficiently small circle centered at a., 4. Evaluate the following integrals along different simple closed curves not, passing through 0 and ±1., +, +, ez − 1, ez, dz, (ii), dz., (i), 2, 2, 2, C z (z − 1), C z (1 − z ), 5. Evaluate, the following integrals., +, sin z, (a), dz, where n is the integer nearest, 2, n, |z|=1/2 1 + z + z + · · · + z, your age, +, 2, ez π cot πz dz., (b), |z|=5/2
Page 339 :
330, , 9 Laurent Series and the Residue Theorem, , 6. For φ ∈ (0, π) and n ∈ N, show that, +, sin nφ, zn, 1, ., dz =, 2πi |z|=2 1 − 2z cos φ + z 2, sin φ, 7. For each integer n, evaluate, +, +, tan πz dz, (b), (a), |z|=n, , cot πz dz., , |z|=n+1/2, , 8. Evaluate, +, (i), , +, ez, dz, (ii), (2z − 1)ez/(z−1) dz., 2, z(2z, +, 1), |z|=1, |z|=2, 9. Let f (z) = z 4 + 6z 2 + 13. Find the residue of z 2 /f (z) at the zeros of, f (z) = 0 which lie in the upper half-plane {w ∈ C : Re w > 0}., 10. Using the concept of the residue at the point at infinity, deduce the, fundamental theorem of algebra., 11. Use contour integration to evaluate, + ∞, + ∞, dx, x2, (b), dx, (a), 1 + x6, 1 + x4, 0, 0, + ∞, + ∞, dx, x, (c), (d), dx, 2, 2, 2, −∞ 1 + x + x, −∞ (x + 2x + 2), + ∞ 2, + ∞, cos ax, x +1, dx, dx, (f), (e), 2, 1, +, x, +, x, x4 + 1, −∞, 0, + ∞, + ∞, x2, x2, dx, (h), dx., (g), (x2 + 4)2 (x2 + 9), (x2 + 1)(x2 + 4), 0, 0, 12. Let C be the*rectangle having vertices at 0, R, R + ic, ic. By considering, 2, the integral C e−z dz, evaluate, + ∞, 2, e−x cos(2cx) dx., 0, , 13. (a) By integrating (eiz −1)/z along the contour of Figure 9.4, show that, + ∞, π, sin x, dx = ., x, 2, 0, (b) By integrating (e2iz − 1)/z 2 along the contour of Figure 9.5, show, that, + ∞, π, sin2 x, dx = ., 2, x, 2, 0, (c) By integrating (1 + 2iz − eiz )/z 2 along the contour of Figure 9.4,, show that, + ∞, π, sin2 x, dx = ., 2, x, 2, 0
Page 340 :
9.4 Argument Principle, , 331, , 14. If t = ±1 is real, show that, + 2π, 0, , 2π, dθ, ., = 2, 1 − 2t cos θ + t2, |t − 1|, , 15. Show that, + 2π, + π/2, π, cos θ, dθ, π, = √, (a), dθ = −, (b), 2, 5, +, 4, cos, θ, 3, 1, +, sin, θ, 2 2, 0, 0, + 2π, dθ, 2π, =, (a, b > 0)., (c), ab, a2 sin2 θ + b2 cos2 θ, 0, , 9.4 Argument Principle, Suppose f (z) is a nonconstant function analytic at z0 with f (z0 ) = 0. Then,, by Corollary 8.45, there exists a neighborhood of z0 that contains no other, zeros of f (z). Thus we may express f (z) as, f (z) = (z − z0 )k F (z), , (k a positive integer),, , where F (z) is analytic at z0 with F (z0 ) = 0. Thus, F (z) = 0 in the neighborhood of z0 or on its boundary C. Note that, f (z) = (z − z0 )k−1 [kF (z) + (z − z0 )F (z)],, has a zero of order k − 1 at z0 and, k, f (z), F (z), =, +, f (z), z − z0, F (z), , (9.27), , so that, at each zero of f of order k, f (z)/f (z) has a simple pole with residue, k. Thus, the residue theorem gives, +, f (z), 1, dz = k,, 2πi C f (z), the order of the zero of f (z). The expression f (z)/f (z) is called the logarithmic derivative of f (z) because it is the derivative of log f (z) at all points, where f (z) is analytic and nonzero., Next suppose that f (z) is analytic inside and on a simple closed contour, C with no zeros on C. By Theorem 8.47, f (z) has at most a finite number of, zeros inside C. Let the zeros be at z1 , z2 , . . . , zn with orders α1 , α2 , . . . , αn ,, respectively. Then, f (z) = (z − z1 )α1 (z − z2 )α2 · · · (z − zn )αn F (z),, , (9.28), , where F (z) has no zeros inside or on C. Forming the logarithmic derivative, in (9.28), we obtain
Page 341 :
332, , 9 Laurent Series and the Residue Theorem, , F (z), f (z) αj, =, ., +, f (z), z − zj, F (z), j=1, n, , (9.29), , An integration of (9.29) leads to, 1, 2πi, , +, C, , , , 1, f (z), dz =, f (z), 2πi, =, , n, , , ⎛, , +, , ⎝, , C, , , αj, , j=1, , n, , j=1, , 1, 2πi, , ⎞, 1, αj ⎠, dz +, z − zj, 2πi, , +, C, , dz, z − zj, , +, C, , F (z), dz (9.30), F (z), , , ., , For each point zj inside C, construct a circle Cj contained in C having center, at zj and containing no other zero of f (z) (see Figure 9.9). Then for each zj ,, ,, ,, 1, dz, dz, 1, =, = 1., (9.31), 2πi C z − zj, 2πi Cj z − zj, An application of (9.31) to (9.30), or a direct application of the residue theorem to (9.29) yields, 1, 2πi, , +, , , f (z), dz =, αj ., f (z), j=1, n, , C, , (9.32), , Thus, we have, Theorem 9.38. If f is analytic inside and on a simple closed contour C with, no zeros on C, then, +, 1, f (z), dz = N, 2πi C f (z), where N is the number of zeros of f (z) inside C. In determining N , zeros are, counted according to their order or multiplicities., , Figure 9.9.
Page 342 :
9.4 Argument Principle, , Thus the expression, 1, 2πi, , +, C, , 333, , f (z), dz, f (z), , may be viewed as a “counting function” for the zeros of f (z) inside C, where, a zero of multiplicity k is counted k times., In Section 9.2 we have shown that a function f (z) having a pole of order, n at z0 may be expressed as, f (z) =, , F (z), ,, (z − z0 )n, , where F (z) is analytic at z = z0 with F (z0 ) = 0. As can be seen in (9.27), if, f (z) has a zero at z0 , then f (z)/f (z) has a simple pole at z0 ., Equation (9.32) may now be generalized in the following manner:, Theorem 9.39. (Argument Principle) Let f (z) be analytic inside and on, a simple closed contour C except for a finite number of poles inside C, and, suppose f (z) = 0 on C. If Nf and Pf are, respectively, the number of zeros, (a zero of order k being counted k times) and poles (again with multiplicity), inside C, then, +, 1, f (z), dz = Nf − Pf ., 2πi C f (z), Proof. Suppose the zeros of f (z) are z1 , · · · , zn with multiplicity α1 , . . . , αn, and the poles of f (z) are w1 , . . . , wm with multiplicity β1 , . . . , βm . Then f (z), may be written as, f (z) =, , (z − z1 )α1 · · · (z − zn )αn, F (z),, (z − w1 )β1 · · · (z − wm )βm, , where F (z) is analytic with no zeros or poles inside or on C. Forming the, logarithmic derivative, we have, βj, F (z), f (z) αj, =, ., −, +, f (z), z − zj j=1 z − wj, F (z), j=1, n, , m, , Integrating (9.33), we obtain, +, +, +, n, m, , , f (z), αj, dz, βj, dz, 1, dz =, −, ., 2πi C f (z), 2πi C z − zj j=1 2πi C z − wj, j=1, , (9.33), , (9.34), , Next enclose each zero and pole of f (z) with disjoint circles containing no, other zeros or poles. Then, just as we went from (9.30) to (9.32), so may we, go from (9.34) to, +, n, m, , , f (z), 1, dz =, αj −, βj = Nf − Pf ,, (9.35), 2πi C f (z), j=1, j=1, and the proof is complete.
Page 343 :
334, , 9 Laurent Series and the Residue Theorem, , Examples 9.40. Let us now illustrate Theorem 9.39 by simple examples., (i) Suppose that f is given by, f (z) =, , (z + 1)(z + 7)5 (z − i)2, (z 2 − 2z + 2)4 (z + i)8 (z − 5i)8, , and C = {z : |z| = 2}. Then, examining the numerator of f (z) shows, that inside C, f has a simple zero at z = −1, and a zero of order 2 at, z = i. Therefore, the number N of the zeros of f inside C is, N = 1 + 2 = 3., Similarly, as z 2 − 2z + 2 = (z − 1)2 + 1 = 0 implies that z = 1 ± i, inside, C, f has a pole at z = 1 + i (order 4), z = 1 − i (order 4) and z = −i, (order 8). Thus, the number P of the poles of f inside C is, P = 4 + 4 + 8 = 16., According to Theorem 9.39,, +, f (z), dz = 2πi(N − P ) = −26πi., |z|=2 f (z), (ii) Let us use the Argument principle to evaluate, +, z+i, I=, dz, C = {z : |z + 1 + i| = 2}., 2, C z + 2iz − 4, Note that this integral may be evaluated either by the Cauchy integral, formula or the residue theorem. We rewrite I as, +, f (z), 1, dz, f (z) = (z + i)2 − 3., I=, 2 C f (z), √, Note √, that the zeros of f are given by, √ z = ± 3 − i. We observe that, −i + 3 lies outside C, while z = − 3 − i lies inside C. Consequently,, by Theorem 9.39, we have, I=, , 1, (2πi) = πi., 2, , (iii) We easily see that, +, +, 1, 1, dz, 3, =, dz = (2πi), 3 |z|=2 3z + 4, 3, |z|=2 3z + 4, , (with f (z) = 3z + 4).
Page 344 :
9.4 Argument Principle, , +, (iv) To evaluate I =, |z|=2, , z+2, dz, we may rewrite it as, z(z + 1), +, , +, I=, , 335, , 1, 2, z2 + z3, 1, 1, |z|=2 z + z 2, , dz = −, , |z|=2, , f (z), dz,, f (z), , where, , 1, z+1, 1, + 2 =, ., z, z, z2, Then, by the Argument principle,, f (z) =, , I = −2πi[Nf − Pf ] = −2πi[1 − 2] = 2πi., This can be checked by using partial fractions and Cauchy’s integral, formula:, , , +, +, +, 1, 1, z+1+1, dz, dz =, +, −, I=, dz, z, z+1, |z|=2 z(z + 1), |z|=2 z, |z|=2, •, = 2πi + 2πi − 2πi = 2πi., It seems strange indeed that (9.35) is always an integer regardless of the, function f (z) or the closed contour C. This phenomenon is based on properties, of the logarithm. Suppose that f (z) is analytic and nonzero for all z on a, simple closed contour C. Set, log f (z) = ln |f (z)| + i arg f (z),, where a fixed branch for the logarithm is chosen. Then, +, +, f (z), dz =, d(log f (z)) = ln |f (z)| + i arg f (z), C, C f (z), C, , ., , (9.36), , C, , Since the initial and terminal points of the closed contour C must coincide,, ln |f (z)| |C = 0. Hence (9.36) simplifies to, +, f (z), dz = i arg f (z) ., (9.37), C, C f (z), Thus the value of the integral depends only on the net change in the argument, of f (z) as z traverses the contour C., Now the image of the simple closed contour C under f (z) is a closed, contour C , which need not be simple. We illustrate two cases:, Case 1: If f (z) has no zero inside C, then C does not surround the origin., Therefore, arg f (z) returns to its original value as f (z) traverses the contour, C (see Figure 9.10). Let z0 ∈ C be mapped to w0 ∈ C . As z0 traverses the, contour C once in the positive direction, w0 traverses C an integer number, of times in the positive or negative direction. However, the number
Page 345 :
336, , 9 Laurent Series and the Residue Theorem, , u, , Figure 9.10., , arg f (z), z=z0, , = arg f (z0 ) = arg w0, , does not change as z0 travels once or several times along C. For example, in, Figure 9.10, arg w0 increases to arg A (up to the point A) then decreases and, when w0 returns to its initial position, arg w0 returns to its initial value. This, means that the net change in arg f (z), as z traverses the contour C, is zero., That is,, +, +, +, dw, df (z), f (z), =, dz =, dz = i arg f (z), = 0., C, Γ =f (C) w, C f (z), C f (z), Case 2: If f (z) does have zeros inside C, then C must wrap around the, origin (Why?). Each time that C winds around the origin (in the positive, sense), the argument of f (z) is increased by 2π. In Figure 9.11, we show a, simple closed contour C being mapped by f (z) = z 2 onto a closed contour C , that twice winds around the origin. When z returns to its initial point on C,, arg f (z) has increased by 4π along C ., , Figure 9.11.
Page 346 :
9.4 Argument Principle, , 337, , Note that we are not concerned with whether or not the origin is inside, the simple closed contour C. The origin only plays a critical role relative to, the image curve C . The net change in arg f (z) as z traverses a contour C is, called the variation of arg f (z) along C, and is denoted by ΔC arg f (z). This, notation allows us to write (9.37) as, +, f (z), dz = iΔC arg f (z)., C f (z), Hence the conclusion of Theorem 9.39 can be expressed as, 1, ΔC arg f (z) = Nf − Pf ., 2π, , (9.38), , Geometrically, the identity in (9.38) is known as the Argument Principle., Remark 9.41. For each zero of f (z) inside C, the curve C winds once around, the origin in the positive sense, whereas for each pole of f (z) inside C, the, curve C winds once around the origin in the negative sense. To prove this, we, need a careful definition of winding number (see Ahlfors [A] and Ponnusamy, •, [P1])., The following two lemmas are consequences of the argument principle., Lemma 9.42. Suppose f (z) and g(z) are analytic inside and on a simple, closed contour C with f (z) and g(z) having no zeros on C. Then, ΔC arg f (z)g(z) = ΔC arg f (z) + ΔC arg g(z)., Proof. Let f (z) and g(z) have N1 and N2 zeros respectively inside C. Then,, by the argument principle,, 1, ΔC arg f (z) = N1, 2π, , and, , 1, ΔC arg f (z) = N2 ., 2π, , But f (z)g(z) has N1 + N2 zeros inside C. Hence, 1, ΔC arg f (z)g(z) = N1 + N2 ,, 2π, and the proof is complete., Lemma 9.43. Suppose h(z) is analytic on a simple closed contour C with, |h(z)| < 1 for all z on C. Then ΔC arg(1 + h(z)) = 0., Proof. The simple closed contour C is mapped by w = F (z) = 1 + h(z) onto a, closed contour C contained in the disk |w−1| < 1 (see Figure 9.12). Since this, disk is in the right half-plane, we may cut the plane along the negative real, axis to obtain a branch (the principle branch) for arg F (z) as F (z) traverses, C . Thus, 1, 1, ΔC arg F (z) =, ΔC arg(1 + h(z)) = 0., 2π, 2π
Page 347 :
338, , 9 Laurent Series and the Residue Theorem, , Figure 9.12., , Remark 9.44. The function F (z) = 1+h(z) need not be analytic for z inside, C in order for a branch of log F (z) to exist for z on C. In view of the argument, principle, we have merely shown that F (z) has the same number of zeros and, poles inside C. For example, h(z) = 1/(2z) satisfies the inequality |h(z)| < 1, on the unit circle. The function, 1 + 2z, 2z, , F (z) = 1 + h(z) =, , has one simple zero (at z = −1/2) and simple pole (at the origin)., , •, , Theorem 9.45. (Rouché’s Theorem) Suppose f (z) and g(z) are analytic, inside and on a simple closed contour C, with |g(z)| < |f (z)| on C. Then, f (z) + g(z) has the same number of zeros as f (z) inside C., Proof. By hypothesis, |g(z)| < |f (z)| on C, which implies that on C, |f (z)| > 0 and |f (z) + g(z)| ≥ |f (z)| − |g(z)| > 0., Thus, f and f + g are analytic inside and on C with f and f + g having no, zeros on C. Since f and g are analytic, Pf = 0 = Pf +g . If we write, f + g = f (1 + g/f ) := f φ,, then, , , , g(z), (f + g) (z) = f (z) 1 +, f (z), , , , , so that, , , , , , , f (z), (f + g) (z), =, +, (f + g)(z), f (z), , 1+, 1+, , , , g(z), f (z), , , , g(z), + f (z) 1 +, f (z), , , g(z), f (z), , =, , , , f (z) φ (z), +, ., f (z), φ(z), , Let Nf and Nf +g denote the number of zeros of f and f + g respectively on, the domain which is bounded by C. By the Argument Principle,
Page 348 :
9.4 Argument Principle, , Nf +g − Nf =, , 1, 2πi, , + , C, , (f + g) (z) f (z), −, (f + g)(z), f (z), , , dz =, , 1, 2πi, , +, C, , 339, , φ (z), dz., φ(z), , Since |g(z)/f (z)| < 1 on C, φ(z) lies in the disc |w − 1| < 1 for z ∈ C. Taking, a branch of the logarithm on the simply connected domain |w − 1| < 1 (which, does contain the origin), we have, d, φ (z), =, log(φ(z))., φ(z), dz, Thus, the integral on the right is zero. That is, Nf +g = Nf ., Alternatively, by Lemma 9.42,, , , g(z), ΔC arg(f (z) + g(z)) = ΔC arg f (z) + ΔC arg 1 +, ., f (z), , (9.39), , Since |g(z)/f (z)| < 1 on C, Lemma 9.43 may be applied to obtain, , , g(z), = 0., ΔC arg 1 +, f (z), Hence (9.39) reduces to, 1, 1, ΔC arg(f (z) + g(z)) =, ΔC arg f (z)., 2π, 2π, , (9.40), , Then, by (9.40) and the argument principle, the theorem follows., The proof of Rouché’s theorem was “geometric” in character. We now give, an “analytic” proof., Alternate proof of Rouché’s Theorem. Let {φt (z)} be a family of functions defined by, φt (z) = f (z) + tg(z) (0 ≤ t ≤ 1)., Then, for each t, φt is analytic inside and on C having no poles inside or on, C. Also, since |f (z)| < |g(z)| on C, we have, |φt (z)| = |f (z) + tg(z)| ≥ |f (z)| − |g(z)| > 0, , for z ∈ C, , and so φt does not have a zero on C. Since |f (z)| − |g(z)| is continuous on the, compact set C, it attains a minimum, say m. Thus for all z on C,, |φt (z)| ≥ m > 0, Define, h(t) =, , 1, 2πi, , +, C, , (0 ≤ t ≤ 1)., , (9.41), , φt (z), dz., φt (z), , Observe that h(t) denotes the number of zeros of φt (z) inside C. We want to, show that h(0) = h(1). Now, given any points t1 and t2 in [0, 1], we have
Page 349 :
340, , 9 Laurent Series and the Residue Theorem, , , + , f + t2 g , f + t1 g , 1, − , dz, 2πi C f + t2 g, f + t1 g, +, 1, (t2 − t1 )(g f − f g), =, dz., 2πi C (f + t2 g)(f + t1 g), , h(t2 ) − h(t1 ) =, , (9.42), , Since f (z), g(z), f (z), and g (z) are analytic on C, they are bounded, and, we may assume that, |f (z)|, |g(z)|, |f (z)|, |g (z)| ≤ M., , (9.43), , Denote the length of C by L. Then, from (9.41), (9.42), and (9.43), we get, |h(t2 ) − h(t1 )| ≤, , 1 |t2 − t1 |2M 2, L = K|t2 − t1 |,, 2π, m2, , where K is constant independent of t1 and t2 , so that h(t) is a continuous, function on [0, 1]. Since h(t) is integer-valued, it follows (by the intermediate, value theorem) that h(t) is constant on [0, 1]. In particular, h(0) = h(1) where, h(0) and h(1) are, respectively, the number of zeros of φ0 (z) = f (z) and the, number of zeros of φ1 (z) = f (z) + g(z) inside C. The proof is complete., Remark 9.46. In Rouché’s theorem, the condition |g(z)| < |f (z)| on C cannot be relaxed to |g(z)| ≤ |f (z)|. This is seen by setting, g(z) = −f (z)., Then f (z) + g(z) ≡ 0 inside C regardless of the number of zeros of f (z)., , •, , Corollary 9.47. Let g be analytic for |z| ≤ 1 and |g(z)| < 1 for |z| = 1. Then, g has a unique fixed point in |z| < 1., Proof. For |z| = 1, |g(z)| < | − z| = 1. By Rouche’s theorem, we have g(z) − z, has exactly one zero in |z| < 1 and the conclusion follows., We ask what happens if we replace |g(z)| < 1 with |g(z)| ≤ 1 in Corollary, 9.47? If, z−a, (0 < |a| < 1),, g(z) =, 1 − az, then g is analytic for |z| ≤ 1 and |g(z)| = 1 for |z| = 1. Moreover,, g(|z| < 1) ⊂ {w : |w| < 1}, and we easily see that, , g(z) = z ⇐⇒ z 2 = a/a, , which shows that, for each a with 0 < |a| < 1, g(z) = z has no solution in, |z| < 1., Next we ask: what happens if we simply assume that f is analytic for, |z| < 1 and |f (z)| < 1 for |z| < 1? Of course, f (z) = z shows that every point
Page 350 :
9.4 Argument Principle, , 341, , of |z| < 1 is a fixed point. Suppose that f (z) ≡ z. Can f have more than one, fixed point in |z| < 1?, To answer this, we suppose that f has two fixed points, say a and b, in, |z| < 1. Consider, a−z, ., φ(z) =, 1 − az, Then, φ(a) = 0 and φ is its own inverse. Set φ(b) = α. Then α = 0, because, φ is one-to-one. Define g = φ ◦ f ◦ φ−1 and see that, g(0) = φ ◦ f ◦ φ−1 (0) = φ(f (a)) = φ(a) = 0, g(α) = φ ◦ f ◦ φ−1 (α) = φ(f (b)) = φ(b) = α., By Schwarz’s lemma, |g(z)| ≤ |z|. But, because equality is attained at an, interior point α, we have g(z) = eiη z for some real constant η. The condition, g(α) = α shows that g must be the identity function. Thus, f (z) = z which, is a contradiction., As a first application of Rouché’s theorem, we prove, Theorem 9.48. (Hurwitz’s Theorem) Let {fn (z)} be a sequence of functions analytic inside and on the simple closed contour C, and suppose {fn (z)}, converges uniformly to f (z) inside and on C. If f (z) has no zeros on C, then, the number of zeros of f (z) inside C is equal to the number of zeros of fn (z), inside C for sufficiently large n., Proof. First note that according to Theorem 8.16, the limit function f (z) is, analytic inside and on C. Let m > 0 denote the minimum of |f (z)| on C. By, the uniform convergence of {fn (z)} on C, we have for n > N (m) that, |fn (z) − f (z)| < m ≤ |f (z)|, on C. Hence by Rouché’s theorem, the number of zeros of f (z) inside C equals, the number of zeros of, f (z) + (fn (z) − f (z)) = fn (z), , (n > N )., , Rouché’s theorem furnishes us with yet another proof for the fundamental, theorem of algebra., Theorem 9.49. (Fundamental Theorem of Algebra) If, Pn (z) = a0 + a1 z + · · · + an−1 z n−1 + an z n, , (an = 0), , is a polynomial of degree n, then it has n zeros in C., Proof. Note that for z = 0,, a0 , 1 an−1, Pn (z), ., +, ·, ·, ·, +, =, 1, +, an z n, an, z, zn
Page 351 :
342, , 9 Laurent Series and the Residue Theorem, , Then for |z| = r > 1,, , , |a0 | 1, |an−1 |, Pn (z), + ··· + n, −1 ≤, an z n, r, r, |an |, |a0 | + |a1 | + · · · + |an−1 |, ≤, |an |r, < 1 if r > max{(|a0 | + · · · + |an−1 |)/|an |, 1}., That is,, |Pn (z) − an z n | < |an z n |, , for |z| = r., , Since an z n has n zeros (all at the origin) inside |z| = r, so does the function, (Pn (z) − an z n ) + an z n = Pn (z)., Remark 9.50. This proof of the fundamental theorem is more satisfying than, the previous proofs. First, we get directly that the polynomial has n roots (as, opposed to “at least one root”). Second, and more important, we get a bound, on the modulus of the roots in terms of the coefficients. We know that all the, roots lie in the disk, |z| ≤, , |a0 | + |a1 | + · · · + |an−1 |, |an |, , (|z| > 1)., , Rouché’s theorem will frequently be an aid in approximating the location of, zeros for an analytic function., •, Example 9.51. Let us use Rouche’s theorem to determine the number of, zeros of the polynomial p(z) = z 10 − 6z 9 − 3z + 1 inside the unit circle |z| = 1., To do this, we set, p(z) = f (z) + g(z),, where f (z) = −6z 9 + 1 and g(z) = z 10 − 3z. Then for |z| = 1,, |f (z)| = | − 6z 9 + 1| ≥ |6z 9 | − 1 = 6 − 1 = 5, and, , |g(z)| = |z 10 − 3z| ≤ |z|10 + 3|z| = 4 < 5 ≤ |f (z)|., , By Rouche’s theorem, f (z) and p(z) have the same number of zeros inside, |z| = 1. But f (z) has nine zeros inside the unit circle |z| = 1. Therefore, p(z), has nine zeros in |z| < 1., •, Example 9.52. Let us show that all five roots of the polynomial, P (z) = z 5 + 6z 3 + 2z + 10, lie in the annulus 1 < |z| < 3., To see this, we let f (z) = z 5 + 6z 3 + 2z and g(z) = 10. On |z| = 1,
Page 352 :
9.4 Argument Principle, , 343, , |f (z)| ≤ |z 5 | + 6|z 3 | + 2|z| = 9 < |g(z)| = 10., Hence P (z) = f (z) + g(z) has the same number of zeros in |z| < 1 as does, g(z), namely none. Observe that, |P (z)| ≥ 10 − |z 5 + 6z 3 + 2z| ≥ 10 − |z|5 − 6|z|3 − 2|z| = 1 on |z| = 1, and so P (z) can have no zeros on the unit circle |z| = 1., Next let f (z) = z 5 and g(z) = 6z 3 + 2z + 10. On |z| = 3,, |g(z)| ≤ 6(33 ) + 2(3) + 10 < |f (z)| = 35 ., Thus all the zeros of P (z) must lie in |z| < 3, that is, in the annulus 1 < |z| < 3., By setting f (z) = 6z 3 and g(z) = z 5 + 2z + 10, we can further show that, three of the roots of P (z) lie in the annulus 1 < |z| < 2 and, consequently,, •, that the other two lie in the annulus 2 < |z| < 3., Example 9.53. We easily show that all the roots of, z 5 − 3z 2 − 1 = 0, lie inside the circle |z| = 22/3 and that two of its roots lie inside the circle, |z| = 3/4. To do this we first set, f (z) = z 5, , and g(z) = −3z 2 − 1., , Then, on |z| = 22/3 , |f (z)| = |z|5 = 210/3 and, |g(z)| ≤ 3|z|2 + 1 = 3(24/3 ) + 1 < 210/3 = |f (z)|,, showing that f and f + g have the same number of zeros inside the circle, |z| = 22/3 . But f has five zeros inside |z| = 22/3 . Thus f + g and hence all the, roots of given equation lie in |z| < 22/3 ., Also, on |z| = 3/4, we have |f (z)| = |z|5 = (3/4)5 and, |g(z)| ≥ 3|z|2 − 1 = 3(3/4)2 − 1 = 11/16 > (3/4)5 = |f (z)|., It follows that g and f + g have, √ the same number of zeros in |z| < 3/4. But, g has two zeros at z = ±i/ 3 which lies inside the circle |z| = 3/4. Hence, the given equation has two roots inside |z| = 3/4. Consequently, the given, •, equation has three zeros in 3/4 ≤ |z| < 22/3 ., Example 9.54. Consider f (z) = z 2 + 7z + 12 − c. Then for |z| ≤ 1,, |z 2 + 7z + 12| = |(z + 3)(z + 4)| ≥ 2(3) = 6., Therefore, if |c| < 6, then f (z) = 0 in the unit disc |z < 1., As a final application of Rouché’s theorem, we prove, , •
Page 353 :
344, , 9 Laurent Series and the Residue Theorem, , Theorem 9.55. (Open Mapping Theorem) A nonconstant analytic function maps open sets onto open sets., Proof. Suppose f (z) is analytic at z = z0 . We must show that the image of, every sufficiently small neighborhood of z0 in the z plane contains a neighborhood of w0 = f (z0 ) in the w plane. Choose δ > 0 such that the function, f (z) − w0 is analytic in the disk |z − z0 | ≤ δ and contains no zeros on the, circle |z − z0 | = δ. This is possible in view of Corollary 8.48. Let m be the, minimum value of |f (z) − w0 | on the circle |z − z0 | = δ. We will show that the, image of the disk |z − z0 | < δ under f (z) contains the disk |w − w0 | < m (see, Figure 9.13)., , y, , f (z), , v, w1, , z0, , m w0, , δ, x, , u, Figure 9.13., , Choose w1 in the disk |w − w0 | < m. Then on the circle |z − z0 | = δ we, have, |w0 − w1 | < m ≤ |f (z) − w0 |., Hence by Rouché’s theorem,, (f (z) − w0 ) + (w0 − w1 ) = f (z) − w1, has the same number of zeros in |z − z0 | < δ as does f (z)− w0 . Since f (z)− w0, has at least one zero (at z0 ), the function f (z) − w1 has at least one zero., That is, f (z) = w1 at least once. Since w1 is arbitrary, the image of the disk, |z − z0 | < δ must contain all points in the disk |w − w0 | < m., Corollary 9.56. A nonconstant analytic function maps a domain onto a domain., Proof. Recall that a domain is an open connected set. In view of the theorem,, we need only show that an analytic function maps connected sets onto connected sets. But this follows from Exercise 2.46(13) since an analytic function, is continuous., The open mapping theorem provides a short proof of the maximum modulus theorem.
Page 354 :
9.4 Argument Principle, , y, , 345, , v, f(z ' ), , δ z', z0, , f(z 0), 0, , x, , u, , Figure 9.14., , Suppose f (z) is analytic in a domain D and that z0 is a point in D. If, f (z) is not constant, then the image of some disk |z − z0 | < δ contains a, disk |w − f (z0 )| < m in the w-plane. If f (z0 ) = Reiθ0 , then to each , that, f (z ) = (R + )eiθ0 (see Figure 9.14). Thus,, |f (z )| = R + > |f (z0 )| = R,, so that z0 is not a maximum for |f (z)|., We end the section with the following corollary which has been proved, earlier by a different method (see Theorem 5.37)., Corollary 9.57. If f is analytic in a domain D and if any one of Re f , Im f ,, |f |, or Arg f is constant, then f is also constant., Proof. By hypothesis, f (D) would be a subset of either the real axis, or imaginary axis, or a circle or a line with constant argument, respectively. Note, that none of them forms an open set. The conclusion follows from the open, mapping theorem., Questions 9.58., 1. Can f (z) be analytic in a deleted neighborhood of z0 even when the, limit limz→z0 (z − z0 )n f (z) does not exist for any integer n?, 2. What is the significance of the constant 2πi?, 3. How do the properties of ΔC arg f (z) and log f (z) compare?, 4. Can Rouché’s theorem be used to locate the quadrants of zeros for an, analytic function?, 5. Can Rouché’s theorem be extended to the case when there are poles, inside the contour?, 6. Does a nonconstant continuous function map open sets onto open sets?, 7. Does an analytic function map closed sets onto closed sets?, *, , (z), 8. Let f be an entire function such that |z|=R ff (z), dz = 0 for all R > 200., Is f a constant?
Page 355 :
346, , 9 Laurent Series and the Residue Theorem, , Exercises 9.59., 1. If f (z) is analytic inside and on the simple closed contour C, and f (z) =, a on C, show that the number of times f (z) assumes the value a inside, C is given by, +, 1, f (z), dz., 2πi C f (z) − a, 2. Let f (z) be analytic inside and on a simple closed contour C except for, a finite number of poles inside C. Denote the zeros by z1 , . . . , zn (none, of which lies on C) and the poles by w1 , . . . , wm . If g(z) is analytic, inside and on C, prove that, 1, 2πi, , +, g(z), C, , n, m, , , f (z), dz =, g(zj ) −, g(wj ),, f (z), j=1, j=1, , where each zero and pole occurs as often in the sum as is required by, its multiplicity., 3. If P (z) = a0 + a1 z + · · · + an z n , evaluate, +, zP (z), 1, dz, 2πi |z|=R P (z), for large values of R., 4. Using the argument principle, prove the Fundamental Theorem of Algebra., 5. If f (z) is analytic at z0 , show that f (z) has a zero of order k at z0 if, and only if 1/f (z) has a pole of order k at z0 ., 6. If f (z) is analytic and nonzero in the disk |z| < 1, prove that for 0 <, r<1, , , + 2π, 1, exp, log |f (reiθ )| dθ = |f (0)|., 2π 0, 7. Show that the polynomial z 4 + 4z − 1 has one root in the disk |z| < 1/3, and the remaining three roots in the annulus 1/3 < |z| < 2., 8. Find the number of roots of the equation z 4 − 8z + 10 = 0 in the unit, disk |z| < 1 and in the annulus 1 < |z| < 3, respectively., 9. Show that there exists one root in |z| < 1, and three roots in |z| < 2 for, the equation z 4 − 6z + 3 = 0., 10. If a > e, show that the equation ez = az n has n roots inside the unit, circle. When n = 2, show that both roots are real., 11. If a > 1, prove that f (z) = z + e−z takes the value a at exactly one, point in the right half-plane., 12. Show that the equation z 3 + iz + 1 = 0 has neither a real root nor a, purely imaginary root., 13. Show that the number of roots of the equation z 4 − 6z + 1 = 0 in the, annulus 1 < |z| < 2 is 3.
Page 356 :
9.4 Argument Principle, , 347, , 14. Let F1 (z) = z 5 + z + 16, F2 (z) = z 7 − 5z 3 − 12 and F3 (z) = z 7 + 6z 3 + 12., Determine whether all zeros of these functions lie in the annulus 1 <, |z| < 2., 15. Show that the polynomial z 3 − z 2 + 4z + 5 has all its roots in the disk, |z| < 3., 16. How many roots of the equation z 4 + z 3 + 1 = 0 have modulus between, 3/4 and 3/2?, 17. Show that, however small R, all the zeros of the function, 1+, , 1, 1, 1, +, + ··· +, z, 2!z 2, n!z n, , lie in the disk |z| < R, if n is sufficiently large., 18. Suppose that f (z) is analytic for |z| ≤ 1 such that |f (z)| < 1 for |z| = 1., Show that f (z) = z n has exactly n solutions in |z| < 1., 19. Suppose {fn (z)} is a sequence of analytic functions that converge uniformly to f (z) on all compact subsets of a domain D. Let fn (zn ) = 0, for every n, where each zn belongs to D. Show that every limit point of, {zn } that belongs to D is a zero of f (z)., 20. Suppose that f (z) is analytic at z0 and that f (z) − f (z0 ) has a zero, of order n at z0 . Show that there exist neighborhoods N (z0 ; δ) and, N (f (z0 ); ) such that each point in N (f (z0 ); ) is the image of at least, one and at most n distinct points in N (z0 ; δ)., 21. Suppose f (z) is analytic at z0 with f (z0 ) = 0. Show that there exists, an analytic function g(z) such that f (g(z)) = z in some neighborhood, of z0 . This is known as the inverse function theorem.
Page 357 :
10, Harmonic Functions, , In Chapter 5, we saw that if an analytic function has a continuous second, derivative, then the real (or imaginary) part of the function is harmonic. In, Chapter 8, it was shown that all analytic functions are infinitely differentiable, and in particular, have continuous second derivatives. Thus, the real part of, an analytic function is always harmonic., In this chapter, we examine the extent to which the converse is true. In, simply connected domains, we show that every harmonic function is the real, part of some analytic function. This result enables us to prove several theorems, for harmonic functions that are analogous to theorems for analytic functions., In particular, an analog to Cauchy’s integral formula, known as Poisson’s, integral formula, gives a method for determining the values of a harmonic, function inside a disk from the behavior at its boundary points., , 10.1 Comparison with Analytic Functions, Recall that a continuous real-valued function u(x, y), defined and single-valued, in a domain D, is said to be harmonic in D if it has continuous first and second, partial derivatives that satisfy Laplace’s equation, uxx + uyy = 0., In Section 5.3, we illustrated how the Cauchy–Riemann equations might be, used to construct a function v(x, y) conjugate to a given harmonic function, u(x, y); that is, a function v(x, y) was found for which f (z) = u(x, y) +, iv(x, y) = u(z) + iv(z) was analytic. The method entailed finding all functions v(z) satisfying the two conditions, ux = vy ,, , uy = −vx ., , *, This method was successful when the partial integration vy dy could explicitly be solved. We now give general conditions for the existence of an analytic
Page 358 :
350, , 10 Harmonic Functions, , function whose real part is a prescribed harmonic function. First note that in, view of the Cauchy–Riemann equations, the derivative of any analytic function f (z) = u(z) + iv(z) may be expressed as, f (z) = ux (z) − iuy (z)., Hence we can find f (by integration) directly from u. The details follow., Theorem 10.1. If u is harmonic on a simply connected domain D, then there, exists an analytic function on D whose real part equals u., Proof. Set g(z) = ux (z) − iuy (z) := U (z) + iV (z), z ∈ D. Then by Laplace’s, equation,, Ux − Vy = uxx − (−uyy ) = 0., , (10.1), , Since the mixed partial derivatives of u(z) are continuous in D,, Uy + Vx = (ux )y + (−uy )x = 0., , (10.2), , But (10.1) and (10.2) are the Cauchy–Riemann equations for g = U + iV ., Noting that Ux , Uy , Vx , Vy are all continuous, we may apply Theorem 5.17 to, establish the analyticity of g(z) in D., Next choose any point z0 in D, and set, + z, F (z) =, g(ζ) dζ., z0, , Then, by Corollary 8.15, F (z) is analytic in D with, F (z) = g(z) = ux (z) − iuy (z)., Observe that the derivative of F (z) may also be expressed as, F (z) =, , ∂, ∂, Re F (z) − i Re F (z)., ∂x, ∂y, , Thus u(z) and Re F (z) have the same first partial derivatives in D, so that, Re F (z) = u(z) + c, , (c a real constant)., , Hence, the function, +, , z, , (ux (ζ) − iuy (ζ)) dζ − c, , f (z) = F (z) − c =, z0, , is analytic in D with Re f (z) = u(z)., Corollary 10.2. If u is harmonic on a simply connected domain D, then, there exists an analytic function on D whose imaginary part equals u.
Page 359 :
10.1 Comparison with Analytic Functions, , 351, , Proof. By Theorem 10.1, there exists an analytic function h(z) such that, Re h(z) = u(z). But then f (z) = ih(z) is analytic with Im f (z) = Re h(z) =, u(z)., Example 10.3. Let u(x, y) = sin x cosh y + cos x sinh y + x2 − y 2 + 2xy. It can, be easily seen that u is harmonic in C. Following the proof of Theorem 10.1,, f (z) = ux − iuy = cos x cosh y − sin x sinh y + 2x + 2y, −i (sin x sinh y + cos x cosh y − 2y + 2x)., As cos(iy) = cosh y and −i sin(iy) = sinh y, we can simplify the last equation, and obtain, f (z) = (1 − i)(cos z + 2z)., Thus, f (z) = (1 − i)(sin z + z 2 ) + c., , •, , The requirement in Theorem 10.1 that the domain be simply connected is, essential. For example, the function, , u(z) = u(x, y) = ln x2 + y 2 = ln |z|, is harmonic in the punctured plane C \{0}. Each point in C \{0} has a neighborhood where log z has a single-valued analytic branch. In other words, we, say that u(z) is locally the real part of an analytic function as guaranteed, by Theorem 10.1. Therefore, u(z) = ln |z|, being the real part of an analytic, function, is harmonic in such neighborhoods. We also know that the principal, logarithm Log z defined by, Log z = ln |z| + iArg z, is analytic in the cut plane D = C \(−∞, 0]. Now if some function, f (z) = ln |z| + iv(z), were analytic throughout the punctured plane C \{0}, then g defined by, g(z) = f (z) − Log z, would be analytic in the slit plane D = C \(−∞, 0]. Since g(z) is purely, imaginary in D, an application of the Cauchy–Riemann equations shows that, g(z) must be constant in D. Thus, any function analytic in D whose real part, is ln |z| must be of the form, u(z) + iv(z) = Log z + ic,, where c is a real constant. It follows that v(z) = Arg z + c. But then, lim v(−1 + iy) = y→0, lim Arg (−1 + iy) + c = π + c, , y→0, y>0, , y>0
Page 360 :
352, , 10 Harmonic Functions, , and, lim Arg (−1 + iy) + c = −π + c, lim v(−1 + iy) = y→0, , y→0, y<0, , y<0, , which means that v is discontinuous at −1, a contradiction. An argument, similar to this shows that v is not continuous at all points in the negative, real axis (−∞, 0]. Thus, there is no hope for defining an analytic function in, C \{0} whose real part is u(z) = ln |z|. Hence, a harmonic function need not, have an analytic completion in a multiply connected domain., In view of Theorem 10.1, we may now modify some of the theorems in, Chapter 8 to obtain harmonic analogs. Our next theorem is the harmonic, analog of Liouville’s theorem., Theorem 10.4. A function harmonic and bounded in C must be a constant., Proof. Suppose u(z) is harmonic and bounded in the plane. Theorem 10.1, guarantees the existence of an entire function f (z) whose real part is u(z)., But then, g(z) = ef (z), is an entire function too. Since |g(z)| = eu(z) , g(z) is also bounded in the plane., By Liouville’s theorem g(z), and hence u(z) = ln |g(z)|, must be constant., Clearly, Theorem 10.4 may be restated in a general form as follows:, Theorem 10.5. If the real or imaginary part of an entire function is bounded, above or below by a real number M , then the function is a constant., We now prove an analog to Gauss’s mean-value theorem for analytic functions. This is one of the fundamental facts about harmonic functions, called, the mean value property of harmonic functions., Theorem 10.6. (Mean Value Property) Suppose u(z) is harmonic in a domain containing the disk |z − z0 | ≤ R. Then, u(z0 ) =, , 1, 2π, , +, 0, , 2π, , u(z0 + Reiθ ) dθ., , Proof. Let f (z) be a function analytic in |z − z0 | ≤ R whose real part is u(z)., By Gauss’s mean-value theorem,, + 2π, 1, f (z0 + Reiθ ) dθ., f (z0 ) =, 2π 0, The result follows upon taking real parts of both sides., The right-hand side of the last formula gives in particular that the mean, (or average) value u on the circle |z − z0 | = R is simply the value of u at the, center of the circle |z − z0 | = R. In Section 10.2, we shall consider a similar
Page 361 :
10.1 Comparison with Analytic Functions, , 353, , expression for a point of the disk |z − z0 | < R other than the center. We, have shown that the behavior of a harmonic function on the boundary of a, closed and bounded region determines the behavior of the harmonic function, throughout the region. For instance, a harmonic function u in the unit disk, |z| < 1 that extends continuously to |z| ≤ 1 is completely determined by, its values on the boundary |z| = 1. The explicit formula for the value of, u for each point in |z| < 1 is given by the Poisson integral formula for a, harmonic function and this is the subject of the discussion in Section 10.2., Unlike the situation for analytic functions, this result cannot be improved to, an arbitrary sequence of points in the region. For instance, the nonconstant, function u(z) = x is harmonic in the plane with u(z) ≡ 0 on the imaginary, axis. Hence, “analytic” cannot be replaced with “harmonic” in the statement, of Theorem 8.47. That is, even if u(z) is harmonic in a domain D, u(zn ) ≡ 0,, and zn → z0 in D, we are not guaranteed that u(z) ≡ 0 in D. Thus, the, analog of the identity principle (see Theorem 8.48) for analytic functions does, not hold for harmonic functions. However, we can salvage the following:, Theorem 10.7. If u(z) is harmonic in a domain D and constant in the neighborhood of some point in D, then u(z) is constant throughout D., Proof. Let A be the set of all points z0 in D for which u(z) is constant in, some neighborhood of z0 . Clearly A is a nonempty open set. To prove that, A = D, it suffices to show that B = D \A is open, for then B would have to, be empty in order for D to be connected., Suppose B is not open. Then for a point z0 in B and an > 0 there is, a point z1 in A such that z1 ∈ N (z0 ; ) ⊂ D. Since A is open, we can find, a δ > 0 sufficiently small so that N (z1 ; δ) ⊂ N (z0 ; ) ∩ A. Now construct an, analytic function f (z) such that, Re f (z) = u(z) for all z in N (z0 ; )., Since u(z) is constant in N (z1 ; δ), f (z) = 0 for z in N (z1 ; δ). An application, of Theorem 8.47 to f (z) shows that f (z) ≡ 0 throughout N (z0 ; ). Then,, by Theorem 5.9, f (z) is constant in N (zo ; ). Hence, u(z) = Re f (z) is also, constant in N (z0 ; ), contradicting the assumption that z0 ∈ B., Example 10.8. Suppose that u(z) is harmonic in a domain D such that the, set {z ∈ D : ux (z) = 0 = uy (z)} has a limit point in D. Then we can easily, show that u(z) is a constant throughout D., To see this, we define, F (z) = ux (z) − iuy (z),, , z ∈ D., , Then F is analytic in D and the set {z ∈ D : F (z) = 0} has a limit point, in D. By the uniqueness theorem for analytic functions (see Theorem 8.47),, F (z) ≡ 0 in D and so, ux (z) = 0 = uy (z) on D, i.e., u is a constant.
Page 362 :
354, , 10 Harmonic Functions, , Analogous to the maximum and minimum modulus theorems for analytic, functions are the maximum and minimum principles for harmonic functions., The fact that a harmonic function is locally the real part of an analytic function produces a number of important results. One of them is the maximum, principle., Theorem 10.9. (Maximum Principle for Harmonic Functions) A nonconstant harmonic function cannot attain a maximum or a minimum in a domain., Note that a harmonic function u(z) attains a maximum at a point z0 if and, only if the harmonic function −u(z) attains a minimum at z0 . So the minimum, principle can be derived directly from the maximum principle. This result has, several proofs., Proof. The maximum modulus theorem for analytic functions is a direct consequence of Gauss’s mean-value theorem and the fact that an analytic function, is continuous. Similarly, we may deduce the maximum principle for harmonic, functions from the mean-value principle for harmonic functions (Theorem, 10.6). Indeed, we assume that u(z) attains the maximum at z0 ∈ D. Then,, for each r with 0 < r ≤ dist (z0 , D), Theorem 10.6 gives, + 2π, 1, (u(z0 ) − u(z0 + Reiθ )) dθ = 0., 2π 0, Since u(z0 ) − u(z0 + Reiθ ) is a continuous function of θ and is nonnegative,, we have, u(z0 ) = u(z0 + Reiθ ) for 0 ≤ θ ≤ 2π., Thus, u(z) = u(z0 ) for all z in some neighborhood N (z0 ; δ). Hence, u(z) =, u(z0 ) on D (see Theorem 10.7)., For a second proof, we assume that u(z) is a nonconstant function harmonic in a domain D. Given z0 in D, construct a function f (z) = u(z) + iv(z), that is analytic in some neighborhood N (z0 ; δ) of z0 ., We set g(z) = ef (z) , and note that |g(z)| = eu(z) . If z0 were a maximum, for u(z) in this neighborhood, then z0 would be a maximum for |g(z)|. By, the maximum modulus theorem for analytic functions, the function g must, be constant on N (z0 ; δ). Therefore, u is constant on N (z0 ; δ) and hence on, D, which contradicts the assumption that u is nonconstant. The proof is, complete., Alternatively, one could use the open mapping theorem (Theorem 9.55)., Then it follows that there exists an > 0 such that N (f (z0 ); ) is contained in, the image of N (z0 ; δ) under f (z). In particular, for some point z1 ∈ N (z0 ; δ), we have Re f (z1 ) = u(z0 ) + /2. Thus, z0 is not a maximum of u(z) in D., Observe that min{|f (z)| : z ∈ D} may be attained at an interior point of, D without the analytic function f on D being constant. For example, consider, f (z) = z, for |z| < 1. Then, for |z| ≤ r (r < 1),
Page 363 :
10.1 Comparison with Analytic Functions, , 355, , |f (z)| = |z| ≥ 0 = |f (0)|, so that the minimum modulus of f (z) is attained at the interior point z = 0., However, the maximum of |f (z)| on |z| ≤ r is attained at z = r which is a, boundary point of |z| < r., The minimum principle for harmonic functions is actually stronger than, the minimum modulus theorem for analytic functions. The hypothesis that the, function be nonzero in the domain is unnecessary for harmonic functions. Of, course, a harmonic function can assume negative values in a domain, whereas, the modulus of an analytic function cannot., Corollary 10.10. Suppose u(z) is harmonic in a bounded domain D whose, boundary is the closed contour C. If u(z) is continuous in D ∪ C, with u(z) ≡, K(K a constant) on C, then u(z) ≡ K throughout D., Proof. Since D ∪ C forms a compact set, u(z) must attain a maximum and, minimum. By Theorem 10.9, the maximum and minimum cannot occur in D., Thus, they must occur on C. But this means that max u(z) = min u(z) = K., Hence, u(z) ≡ K throughout D., The boundedness of D in Corollary 10.10 is essential. The domain, {z : Re z > 0} has the boundary {z : Re z = 0}. The function u(z) = x, is continuous for Re z ≥ 0 with u(z) ≡ 0 on the boundary. But u(z) = 0 for, Re z > 0., Corollary 10.11. Suppose u1 (z) and u2 (z) are harmonic in a bounded domain D whose boundary is the closed contour C. If u1 (z) and u2 (z) are, continuous in D ∪ C, with u1 (z) ≡ u2 (z) on C, then u1 (z) ≡ u2 (z) throughout D., Proof. Set u(z) = u1 (z) − u2 (z) and apply Corollary 10.10., Example 10.12. Suppose that f (z) is an entire function such that f (z) is, real on the unit circle |z| = 1. Then f (z) is constant., To see this, we set f = u + iv. By assumption, v(z) = 0 on |z| = 1., By Corollary 10.10, v(z) = 0 for |z| < 1. Hence, f (z) is real for |z| < 1,, i.e., f (|z| < 1) ⊆ R. By the open mapping theorem, f must be constant, for |z| < 1. By the uniqueness theorem for analytic functions, f must be a, constant throughout C., There is an interesting relationship between the maximum modulus of an, analytic function and the maximum of its real part., Theorem 10.13. (Borel–Carathéodory) Suppose f (z) is analytic in the, disk |z| ≤ R. Let M (r) = max|z|=r |f (z)| and A(r) = max|z|=r Re f (z). Then, for 0 < r < R,, R+r, 2r, M (r) ≤, A(R) +, |f (0)|., R−r, R−r
Page 364 :
356, , 10 Harmonic Functions, , Proof. If f (z) is constant (say f (z) = k), then the right-hand side is bounded, below by, R+r, −2r, |k| +, |k| = |k| = M (r),, R−r, R−r, and the result follows. Hence, we may assume that f (z) is nonconstant., If f (0) = 0, then by Theorem 10.9, A(R) > A(0) = 0. Since, Re {2A(R) − f (z)} ≥ A(R) > 0, for |z| ≤ R, and, |2A(R) − f (z)|2 ≥ |f (z)|2 + 4A(R)[A(R) − Re f (z)] ≥ |f (z)|2 ,, the function, g(z) =, , f (z), 2A(R) − f (z), , is analytic and |g(z)| ≤ 1 for |z| ≤ R. Then by Schwarz’s lemma,, max |g(z)| ≤ r/R., , |z|=r, , But, |f (z)| =, , 2A(R)g(z), 2A(R)r/R, 2rA(R), ≤, =, ,, 1 + g(z), 1 − r/R, R−r, , (10.3), , and the result follows when f (0) = 0., Finally, if f (0) = 0, we apply (10.3) to f (z) − f (0). This leads to, |f (z) − f (0)| ≤, , 2r, 2r, max Re {f (z) − f (0)} ≤, (A(R) + |f (0)|)., R − r |z|=r, R−r, , Thus, |f (z)| ≤, , 2r, R+r, 2r, (A(R) + |f (0)|) + |f (0)| =, A(R) +, |f (0)|,, R−r, R−r, R−r, , and the theorem is proved., Theorem 10.13 may be used to generalize both Theorem 8.35 and Theorem, 10.4 as follows., Theorem 10.14. Suppose f (z) is an entire function and that Re f (z) ≤ M rλ, for |z| = r ≥ r0 and for some nonnegative real number λ. Then f (z) is a, polynomial of degree at most [λ]., Proof. Set R = 2r in Theorem 10.13. Then, |f (z)| ≤, , 2r + r, 2r, A(2r) +, |f (0)| ≤ 2(2r)λ M + 3|f (0)| ≤ M1 rλ ,, 2r − r, 2r − r, , for M1 sufficiently large. The result now follows from Theorem 8.35.
Page 365 :
10.1 Comparison with Analytic Functions, , 357, , Questions 10.15., 1. When can we say ln |f (z)| is harmonic? Where is it harmonic?, 2. In the proof of Theorem 10.1, where did we use the fact that the domain, was simply connected?, 3. What theorems are valid for disks but not for a simply connected domain?, 4. Where was continuity of the second partial derivatives for harmonic, functions important?, 5. Can a nonconstant function harmonic in the plane omit more than one, real value?, 6. Let f = u + iv be analytic in a domain D. Is uxx harmonic in D?, 7. Can the maximum modulus theorem for analytic functions be proved, using the maximum principle for harmonic functions?, 8. Suppose a function is harmonic in a domain D and continuous on its, boundary C. Must the function be continuous in D ∪ C?, 9. For a harmonic function u in a domain D which vanishes in an open, subset of D, does u vanish identically in D?, 10. Is there a relationship between the coefficients of an analytic function, and the maximum of its real part?, 11. Why is Theorem 10.14 a generalization of Theorem 8.35 and Theorem, 10.4?, 12. Is every harmonic function an open mapping?, 13. Let Ω be a domain and u ∈ C 3 (Ω). If u is harmonic on Ω, must ux be, harmonic on Ω? Must uy be harmonic on Ω?, Note: C k (Ω) denotes the set of all functions u whose partial derivatives, of order k all exist and are continuous on Ω., 14. What is the average value of the harmonic function u(x, y) = xy on the, circle (x − 2)2 + (y + 1)2 = 1?, 15. Let u(z) be harmonic on the disk |z| < r such that ux (z) = 0 on |z| < r., What can we conclude about u?, 16. Let u be harmonic for |z| < 1. Suppose that {zn }n≥1 is a sequence of, complex numbers not equal to z0 such that zn → z0 in |z| < 1 and, u(zn ) = 0 for n ∈ N. Must u be identically zero? If not, under what, additional assumption, do we get u ≡ 0?, 17. Must a product of two harmonic functions u and v be harmonic?, 18. Suppose that u is harmonic in a domain D and v is its harmonic conjugate. Must uv be harmonic on D? Must u2 be harmonic on D?, 19. We know that u(z) = ln |z| is harmonic in the annulus D = {z : 1 <, |z| < 2}. Can u(z) have a harmonic conjugate on D?, Exercises 10.16., 1. Show that a function harmonic in a domain must have partial derivatives, of all orders.
Page 366 :
358, , 10 Harmonic Functions, , 2. If u(z) is nonconstant and harmonic in the plane, show that u(z) comes, arbitrarily close to every real value., 3. Prove the minimum principle directly by each of the three methods in, which the maximum, principle was proved., *π, 4. Show that 0 ln sin θ dθ = −π ln 2 by applying the mean-value principle, to ln |1 + z| for |z| ≤ r < 1, and then letting r → 1., 5. Suppose f (z) and g(z) are analytic inside and on a simple closed contour, C, with Re f (z) = Re g(z) on C. Show that f (z) = g(z) + iβ inside C,, where β is a real constant., 6. Generalize the previous exercise by showing that the conclusion still, holds if it is only assumed that f (z) and g(z) are analytic inside C and, continuous in the region consisting of C and its interior., 7. If u(z) is harmonic and bounded in the punctured disk 0 < |z − z0 | < R,, show that limz→z0 u(z) exists., 8. Suppose u1 (z) and u2 (z) are harmonic in a simply connected domain, D, with u1 (z)u2 (z) ≡ 0 in D. Prove that either u1 (z) ≡ 0 or u2 (z) ≡ 0, in D., 2, , 1+z, is harmonic in the unit disk, 9. It is easy to see that u(z) = Im 1−z, , 10., 11., 12., 13., , |z| < 1 and limr→1− u(reiθ ) = 0 for all θ. Why does this not contradict, the maximum principle for harmonic functions? Is u continuous on |z| =, 1?, Does there exist a harmonic function in |z| < 1 taking the value 1, everywhere on |z| = 1? Is your solution unique?, Does there exist a harmonic function on the strip {z : 0 < Re z < 1}, with u(x, 0) = 0 and u(x, 1) = 1? Is your solution unique?, If u(z) = u(x, y) is harmonic in the plane with u(z) ≤ |z|n for every z,, show that u(z) is a polynomial in the two variables x and y., Suppose that f (z) is analytic in the disk |z| ≤ R, and let A(r) =, max|z|=r| Re f (z). Prove that for r < R,, max, , |z|=r, , 2n+2 R, |f (n) (z)|, ≤, {A(r) + |f (0)|}., n!, (R − r)n+1, , 10.2 Poisson Integral Formula, In this section, we shall attempt to find a harmonic analog to Cauchy’s integral, formula. If f is analytic inside and on a simple closed contour C, then, +, f (ζ), 1, dζ, (10.4), f (z) =, 2πi C ζ − z, at all points z inside C. We would like to find an expression for Re f at points, inside C in terms of the values of Re f on C. Unfortunately, the expression
Page 367 :
10.2 Poisson Integral Formula, , , Re, , 1, 2πi, , +, C, , 359, , f (ζ), dζ, ζ −z, , simplifies into one involving both Re f and Im f on C., If, however, the integral of (10.4) is transformed into one of the form, *b, φ(t), dt, where φ(t) is a complex-valued function of a real variable t, then, a, +, , +, , b, , φ(t) dt =, , Re, a, , b, , Re φ(t) dt., a, , Recall that we performed this kind of transformation when proving the meanvalue principle for harmonic functions. This enabled us to determine the value, of a harmonic function at the center of a circle based on its values on the circumference. By (10.4), we have the so-called mean value property for analytic, functions:, + 2π, 1, f (a + reiφ ) dφ, 0 < r < dist (a, C) = R, f (a) =, 2π 0, for a inside C. The value f (a) of f at the center a of the disk |z − a| < r, is expressed by the integration of f over the boundary circle |z − a| = r of, this disk. Note that f (a) is the same for all r in the interval (0, R). We wish, to obtain similar expression for a point of the disk |z − a| < r other than, the center. But an analog to the Cauchy integral formula for the circle is an, expression for the harmonic function at all points inside the circle in terms of, its values on the circle., Lemma 10.17. (Poisson Integral Formula for Analytic Functions) Suppose f (z) is analytic in a domain containing the closed unit disk |z| ≤ 1., Then for |z| < 1, we have, +, dζ, 1 − |z|2, 1, (10.5), f (ζ) ,, f (z) =, 2π |ζ|=1 |ζ − z|2, iζ, or equivalently,, 1, f (z) =, 2π, , +, 0, , 2π, , 1 − |z|2, f (eiφ ) dφ., |eiφ − z|2, , Proof. By Cauchy’s integral formula, we have, +, + 2π, 1, f (ζ), ζf (ζ), 1, dζ =, dφ, f (z) =, 2πi |ζ|=1 ζ − z, 2π 0 ζ − z, , (|z| < 1)., , (10.6), , (10.7), , If z = 0, the result follows from Gauss’s mean-value theorem. So we may, suppose that z = 0, and set, z ∗ = 1/z
Page 368 :
360, , 10 Harmonic Functions, , which is the reflection of z in the unit circle. The point z ∗ , which lies on the, ray from the origin through z, is outside the unit circle |ζ| = 1. Hence (as, z ∗ = 1/z), for |z| < 1, +, +, 1, f (ζ), zf (ζ), 1, dζ., (10.8), dζ = −, 0=, 2πi |ζ|=1 ζ − z ∗, 2πi |ζ|=1 1 − ζz, Subtracting (10.8) from (10.7), we get, , , +, 1, 1, z, +, f (z) =, f (ζ) dζ., 2πi |ζ|=1 ζ − z, 1 − ζz, , (10.9), , We can simplify (since |ζ| = 1) to, 1 − |z|2, 1 − |z|2, 1 − |z|2 1, z, 1, +, =, =, ., =, ζ −z, 1 − ζz, (ζ − z)(1 − ζz), |ζ − z|2 ζ, (ζ − z)(ζ − z)ζ, Using the last equality, (10.9) gives (10.5). Equation (10.6) follows if we let, ζ = eiθ in (10.5)., The following general result is a consequence of Lemma 10.17., Theorem 10.18. (Poisson Integral Formula for analytic functions) Suppose f (z) is analytic in a domain containing the closed disk |z − a| ≤ R. Then, for |z − a| < R, we have, +, dζ, R2 − |z − a|2, 1, ,, f (ζ), f (z) =, 2, 2π |ζ−a|=R, |ζ − z|, i(ζ − a), or equivalently,, f (z) =, , 1, 2π, , +, 0, , 2π, , R2 − |z − a|2, f (a + Reiφ ) dφ., |Reiφ − (z − a)|2, , (10.10), , Proof. By the change of variable w = (z − a)/R, it reduces to the case where, R = 1 and a = 0., In particular, for a = 0, the formula reduces to, + 2π, 1, R2 − r 2, f (Reiφ ) dφ., f (reiθ ) =, 2π 0 |Reiφ − reiθ |2, The expression (with ζ = Reiφ , z = reiθ and r < R), , , ζ +z, R2 − r 2, |ζ|2 − |z|2, = 2, =, Re, P (z, ζ) =, 2, |ζ − z|, ζ −z, R − 2rR cos(θ − φ) + r2, is known as the Poisson kernel for the disk |z| < R. Note that the Poisson, kernel is bounded above by
Page 371 :
10.2 Poisson Integral Formula, , 363, , Remark 10.24. We know in general (Exercise 10.16(5)) that an analytic, function is determined to within an imaginary constant by its real part. In, the case that the function f (z) = u(z) + iv(z) is analytic in the disk |z| ≤ R,, Theorem 10.22 gives this relationship explicitly., •, As we shall now see, the conclusion of Theorem 10.19 is valid under less, stringent conditions. It need not be assumed that u(z) is harmonic on the, circle |z| = R. The proof requires an acquaintance with the notion of uniform, convergence. Again it suffices to deal with R = 1, since the general case follows, from a simple transformation., Theorem 10.25. Suppose u(z) is harmonic in the open disk |z| < 1 and, continuous on the closed disk |z| ≤ 1. Then for z = reiθ , r < 1, we have, + 2π, 1, 1 − r2, iθ, u(eiφ ) dφ., u(re ) =, 2π 0 1 − 2r cos(θ − φ) + r2, Proof. Let f (z) = u(z) + iv(z) be analytic for |z| < 1, and let {tn } be an, increasing sequence of positive real numbers approaching 1. Then for each n,, define, fn (z) = f (tn z), un (z) = u(tn z), and vn (z) = v(tn z)., Clearly, vn (0) = v(0) for each n and, un (z) = Re f (tn z), and vn (z) = Im f (tn z)., As u(tn z) is harmonic in the closed disk |z| ≤ 1, we obtain that f (tn z) is, analytic in the closed disk |z| ≤ 1 (since 1/tn > 1), and so Theorem 10.21 is, applicable for fn . Thus, for each fixed z with |z| < 1,, + 2π iφ, 1, e +z, un (eiφ ) dφ + ivn (0)., fn (z) =, 2π 0 eiφ − z, Since fn (z) is continuous at z (|z| < 1) and tn z → z as n → ∞,, lim fn (z) = lim f (tn z) = f (z),, , n→∞, , n→∞, , |z| < 1., , The proof will be completed by verifying that, + 2π iφ, + 2π iφ, e +z, e +z, iφ, u (e ) dφ →, u(eiφ ) dφ., iφ − z n, e, eiφ − z, 0, 0, , (10.15), , (Recall that vn (0) = v(0)). It suffices to show that the difference, +, + 2π iφ, 1 + r 2π, e +z, iφ, iφ, (u, un (eiφ ) − u(eiφ ) dφ, (e, ), −, u(e, )), dφ, ≤, n, iφ − z, e, 1, −, r, 0, 0, can be made arbitrarily small. Note that, u(z), being continuous on the compact set |z| ≤ 1, is uniformly continuous on |z| ≤ 1. So
Page 372 :
364, , 10 Harmonic Functions, , un (eiφ ) = u(tn eiφ ) → u(eiφ ), uniformly with respect φ, 0 ≤ φ ≤ 2π. Consequently, the expression on the, last integral converges to zero as n → ∞. Thus, for |z| < 1, we have, 1, f (z) =, 2π, , +, , 2π, , 0, , eiφ + z, u(eiφ ) dφ + iv(0)., eiφ − z, , Equating the real part on both sides, we have the desired result., Remark 10.26. The uniform continuity of u(z) (|z| ≤ 1) enabled us to show, that the sequence un (z) = u(tn z) converged uniformly to u(z) (|z| ≤ 1). Thus,, •, the validity of (10.15) is a consequence of Theorem 8.11., By a simple transformation, Theorem 10.25 shows that a function, harmonic for |z| < R and continuous for |z| ≤ R, has the property that its values, inside the disk are determined by its values on the boundary. Suppose, instead, that we start with a real-valued function F (θ) continuous on the circle, |z| = R. Does there exist a function u(z) harmonic in the disk |z| < R having prescribed boundary values? More generally, the Dirichlet problem deals, with the following question: Given a domain D, and a function F : ∂D → R,, does there exist a function u that is harmonic in D such that u = F on the, boundary ∂D? The solution to this problem has immediate applications in, fluid mechanics. Our next theorem solves the Dirichlet problem for the disk., Theorem 10.27. (Schwarz’s Theorem) Let F be a continuous function of a, real variable defined on the unit circle |ζ| = 1. Then the real-valued function, u(z) defined by, u(z) =, , 1, 2π, , +, , 2π, , P (z, eiφ )F (eiφ ) dφ, , (|z| < 1), , 0, , is harmonic in the disk |z| < 1, and for each fixed t, 0 ≤ t ≤ 2π,, lim u(z) := lim− u(reiθ ) = F (eit ), , z→eit, , r→1, θ→t, , (|z| < 1)., , ( In addition, if we let u(z) = F (z) for |z| = 1, then u(z) becomes continuous, for |z| ≤ 1 )., Proof. First we verify that the function u defined in the statement is harmonic, in the disk |z| < 1. To see this, we may rewrite, , , + 2π iφ, 1, e +z, iφ, F, (e, ), dφ, u(z) = Re, 2π 0 eiφ − z, ', &, +, dζ, 1, ζ +z, F (ζ), = Re, 2π |ζ|=1 ζ − z, iζ
Page 375 :
10.2 Poisson Integral Formula, , The second term on the right can be made less than any, 1. Thus, there exists a δ > 0 such that, u(z) − F (eit ) < 2, , 367, , > 0 for r close to, , whenever z with |z| < 1 and |z − eit | < δ ., , Thus, limz→eit u(z) = F (eit )., Remark 10.28. A slight modification in the above proof shows that any, function satisfying the conditions of the theorem must be continuous on the, closed disk |z| ≤ 1. In view of Corollary 10.11, the function u(z) of Theorem, 10.19 (with R = 1) is the only function that can satisfy the conditions of the, theorem. Using the Riemann mapping theorem which will be proved in Chapter 11, we see that Dirichlet’s problem can be solved for simply connected, •, domain D., A simple translation applied to Theorem 10.27 leads to a general result, which we formulate as follows., Theorem 10.29. Let F (φ) := F (Reiφ ) be a continuous function of the real, variable φ, 0 ≤ φ ≤ 2π, with F (0) = F (2π). Then the function u(z) defined, by, + 2π, 1, P (z, Reiφ )F (Reiφ ) dφ (|z| < R), u(z) =, 2π 0, satisfies the following conditions:, (i) u(z) is harmonic in the disk |z| < R., (ii) For each fixed t, 0 ≤ t ≤ 2π,, lim u(z) = F (Reit ), , z→Reit, , (|z| < R)., , Remark 10.30. By requiring in Theorem 10.29 only that the function F (φ), be sectionally continuous, the conclusion (i) still holds with the restriction, that limr→R u(reiθ ) = F (θ) only at the points of continuity for F . The proof, is identical. Again, the analog of Theorem 10.27 for |z − z0 | ≤ R follows, •, routinely as well., Example 10.31. Suppose that we wish to find a real-valued function u harmonic in the open first quadrant D = {z = x + iy : x, y > 0}, continuous on, D \ {0} and u(x, 0) = 5 for x > 0 and u(0, y) = 3 for y > 0., To do this, we may consider, Log z = ln |z| + iArg z on Dπ = C \ (−∞, 0]., Then v(x, y) = Arg z is harmonic on Dπ ,, v(x, 0) = 0 for x > 0 and v(0, y) = π/2 for y > 0., To obtain u satisfying the desired properties, we define
Page 376 :
368, , 10 Harmonic Functions, , u(x, y) = av(x, y) + b., To find a and b, we set y = 0 and obtain, 5 = u(x, 0) = av(x, 0) + b = a(0) + b, i.e., b = 5., Setting x = 0, we have, 3 = u(0, y) = av(0, y) + b = a(π/2) + b,, , i.e., a = −4/π., , The desired function is then, u(x, y) = −(4/π)Arg z + 5., This problem can be also solved by using the Poisson integral formula for the, half-plane (see Exercise 10.35(10))., Remark 10.32. As remarked before, the results of this section that are stated, for unit disks could be stated for arbitrary disks. To illustrate, suppose u(z) is, harmonic in a domain containing the disk |z − z0 | ≤ R. Setting z − z0 = reiθ ,, the conclusion of Theorem 10.19 remains valid for any point inside the circle, •, |z − z0 | = R., Example 10.33. Solve the Dirichlet problem:, uxx + uyy = 0,, , −∞ < x < ∞, y > 0,, , subject to u(x, 0) = 0 for |x| > 1 and u(x, 0) = x for x ∈ (−1, 1), see Figure 10.2. According to Exercise 10.35(10), +, t dt, y 1, u(x, y) =, π −1 (x − t)2 + t2, and a simple computation gives, , , (x − 1)2 + y 2, y, x, x+1, x−1, − tan−1, ln, ., tan−1, +, u(x, y) =, π, y, y, 2π (x + 1)2 + y 2, , 0, , 1, , 0, , 0, x, , Figure 10.2., , 1, , 0, , •
Page 377 :
10.2 Poisson Integral Formula, , 369, , Questions 10.34., 1. What properties of the point z ∗ , chosen in the proof of Theorem 10.19,, made the proof work?, 2. Can Theorem 10.14 be proved using Poisson’s formula?, 3. If F (φ), defined on the circle |z| = R, is continuous at all but a finite, number of points, is there a unique function u harmonic for |z| < R that, approaches F (φ) as z approaches the boundary?, 4. Can the solution to the Dirichlet problem for the disk be used to solve, a Dirichlet problem for different regions?, 5. What is the relationship between Theorem 10.6 and Theorem 10.19?, Exercises 10.35., 1. If f (z) is a continuous function on |z| = 1, show that F (z) defined by, 1, F (z) =, 2πi, , +, 0, , 2π, , f (ζ), dζ, ζ −z, , is analytic for |z| < 1., 2. (a) For ρ = Reiφ , z = reiθ (r < R), show that, ∞ n, , ρ+z, r, =1+2, ein(θ−φ) ., ρ−z, R, n=1, , (b) Conclude that, ∞ n, , R2 − r 2, r, =, 1, +, 2, cos nα, R2 − 2rR cos α + r2, R, n=1, , (α real)., , 3. Use the previous exercise to find an alternate expression for the conclusion of Theorem 10.19., 4. Show that, + 2π, sin(θ − φ), dφ = 0., 2, R − 2rR cos(θ − φ) + r2, 0, 5. If u(z) is harmonic for |z| > R and continuous for |z| ≥ R, show that, for ρ = Reiφ , z = reiθ (r > R),, u(z) = −, , 1, 2π, , +, , 2π, , Re, 0, , ρ+z, u(Reiθ ) dφ., ρ−z, , 6. Find a function u(z) harmonic in the disk |z| < R for which, , 0 if 0 < θ < π, lim u(reiθ ) =, 1 if π < θ < 2π., r→R
Page 378 :
370, , 10 Harmonic Functions, , 7. Show that the function, 2, 2r sin θ, tan−1, (r < 1), π, 1 − r2, is harmonic for |z| < 1 and satisfies the boundary conditions, , 1 if 0 < θ < π, lim u(reiθ ) =, −1 if π < θ < 2π., r→1, u(reiθ ) =, , 8. Set F (θ) = θ/2, 0 ≤ θ ≤ 2π. Show that the function, u(reeiθ ) = tan−1, , r sin θ, 1 + r cos θ, , is harmonic for |z| < 1, and that limr→1 u(reiθ ) = F (θ) for all θ. Can, you derive this from the Poisson integral formula?, 9. Suppose f = u + iv is analytic and bounded on the real line and the, upper half-plane. Show that for z = x + iy, y > 0, we have, +, +, y ∞, y ∞, u(t, 0), v(t, 0), dt,, v(z), =, dt., u(z) =, 2, 2, π −∞ (t − x) + y, π −∞ (t − x)2 + y 2, These are called the Poisson integral formula for u and v in the upper, half-plane., 10. Using the previous exercise, formulate and solve a Dirichlet problem for, a half-plane. More precisely, prove the following: If F (x) is a continuous, function on R, then show that the function u(x, y) defined by, +, y ∞, F (t), u(x, y) =, dt, π −∞ (x − t)2 + y 2, is a solution of the Dirichlet problem in the upper half-plane Im z > 0, with the boundary condition u(x, 0) = F (x) for x ∈ R., 11. Suppose that f = u + iv is analytic for |z| < 1. Show that for |z| <, r (0 < r < 1),, +, 1, 2, f (n) (z), =, u(ζ) dζ., n!, 2πi |ζ|=r (ζ − z)n+1, 12. Find a harmonic function u on the upper half-plane Im z > 0 such that, u(x, 0) = 0 for x > 0 and u(x, 0) = 1 for x < 0., 13. Find a harmonic function u on the upper half-plane Im z > 0 such that, (a) u(x, 0) = 1 for x < −1, (b) u(x, 0) = 2 for −1 < x < 1, (c) u(x, 0) = 3 for x > 1., 14. Find a function u(z) harmonic for |z| < 1 such that, , 1 if 0 < θ < π, lim u(reiθ ) =, 0 if π < θ < 2π., r→1, 15. Suppose that f = u + iv is entire and z −1 Re f (z) → 0 as z → ∞. Show, that f is a constant.
Page 379 :
10.3 Positive Harmonic Functions, , 371, , 10.3 Positive Harmonic Functions, As an application of Poisson’s integral formula, we prove, Theorem 10.36. (Harnack’s Inequality) Suppose u(z) is harmonic in the, disk Δ(z0 ; R) = {z : |z − z0 | < R}, with u(z) ≥ 0 for all z ∈ Δ(z0 ; R). Then, for every z in this disk, we have, u(z0 ), , R − |z − z0 |, R + |z − z0 |, ≤ u(z) ≤ u(z0 ), ., R + |z − z0 |, R − |z − z0 |, , Proof. Fix z ∈ Δ(z0 ; R), and let s < R. Then, for every s with s < R, the, Poisson integral formula given by Theorem 10.18 leads to, + 2π, 1, s2 − |z − z0 |2, u(z) =, u(z0 + seiφ ) dφ, (10.17), 2π 0 |seiφ − (z − z0 )|2, for every z ∈ Δ(z0 ; s). Using the positivity of u(z) and the inequality, s2 − |z − z0 |2, s + |z − z0 |, s − |z − z0 |, ≤, ≤, s + |z − z0 |, |seiφ − (z − z0 )|2, s − |z − z0 |, we get, from (10.17),, s + |z − z0 |, s − |z − z0 |, u(z0 ) ≤ u(z) ≤, u(z0 ), s + |z − z0 |, s − |z − z0 |, because, by the mean-value property (i.e., (10.17) for z = z0 ),, u(z0 ) =, , 1, 2π, , +, 0, , 2π, , u(z0 + seiφ ) dφ., , Since the last inequalities are valid whenever |z − z0 | ≤ s < R, these inequalities continues to hold when s approaches R., Using Harnack’s inequality we can present an alternate proof of Liouville’s, theorem for harmonic functions (Theorem 10.4) in the following form., Corollary 10.37. If u is harmonic in C and is bounded above (or below),, then u is constant., Proof. It suffices to prove for u(z) ≥ 0 in C. Fix z (|z| = r) and let R > r. By, Harnack’s inequality, R+r, R−r, u(0) ≤ u(reiθ ) ≤, u(0)., R+r, R−r, Letting R → ∞, we see that u(z) ≤ u(0) so that u attains its maximum at, z = 0 and therefore, u is constant for |z| < r and hence in C.
Page 380 :
372, , 10 Harmonic Functions, , It was shown (see Theorem 10.6) that every harmonic function satisfies, the mean-value property. For continuous functions, the converse is also true., Theorem 10.38. Suppose u(z) is a real-valued continuous function such that, for each point z0 in a domain D,, + 2π, 1, u(z0 ) =, u(z0 + reiθ ) dθ, 2π 0, whenever the disk |z − z0 | ≤ r is contained in D. Then u(z) is harmonic, throughout D., Proof. Choose a point z0 ∈ D and r > 0 such that |z − z0 | ≤ r is contained in, D. As a consequence of Theorem 10.29, there exists a function u1 (z) harmonic, for |z − z0 | < r, continuous for |z − z0 | ≤ r, and equal to u(z) on the circle, |z − z0 | = r. Since u1 (z) − u(z) is a continuous function that satisfies the, mean-value property, the first proof of Theorem 10.9 shows that u1 (z) − u(z), attains both its maximum and minimum on the boundary. Because, u1 (z) − u(z) ≡ 0, , on |z − z0 | = r,, , it follows that u1 (z) ≡ u(z) for |z − z0 | < r. Hence u(z) is harmonic in a, neighborhood of z0 . Since z0 was arbitrary, u(z) is harmonic in D., Thus, a necessary and sufficient condition for a continuous function to be, harmonic in a domain is that it satisfies the mean-value property at each point, in the domain. As an application, we prove the following analog to Theorem, 8.16., Theorem 10.39. Suppose {un (z)} is a sequence of real-valued harmonic, functions that converges uniformly on all compact subsets of a domain D, to a function u(z). Then u(z) is harmonic throughout D., Proof. Since un (z) is continuous for each n, the continuity of u(z) is a consequence of Theorem 6.26. Given z0 ∈ D and a disk |z − z0 | ≤ r contained in, D, we have for each n that, + 2π, 1, un (z0 ) =, un (z0 + reiθ ) dθ., 2π 0, By Theorem 8.11,, 1, n→∞ 2π, , +, , u(z0 ) = lim un (z0 ) = lim, n→∞, , 1, =, 2π, , +, , 0, , 2π, , un (z0 + reiθ ) dθ, , 2π, 0, , u(z0 + reiθ ) dθ., , Thus, u has the mean-value property. The result now follows from Theorem, 10.38.
Page 381 :
10.3 Positive Harmonic Functions, , 373, , Harnack’s inequality leads to a theorem concerning sequences of harmonic, functions., Theorem 10.40. (Harnack’s Principle) Suppose {un (z)} is a sequence of, real-valued harmonic functions defined in a domain D, and that un+1 (z) ≥, un (z) for each z ∈ D and each n. If {un (z)} converges for at least one point, in D, then {un (z)} converges for all points in D. Furthermore, the convergence is uniform on compact subsets of D, and the limit function is harmonic, throughout D., Proof. We may assume that un (z) ≥ 0; for if not, the theorem can be proved, for the nonnegative sequence {un (z) − u1 (z)}. By the monotonicity property,, for each z in D, either {un (z)} converges or approaches ∞. Let, A = {z ∈ D : un (z) → ∞}, and B = {z ∈ D : un (z) converges}., Given z0 ∈ D, choose a disk |z − z0 | ≤ R contained in D. Then for all z, satisfying |z − z0 | ≤ R/2, Harnack’s inequality gives, R − R/2, R + R/2, 1, un (z0 ) =, un (z0 ) ≤ un (z) ≤, un (z0 ) = 3un (z0 )., 3, R + R/2, R − R/2, , (10.18), , If un (z0 ) → ∞, the left hand inequality of (10.18) shows that un (z) → ∞ for, |z − z0 | ≤ R/2. If {un (z0 )} converges, the right hand inequality shows that, {un (z)} converges for |z − z0 | ≤ R/2. Hence, A and B are both open sets,, with A ∪ B = D. Since the domain D is connected, either A = ∅ or B = ∅., By hypothesis, there is at least one point in B. Thus B = D, and {un (z)}, converges for all z in D., Next we must show that {un (z)} converges uniformly on compact subsets, of D. Applying Harnack’s inequality to un+p (z) − un (z), we get as in (10.18),, un+p (z) − un (z) ≤ 3[un+p (z0 ) − un (z0 )], , (10.19), , for |z − z0 | ≤ R/2 and p = 1, 2, . . . . By the Cauchy criterion,, un+p (z0 ) − un (z0 ) <, , (n > N ( ))., , Hence from (10.19), we see that {un (z)} converges uniformly in some neighborhood of z0 . Since z0 was arbitrary, to every point in D there corresponds, a neighborhood in which the convergence of {un (z)} is uniform., Now let K be a compact subset of D. For each point of K, construct a, neighborhood in which {un (z)} converges uniformly. By the Heine–Borel theorem, finitely many such neighborhoods cover K. But a sequence converging, uniformly on finitely many different sets must converge uniformly on their, union. Therefore, {un (z)} converges uniformly on K., Finally, it follows from Theorem 10.39 that the limit function is harmonic, throughout D.
Page 382 :
374, , 10 Harmonic Functions, , Remark 10.41. According to Harnack’s principle, boundedness of the sequence {un (z)} at one point forces the boundedness for all other points of D., Further, the contrapositive of the theorem says that if the sequence approaches, ∞ at one point in the domain, then it approaches ∞ at all points. That this, can actually happen is seen by considering the sequence un (z) = x + n, which, is harmonic in every domain and satisfies the conditions of the theorem. •, We turn now to the class of analytic functions with positive real part in, the disk |z| < 1, and apply our knowledge of harmonic functions. According, to Harnack’s inequality, if u(z) is harmonic and positive for |z| < 1 with, u(0) = 1, then, 1 + |z|, (|z| < 1)., u(z) ≤, 1 − |z|, Consider the following generalization to analytic functions., Theorem 10.42. Suppose f (z) is analytic for |z| < 1 with f (0) = 1. If, Re f (z) > 0 for |z| < 1, then, |f (z)| ≤, , 1 + |z|, 1 − |z|, , (|z| < 1)., , Proof. Set Re f (z) = u(z). In view of (10.11), we may write, + 2π, 1, Reiφ + z, f (z) =, u(Reiφ ) dφ (|z| < R < 1)., 2π 0 Reiφ − z, Hence,, |f (z)| ≤, , R + |z| 1, R − |z| 2π, , +, , 2π, , u(Reiφ ) dφ =, 0, , R + |z|, R + |z|, u(0) =, ., R − |z|, R − |z|, , −, , By letting R → 1 , the result is obtained., Here is an alternate proof which relies on Schwarz’s inequality (Schwarz’s, lemma) rather than on Harnack’s inequality. If Re f (z) > 0, then the function, g(z) =, , f (z) − 1, f (z) + 1, , (10.20), , satisfies |g(z)| < 1 for |z| < 1. Since g(0) = 0, it follows from Schwarz’s, inequality that |g(z)| ≤ |z| for |z| < 1. Solving for f (z) in (10.20), we get, f (z) =, But, |f (z)| ≤, and the proof is complete., , 1 + g(z), ., 1 − g(z), , 1 + |g(z)|, 1 + |z|, ≤, ,, 1 − |g(z)|, 1 − |z|
Page 383 :
10.3 Positive Harmonic Functions, , 375, , Clearly, Theorem 10.42 is a generalization of Harnack’s inequality because, Re f (z) ≤ |f (z)|., Remark 10.43. The assumption f (0) = 1 does not restrict the generality of, the inequality. For if Re f (z) > 0, then the theorem can be applied to the, function, f (z) − iIm f (0), h(z) =, ,, Re f (0), which satisfies the conditions Re h(z) > 0, h(0) = 1. Also, if we had assumed, only that Re f (z) ≥ 0 for |z| < 1, we could have deduced from the open, mapping theorem that Re f (z) > 0 for |z| < 1., •, Our next theorem may also be proved by methods that rely on harmonic, functions or on Schwarz’s lemma., ∞, Theorem 10.44. Suppose f (z) = 1 + n=1 an z n is analytic for |z| < 1. If, Re f (z) > 0 for |z| < 1, then |an | ≤ 2 for every n., Proof. Set f (z) = u(reiθ ) + iv(reiθ ), with an = αn + iβn . Then, u(reiθ ) = 1 + Re, , ∞, , , m imθ, , am r e, , =1+, , m=1, , ∞, , , (αm cos mθ − βm sin mθ)rm ., , m=1, , This series converges uniformly on the circle |z| = r < 1. By Theorem 8.11,, we may multiply by cos nθ or sin nθ and then integrate term-by-term. Since, + 2π, + 2π, cos nθ cos mθ dθ =, sin nθ sin mθ dθ = 0, 0, , for n = m and, , 0, , +, , 2π, , cos nθ sin mθ dθ = 0, 0, , for all n and m, we have the identities, +, +, 1 2π, 1 2π, u(reiθ ) cos nθ dθ =, αn rn cos2 nθ dθ = αn rn ,, π 0, π 0, 1, π, , +, , 2π, , u(reiθ ) sin nθ dθ =, 0, , 1, π, , +, 0, , 2π, , −βn rn sin2 nθ dθ = −βn rn ., , Multiplying (10.22) by −i and adding to (10.21), we obtain, an rn = (αn + iβn )rn =, Thus,, , 1, π, , +, 0, , 2π, , u(reiθ )e−inθ dθ., , (10.21), , (10.22)
Page 384 :
376, , 10 Harmonic Functions, , |an |rn ≤, , 1, π, , +, , 2π, , 0, , u(reiθ )|e−inθ | dθ =, , 1, π, , +, , 2π, , u(reiθ ) dθ., 0, , By the mean-value property,, +, 1 2π, u(reiθ ) dθ = 2u(0) = 2., π 0, Hence, |an |rn ≤ 2. Letting r → 1− , the result follows., As an alternate proof, consider, g(z) =, , f (z) − 1, f (z) + 1, , which is analytic in the disk |z| < 1, with g(0) = 0 and |g(z)| < 1. By Exercise, 8.73(10), |g (0)| ≤ 1. But g (0) = a1 /2, so that |a1 | ≤ 2., We will now show that |an | ≤ 2, for arbitrary n by constructing a new, function of the form 1 + an z + · · · , which satisfies the conditions of the, theorem. In view of the identity, , n, , n if m is a multiple of n, e(2kπi)m/n =, 0 otherwise,, k=1, , we can verify (do it!) that the function, 1, f (e2kπi/n z 1/n ) = 1 + an z + · · ·, n, n, , h(z) =, , k=1, , is analytic for |z| < 1, and has positive real part. Therefore, |an | ≤ 2 and the, proof is complete., The function, , 1+z, 1−z, maps the circle |z| = 1 onto the imaginary axis and the disk |z| < 1 onto the, right half-plane. This function shows that equality holds in the previous two, theorems. That is,, 1 + |z|, Re f (z) =, 1 − |z|, f (z) =, , when z is a positive real number, and, f (z) =, , ∞, ∞, , , 1+z, zn = 1 + 2, zn., = (1 + z), 1−z, n=0, n=0, , Questions 10.45., 1. Can the mean-value property hold for discontinuous functions?
Page 385 :
10.3 Positive Harmonic Functions, , 377, , 2. If two continuous functions satisfy the mean-value property, does their, sum? Their product?, 3. Suppose {un (z)} is a sequence of harmonic functions having harmonic, conjugates {vn (z)}. If {un (z)} converges uniformly in a region, does, {vn (z)} also converge uniformly?, 4. Is the conclusion of Theorem 10.40 valid if the hypothesis un+1 (z) ≥, un (z) is replaced with un+1 (z) ≤ un (z)?, 5. What kind of generalizations of Theorem 10.42 can you prove?, 6. Why are some theorems valid for compact subsets of a domain but not, for the whole domain?, 7. Where was the positivity of Re f (z) used in the first proof of Theorem, 10.44?, 8. What is the relationship between Schwarz’s inequality and Harnack’s, inequality?, Exercises 10.46., 1. Suppose {un (z)} is a sequence of functions harmonic in a domain D,, and that un+1 (z) ≥ un (z) for each z ∈ D and each n. If un (z0 ) → ∞, for some z0 ∈ D, show that un (z) → ∞ uniformly on compact subsets., That is, given a compact subset C and a real number M , show that, un (z) ≥ M for n > N and all z ∈ C., 2. Let K be a compact subset of a domain D. Given z0 ∈ D, show that, there exist real constants A and B (depending on z0 , K, and D) such, that, A · u(z0 ) ≤ u(z) ≤ B · u(z0 ), for all z in K and all functions u(z) harmonic in D., 3. A continuous real-valued function u(z) is said to be subharmonic in a, domain in D if, + 2π, 1, u(z0 ) ≤, u(z0 + reiθ ) dθ, 2π 0, for every disk |z − z0 | ≤ r contained in D. Show that a nonconstant, subharmonic function cannot attain a maximum in a domain. Can it, attain a minimum?, 4. Suppose f (z) is analytic with Re f (z) > 0 for |z| < 1. If f (0) = 1, then, apply Theorem 10.42 to 1/f (z) to show that, |f (z)| ≥, , 1 − |z|, ., 1 + |z|, , Can this be deduced from Harnack’s inequality?, 5. Suppose g(z) is analytic for |z| < 1 with g(0) = 1. If Re g(z) > α, show, that, 1 + (1 − 2α)|z|, (|z| < 1)., |g(z)| ≤, 1 − |z|
Page 386 :
378, , 10 Harmonic Functions, , 6. Under the assumptions of the previous exercise, show that, 1 − (1 − 2α)|z|, (|z| < 1)., 1 + |z|, ∞, 7. Suppose that g(z) = 1 + n=1 an z n is analytic for |z| < 1, with, Re g(z) > α. Show that |an | ≤ 2(1 − α) for every n., 8. Suppose that g(z) is analytic for |z| < 1 with g(0) = a > 0. If Re g(z) > 0, in |z| < 1, then show that, |g(z)| ≥, , g(z) − a, ≤ |z| for |z| < 1, and |g (0)| ≤ 2a., g(z) + a, 9. Suppose that g(z) is analytic for |z| < 1, and g(0) = 0. If Re g(z) <, a (a > 0), show that, |g(z)| ≤, , 2a|z|, for |z| < 1., 1 − |z|, , Is |g (0)| ≤ 2a?, 10. Show that equality holds in Theorem 10.42 and Theorem 10.44 if and, only if f (z) is of the form, f (z) =, , 1 + eiθ0 z, (θ0 real)., 1 − eiθ0 z
Page 387 :
11, Conformal Mapping and the Riemann, Mapping Theorem, , Our study of mapping properties in Chapters 3 and 4 was limited because, derivatives had not yet been introduced. That remedied, we look anew at, some old functions. We shall see that the derivative relates the angle between, two curves to the angle between their images. In addition, the derivative will, be seen to measure the “distortion” of image curves., Analytic functions mapping disks and half-planes onto disks and halfplanes, disks onto the interior of ellipses, etc., have previously been constructed. The major result of this chapter, known as the Riemann mapping, theorem, tells us that there is nearly always an analytic function that maps a, given simply connected domain onto another given simply connected domain., This is a very powerful result and is used in a wide range of mathematical settings. Our method of proof relies on normal families, a concept that enables us, to extract limit functions from families of functions. Recall how we previously, had extracted limit points from sequences of points (Bolzano–Weierstrass theorem)., , 11.1 Conformal Mappings, Any straight line in the plane that passes through the origin may be parameterized by σ(s) = seiα , where s traverses the set of real numbers and α is the, angle−measured in radians−between the positive real axis and the line. More, generally, a straight line passing through the point z0 and making an angle α, with the real axis can be expressed as σ(s) = z0 + seiα , s real., Suppose now that a function f is analytic on a smooth (parameterized), curve z(t), t ∈ [a, b]. Then the image of z(t) under f is also a smooth curve, whose derivative is given by f (z(t))z (t). A smooth curve is characterized, by having a tangent at each point. So, we interpret z (t) as a vector in the, direction of the tangent vector at the point z(t). Our purpose is to compare, the inclination of the tangent to the curve at a point with the inclination of, the tangent to the image curve at the image of the point.
Page 388 :
380, , 11 Conformal Mapping and the Riemann Mapping Theorem, , Let z0 = z(t0 ) be a point on the curve z = z(t). Then the vector z (t0 ) is, tangent to the curve at the point z0 and arg z (t0 ) is the angle this directed, tangent makes with the positive x-axis. Suppose that w = w(t) = f (z(t)),, with w0 = f (z0 ). For any point z on the curve other than z0 , we have the, identity, f (z) − f (z0 ), w − w0 =, (z − z0 )., z − z0, Thus,, arg(w − w0 ) = arg, , f (z) − f (z0 ), + arg(z − z0 ), z − z0, , ( mod 2π ),, , (11.1), , where it is assumed that f (z) = f (z0 ) so that (11.1) has meaning. Note that, arg(z − z0 ) is the angle in the z plane between the x axis and the straight line, passing through the points z and z0 , while arg(w − w0 ) is the angle in the w, plane between the u axis and the straight line passing through the points w, and w0 . Hence as z approaches z0 along the curve z(t), arg(z − z0 ) approaches, a value θ, which is the angle that the tangent to the curve z(t) at z0 makes, with the x axis. Similarly, arg(w − w0 ) approaches a value φ, the angle that, the tangent to the curve f (z(t)) at w0 makes with the u axis., Suppose f (z0 ) = 0 so that arg f (z0 ) has meaning. Then taking limits in, (11.1), we find (mod 2π) that, φ = arg f (z0 ) + θ,, , or, , arg w (t0 ) = arg f (z0 ) + arg z (t0 )., , (11.2), , That is, the difference between the tangent to a curve at a point and the, tangent to the image curve at the image of the point depends only on the, derivative of the function at the point (see Figure 11.1)., For instance, consider f (z) = z 2 . Then f (z) = 0 on C \{0}. Choose z0 =, 1 + i. Then f (z0 ) = 2(1 + i) so that, arg f (z0 ) = (π/4) + 2kπ., , Figure 11.1. The direction of the tangent line at z(t)
Page 389 :
11.1 Conformal Mappings, , 381, , To verify the angle of rotation of a particular curve, we consider a simple, curve C passing through z0 :, C : z(t) = t(1 + i), t ∈ R., Clearly, π/4 is the angle which the curve C makes with the x axis. The image, of C under f (z) = z 2 = (x2 − y 2 ) + i(2xy) is given by w(t) = 0 + 2t2 i. Thus,, the angle of rotation at 1 + i is π/2 which corresponds to the case k = 0., If two smooth curves intersect at a point, then the angle between these, two curves is defined as the angle between the tangents to these curves at the, point. We can now state, Theorem 11.1. Suppose f (z) is analytic at z0 with f (z0 ) = 0. Let C1 : z1 (t), and C2 : z2 (t) be smooth curves in the z plane that intersect at z0 =: z1 (t0 ) =:, z2 (t0 ), with C1 : w1 (t) and C2 : w2 (t) the images of C1 and C2 , respectively., Then the angle between C1 and C2 measured from C1 to C2 is equal to the, angle between C1 and C2 measured from C1 to C2 ., Proof. Let the tangents to C1 and C2 make angles θ1 and θ2 , respectively, with, the x axis (see Figure 11.2). Then the angle between C1 and C2 is θ2 − θ1 ., , Figure 11.2. The curves C1 and C2 intersect at angle α, , According to (11.2), the angle between C1 and C2 , which is the angle between, the tangent vectors f (z0 )z1 (t0 ) and f (z0 )z2 (t0 ), of the image curves is, θ2 + arg f (z0 ) − (θ1 + arg f (z0 ) ) = θ2 − θ1 ,, and the theorem is proved., A function that preserves both angle size and orientation is said to be, conformal. Theorem 11.1 says that an analytic function is conformal at all, points where the derivative is nonzero. We have already discussed a number, of examples of conformal maps without referring to the name “conformal”.
Page 390 :
382, , 11 Conformal Mapping and the Riemann Mapping Theorem, , For instance, f (z) = ez maps vertical and horizontal lines into circles and, orthogonal radial rays, respectively., A function that preserves angle size but not orientation is said to be isogonal. An example of such a function is f (z) = z. To illustrate, z maps the, positive real axis and the positive imaginary axis onto the positive real axis, and the negative real axis respectively (see Figure 11.3). Although the two, curves intersect at right angles in each plane, a “counterclockwise” angle is, mapped onto a “clockwise” angle., , Figure 11.3., , Suppose f (z) is analytic at z0 and f (z0 ) = 0. When z is near z0 , there is, an interesting relationship concerning the distance between the points z and, z0 and the distance between their images. Note that, f (z) = f (z0 ) + f (z0 )(z − z0 ) + (z)(z − z0 ), where (z) → 0 as z → z0 . Thus for z close to z0 ,, f (z) ≈ f (z0 )z + (−f (z0 )z0 + f (z0 )), so that we may approximate f (z) by the linear function. Also,, |f (z) − f (z0 )| ≈ |f (z0 )| |z − z0 | ., , (11.3), , In view of (11.3), “small” neighborhoods of z0 are mapped onto roughly the, same configuration, magnified by the factor |f (z0 )|, see Figure 11.4. Hence,, f (z0 ) plays two roles in determining the geometric character of the image., According to (11.2), arg f (z0 ) measures the rotation; according to (11.3),, |f (z0 )| measures (for points nearby) the magnification or distortion of the, image., An interesting comparison can now be made between the derivatives of real, and complex functions. For real differentiable functions, the nonvanishing of, the derivative is sufficient to guarantee that the function is one-to-one on an, interval. This is not the case for complex functions on a domain. Even though
Page 392 :
384, , 11 Conformal Mapping and the Riemann Mapping Theorem, , The vanishing of a derivative does not preclude the possibility of a real, function being one-to-one. Although the derivative of f (x) = x3 is zero at the, origin, the function is still one-to-one on the real line. That this cannot occur, for complex functions is seen by, Theorem 11.3. If f (z) is analytic and one-to-one in a domain D, then, f (z) = 0 in D, so that f is conformal on D., Proof. If f (z) = 0 at some point z0 in D, then, f (z) − f (z0 ) =, , f (z0 ), (z − z0 )2 + · · ·, 2!, , has a zero of order k (k ≥ 2) at z0 . Since zeros of an analytic function are, isolated, there exists an r > 0 so small that both f (z) − f (z0 ) and f (z) have, no zeros in the punctured disk 0 < |z − z0 | ≤ r. Let g(z) := f (z) − f (z0 ),, C = {z : |z − z0 | = r} and, m = min |g(z)|., z∈C, , Then, g has a zero of order k (k ≥ 2) and m > 0. Let b ∈ C be such that, 0 < |b − f (z0 )| < m. Then, as m ≤ |g(z)| on C,, |f (z0 ) − b| < |g(z)| on C., It follows from Rouche’s theorem that g(z) and, g(z) + (f (z0 ) − b) = f (z) − b, have the same number of zeros inside C. Thus, f (z) − b has at least two zeros, inside C. Observe that none of these zeros can be at z0 . Since f (z) = 0 in the, punctured disk 0 < |z − z0 | ≤ r, these zeros must be simple and so, distinct., Thus, f (z) = b at two or more points inside C. This contradicts the fact that, f is one-to-one on D., We sum up our results for differentiable functions. In the real case, the, nonvanishing of a derivative on an interval is a sufficient but not a necessary, condition for the function to be one-to-one on the interval; whereas in the, complex case, the nonvanishing of a derivative on a domain is a necessary but, not a sufficient condition for the function to be one-to-one on the domain., An analytic function f : D → C is called locally bianalytic at z0 ∈ D, if there exists a neighborhood N of z0 such that restriction of f from N, onto f (N ) is bianalytic. Clearly, a locally bianalytic map on D need not be, bianalytic on D, as the example f (z) = z n (n > 2) on C \{0} illustrates., Combining Theorem 11.2 and Theorem 11.3 leads to the following criterion, for local bianalytic maps., Theorem 11.4. Let f (z) be analytic in a domain D and z0 ∈ D. Then f is, bianalytic at z0 iff f (z0 ) = 0.
Page 393 :
11.1 Conformal Mappings, , 385, , A sufficient condition for an analytic function to be one-to-one in a simply, connected domain is that it be one-to-one on its boundary. More formally, we, have, Theorem 11.5. Let f (z) be analytic in a simply connected domain D and on, its boundary, the simple closed contour C. If f (z) is one-to-one on C, then, f (z) is one-to-one in D., Proof. Choose a point z0 in D such that w0 = f (z0 ) = f (z) for z on C., According to the argument principle, the number of zeros of f (z) − f (z0 ) in D, is given by (1/2π) C {f (z) − f (z0 )}. By hypothesis, the image of C must be a, simple closed contour, which we shall denote by C (see Figure 11.5). Thus the, net change in the argument of w − w0 = f (z) − f (z0 ) as w = f (z) traverses, the contour C is either +2π or −2π, according to whether the contour is, traversed counterclockwise or clockwise. Since f (z) assumes the value w0 at, least once in D, we must have, 1, 2π, , C, , {f (z) − f (z0 )} =, , 1, 2π, , C, , {w − w0 } = 1., , That is, f (z) assumes the value f (z0 ) exactly once in D., , Figure 11.5., , This proves the theorem for all points z0 in D at which f (z) = f (z0 ), when z is on C. If f (z) = f (z0 ) at some point on C, then the expression, C {f (z) − f (z0 )} is not defined. We leave for the reader the completion of, the proof in this special case., In the proof of Theorem 11.1, we relied on the nonvanishing of the derivative. In Theorem 11.2, we see that every analytic function is locally one-to-one, at points where the derivative is nonvanishing. More generally, it can be shown, that if f is analytic at z0 and f has a zero of order k at z0 , then f is locally, (k + 1)-to-one. For example, if f (z) = z 2 , then f (z) has a zero of order 1 at, the origin and hence, it is two-to-one in any neighborhood of the origin.
Page 394 :
386, , 11 Conformal Mapping and the Riemann Mapping Theorem, , We now examine the behavior of an analytic function in a neighborhood, of a critical point, a point where the derivative vanishes. First we note that, the angle of intersection of two smooth curves at a critical point of an analytic, function is not the same as the angle of intersection of their images under f . If, f (z) is analytic and f (z) has a zero of order k − 1 at z = z0 , then f (j) (z) = 0, for j = 1, · · · , k − 1 and so we may write, f (z) = f (z0 ) + ak (z − z0 )k + ak+1 (z − z0 )k+1 + · · · ., Thus, f (z) − f (z0 ) = (z − z0 )k g(z), where g(z) is analytic at z0 and g(z0 ) =, ak = 0. Consequently,, arg[f (z) − f (z0 )] = k arg(z − z0 ) + arg g(z)., , (11.4), , Suppose θ is the angle that the tangent to a smooth curve C at z0 makes with, the x axis, and φ is the angle that the tangent to the image C of the curve, C at f (z0 ) makes with the u axis. If z approaches z0 along the curve C, then, w = f (z) approaches w0 = f (z0 ) along the curve C , and so (11.4) yields, φ = kθ + arg g(z0 )., , (11.5), , Observe that (11.5) reduces to (11.2) in the special case when k = 1. In, general, the tangent to an image curve depends on the tangent to the original, curve as well as on the order and argument of the first nonzero derivative at, the point in question. Just as (11.2) led to Theorem 11.1, so (11.5) leads to, Theorem 11.6. Suppose f (z) is analytic at z0 , and that f (z) has a zero of, order k − 1 at z0 . If two smooth curves in the domain of f intersect at an, angle θ, then their images intersect at an angle kθ., Proof. Suppose that the tangents to the two curves make angles θ1 and θ2, with respect to the real axis. Then θ = θ2 − θ1 is the angle between the two, curves. According to (11.5), the angle φ between their images is given by, φ = kθ2 + arg g(z0 ) − (kθ1 + arg g(z0 )) = kθ,, , g(z0 ) =, , f (k) (z0 ), ., k!, , Combining Theorems 11.1 and 11.6, we see that an analytic function is, conformal at a point if and only if it has a nonzero derivative at the point., Thus, an analytic function f is conformal on a domain D iff f (z) = 0 on D., It now pays to reexamine bilinear transformations, studied in Chapter 3,, from a conformal mapping point of view. Recall that the transformation, w = f (z) =, , az + b, cz + d, , (ad − bc = 0), , (11.6), , represents a one-to-one continuous mapping from the extended plane onto, itself, with f (−d/c) = ∞ and f (∞) = a/c. Since f (z) = 0 (ad − bc = 0), the, mapping is conformal for all finite z, z = −d/c.
Page 395 :
11.1 Conformal Mappings, , 387, , As we have seen, a circle or a straight line is mapped onto either a circle, or a straight line, depending on which point is mapped onto the point at ∞., For instance, the inversion transformation w = 1/z maps straight lines not, passing through the origin onto circles. In particular, the lines y = x + 1 and, y = −x + 1 are mapped, respectively, onto the circles, , , 2 , 2 , 2 , 2 , 2, 2, 1, 1, 1, 1, 1, 1, u+, + v+, = √, and, u−, + v+, = √, ., 2, 2, 2, 2, 2, 2, At first glance, Figure 11.6 is somewhat misleading. It shows a pair of straight, lines that intersect at one point being mapped onto a pair of circles that, intersect at two points. It should not be forgotten, however, that these straight, lines also intersect at ∞. For both lines, the point (0, 1) is mapped onto the, point (0, −1) while the point at ∞ is mapped onto the origin. The two lines, intersect at right angles at (0, 1) as do the two circles at (0, −1). This is in, harmony with Theorem 11.1., , Figure 11.6., , But at what angle do the two lines intersect at ∞? We need the following, definition: Two smooth curves in the extended plane are said to intersect at, an angle α at ∞ if their images under the transformation w = 1/z intersect, at an angle α at the origin. Since the two circles in Figure 11.6 intersect at, right angles at the origin, the lines y = x + 1 and y = −x + 1 intersect at right, angles at ∞., With this definition, we can show that all transformations of the form, (11.6) are conformal at ∞. There are two cases to consider., Case 1: Let c = 0. The behavior of f at ∞ is determined from the behavior, of f (1/z) at 0 in (11.6). Thus we consider, , bz + a, a/z + b, 1, =, ., =, g(z) = f, z, c/z + d, dz + c, Since g (0) = (bc − ad)/c2 = 0, it follows that g(z) is conformal at ζ = 0. But, this means that f (z) is conformal at z = ∞.
Page 396 :
388, , 11 Conformal Mapping and the Riemann Mapping Theorem, , Case 2: Let c = 0. Then (11.6) is linear, and maps z = ∞ onto w = ∞., So we need to consider the expression h(z) = 1/f (1/z) in (11.6):, w = h(z) =, , dz, ., bz + a, , Since h (0) = d/a = 0, h(z) is conformal at z = 0; that is, f (z) is conformal at, z = ∞. Hence, a bilinear transformation is a one-to-one conformal mapping, of the extended plane onto itself., Recall from Chapter 4 that the exponential function ez maps lines parallel, to the y axis onto circles centered at the origin and lines parallel to the x axis, onto rays emanating from the origin. From elementary geometry we know that, these two image curves must intersect at right angles (see Figure 11.7)., , Figure 11.7., , Finally, consider the function w = cos z, which maps lines parallel to the, y axis onto ellipses and lines parallel to the x axis onto hyperbolas. According to Theorem 11.1, these conic sections must intersect at right angles (see, Figure 11.8)., , Figure 11.8.
Page 397 :
11.1 Conformal Mappings, , 389, , Questions 11.7., 1., 2., 3., 4., 5., 6., , 7., 8., 9., 10., 11., 12., , What is meant by a tangent to a point on a straight line?, Was it necessary to require the curves in Theorem 11.1 to be smooth?, Can nonanalytic functions be conformal?, What kind of functions are isogonal?, Why does the derivative play such a central role?, If a function is one-to-one in some neighborhood of each point in a, domain, why does this not mean that the function is one-to-one in the, domain?, If f is conformal on a domain D, is f always one-to-one on D?, If f is conformal on a domain D which is symmetric with respect to the, real axis, is f (z) conformal on D?, What is the relationship between conformal and one-to-one?, At what angle do parallel lines intersect at ∞?, How might we define a function to be analytic at ∞?, Is the sum of conformal maps conformal? The product? The composition?, , Exercises 11.8., 1. Given a complex number z0 and an > 0, show that there exists a, function f (z) analytic at z0 with f (z0 ) = 0 and such that f (z) is not, one-to-one for |z − z0 | < . Does this contradict Theorem 11.2?, 2. Show that z 2 is one-to-one in a domain D if and only if D is contained, in a half-plane whose boundary passes through the origin., 3. Find points at which the mapping defined by f (z) = nz + z n (n ∈ N) is, not conformal., 4. Prove that two smooth curves intersect at an angle α at ∞ if and only, if their images under stereographic projection (see Section 2.4) intersect, at an angle α at the north pole., 5. Show that f (z) and f (z) are both isogonal at points where f (z) is, analytic with nonzero derivative., 6. If two straight lines are mapped by a bilinear transformation onto circles, tangent to each other, show that the two lines must be parallel. Is the, converse true?, 7. Find the radius of the largest disk centered at the origin in which w = ez, is one-to-one. Is the radius different if the disk is centered at an arbitrary, point z0 ?, 8. For f (z) = ez , find arg f (z). Use this to verify that lines parallel to the, y axis and x axis map, respectively, onto circles and rays., 9. Suppose f (z) is analytic at z0 with f (z0 ) = 0. Prove that a “small”, rectangle containing z0 and having area A is mapped onto a figure whose, area is approximately |f (z0 )|2 A., 10. Either directly or by making use of Theorem 11.5, show that the function, w = z n maps the ray arg z = θ (0 ≤ θ < 2π/n) onto the ray arg z = nθ.
Page 398 :
390, , 11 Conformal Mapping and the Riemann Mapping Theorem, , 11. If f (z) is nonconstant and analytic in a domain D, show that f (z) = 0, for only a countable number of points in D. Thus conclude that f (z), is locally one-to-one and conformal at all but a countable number of, points in D., 12. Show that f (z) = z + 1/z is conformal except at z = ±1. With this in, mind, review its mapping properties from Chapter 3., , 11.2 Normal Families, We have previously seen significant differences between pointwise and uniform, continuity as well as between pointwise and uniform convergence. Once again, we encounter the contrast between local and global properties. This time, we, shall require a uniformity to hold over a set consisting of a family of functions., A family F of functions is said to be uniformly bounded on a set A if there, exists a real number M such that |f (z)| ≤ M for all f ∈ F and all z ∈ A., Certainly the uniform boundedness of a family implies that each member, of the family is bounded. On the other hand, each member of the sequence, {fn (z)} of functions fn (z) = nz is bounded in the disk |z| ≤ R, but there is, no bound that works for every member of the family., A family F of functions is said to be locally uniformly bounded on a set A, if to each z ∈ A there corresponds a neighborhood in which F is uniformly, bounded. The sequence fn (z) = 1/(1 − z n ) is locally uniformly bounded, but, not uniformly bounded in the disk |z| < 1. We have the following characterization:, Theorem 11.9. A family F of functions is locally uniformly bounded in a, domain D if and only if F is uniformly bounded on each compact subset, of D., Proof. Let F be locally uniformly bounded and suppose K is a compact subset, of D. For each point in K, choose a neighborhood in which F is uniformly, bounded. This provides an open cover for K. According to the Heine–Borel, theorem, there exists a finite subcover, n of K. That is, there are finitely many, zi ∈ K and i > 0 such that K ⊂ i=1 N (zi ; i ), where |f (z)| ≤ Mi for all, f ∈ F and all z ∈ N (zi ; i ). Then F is uniformly bounded on K, having for, a bound M = max{M1 , M2 , . . . , Mn }., The converse is immediate from the fact that the closure of a neighborhood, of a point is a compact set., By restricting ourselves to locally uniformly bounded families of analytic, functions, we can obtain additional information., Theorem 11.10. Suppose F is a family of locally uniformly bounded analytic functions in a domain D. Then the family F (n) , consisting of the nth, derivatives of all functions in F, is also locally uniformly bounded in D.
Page 399 :
11.2 Normal Families, , 391, , Proof. It suffices to prove this when n = 1, since then the result may be reapplied successively to each new class. Suppose for some z0 in D that |f (z)| ≤ M, for each f ∈ F and all z inside or on the circle C : |z − z0 | = r contained in, D. Then for z in the smaller disk |z − z0 | ≤ r/2, Cauchy’s integral formula, yields, +, 1, f (ζ), f (z) =, dζ, 2πi C (ζ − z)2, and so, as |ζ − z| ≥ |ζ − z0 | − |z − z0 | ≥ r − r/2 = r/2,, +, 1, 4M, , ., |f (ζ)| |dζ| ≤, |f (z)| ≤, 2π(r/2)2 C, r, This shows that the family F is locally uniformly bounded at z0 . Since z0, was arbitrary, the proof is complete., We next extend the concept of uniform continuity. A family F of functions, is said to be equicontinuous in a region R if for every > 0 there exists a, δ > 0 such that |f (z1 ) − f (z0 )| < for all f ∈ F and all points z0 , z1 ∈ R, satisfying |z1 −z0 | < δ. Observe that each member of an equicontinuous family, is uniformly continuous. That is, for an equicontinuous family we can find a, δ = δ( ) that works for all points in the set as well as for all functions in the, family., It is possible for each member of a family to be uniformly continuous, without the family being equicontinuous. To see this, set fn (z) = nz. Each fn, is uniformly continuous on |z| ≤ R because, |fn (z1 ) − fn (z0 )| = n|z1 − z0 | <, whenever |z1 − z0 | < /n = δ. But a δ cannot be chosen that works for all n., Hence the sequence {nz} is not equicontinuous on |z| ≤ R., There is an important relationship between locally uniformly bounded and, equicontinuous families of analytic functions., Theorem 11.11. If F is a locally uniformly bounded family of analytic functions in a domain D, then F is equicontinuous on compact subsets of D., Proof. We prove the theorem in the special case that K is a closed disk contained in D. The proof for general compact subsets of D is similar to the, proof of Theorem 11.9, and is left for the reader. By Theorem 11.10, the family F , consisting of the derivatives of functions in F, is also locally uniformly, bounded. In view of Theorem 11.9. we may therefore assume that |f (z)| ≤ M, for all f ∈ F and all z ∈ K. Then for z0 , z1 ∈ K, we have, + z1, f (z) dz ≤ M |z1 − z0 |,, |f (z1 ) − f (z0 )| =, z0, , where the path from z0 to z1 is taken to be the straight line segment. By, choosing δ = /M ( arbitrary), we see that the family F is equicontinuous, on the disk K.
Page 400 :
392, , 11 Conformal Mapping and the Riemann Mapping Theorem, , Remark 11.12. The converse of Theorem 11.11 is not true. The sequence, fn (z) = z + n is equicontinuous on all compact subsets of the plane. In fact, fn (z1 ) − fn (z0 ) = z1 − z0 for each n, so that δ = may be chosen. However,, {fn (z)} is not uniformly bounded in any neighborhood in C., •, In Chapter 2, we showed that every bounded sequence of complex numbers contains a convergent subsequence. Our goal in this section is to obtain, analogous results for sequences of functions. It is not clear, at this point, what, form of convergence is most reasonable or most applicable. To help clarify the, situation, we need the following definition. A family F of functions is said to, be normal in a domain D if every sequence {fn } in F contains a subsequence, {fnk } that converges uniformly on each compact subset of D., As an example, the family consisting of the sequence {z n } is normal in, the domain |z| < 1. In fact, the sequence itself converges uniformly to zero, on every compact subset of |z| < 1. Note, however, that neither the sequence, nor any subsequence converges uniformly in the whole domain., Just as a bounded sequence may contain different subsequences that converge to different limits, so may a normal family contain different sequences, that converge uniformly on compact subsets to different functions. To illustrate, set, , z n if n odd,, fn (z) =, 1 − z n if n even., Then {f2n+1 } converges uniformly to 0 and {f2n } converges uniformly to 1, on all compact subsets of |z| < 1., A set of points E is said to be dense in a set A if every neighborhood of each, point in A contains points of E. Every domain in the plane contains a dense, sequence of points (for example, the set of points in the domain having both, coordinates rational is countable, and so may be expressed as a sequence)., Before proving the major result of this section, we need the following:, Lemma 11.13. Suppose {fn (z)} is a sequence of analytic functions that is, locally uniformly bounded in a domain D. If {fn (z)} converges at all points, of a dense subset of D, then it converges uniformly on each compact subset, of D., Proof. Given a compact set K contained in D, we wish to show that the, sequence {fn (z)} converges uniformly on K. By Theorem 11.11, {fn (z)} is, equicontinuous on K. Thus to each > 0, there corresponds a δ > 0 such that, |fn (z) − fn (z )| < /3, , for |z − z | < δ,, , (11.7), , where z, z are any points in K and n is arbitrary. Since K is compact, finitely, many, say p, neighborhoods of radius δ/2 cover K. In each of these p neighborhoods, choose a point zk (k = 1, 2, . . . , p) from the dense subset of K, at, which {fn } converges. Next choose n and m large enough so that
Page 401 :
11.2 Normal Families, , |fn (zk ) − fm (zk )| < /3 for k = 1, 2, . . . , p., , 393, , (11.8), , In view of (11.7) and (11.8), we see that, to each z in K, there corresponds a, zk in K such that, |fn (z) − fm (z)| ≤ |fn (z) − fn (zk )| + |fn (zk ) − fm (zk )| + |fm (zk ) − fm (z)|, < ., Hence the sequence {fn (z)} is uniformly Cauchy on K, and must therefore, converge uniformly on K., Note the lemma concludes that {fn (z)} is a normal family in D. We will, now show, by a diagonalization process, that this conclusion is true without, the assumption that the sequence converges on a dense subset., Theorem 11.14. (Montel’s Theorem) If F is a locally uniformly bounded, family of analytic functions in a domain D, then F is a normal family in D., Proof. Given a sequence {fn } of functions in F, we must show that some, subsequence of {fn } converges uniformly on compact subsets. Choose any, sequence of points {zk } that is dense in D. According to Lemma 11.13, it, suffices to construct a subsequence of {fn } that converges at each point of, the sequence {zk }. By hypothesis, the sequence {fn (z1 )} of complex numbers, is bounded. Hence by the Bolzano–Weierstrass property (see Theorem 2.17),, there exists a subsequence of {fn }, which we shall denote by {fn,1 }, that, converges at z1 . But the sequence of {fn,1 (z2 )} of points is also bounded. Thus, there is a subsequence {fn,2 } of {fn,1 } that converges at z2 . Since {fn,2 } is a, subsequence of {fn,1 }, it must also converge at z1 ., Continuing the process, for each positive integer m, we obtain the mth, subsequence {fn,m } of {fn } so that it converges at z1 , z2 , . . . , zm . As seen in, the chart below,, f1,1 (z), f2,1 (z), f3,1 (z), . . . fm,1 (z), . . ., f1,2 (z), f2,2 (z), f3,2 (z), . . . fm,2 (z), . . ., f1,3 (z), f2,3 (z), f3,3 (z),, .., .., .., ., ., ., , . . . fm,3 (z), . . ., .., .., ., ., , f1,m (z), f2,m (z), f3,m (z),, .., .., .., ., ., ., , . . . fm,m (z), . . ., .., .., ., ., , the mth sequence of complex functions converges at zm and all preceding, points of the sequence {zk }. Now consider the sequence {fn,n (z)}, which makes, up the diagonal of the chart. For each fixed m, the sequence {fn,n (zm )},, n ≥ m, is a subsequence of the convergent sequence {fn,m (zm )}, and hence, converges. Therefore, {fn,n (z)} is a subsequence of {fn } that converges at all, points of the sequence {zk }. This completes the proof.
Page 402 :
394, , 11 Conformal Mapping and the Riemann Mapping Theorem, , The Bolzano–Weierstrass theorem guarantees the existence of a limit point, for every bounded infinite set of points, and consequently the existence of a, convergent subsequence for every bounded sequence. Montel’s theorem can be, viewed as an “analytic function” analog to Bolzano’s theorem. It guarantees,, in some sense, the existence of a convergent sequence of functions associated, with every locally uniformly bounded family of analytic functions., Carrying the analogy one step further, both theorems suffer from the same, deficiency. The limit point of Bolzano’s theorem need not be a member of the, set, while the convergent function of Montel’s need not be a member of the, normal family. For example, the sequence {z n } is a normal family in |z| < 1, because it converges uniformly to 0 on all compact subsets of |z| < 1. However,, 0 is not a member of the family {z n }., Recall that a bounded set that contains all its limit points is compact. This, leads to the following definition. A normal family F of functions is said to be, compact if the uniform limits of all sequences converging in F are themselves, members of F., Example 11.15. The family F of functions of the form, f (z) = 1 +, , ∞, , , an z n, , n=1, , that are analytic with positive real part in the disk |z| < 1 is a compact,, normal family. By Theorem 10.42, all functions f ∈ F satisfy the inequality, |f (z)| ≤, , 1 + |z|, 1 − |z|, , (|z| = r < 1)., , Hence F is locally uniformly bounded and, by Montel’s theorem, is normal., To show compactness, suppose a sequence {fn } of functions in F converges, uniformly to a function g. We wish to show that g ∈ F. By Theorem 8.16, g is, analytic in |z| < 1. Since fn (0) = 1 for every n, g(0) = 1. Since Re fn (z) > 0, for every n, Re g(z) ≥ 0 for |z| < 1. But then by the open mapping theorem,, •, we must have Re g(z) > 0 for |z| < 1. Thus g ∈ F, and F is compact., Questions 11.16., 1. What kinds of families of functions are locally uniformly bounded but, not uniformly bounded?, 2. Is the family of polynomials locally uniformly bounded on some set?, 3. If F is a uniformly bounded family of analytic functions, is F (n) also, uniformly bounded?, 4. If a family of functions is uniformly bounded at each point in a domain,, is the family locally uniformly bounded?, 5. Where, in the proof of Theorem 11.7, did we use the fact that the set, K was a disk?
Page 403 :
11.3 Riemann Mapping Theorem, , 395, , 6. What is an important distinction between a dense sequence and a dense, set?, 7. What kinds of normal families have more than one subsequential limit, function?, 8. Can a normal family have infinitely many subsequential limit functions?, Exercises 11.17., 1. Suppose that for each point in a domain D there corresponds a neighborhood in which a family F is equicontinuous. Show that F is equicontinuous on compact subsets of D. Is F equicontinuous in D?, 2. Show that the sequence {nz} is not equicontinuous in any region., 3. If F is locally uniformly bounded family of analytic functions in a domain D, show that F , the family of functions consisting of the derivatives of functions in F, is equicontinuous on compact subsets of D., 4. Suppose F is a normal family of analytic functions in* the disk |z| < 1., z, Let G be the family of functions of the form g(z) = 0 f (ζ) dζ, where, f ∈ F. Show that G is normal in |z| < 1., 5. Show that the sequence {fn (z)} defined by, , z n if n odd, fn (z) =, 1 − z n if n even,, forms a normal family in the disk |z| < 1., ∞, 6. Show that the family of functions of the form f (z) = n=0 an z n , where, |an | ≤ n, is a compact normal family of analytic functions in the disk, |z| < 1., 7. Let F be the family consisting of all functions f (z) that are analytic in, a domain D with |f (z)| ≤ M in D. Show that F is a compact, normal, family in D., , 11.3 Riemann Mapping Theorem, We have already discussed a number of examples of analytic functions between, various domains of the complex plane. In some cases, we have given complete, characterizations for mappings between certain domains such as disks and, half-planes. Also, we know from the open mapping theorem that nonconstant, analytic functions map domains into domains. Now, suppose D1 and D2 are, simply connected domains. Then there is almost always an analytic function, mapping D1 onto D2 . We first discuss a “typical” exception. Suppose D1 is, the whole plane and D2 is the disk |z| < 1. There can be no function analytic, in the plane (entire) that maps onto the (bounded) disk |z| < 1, for, according, to Liouville’s theorem, constant functions are the only entire functions whose, images are contained in the disk. Our major theorem of this section says, that a one-to-one analytic mapping exists between any two simply connected
Page 404 :
396, , 11 Conformal Mapping and the Riemann Mapping Theorem, , domains, neither of which is the whole plane. Before proving this remarkable, (existence) result, we shall need some preliminaries concerning univalent (a, fancy term for one-to-one) functions., Theorem 11.18. Suppose {fn (z)} is a sequence of analytic, univalent functions defined in a domain D and converging uniformly on each compact subset of D to a nonconstant function f (z). Then f (z) is analytic and univalent, in D., Proof. The analyticity of f follows from Theorem 8.16. To prove the univalence of f , assume there are distinct points z0 , z1 in D for which f (z0 ) =, f (z1 ) = a. We can find r > 0 (e.g., r < |z0 − z1 |/2) so small that the closed, disks centered at z0 and z1 with radius r are mutually disjoint and are contained in D. Assume further that f (z) = a on the circles C0 : |z − z0 | = r and, C1 : |z − z1 | = r. This is possible because f is nonconstant. Let, m=, , min, , z∈C0 ∪C1, , |f (z) − a|., , Now choose n sufficiently large so that |fn (z) − f (z)| < m on C0 ∪ C1 . So, on, C0 ∪ C1 ,, |f (z) − a)| > m > |fn (z) − f (z)| for large n., By Rouche’s theorem, the function, fn (z) − a = (fn (z) − f (z)) + (f (z) − a), has at least one zero inside C0 and at least one zero inside C1 . This contradicts, the univalence of fn (z) in D., Note that it is possible for the uniform limit of a sequence of univalent, functions to be constant. For example, the univalent sequence fn (z) = z/n, converges uniformly to f (z) = 0 on any compact subset of C. Thus the uniform, limit of a sequence of univalent functions need not be univalent., Theorem 11.19. Suppose f (z) is analytic and univalent in a domain D, and, that g(z) is analytic and univalent on the image of D under f (z). Then the, composition function g(f (z)) is analytic and univalent in D., Proof. The analyticity of g(f (z)) follows from Theorem 5.6. To show univalence, suppose, g(f (z0 )) = g(f (z1 )) for z0 , z1 ∈ D., By the univalence of g, we have f (z0 ) = f (z1 ). From the univalence of f ,, z0 = z1 and the theorem is proved., Theorem 11.20. Suppose f , mapping a domain D1 onto D2 , is analytic and, univalent in D1 . Then the inverse function g, defined by g(f (z)) = z for all, z ∈ D1 , is an analytic and univalent mapping from D2 onto D1 .
Page 405 :
11.3 Riemann Mapping Theorem, , 397, , Proof. The univalence of g is an immediate consequence of the univalence of, f . To show analyticity, fix a point w0 ∈ D2 . Then w0 = f (z0 ) for a unique, z0 ∈ D1 . Setting w = f (z), we have, z − z0, g(w) − g(w0 ), ., =, w − w0, f (z) − f (z0 ), , (11.9), , Since f maps open sets onto open sets (Theorem 9.55), g is continuous in, D2 . Thus z → z0 as w → w0 . By Theorem 11.3, f (z0 ) = 0. Hence we may, take limits in (11.9) to obtain g (w0 ) = g (f (z0 )) = 1/(f (z0 )). Therefore g is, analytic in D2 , and the theorem is proved., If f and g are analytic and univalent in domains D1 and D2 , respectively,, and map onto the disk |z| < 1, then g −1 (f (z)) is an analytic and univalent, mapping from D1 onto D2 (see Figure 11.9)., , Figure 11.9., , Thus the set of domains that may be mapped analytically and univalently, onto the interior of the unit disk can also be mapped analytically and univalently onto one another., Suppose f is analytic and univalent in D and maps onto |z| < 1. Are there, other functions with the same property? In general, there are infinitely many., To see this, recall from Section 3.2 (see Theorem 3.21) that all functions of, the form, g(z) = eiα, , z − z0, 1 − z0z, , (|z0 | < 1, α real), , (11.10), , map the interior of the unit circle onto itself. Hence the functions g(f (z)) and, f (z) simultaneously map D onto |z| < 1. Our next result suggests conditions, for establishing a unique mapping function., Given a domain D ⊆ C, we define the group of analytic automorphisms of, D as follows: If f : D → D is an analytic function that is one-to-one and onto,, then f (z) is called an analytic/holomorphic automorphism of D. That is, f (z)
Page 406 :
398, , 11 Conformal Mapping and the Riemann Mapping Theorem, , is called a conformal self-mapping of D. The set of all analytic automorphisms, of D form what is called an “automorphism group” (with composition as the, group operation) of D, and is denoted by Aut (D). The Schwarz lemma can, be used to describe the automorphism groups of the upper half-plane, and the, unit disk Δ (see also Theorems 3.18 and 3.21). It is easy to see that, Aut a (D) = {f ∈ Aut (D) : f (a) = a}, forms a subgroup of the group Aut (D). Our next result is a reformulation of, Theorem 3.21 in the language of automorphisms, but the new proof uses the, Schwarz lemma., Theorem 11.21. We have, , , z−a, Aut (Δ) = eiα, : |a| < 1, 0 ≤ α ≤ 2π ., 1 − az, In particular, Aut 0 (Δ) := {f ∈ Aut (Δ) : f (0) = 0} = {eiα z : α real}., Proof. Let a ∈ Δ, and, ϕa (z) =, , a−z, ., 1 − az, , Obviously, ϕa is analytic for |z| < 1/|a| (|a| < 1), ϕa (Δ) ⊆ Δ, and ϕa (∂Δ) =, ∂Δ. Moreover, ϕa is univalent on Δ and (ϕa )−1 = ϕa . Thus, ϕa ∈ Aut (Δ)., Also, the rotation eiθ ϕa (z) (θ ∈ R) belongs to Aut (Δ)., Conversely, let f ∈ Aut (Δ). Then there exists a b ∈ Δ such that f (0) = b., Then F (z) defined by F = ϕb ◦ f is also analytic and univalent in Δ, F maps, Δ onto Δ, and F (0) = 0. By the Schwarz lemma,, |F (z)| ≤ |z| for z ∈ Δ., Since F is analytic and one-to-one on Δ, F −1 exists on Δ. Moreover, F −1, is analytic and one-to-one on Δ with F −1 (0) = 0. We may again apply the, Schwarz lemma to F −1 and obtain |F −1 (w)| ≤ |w| for w ∈ Δ. If we take, w = F (z), we get, |z| ≤ |F (z)| for z ∈ Δ., Hence, |F (z)| = |z|, and so F (z) = λz with |λ| = 1, or, ϕb (f (z)) = λz or f (z) = ϕb (λz)., The desired result follows., Our next result suggests conditions for establishing a unique mapping, function., Lemma 11.22. Suppose f (z) is analytic and univalent in |z| < 1 and maps, the disk onto itself. If f (0) = 0 and f (0) > 0, then f (z) = z.
Page 407 :
11.3 Riemann Mapping Theorem, , 399, , Proof. As Aut 0 (Δ) := {f ∈ Aut (Δ) : f (0) = 0} = {eiα z : α real} and, f (0) > 0, the result follows., Two domains D1 and D2 are said to be conformally equivalent if there, is a bijective analytic function mapping D1 onto D2 . Both the existence and, method of finding it are two important components for conformal mappings., We start with a couple of examples illustrating conformal mappings between, standard simply connected domains. It follows that conformally equivalent, domains are homeomorphic but not the converse., Example 11.23. We are interested in showing that the upper half disk D =, {z : |z| < 1, Im z > 0} and the unit disk Δ = {z : |z| < 1} are conformally, equivalent., Step 1: We consider, 1, ., 1−z, Then we know that f1 transforms the unit disk Δ onto the right half-plane, Re w1 > 1/2. Rewriting, w1 = f1 (z) =, , w1 = f1 (z) =, , 1−z, 1 − x + iy, =, ,, |1 − z|2, |1 − z|2, , we see that Im w1 > 0 iff Im z > 0. Moreover, z = 1 is a pole of f1 (z), the, segment [−1, 1] maps onto the half-line [1/2, ∞) and the upper half circle, {z : |z| = 1, Im z > 0} onto the half-line {w1 : Re w1 = 1/2, Im w1 > 0}., Therefore, f1 maps D onto D1 = {w1 : Re w1 > 1/2, Im w1 > 0}., Step 2: The map w2 = f2 (w1 ) = w1 − 1/2 maps the domain D1 onto the, first quadrant D2 = {w2 : Re w2 > 0, Im w2 > 0}., Step 3: The map w3 = f3 (w2 ) = w22 maps D2 onto the upper half-plane, H = {w3 : Im w3 > 0}., +, , +, 3 −i, Step 4: The map w = f4 (w3 ) = w, w3 +i carries the upper half-plane H onto, the unit disk {w : |w| < 1}. Finally a map f with the desired property is a, composition, , w = f (z) = (f4 ◦ f3 ◦ f2 ◦ f1 )(z) = f4 (f3 (f2 (f1 (z)))), which gives, w = f (z) =, , (1 + z)2 − 4i(1 − z)2, ., (1 + z)2 + 4i(1 − z)2, , •, , Example 11.24. Let D = {z : |z| < 1, |z − 1/2| > 1/2}. Now we want to, find a conformal map of D onto the unit disk Δ. As we can see from the, picture, it suffices to focus on certain key points to understand the sequence, of mappings considered here. If w1 = 1/(1 − z), then z = 1 − 1/w1 and
Page 408 :
400, , 11 Conformal Mapping and the Riemann Mapping Theorem, , Figure 11.10. A conformal map of D onto the strip, , , , |z| < 1, ⇐⇒ Re w1 > 1/2, |z − 1/2| > 1/2 ⇐⇒ Re w1 < 1., , Because of the basic property of Möbius transformations, it follows easily that, f1 maps D onto the strip D1 = {w1 : 1/2 < Re w1 < 1}. A similar explanation, may be provided for other mappings. Finally, the composition, w = f (z) = f4 ◦ f3 ◦ f2 ◦ f1 (z), gives the formula which does the required job, where, w2 = f2 (w1 ) = iπ(w1 − 1/2), w3 = f3 (w2 ) = ew2 , f4 (w3 ) =, , w3 − i, ., w3 + i, , •, , We are now ready to formally state and prove the Riemann mapping theorem which is a classical example of existence theorems., Theorem 11.25. (Riemann Mapping Theorem) Suppose D is a simply, connected domain, other than the whole plane, and z0 is a point in D. Then, there exists a unique function f (z), analytic and univalent in D, which maps, D onto the disk |w| < 1 in such a manner that f (z0 ) = 0 and f (z0 ) > 0., Proof. We first prove the uniqueness of the mapping function f . If g1 and g2, are two functions each of which maps D onto the unit disk |w| < 1 in the, prescribed manner, then h = g2 ◦ g1−1 is an analytic and univalent mapping, of the unit disk |w| < 1 onto itself. Furthermore,, h(0) = g2 (g1−1 (0)) = g2 (z0 ) = 0, and, because g1 (z0 ) > 0 and g2 (z0 ) > 0,, h (0) = g2 (g1−1 (0))(g1−1 ) (0) =, , g2 (z0 ), > 0., g1 (z0 ), , Hence, by Lemma 11.22, h is the identity function. That is, g1 (z) = g2 (z) and, uniqueness is proved.
Page 409 :
11.3 Riemann Mapping Theorem, , 401, , To prove existence of the mapping function, we first show that there is an, analytic and univalent function mapping D into the disk |w| < 1. Since D is, not the whole plane C, there is a point a ∈ C \D. If there is actually a disk, |z − a| < outside of D, then |z − a| > for all points z in D. In this case,, w=, , z−a, , is an analytic and univalent function that maps all points of D into the unit, disk |w| < 1. Thus, the proof follows if D is a bounded domain. However, if D, is unbounded, then it is possible that the complement of D does not contain, any disk. For instance, D might be the plane minus a ray from some point z0, to ∞. This kind of difficulty will be avoided by considering a branch of the, square root function, which maps a domain onto one “half” its size., According to Corollary 7.52, if a ∈ C \D, then there exists an analytic, function φ : √, D → C, called analytic branch of (z − a)1/2 with φ2 (z) = z − a so, that φ(z) = z − a. Furthermore, φ(z) is univalent in D. For if φ(z1 ) = φ(z2 ), for z1 , z2 ∈ D, then, [φ(z1 )]2 = [φ(z2 )]2 , i.e., z1 − a = z2 − a., Now let D = φ(D). Then D is simply connected since D is simply connected. Then the complement of D contains a disk. To see this, we will show, that points b and −b cannot simultaneously be in D . For if they are, then, there exist two points z1 and z2 in D such that φ(z1 ) = b and φ(z2 ) = −b., Now,, φ(z1 ) = −φ(z2 ) =⇒ [φ(z1 )]2 = [φ(z2 )]2, =⇒ z1 − a = z2 − a, i.e., z1 = z2, =⇒ b = −b, i.e., φ(z1 ) = 0 = φ(z2 ), =⇒ z2 = a ∈ C \D,, contradicting the fact that z1 and z2 are distinct., Next choose a point w0 ∈ D and an > 0 so that the disk |w − w0 | < is, contained in D . Then the disk |w + w0 | < is contained in the complement, C \D . Hence the function, ψ(w) =, w + w0, maps D into the unit disk, because |w + w0 | > for all w ∈ D . Therefore,, the composition, f (z) = ψ(φ(z)) =, φ(z) + w0, is analytic and univalent in D and maps D into the unit disk. By a suitable, bilinear transformation (fill in details!), we can transform this function into a, function f0 (z) satisfying the additional conditions f0 (z0 ) = 0 and f0 (z0 ) > 0.
Page 410 :
402, , 11 Conformal Mapping and the Riemann Mapping Theorem, , Let F denote the family of all analytic functions g : D → C such that g(z), is univalent in D, g(z0 ) = 0, g (z0 ) > 0, and satisfies |g(z)| < 1 for all z in D., The family F is nonempty because f0 (z) ∈ F. Certainly the function whose, existence we are determined to prove must also be in the family F. It will be, shown that the desired function has a larger derivative at z0 than any other, function in F. To show the existence of a function in F with a maximum, derivative at z0 , we will rely on the theory of normal families., Since the family F is locally uniformly bounded (in fact, uniformly, bounded) in D, it follows from Theorem 11.14 that F is a normal family., Set, A = lub {g (z0 ) : g ∈ F}., Then, A > 0 because g (z0 ) > 0 for each g ∈ F. But A may be infinite., By the definition of A, there is a sequence {fn } of functions in F such that, fn (z0 ) → A. By the normality of F, there exists a subsequence {fnk } that, converges uniformly on the compact subsets of D to an analytic function, f (z). An application of Corollary 8.18 shows that f (z0 ) = A, so that A is, finite. Since f (z0 ) ≥ f0 (z0 ) > 0, the function f (z) is not constant in D. It, thus follows from Theorem 11.18 that f (z) is univalent and, consequently, a, member of F., We shall now show that this f maps D onto the unit disk, and so it is, the required function. For the sake of obtaining a contradiction we suppose, that f (D) is not the whole unit disk |w| < 1. Then f (z) = α for some α with, |α| < 1. By the definition of analytic branch of square roots, there exists an, analytic function F (z) in D so that, F (z)2 =, , f (z) − α, ., 1 − αf (z), , The univalence of F (z) follows from the univalence of f (z), and the inequality |F (z)| < 1 follows from the inequality |f (z)| < 1. However, F (z) is not, properly normalized. We therefore consider the function, G(z) =, , |F (z0 )| F (z) − F (z0 ), ,, F (z0 ) 1 − F (z0 )F (z), , which satisfies G(z0 ) = 0 and G (z0 ) > 0, so that G(z) ∈ F. Moreover,, G (z0 ) =, , |F (z0 )|, 1 + |α|, = A > A = f (z0 ),, 1 − |F (z0 )|2, 2 |α|, , contradicting the maximality of f (z0 ). Thus f (z) omits no values inside the, unit disk, and the proof is complete., Remark 11.26. Since univalence in a domain guarantees a nonvanishing, derivative, the Riemann mapping theorem shows that any two simply connected domains (neither of which is the plane) are conformally equivalent.
Page 411 :
11.3 Riemann Mapping Theorem, , 403, , In the proof of Theorem 11.25, we assumed that an analytic, univalent, function maps simply connected domains onto simply connected domains. In, elementary topology, it is proved that the one-to-one continuous image of a, simply connected domain cannot be multiply connected. Thus, we conclude, that no simply connected domain can be conformally equivalent to a multiply, connected domain., •, Remark 11.27. Recall that a bilinear transformation maps circles and, straight lines onto circles and straight lines. Hence any conformal mapping, of a domain, other than a disk or a half-plane, onto the interior of the unit, circle must be accomplished by a function other than a bilinear transformation. Furthermore, by the uniqueness property of the Riemann mapping, theorem, no univalent function other than a bilinear transformation can map, a disk or a half-plane onto the interior of the unit circle., At this point, we must reflect on a sobering thought. The Riemann mapping theorem, like many existence theorems, has the drawback of not furnishing much insight into the actual construction. Therefore, given two “unfamiliar” simply connected domains, we must plod along as before to develop, •, techniques for determining an appropriate mapping function., Remark 11.28. The mapping of the interior of an arbitrary polygon onto, the interior of the unit circle, whose existence is guaranteed by the theorem,, can be found explicitly. This is accomplished in several stages. The Schwarz–, Christoffel formula gives an analytic and univalent mapping of the upper, half-plane onto the interior of an arbitrary polygon. For a complete discussion, of the Schwarz–Christoffel transformation, we refer the reader to Nehari [N]., Composing the inverse of such a mapping with a bilinear transformation from, the upper half-plane onto the open unit disk (see Section 3.3) gives the desired, •, mapping., Example 11.29. Let f : Ω → Ω be analytic in a simply connected domain Ω, (= C) having a fixed point in Ω. Then it can easily be shown that |f (a)| ≤ 1,, and if |f (a)| = 1, then f is actually a homeomorphism from Ω onto Ω., The Riemann mapping theorem assures the existence of a bijective conformal map φ : Ω → Δ such that φ(a) = 0. Then we see that g defined, by, g(z) = φ ◦ f ◦ φ−1 (z), maps Δ into Δ and satisfies the hypothesis of the Schwarz lemma. Now, we, easily see that g (0) = f (a) and so |f (a)| ≤ 1, because |g (0)| ≤ 1. Moreover,, |f (a)| = 1 =⇒ |g (0)| = 1, =⇒ g(z) = eiα z, =⇒ φ ◦ f ◦ φ, , −1, , =⇒ f (z) = φ, , (by the Schwarz lemma), , (z) = eiα z, , −1, , (eiα φ(z))
Page 412 :
404, , 11 Conformal Mapping and the Riemann Mapping Theorem, , which implies that f must be a bijective mapping from Ω onto Ω, because, •, φ : Ω → Δ and φ−1 : Δ → Ω are bijective maps., Questions 11.30., 1. Must the convergence be uniform in Theorem 11.18 in order for the, conclusion to be valid?, 2. Are there conformal mappings from multiply connected domains onto, multiply connected domains?, 3. If f (z) is analytic and conformal in a domain D1 and maps D1 onto D2 ,, are D1 and D2 conformally equivalent?, 4. What other initial conditions could we have prescribed in the Riemann, mapping theorem to guarantee uniqueness?, 5. Does there exist a one-to-one conformal mapping from the unit disk, onto the disk minus the origin?, 6. If two domains are conformally equivalent, what can be said about their, boundaries?, 7. Does there always exist an analytic function which maps a simply connected domain Ω(= C) into the unit disk |z| < 1?, 8. Let Ω (= C) be a simply connected domain and let F be the set of all, one-to-one analytic functions which map Ω into the unit disk |z| < 1,, and a ∈ Ω. If f ∈ F and is not onto, is there a function g ∈ F such that, |g (a)| > |f (a)|?, 9. Are the plane C and the unit disk |z| < 1 conformally equivalent? Are, they homeomorphic?, 10. Are the plane C and the upper half-plane Im w > 0 conformally equivalent? Are they homeomorphic?, 11. In the statement of the Riemann mapping theorem, why do we require, the domain D to be a proper subset of C? Does the theorem still hold, if we remove that assumption?, 12. Does the proof of the Riemann mapping theorem use the fact that every, nonvanishing analytic function in a simply connected domain D admits, analytic square root function in D?, 13. Where, in the proof of the Riemann mapping theorem, did we require, the domain to be simply connected?, 14. Why was it necessary to first show that some function mapped the, domain into the unit disk?, 15. Why does the function G(z), constructed in the proof of the Riemann, mapping theorem, work?, 16. What is a conformal map between the upper half-plane H+ =, {z : Im z > 0} and C \ [0, ∞)?, 17. What is a conformal map between the right half-plane D1 =, {z : Re z > 0} and D2 = {z : |Arg z| < π/8}?, 18. What is a conformal map between the strip D1 = {z : 0 < Im z < π/2}, and the upper half-plane H+ ?
Page 413 :
11.4 The Class S, , 405, , 19. What is a conformal map between the strip D1 = {z : 0 < Im z < α}, and the upper half-plane H+ ?, 20. What is a conformal map between the infinite strip |Re z| < π/2 and, the unit disk |w| < 1?, 21. What is a conformal map between the unit disk |z| < 1 and C \ Δ?, Exercises 11.31., 1. Suppose f (z) is analytic at z0 with f (z0 ) = 0. Show that there exist, neighborhoods U and V of z0 and f (z0 ), respectively, such that f (z) is, a univalent mapping from U onto V ., 2. Show that the plane is not conformally equivalent to the upper halfplane. More generally, show that the plane is only conformally equivalent, to itself., 3. Let D1 = {z : 0 < Re z, Im z < ∞} and D2 = {w : Im w > 0} be, the open first quadrant and the upper half-plane, respectively. By the, Riemann mapping theorem D1 and D2 are conformally equivalent. Show, that f (z) = z 2 does this job., 4. Let D1 = {z : |Re z| < π/2} and D2 = {w : Re w > 0}. Show that, f : D1 → D2 given by f (z) = eiz is conformal., 5. Even though the interior of a square can be mapped conformally onto, the interior of a circle, show that no square can be mapped conformally, onto a circle., 6. Let D1 be the annulus 0 < r1 < |z| < R1 and D2 be the annulus, 0 < r2 < |z| < R2 . If, R2, R1, =, ,, r1, r2, construct an analytic and univalent function that maps D1 onto D2 ., 7. Suppose D1 and D2 are conformally equivalent, and that D2 and D3 are, conformally equivalent. Show that D1 and D3 are conformally equivalent., , 11.4 The Class S, We continue our investigation of univalent functions—a specialized topics in, complex analysis. Analytically, a univalent function has a nonvanishing derivative (Theorem 11.3); geometrically, a univalent function maps simple curves, onto simple curves., Functions that are both analytic and univalent have a nice property of, mapping simply connected domains onto simply connected domains. By the, Riemann mapping theorem, we can associate a univalent function defined in, an arbitrary simply connected domain (other than the whole plane) with one, defined in the unit disk. Therefore, we shall restrict the domain on which these, functions are defined to the disk |z| < 1. Our results will have a nicer form
Page 414 :
406, , 11 Conformal Mapping and the Riemann Mapping Theorem, , if we also assume that the function has a zero (hence its only zero) at the, origin and that its derivative is equal to one at the origin. Since the derivative, of a univalent function never vanishes, every univalent function h(z) may be, reduced to a function of this form by replacing it with, f (z) =, , h(z) − h(0), ., h (0), , We shall denote by S the class of all functions f (z) that are analytic and, univalent in the unit disk |z| < 1, and are normalized by the conditions, f (0) = 0 and f (0) = 1. Thus a function f (z) in S has the power series, representation, f (z) = z + a2 z 2 + a3 z 3 + · · ·, , (|z| < 1)., , We shall denote by T the class of all functions of the form, g(z) = z + b0 +, , b2, b1, + 2 + ···, z, z, , that are analytic and univalent in the domain |z| > 1. The following relationship will enable us to deduce information about S from information about, T., Theorem 11.32. If f (z) ∈ S, then 1/f (1/z) ∈ T ., Proof. First suppose 1/f (1/z1 ) = 1/f (1/z2 ) (|z1 | > 1, |z2 | > 1). Then, f (1/z1 ) = f (1/z2 ), where |1/z1 | < 1 and |1/z2 | < 1. The univalence of, 1/f (1/z) (|z| > 1) now follows from the univalence of f (z) (|z| < 1). The, analyticity of 1/f (1/z) will be a consequence of the analyticity of f (z) if we, can show that f (1/z) = 0 for |z| > 1. If f (1/z0 ) = 0 for 0 < |1/z0 | < 1, then, f (0) = f (1/z0 ) = 0, contradicting the univalence of f (z) for |z| < 1. Hence, 1/f (1/z) ∈ T , and the proof is complete., The next theorem, because of its proof rather than its statement, is known, as the area theorem., Theorem, 11.33. If g(z) = z + b0 + (b1 /z) + (b2 /z 2 ) + · · · is in T , then, ∞, 2, n=1 n|bn | ≤ 1., Proof. The univalent function g(z) maps the circle |z| = r > 1 onto a simple, closed contour C. Set g(z) = u(z) + iv(z). The area of the region R enclosed, by C, denoted by A(r), is, ++, A(r) =, du dv., R, , Note that A(r) > 0 for each r > 1. If we now let P (u, v) = −v/2 and, Q(u, v) = u/2, an application of Green’s theorem yields
Page 415 :
11.4 The Class S, , A(r) =, , 1, 2, , +, u dv − v du =, C, , 1, 2, , +, , 2π, , 0, , , , ∂u, ∂v, −v, u, dθ,, ∂θ, ∂θ, , 407, , (11.11), , where A(r) > 0. By Exercise 5.2(13), we have g (z) = (1/iz)(∂g/∂θ). To, evaluate the line integral of (11.11), consider the integral, , , +, +, ∂v, 1 2π, 1 ∂u, 1, , +i, iz dθ, (11.12), g(z)g (z) dz =, (u − iv), 2 |z|=r, 2 0, iz ∂θ, ∂θ, , , , , +, +, 1 2π, ∂v, ∂u, i 2π, ∂u, ∂v, =, +v, −v, dθ +, dθ,, u, u, 2 0, ∂θ, ∂θ, 2 0, ∂θ, ∂θ, whose imaginary part corresponds to A(r). In order to simplify (11.12), we, write, , , , +, +, ∞, ∞, , , , −m, −n−1, 1−, dz,, g(z)g (z) dz =, z+, bm (z), nbn z, |z|=r, , and note that, , |z|=r, , +, , m=0, , −m −n−1, , (z), , z, , |z|=r, , n=1, , , dz =, , 2πir−2m if n = m,, 0 if n = m., , This leads to the identity, ∞, +, +, +, 2 −2n, 1, 1, 1, , n=1 n|bn | r, dz, g(z)g (z) dz =, z dz −, 2 |z|=r, 2 |z|=r, 2 |z|=r, z, , , ∞, , n|bn |2, 2, = πi r −, ., r2n, n=1, Therefore (11.12) is purely imaginary, and, 1, A(r) =, 2, , +, 0, , 2π, , , , , , ∞, 2, , ∂u, ∂v, n|b, |, n, −v, ., u, dθ = π r2 −, ∂θ, ∂θ, r2n, n=1, , (11.13), , Since A(r) > 0, we have, r2 −, , ∞, , n|bn |2, >0, r2n, n=1, , (r > 1)., , (11.14), , But (11.14) is valid for every r > 1 so that the result follows upon letting, r → 1+ ., Remark 11.34. According to (11.13), the area enclosed by the image of the, circle |z| = r is at most πr2 (the area enclosed by the circle), with equality, only for g(z) = z + b0 . Furthermore, equality in the conclusion of the theorem, holds if and only if the area enclosed by the image of |z| = r > 1 becomes, •, arbitrarily small as r → 1.
Page 416 :
408, , 11 Conformal Mapping and the Riemann Mapping Theorem, , Remark 11.35. If b1 = 1, then bn = 0 for n > 1. Recall that the properties of g(z) = z + 1/z were extensively studied in Section 3.3. In particular,, this function was shown to map |z| = r > 1 onto an ellipse, and the ellipse, approaches the linear segment [−2, 2] as r approaches 1., •, The coefficient bound for functions in T , as expressed by the area theorem,, will enable us to obtain a coefficient bound for functions in S. But first we, need the following:, , Lemma 11.36. If f (z) ∈ S, then z f (z 2 )/z 2 ∈ S., ∞, Proof. Set f (z) = z + n=2 an z n . Then, f (z 2 ) = z 2 [1 + a2 z 2 + a3 z 4 + · · · ] := z 2 h(z),, where h(z) is analytic and never vanishes in the unit disk. Therefore, choosing, a branch of (h(z))1/2 with (h(0))1/2 = 1, we see that g(z) defined by, /, , f (z 2 ), g(z) = z, =, z, 1 + a2 z 2 + a3 z 4 + · · ·, (11.15), z2, is analytic with g(0) = 0 and g (0) = 1. To prove that g(z) is univalent,, suppose g(z1 ) = g(z2 ). Then f (z12 ) = f (z2 2 ), and the univalence of f (z) shows, that z12 = z22 , that is, z1 = ±z2 . But from (11.15), we see that g(z) is an odd, function. Hence, z1 = −z2 implies g(z1 ) = −g(z2 ), which is a contradiction, unless z1 = z2 = 0. Therefore z1 = z2 , thus establishing the univalence of, g(z)., , , Remark 11.37. It was necessary to write z f (z 2 )/z 2 instead of f (z 2 ), because f (z 2 ) has a zero at the origin, which makes the expression, , , f (z 2 ) = e(1/2) Log f (z, , 2, , ), , •, , meaningless., , Theorem 11.38. If f (z) = z + a2 z 2 + · · · is in S, then |a2 | ≤ 2., , Proof. By Lemma 11.36, g(z) = z f (z 2 )/z 2 ∈ S. We can verify from the, expansion in (11.15) that g (0) = 3a2 . Thus we may write, g(z) = z +, , a2 3, z + ··· ., 2, , In view of Theorem 11.32, the Laurent expansion for 1/g(1/z) shows that, 1, a2 1, 1, =, =z−, + ··· ∈ T ., g(1/z), (1/z)[1 + (a2 /2)z 2 + · · · ], 2 z, Applying Theorem 11.33, we find that |a2 /2|2 ≤ 1, i.e., |a2 | ≤ 2.
Page 417 :
11.4 The Class S, , 409, , Remark 11.39. Retracing the steps in the proof, we can determine when, iα, equality holds. For if a2 = 2eiα , α real,, then 1/g(1/z) = z − e /z. But this, iα 2, 2, 2, means that g(z) = z/(1 − e z) = z f (z )/z , so that, f (z) =, , z, = z + 2eiα z 2 + 3e2iα z 3 + · · · ., (1 − eiα z)2, , (11.16), , For each α ∈ R, this function is known as the Koebe function Moreover, it is, easy to verify that the functions f maps |z| < 1 onto the w plane cut along, •, the ray with constant argument from − 14 e−iα to ∞., The functions in (11.16) are extremal for Theorem 11.38 in the sense that, there is equality on the bound for the second coefficient. Impressed by the, fact that the Koebe function appears in many problems concerning the class, S, Bieberbach asked whether we always have |an | ≤ n. This give rise to the, famous, ∞, Bieberbach Conjecture. If f (z) = z + n=2 an z n is in S, then |an | ≤ n, for every n., Theorem 11.38 proves the conjecture for n = 2. Although stated in 1916,, the conjecture was verified only for the values of n up to n = 7 until Louis, de Branges proved the whole conjecture in 1985. For all n the maximization, of |an | is achieved only by the Koebe function. A large amount of research in, the theory of univalent functions is centered on the Bieberbach conjecture., The result for n = 2 can be used to prove the following elegant theorem, which shows that this mapping property is, in a sense, extremal., Theorem 11.40. If f (z) ∈ S and f (z) = c for |z| < 1, then |c| ≥ 14 ., Proof. Set f (z) = z + a2 z 2 + · · · . Since f (z) = c, the function, , , 1, cf (z), = z + a2 +, z2 + · · ·, g(z) =, c − f (z), c, is also in S. Applying Theorem 11.38 to g(z), we get |a2 + (1/c)| ≤ 2. Thus,, |1/c| − |a2 | ≤ |(1/c) + a2 | ≤ 2. Now, applying Theorem 11.38 to f (z), we have, |1/c| ≤ 2 + |a2 | ≤ 4, and the result follows., Remark 11.41. Theorem 11.40 is known as a covering theorem or Koebe onequarter theorem. It says that every function in S maps the unit disk |z| < 1, onto a domain in the w plane that contains the disk |w| < 14 . This result has, a lot of interesting applications in many other parts of complex analysis. By, the inverse function theorem (also by the open mapping theorem), f (|z| < 1), contains an open neighborhood of the origin (since f (0) = 0 and f (0) = 0)., •, The Koebe 14 –theorem actually estimates the size of this neighborhood., Finally, we end the section with the following results which provides a, sufficient condition for an analytic function to be univalent.
Page 418 :
410, , 11 Conformal Mapping and the Riemann Mapping Theorem, , Theorem 11.42. If f (z) is analytic in a convex domain D, and Re f (z) > 0, in D, then f (z) is univalent in D., Proof. Choose distinct points z0 , z1 ∈ D. Then the straight line segment z =, z0 + t(z1 − z0 ), 0 ≤ t ≤ 1, must lie in D. Integrating along this path, we get, +, f (z1 ) − f (z0 ) =, , z1, , z0, , f (z) dz =, , +, 0, , 1, , f (z0 + t(z1 − z0 )) (z1 − z0 ) dt., , Dividing by z1 − z0 and taking real parts, we have, , + 1, f (z1 ) − f (z0 ), Re, f (z0 + t(z1 − z0 )) dt, = Re, z 1 − z0, 0, , > 0., , Thus f (z1 ) = f (z0 ), and f (z) is univalent in D., Questions 11.43., 1. What kind of results could have been obtained in this section if the, functions had not been normalized?, 2. What was the importance of the class T ?, 3. Why was a bound on |a2 | so useful?, 4. Can |a2 | = 2 if f (z) is a bounded function in S?, 5. Why is the Koebe function extremal for so many theorems?, 6. For each n, are we guaranteed the existence of a function in S for which, the absolute value of its nth coefficient is at least as large as the absolute, value of the nth coefficient for any other function in S?, Exercises 11.44., 1. Give an example of a function that is univalent but not analytic in the, disk |z| < 1., 2. (a) If f (z) ∈ S, show that for any nonzero complex number t, |t| ≤ 1,, the function f (tz)/t ∈ S., 2, / S., (b) If f (z) = z/(1, ∞− z) kand |t0 | > 1, show that f (t0 z)/t0 ∈, in S, show that, for each integer n, there, 3. If f (z) = z + k=2 ak z is , ∞, exists a function g(z) = z + k=2 bk z k in S such that bn = |an |., 4. For α real, verify that z/(1 − eiα z)3 is univalent in |z| < 12 , but in no, larger disk centered at the origin., 5. If f (z) ∈ S, show that z(f (z k )/z k )1/k ∈ S for every positive integer k., 6. Let f (z) be analytic, D and suppose C is a closed contour, * in a domain, , f, (z)f, (z), dz, is, in D. Prove that, ∞ purely imaginary., ∞ C, 7. If f (z) = z + n=2 an z n and n=2 n|an | ≤ 1,show that f (z) ∈ S., ∞, ∞, 8. If f (z) = z − n=2 |an |z n is in S, show that n=2 n|an | ≤ 1.
Page 419 :
12, Entire and Meromorphic Functions, , We begin this chapter by defining infinite products, and show that their convergence properties are similar to those of infinite series. Just as infinite series, were used as tools to develop power series expansions for analytic functions, so, infinite products may be used as tools to develop product expansions for analytic functions. As we shall see, a comparison of product and series expansions, enables us to determine some interesting identities., Given any sequence {an } tending to ∞, we show that there exists an entire, function whose only zeros are at {an }. In one sense, this theorem is nicer than, the Riemann mapping (existence) theorem of Section 11.3, because we can, actually construct the entire function. We next consider an arbitrary sequence, {bn } tending to ∞, and show that a function can always be found that has, poles at {bn } and is analytic otherwise. In reading this chapter, it is worth, keeping in mind the similarities between properties of zeros and poles., , 12.1 Infinite Products, An, .∞ infinite product is an expression of the form u1 u2 u3 · · · (denoted by, n=1 un ), where the un are complex numbers. By analogy with infinite, .nseries,, we are tempted to say that an infinite product converges if limn→∞ ( k=1 uk ), exists. However, such a definition would be incomplete, because the vanishing, of one term would necessitate the convergence of the infinite product regardless of the behavior of the other factors. This certainly is not in keeping with, the spirit of “limit” definitions. We shall thus assume that no factor of an, infinite product vanishes. Even so, we can (unlike the case for finite products), have an infinite ., product be zero although none of its factors is zero. For inn, stance, limn→∞ k=1 (1/k) = 0. Such an occurrence (the reasons for which, will become, .∞ evident later on) we also wish to avoid. We say that an infinite, if and only if there is an N such that uk = 0 for, product n=1 un converges, .n, all k ≥ N , limn→∞ k=N uk exists and is nonzero. An infinite product that, does not converge is said to diverge. Moreover, if convergence condition holds
Page 420 :
412, , 12 Entire and Meromorphic Functions, , but, .∞ finitely many uk ’s are equal to zero, then we say that the infinite product, case, the value of the product is set as 0. If, n=1 un converges to zero. In this ., n, uk = 0 for all k ≥ 1, and limn→∞ k=1 uk = 0, then we say that the infinite, product diverges to zero., To help clarify the definition of an infinite product, we make several simple, observations. When discussing convergence, we need only to consider, infinite, .n, products whose factors are all nonzero. Now, we set Pn = k=1 u., k , where, ∞, uk = 0 for all k ≥ 1. Here Pn is called the n-th partial product. If n=1 un, converges, then the sequence {Pn } approaches some nonzero value P . Thus,, un =, , Pn, P, = 1 as n → ∞., →, Pn−1, P, , It, is therefore convenient to express a convergent infinite product in the form, .∞, n=1 (1 + an ), where an → 0. We formulate, Theorem 12.1..(Necessary condition for convergence of a product) If the, ∞, infinite product n=1 (1+an ) converges, then an → 0 as n → ∞, and an = −1, for at most finitely many n., As is the case with infinite series, the convergence of ., the sequence {an }, ∞, to 0 is not sufficient for the convergence of the product n=1 (1 + an ). For, example,, , n , , n+1, 2 3 4, 1, = n + 1 → ∞,, Pn =, = · · ···, 1+, k, 1 2 3, n, k=1, , so that, , .∞, , n=1 (1, , Pn =, , + 1/n) diverges; similarly,, , n+1, , , 1−, , k=2, , 1, k, , , =, , n−1, n, 1, 1 2, · ···, ·, =, → 0,, 2 3, n, n+1, n+1, , .∞, , so that n=2 (1 − 1/n) diverges., It certainly seems, between, .comparison, ∞natural to investigate further the, ∞, a, and, the, infinite, product, (1, +, a, )., In the, the infinite series, n, n, n=1, n=1, special case that an ≥ 0 for all n, the following relationship is particularly, nice., .∞, Theorem, 12.2. If an ≥ 0, the product n=1 (1 + an ) converges if and only if, ∞, the series n=1 an converges., Proof. First note that, Sn := a1 + a2 + · · · + an ≤ Pn := (1 + a1 )(1 + a2 ) · · · (1 + an ),, since all the terms are nonnegative. Also, ex ≥ 1 + x for nonnegative x. Thus, we have the double inequality
Page 421 :
12.1 Infinite Products, , 413, , a1 + a2 + · · · + an ≤ (1 + a1 )(1 + a2 ) · · · (1 + an ) ≤ ea1 ea2 · · · ean ., That is,, Sn ≤ Pn ≤ eSn ., , (12.1), , ∞, , If n=1 an converges to the real number S, then Pn is an increasing sequence, that is bounded above (by eS ), and hence must converge. Conversely,, ∞if Pn, converges to P , then the left-hand inequality of (12.1) shows that n=1 an, converges to a value no greater than P ., p, For, ∞instance, if an = O(1/n ) as n → ∞, then we know that the sea, converges, for, p, >, 1 and diverges for p ≤ 1. Consequently,, ries, .∞ n=1 n p, (1, +, 1/n, ), converges, for, p, >, 1 and diverges for p ≤ 1., n=1, As we shall see in the next theorem, similar results are obtained when the, terms of the product are negative., .∞, Theorem, 12.3. If an ≥ 0, an = 1, then n=1 (1 − an ) converges if and only, ∞, if n=1 an converges., ∞, Proof. First suppose that n=1 an converges. By the Cauchy criterion, there, exists N such that, 1, aN + aN +1 + · · · + an <, 2, and 0 ≤ an < 1 for all n ≥ N . We have, , (1 − aN )(1 − aN +1 ) = 1 − aN − aN +1 + aN aN +1 ≥ 1 − aN − aN +1 >, , 1, ., 2, , It can be shown by induction that for n ≥ N ,, n, , , (1 − ak ) ≥ 1 −, , k=N, , Write, Pn =, , n, , k=1, , n, , , ak >, , k=N, , (1 − ak ) = PN −1, , n, , , 1, ., 2, , (12.2), , (1 − ak )., , k=N, , Therefore, Pn /PN −1 is a decreasing sequence (since 0 < 1−an ≤ 1 for n ≥ N ), and has a lower bound. Thus, we get from (12.2) that Pn /PN −1, .∞approaches, a value P , 12 ≤ P ≤ 1. Thus Pn → PN −1 P = 0, and hence k=1 (1 − ak ), converges., ∞, To prove the converse, we suppose that n=1 an diverges. If an → 0, then, 1 − an → 1 and the product diverges. So we may assume, without loss of, generality, that an → 0. Then 0 ≤ an < 1 for n ≥ N . From the identity, 4, , 2, x3, x, x5, x, −x, −, −, +, + ··· ,, e =1−x+, 2!, 3!, 4!, 5!
Page 422 :
414, , 12 Entire and Meromorphic Functions, , we see that 1 − x ≤ e−x for 0 ≤ x < 1, because all the terms in parenthesis, are nonnegative. (Alternatively, it suffices to note that f (x) = (1 − x)ex is, decreasing on [0, ∞) so that f (x) ≤ f (0) = 1). Hence, , , n, n, n, , , , 0≤, (1 − ak ) ≤, exp(−ak ) = exp −, ak, (n ≥ N )., k=N, , k=N, , k=N, , ∞, , .∞, Letting n →, .∞∞, the divergence of k=N ak shows that k=N (1 − ak ) = 0., Therefore k=1 (1 − ak ) diverges, and the proof is complete., Remark 12.4. If a product were allowed to converge to 0, Theorem 12.3, •, would be false, as can be seen by letting an = 1/(n + 1)., on {an } are relaxed, the comparison between, ∞, .∞When the restrictions, n=1 (1+an ) and .n=1 an is less straightforward. In the, ∞exercises, we give an, ∞, (1+a, ), diverges, even, though, an converges, and, example for which n=1, n, n=1, .∞, ∞, an example for which n=1 (1 + an ) converges even though n=1 an diverges., In relating infinite products to infinite, .∞ series in the general case, we will, make use of the complex logarithm. If n=1 (1 + an ) = P = 0, then, &∞, ', , log, (1 + an ) = log P., n=1, , ∞, However, this does not mean that the series n=1 log(1 + an ) converges to, log P . Suppose, n, , (1 + ak ) = |Pn |ei arg Pn ., Pn =, k=1, , Then |Pn | → |P |; but all we can say about arg Pn is that, arg Pn → arg P (mod 2π)., .∞, In , order to compare the convergence of n=1 (1 + an ) with the convergence, ∞, of n=1 log(1 + an ), it is necessary to deal with multiple-valued properties, of the logarithm. The key step in the proof of our next theorem consists of, showing that, for a convergent product, the arguments of the partial products, eventually cluster about a fixed point when the same branch of arg(1 + ak ) is, chosen for each k., .∞, Theorem 12.5. For an complex,, ∞an = −1, the product n=1 (1 + an ) converges if and only if the series n=1 Log (1 + an ) converges., .n, Proof. Set Pn = k=1 (1 + ak ), and write, log Pn = ln |Pn | + i arg Pn =, , n, , k=1, , Log (1 + ak ) =: Sn ,, , (12.3)
Page 423 :
12.1 Infinite Products, , 415, , where for each k branch of the logarithm has been chosen that satisfies, −π < Im Log (1 + ak ) = Arg (1 + ak ) ≤ π. This determines the branch of, log Pn ., Suppose that Pn → P = 0. Then, ln |Pn | → ln |P | and arg Pn → arg P (mod 2π)., Thus there exists a sequence of real numbers { n },, integers {mn } such that for all n,, arg Pn = arg P + 2πmn +, , n, , → 0, and a sequence of, , n., , (12.4), , We will show that mn is constant (say m) for n sufficiently large. To see this,, by (12.3), we consider the difference, log Pn+1 − log Pn = Log (1 + an+1 ), so that, arg Pn+1 − arg Pn = Arg (1 + an+1 ) = 2π(mn+1 − mn ) +, , n+1, , −, , n., , Since an → 0, we have Arg (1 + an ) → 0. Hence for all n > N ,, 2π|mn+1 − mn | ≤ |Arg (1 + an+1 )| + |, , n+1 |, , + | n | < 2π., , Therefore mn+1 = mn = m for n > N , and from (12.4) we see that, arg Pn → arg P + 2mπ., ∞, In view of (12.3), it now follows, nthat n=1 Log (1 + an ) converges., Conversely, suppose Sn = k=1 Log (1 + an ). Then, we have, n, , eSn = e, , k=1, , Log (1+ak ) = e Log (1+a1 ) · · · e Log (1+an ) = P ., n, , (12.5), , Since the exponential function ez is continuous, Sn → S implies that, eSn → eS as n → ∞., .∞, Thus, letting n → ∞ in (12.5) we find that k=1 (1 + ak ) = eS = 0., Theorem 12.5 conveys that any question deal about infinite products can, be translated into a question, .∞ about infinite series by taking logarithms., The infinite, product, n=1 (1 + an ), an = −1, is said to be absolutely, .∞, convergent if n=1 (1+|an |) converges. As in the case with series, an absolutely, convergent product is convergent. Before proving this, we need the following., ∞, Lemma 12.6., n=1 an converges absolutely if, ∞For an complex, an = −1,, Log, (1, +, a, ), converges, absolutely., This occurs if and only, and, only, if, n, n=1, .∞, if n=1 (1 + |an |) converges.
Page 424 :
416, , 12 Entire and Meromorphic Functions, , Proof. If either of the two series converges, then there exists an N such that, |an | ≤ 12 for n ≥ N . The Maclaurin expansion, , , ∞, , z3, z4, (−1)n n, z2, Log (1 + z) = z −, +, −, + ··· = z 1 +, z, 2, 3, 4, n+1, n=1, is valid for |z| < 1. In particular, for |z| ≤ 12 ,, ∞, ∞, , 1 1, 1, (−1)n n, z ≤, =, n, n+1, 2 n=1 2, 2, n=1, , which shows that, 3, 1, |z| ≤ | Log (1 + z)| ≤ |z| for |z| ≤ 1/2., 2, 2, Setting z = an , we have for n ≥ N ,, 3, 1, |an | ≤ | Log (1 + an )| ≤ |an |., 2, 2, Hence either both series converge absolutely or neither series converges absolutely., , n+1, .∞ , It is easy to see that n=1 1 + (−1)n, converges to 1 (see Exercise, ∞ 1, 12.171(d) in 12.17) but not absolutely as n=1 n diverges., Remark 12.7. In view of Lemma 12.6, Theorems 12.2 and 12.3 are seen to, •, be special cases of Theorem 12.5., .∞, Example 12.8. Consider the product k=1 (1 + 1/[k(2k + 3)]). Clearly (for, example, by Theorem 12.5 and Lemma 12.6), the product converges. To find, the limit value, we write the kth factor as, , , , k+1, 2k + 1, 1, =, 1+, k(2k + 3), k, 2k + 3, , , so that, Pn = 3, as n → ∞. Similarly, writing, , n+1, 2n + 3, , , →, , 3, 2, , , , , k+1, 2, k+2, =, ,, k(k + 3), k, k+3, .∞, we see that the product k=1 (1 + 1/[k(k + 3)]) converges to 3 because, , , n+1, → 3 as n → ∞., Pn = 3, n+3, 1+
Page 425 :
12.1 Infinite Products, , 417, , Finally, considering the equality, , , , , 2, k, k+3, =, ,, (k + 1)(k + 2), k+1, k+2, .∞, we obtain that the product k=1 (1 − 2/[(k + 1)(k + 2)]) converges to 1/3, because, , , 1 n+3, 1, Pn =, → as n → ∞., 3 n+1, 3, Note that the latter product is the inverse of the former., •, .∞, .∞, Theorem 12.9. If n=1 (1 + an ) converges absolutely, then n=1 (1 + an ), converges. The converse is not true., ∞, .∞, By Theorem 12.2, n=1 |an | conProof. Suppose n=1 (1 + |an |) converges., ∞, verges. Then according to Lemma 12.6, n=1 | Log (1 + an )| converges. Since, an absolutely convergent series is convergent, the result follows from Theorem, 12.5., .∞, The product n=1 (1 + (−1)n+1 /n) converges to 1 but does not converge, absolutely., 1−, , Remark 12.10. The terms of an absolutely convergent product can be rearranged without affecting the convergence or the value of the product. Its, proof is similar to the comparable proof for infinite series, and is left as an, exercise for the reader., •, √, , ∞, k−1, Example 12.11., / k + 1. Then we see that k=1 ak, .∞Consider ak = (−1), converges but k=1 (1 + ak ) diverges, where 1 + ak = 0 for each k. To show, that the product diverges, we let, , n , , (−1)k−1, ., 1+ √, Pn =, k+1, k=1, Then, P2n, , √, √, , √, √, , 2+1, 3−1, 2n + 1, 2n + 1 − 1, √, √, √, √, =, ···, 2n + 1, 2, 3, 2n, √, , √, √, , √, 3+1 3−1, 2n + 1 + 1 2n + 1 − 1, √, √, √, √, ≤, ···, 2n + 1, 2, 3, 2n, , , , , , 2, 4, 2n, √ √, = √ √, ··· √ √, 2 3, 4 5, 2n 2n + 1, 1, 1, 1, , ··· , = , 3/2 5/4, (2n + 1)/2n, 1, = , , , , ., 1, 1, 1, 1+, 1+, ··· 1 +, 2, 4, 2n
Page 426 :
418, , 12 Entire and Meromorphic Functions, , .∞, , (1 + 1/(2k)), diverges to ∞, it follows that P2n → 0 as, k−1, .∞ , √, n → ∞. In particular, k=1 1 + (−1), is divergent, whereas the series, k+1, ∞ (−1)k−1, √, , by alternating series test, converges but not absolutely., •, k=1, k+1, As the product, , k=1, , Examples 12.12. We easily have, , ∞ , , ∞, (−1)k (1 + i), (i), 1+, converges (since k=1 1/k 3 converges), 3, k, k=1 , , ∞, , i, diverges, 1+ , (ii), k(k + 1), k=1, , ∞ , , (−1)k, (iii), 1+ √, diverges, k, k=1, , ∞ , ∞, , , i, Log (1 + i/k) are divergent,, 1+, and the series, (iv) the product, k, k=1, k=1, whereas, 1/2, ∞, ∞ , , , i, 1, =, 1+, 1+ 2, k, k, k=1, k=1, ∞ −2, converges, because the series k=1 k converges., •, Just as we went from series of complex numbers to series of functions,, so may we go from products of complex numbers to products of complex, functions. Also, as in the case of series of complex functions, the concept, of uniform convergence plays an important role in the study of product of, functions., Let {fn (z)}n≥1 be a sequence of functions defined on a region Ω. Then, the infinite product, ∞, , [1 + fn (z)], n=1, , is said to converge uniformly on Ω iff, (i) there exists an, .nN such that fn (z) = −1 for all n > N and all z ∈ Ω, (ii) the sequence k=N +1 [1+fk (z)] converges uniformly on Ω to some P (z),, where P (z) = 0 for all z ∈ Ω., The most useful test for uniform convergence of products is analogous to, the M -test (see Theorem 6.31) which has been used extensively to establish, uniform (and absolute) convergence of series., Theorem 12.13. (M-test for the convergence of a product) Suppose that, is a sequence of functions, such that |fn (z)| ≤ Mn for all z in a region, {fn (z)}, .∞, ∞, then, Ω. If n=1 Mn converges,, n=1 [1 + fn (z)] converges uniformly in Ω. In, .∞, addition, if f (z) = n=1 [1 + fn (z)] and each fn (z) is analytic in Ω, then f (z), is analytic in Ω. Also, f (z0 ) = 0 for some z0 ∈ Ω if and only if fn (z0 ) = −1
Page 427 :
12.1 Infinite Products, , 419, , for some n. The order of the zero of f at z0 is the sum of the order of zeros, of the functions 1 + fn (z) at z0 ., .∞, Proof. According to Theorem 12.2, n=1 [1 + fn (z)] converges absolutely,, hence converges,, for each point in Ω. It suffices to show that the sequence, .n, Pn (z) = k=1 [1 + fk (z)] is uniformly Cauchy in Ω. Note that for any positive, integers m and n (m < n), we have, Pn (z) − Pm (z) =, , n−1, , , [Pk+1 (z) − Pk (z)] =, , k=m, , n−1, , , Pk (z)fk+1 (z)., , (12.6), , k=m, , But for all k,, |Pk (z)| ≤, , ∞, , , (1 + |fn (z)|) ≤ exp, , n=1, , ∞, , ∞, , , |fn (z)| ≤ exp, , n=1, , ∞, , , Mn := eM ., , n=1, , By Theorem 6.26, the series n=1 |fn (z)| converges uniformly in R. Hence,, n−1, choosing m large enough in (12.6) so that k=m |fk+1 (z)| < for all z ∈ Ω, and for all n, we have, |Pn (z) − Pm (z)| ≤, , n−1, , , n−1, , , |Pk (z)| |fk+1 (z)| ≤ eM, , k=m, , |fk+1 (z)| < eM ., , k=m, , Since is arbitrary, it follows that the sequence {Pn (z)} is uniformly Cauchy, in Ω and the proof of the first part is complete., Next suppose that f (z0 ) = 0 for some z0 ∈ Ω. Then by the definition of, the convergence of infinite products, there exists an N such that, FN (z) =, , ∞, , , (1 + fk (z)), , k=N +1, , is nonvanishing at z0 . By the above discussion, FN (z) is the limit of a uniformly convergent sequence of analytic functions. Hence the limit FN (z) is, analytic in Ω. Continuity of FN (z) at z0 shows that FN (z) is nonvanishing in, some neighborhood D(z0 ; δ) of z0 . Now, , , n, ∞, N, , , , f (z) =, (1 + fk (z)) =, (1 + fk (z)) lim, (1 + fk (z)) ., k=1, , k=1, , n→∞, , k=N +1, , Note that the second factor is analytic and nonvanishing on D(z0 ; δ). Therefore, the zero of f (z) and their order arise only from the zeros of the factor, .N, k=1 (1 + fk (z))., Remark 12.14. Showing the sequence {Pn (z)} to be uniformly Cauchy does, not preclude the possibility that Pn (z) → 0 for some z. This is why it was, necessary to show that {Pn (z)} also converged pointwise (to a nonvanishing, function).
Page 428 :
420, , 12 Entire and Meromorphic Functions, , .∞, k, Example 12.15. Consider the product k=0 (1 + z 2 ). Clearly, the series, ∞ 2k, converges absolutely for |z| < 1, and hence the product converges, k=0 z, absolutely for |z| < 1. Now we observe that, (1 − z)P0 (z) = 1 − z 2, 2, , (1 − z)P1 (z) = (1 − z 2 )(1 + z 2 ) = 1 − z 2, .., .., ., ., n, n, n+1, (1 − z)Pn (z) = (1 − z 2 )(1 + z 2 ) = 1 − z 2, and hence, for |z| < 1, we have, (1 − z) lim Pn (z) = lim (1 − z 2, n→∞, , n→∞, , n+1, , ) = 1,, , as desired. More generally, we have, ∞ , z 2k , , R, 1+, =, R, R−z, , for |z| < R., , k=0, , In particular,, ∞ , , , 1+, , k=1, , z 2k , R, , =, , R2, − z2, , R2, , for |z| < R., , •, , Questions 12.16., 1. What are the similarities between infinite series and infinite products?, 2. If a sequence of partial products {Pn } converges, does {log Pn } converge? What about the converse?, 3. In view of ., Theorem 12.2 and Theorem 12.3, is it true, ∞ for real sequences, ∞, {an } that n=1 (1 + an ) converges if and only if n=1 an converges?, 4. How, an infinite product diverge? ., .∞many ways can ., ∞, ∞, 5. If n=1 (1 + an ) and n=1 (1 + bn ) converge, does n=1 (1 + an )(1 + bn ), also, converge?, .∞, .∞, .∞, 6. If n=1 (1 + a., n ) and, n=1 (1 + bn ) diverge, does, n=1 (1 + an )(1 + bn ), ∞, diverge? Can n=1 (1 + an )(1 + bn ) be convergent?, 7. If, α and β are two real!numbers such that α + β = −1, does the product, .∞, α, β, converge?, n=1 1 + n (1 + n), 8. We know that a series converges if and only if every subsequence obtained by deleting a finite number of terms of the series converges. Does, the definition of infinite product assure the analogous statement for infinite, .∞products?, .∞, 9. If . n=1 (1 + an ) converges,, does n=1 |1 + an | converge?, ., ∞, ∞, 10. If n=1 (1 + an ) and n=1 (1 + |an |) diverge, what can we say about the, sequence {an }?
Page 429 :
12.1 Infinite Products, , 421, , .∞, 11. Can n=1 (1+fn (z)) converge uniformly but not absolutely? Absolutely, but not uniformly?, 12. What would we mean by an infinite product satisfying the Cauchy criterion?, .∞, ∞ What is its relation to convergence?, said about n=1 (1 + an z)? How, 13. If n=1, n | converges, what can be ., .|a, ∞, ∞, about n=1 (1 + an z 2 )? How about n=1 (1 + an p(z)), where p(z) is, some polynomial?, 14. How would Theorem 12.5 be proved without making the assumption, that −π < arg(1 + ak ) ≤ π for each k?, Exercises 12.17., 1. Show that, , , , .∞, .∞, 1, 1, 1, (a) n=2 1 − 2 =, =2, (b) n=1 1 +, n, 2, n(n + 2), , , , , .∞, .∞, 2, 1, (−1)n+1, (c) n=2 1 −, =, =1, (d) n=1 1 +, n(n + 1), 3, n, . ∞ n2 − 1, . ∞ n3 − 1, 2, =4, (e) n=3 3, = ., (f) n=2 3, n −4, n +1, 3, 2. Show that the product, , , , , , , 1, 1, 1, 1, 1, 1−, 1+, 1−, 1+, 1−, ···, 2, 2, 3, 3, 4, converges, but not absolutely., ∞, 3. Suppose, {an } is real with |an | < 1. If n=1 an converges, show that, .∞, ∞, 2, n=1 (1 + an ) converges if and only if, n=1 an converges., 4. Set, ⎧, 1, 1, 1, ⎪, ⎪, ⎨ √ + + √ if n even,, n n n n, an =, 1, ⎪, ⎪, if n odd., ⎩−√, n, ∞, ∞ 2, .∞, Show that n=1 (1 + an ) converges but both, n=1 an and, n=1 an, diverge., 5. Suppose that {an } is a decreasing sequence of real numbers, with, lim an = 0., , n→∞, , Show that, verges., 6. Set, , .∞, , n=1 [1, , + (−1)n an ] converges if and only if, ⎧, 1, 1, ⎪, ⎪, if n odd,, ⎨√ +, n n, an =, 1, ⎪, ⎪, if n even., ⎩−√, n, , ∞, n=1, , a2n con-
Page 430 :
422, , 12 Entire and Meromorphic Functions, , .∞, ∞, Show that n=1 (1+an ) converges even though n=1 an diverges. Does, this provide an example of an infinite product that is convergent but, not absolutely?, 7. Show that the following either all converge or all diverge:, ∞, , , |an |,, , n=1, , ∞, , , ∞, , , (1 + |an |),, , n=1, , | log(1 + an )|,, , n=1, , ∞, , , log(1 + |an |)., , n=1, , 8. Determine, the region of convergence, for, .∞, .∞, .∞, n, (a) n=1 (1 + z n ), (b) n=1 (1 − z 2 ), (c) n=2 (1 − n−z ), .∞, .∞, .∞, (d) n=1 cos(z/n), (e) n=1 sin(z/n), (f) n=2 cos(z/2n )., ∞, 9. Suppose an > 0 for every n, and n=1 an converges. Show that, a1, , ∞ , , , 1+, , n=2, , where sn =, , n, k=1, , an, , , , sn−1, , =, , ∞, , , an ,, , n=1, , ak ., , 12.2 Weierstrass’ Product Theorem, Let us now consider the factorization of entire functions. As a first step consider an entire function f which does not vanish in C. Then we may express, f (z) as, f (z) = exp(g(z)),, where g(z) is an entire function. In fact, as f (z)/f (z) is analytic in C,, f (z)/f (z) possesses an anti-derivative g(z) so that, g (z) =, , f (z), f (z), , for some entire function g. Using this, we see that, (f (z)e−g(z) ) = 0,, , i.e., f (z) = c exp(g(z)), , for some constant c. Absorbing the constant c into g(z), we obtain the desired, representation., Next, given a finite set of complex numbers {0, a1 , a2 , . . . , an } (ak = 0, for 1 ≤ k ≤ n), we can always find an entire function (in fact, a polynomial), having zeros at these points of order m, m1 , . . . , mn , respectively. Since such, an entire function is analytic in C with no singularities except at ∞, one such, entire function is the polynomial, mk, n , , z, m, ., 1−, p(z) = z, ak, k=1
Page 431 :
12.2 Weierstrass’ Product Theorem, , 423, , Moreover, if f is an entire function with a finite number of zeros at 0, ak, (ak = 0) for 1 ≤ k ≤ n, of order m, mk (1 ≤ k ≤ n), respectively, then, h(z) =, , f (z), p(z), , has removable singularities at 0, ak (1 ≤ k ≤ n). It follows then that h(z), defines an entire function with no zeros in C. As a consequence,, f (z) = eg(z) p(z) = eg(z) z m, , mk, n , , z, 1−, ., ak, , k=1, , That is, f (z) can be expressed as a product of a polynomial and a nonvanishing, entire function., Finally, the question arises as to whether an entire function can always, be found whose only zeros are at an arbitrarily prescribed sequence of points., Unfortunately, the answer is no. For example, if an entire function has zeros, at 1/n (n ∈ N), then according to Theorem 8.47, the entire function must be, identically zero. More generally, a nonconstant entire function cannot have a, limit point of zeros in C. Consequently, the set of zeros of an entire function, which has infinitely many zeros in C must have ∞ as its only limit point. For, example,, 0 = cos z =, , 0 = sin z =, , eiz + e−iz, =⇒ e2iz = −1 = ei(π+2kπ), 2, =⇒ z = (2k + 1)π/2, k ∈ Z;, eiz − e−iz, =⇒ e2iz = 1 = e2kπi , i.e., z = kπ,, 2, , k ∈ Z,, , and ez − 1 = 0 =⇒ z = 2kπi for k ∈ Z. Thus, in each case, the limit point, of the zeros of cos z, sin z, and ez is ∞. We will show, however, that if the, sequence of zeros of an entire function has no finite limit point, then the, question concerning factorization can be answered in the affirmative., Suppose a sequence {an }n≥1 approaching ∞ is arranged so that, 0 < |a1 | ≤ |a2 | ≤ |a3 | ≤ · · · ., A naive guess for an appropriate entire function is, , ∞ , , z, 1−, ., f (z) =, an, n=1, .∞, Unfortunately, n=1 (1 − z/an ) may diverge. Certainly if f (z) is entire, then, the function has its only zeros at points of the sequence {an } and nowhere, else.
Page 432 :
424, , 12 Entire and Meromorphic Functions, , Example 12.18. Let, first discuss an example with some details. Consider, .us, ∞, the infinite product n=1 (1 − z 2 /n2 ) which has zeros at z = ±n, n ∈ N. Fix, an arbitrary R > 0. If |z| ≤ R, then choose N large enough so that, |z/n| ≤ R/n < 1 for all n ≥ N ,, and so, 1 − z 2 /n2 = 0 for |z| ≤ R and for n ≥ N . In view of this observation,, we may write, , , ∞ , ∞ , , , z2, z2, 1 − 2 = PN −1 (z), 1 − 2 = PN −1 (z)FN (z)., n, n, n=1, n=N, , Note that PN −1 (z) is entire and has zeros only at the points z = ±n (n < N ), and FN (z) is an infinite product with no zeros in |z| ≤ R. Also,, ∞, ∞, ∞, , , , z2, 1, 1, 2, 2, = |z|, ≤R, <∞, 2, 2, 2, n, n, n, n=1, n=1, n=1, , and so, by Theorem 12.13, the infinite product is uniformly convergent for, |z| ≤ R and hence, , on every compact subsets of C. By Theorem 12.13, we, .∞, 2, conclude that n=1 1 − nz 2 is entire, and the infinite product is zero only, •, at z = ±n, n ∈ N., We shall now, ∞determine the restrictions on {an } for which f (z) is entire., Suppose that, n=1 1/|an | converges. Fix an arbitrary R > 0. If |z| ≤ R,, of Theorem 12.13 shows that Pn (z), then |z/an | ≤ R/|an | and an application, .∞, converges uniformly to f (z) = n=1 (1 − z/an ) in |z| ≤ R and hence, on, every compact subset of C. By Theorem 8.16, f (z) must therefore be an, entire function. For example, we can now easily construct, .∞ an entire function, whose only zeros are at 1, 4, 9, 16, . . . . The product n=1 (1 − z/n2 ) is such a, function., However, it is more difficult to construct, an entire function whose zeros, .∞, are at the positive integers. The expression n=1 (1 − z/n) does not represent, an entire function. In fact, setting z = −1 we see that, , n , , 1, = lim (n + 1) = ∞., 1+, lim, n→∞, n→∞, k, k=1, , What, producing” factor. Moreover, if the series, ∞, ∞ is needed is a “convergence, 2, 1/|a, |, diverges, but, 1/|a, n, n | converges, we can modify the above, n=1, n=1, construction and show that, , ∞ , , z, 1−, ez/an, an, n=1, is entire. We will first show that
Page 433 :
12.2 Weierstrass’ Product Theorem, , 425, , ∞ , , z z/n, e, f (z) =, 1−, n, n=1, , is an entire function and the same method can be adapted to solve a more, general problem. We set, 1 + fn (z) = (1 − z/n)ez/n ., To determine the uniform convergence of the product, we need to find an, upper bound for |fn (z)|. We write, 1 + fn (z) = exp ( Log (1 − z/n) + z/n) ., Fix an arbitrary R > 0. If |z| ≤ R, then choose N large enough so that, N ≥ 2R. Then |z/n| ≤ R/n < 1 for all n ≥ N , and so the identity, ∞, , , 1 z k, z , Log 1 −, =−, n, k n, , (12.7), , k=1, , is valid for |z| ≤ R. We get, ∞, ∞ k, , , 1 R, 1 z k, z z, = −, ≤, Log 1 −, +, n, n, k n, 2, n, k=2, , k=2, , R2, 1 R2 /n2, ≤ 2,, =, 2 1 − R/n, n, because R < 2R ≤ N ≤ n. Hence,, |fn (z)| = |exp ( Log (1 − z/n) + z/n) − 1|, ≤ exp (| Log (1 − z/n) + z/n|) − 1, ≤ exp(R2 /n2 ) − 1, ≤ (R2 /n2 ) exp(R2 /n2 ), 2, , 2, , ≤ e(R /n ) = Mn, so that, , ∞, , , |fn (z)| ≤ eR2, , n=N, , According to Theorem 12.5,, Hence, , .∞, , (since ex − 1 ≤ xex for x ≥ 0), , (since R/n < 1), ∞, , 1, <∞, n2, , (N > 2R)., , n=N, , n=1 (1 + fn (z)), , converges uniformly for |z| ≤ R., , ∞, ∞ , N, −1 , , , z z/n, z z/n , z z/n, f (z) =, e, e, e, =, 1−, 1−, 1−, n, n, n, n=1, n=1, n=N, , is an entire function with the prescribed zeros. Another such function may be, obtained just by multiplying f (z) by any nonvanishing entire function.
Page 434 :
426, , 12 Entire and Meromorphic Functions, , .∞, Similarly, f (z) = n=1 (1 + z/n)e−z/n is an entire function with simple, zeros at the negative integers and no other zeros. Consequently, an entire, function having a simple zero at each integer is given by, z, , ∞, ∞ , ∞ , , , z z/n−z/n, z z/n , z −z/n, z , e, e, 1+, e, 1−, 1+, 1−, =z, n, n, n, n, n=1, n=1, n=1, , ∞ , , z2, =z, 1− 2 ., n, n=1, , The rearrangement of the factors in the first equality is justified by the absolute convergence of the infinite products., A more general method for constructing an entire function with prescribed, zeros is indicated by the identity (12.7)., √ For instance, suppose we want an, entire function to have its zeros at z = n, n ∈ N. Since, , , , , 2, 3, z, z, 1, z, 1, z, √, √, −, − ··· ,, Log 1 − √, = −√ −, 3, n, n 2, n, n, the above method shows that, , ∞ , , √, 2, z, 1− √, e(z/ n)+(1/2)(z /n), n, n=1, is such a function. Similarly, an entire function that has simple zeros on the, real axis at points z = ±n1/4 (n ≥ 0) and nowhere else is given by, , ∞ , , 2 √, 4, z2, 1− √, z, ez / n+(1/2)(z /n) ., n, n=1, More generally, if, , ∞, , 1/|an |p+1 converges for some positive integer p, then, , , ∞ , , z, z, 1−, Ep, (12.8), f (z) =, an, an, n=1, , n=1, , is an entire function whose only zeros are at z = an , where, !, Ep (z) = exp z + (1/2)z 2 + · · · + (1/p)z p, and is referred to as the convergence producing factor., As general as (12.8) appears, we still have not accounted for all sequences., For instance, suppose we wish to construct an entire function whose zeros, occur at the points log n (n = 2, 3, 4, . . . ). We cannot use (12.8) because, , ∞, p, n=2 [1/(ln n) ] diverges for all p. Observe that the convergence producing, factors in (12.8) involve a sequence of polynomials, all of degree p. In the, general case, we will not place a uniform bound on the degree of the polynomials. This, in turn, will enable us to construct an appropriate entire function, without regard to the convergence of a series involving its zeros.
Page 435 :
12.2 Weierstrass’ Product Theorem, , 427, , Theorem 12.19. (Weierstrass’s Product Theorem) Given any complex sequence having no finite limit point, there exists an entire function that has, zeros at these points and only these points., Proof. We suppose that the entire function f (z) to be constructed is to have, zeros at {an } so arranged that 0 < |a1 | ≤ |a2 | ≤ |a3 | ≤ · · · . We have assumed,, without loss of generality, that none of the an is zero, for if the k of these points, are zero, then z k f (z) has the desired property., For each n, set, 2, n, z, 1 z, 1 z, +, + ··· +, Pn (z) =, an, 2 an, n an, so that exp(Pn (z)) = En (z/an ), where En (z) is the convergence producing, factor as defined above. We wish to show that the function, , , ∞ , , z, z, 1−, En, f (z) =, an, an, n=1, satisfies the conditions of the theorem. As in the proof of above example, it, suffices to show that, , , , ∞ , , z, Log 1 −, + Pn (z), an, n=1, converges uniformly on an arbitrary compact subset |z| ≤ R of the plane., Choose |an | large enough so that |an | ≥ 2R ≥ 2|z|. Then, we have, , , k, ∞, , z, 1 z, Log 1 −, + Pn (z) = −, an, k an, ≤, ≤, , k=n+1, ∞, , , 1, n+1, , k=n+1, ∞, n, , 1, z, n + 1 an, , z, ≤, an, , n, , z, an, , k=1, , k, , 1, 2k, , 1, ≤ n., 2, , If we set 1 + fn (z) = exp ( Log (1 − z/an ) + Pn (z)), then we have, |fn (z)| ≤ exp(1/2n ) − 1 ≤ (1/2n )e., .∞, According to Theorem 12.5, n=1 (1 + fn (z)) converges uniformly to f (z) for, |z| ≤ R. Since R was arbitrary, the infinite product defines an entire function, with prescribed zeros. The assertion in the theorem about the zeros of f (z), follows from the definition.
Page 436 :
428, , 12 Entire and Meromorphic Functions, , Remark 12.20. The theorem does not exclude the case of multiple zeros. It, •, is certainly possible that ak = ak+1 for some k., Remark 12.21. The function f (z) is by no.means uniquely determined by, ∞, the zeros. For instance, we have seen that n=1 (1 − z/n)ez/n is an entire, function having zeros at the positive integers. In the proof of the theorem, we, have shown that, ∞ , , z (z/n)+(1/2)(z/n)2 + ··· +(1/n)(z/n)n, e, 1−, n, n=1, , is another such function. We can also show that, ∞ , , z (z/n)+(1/2)(z/n)2, ,, 1−, e, n, n=1, , ∞ , , z (z/n)+(1/2)(z/n)2 +(1/3)(z/n)3, ,, 1−, e, n, n=1, , and infinitely many more entire functions also have their only zeros at the, positive integers., •, Suppose two entire functions have the same zeros with the same multiplicities. How do the two functions compare? Since the complex plane C is simply, connected, our characterization is a consequence of Theorem 7.51. We have, Theorem 12.22. If f (z) and g(z) are entire functions whose zeros coincide, in location and in multiplicity, then there exists an entire function φ(z) such, that f (z) = eφ(z) g(z)., Proof. After we cancel the common factors, the function f (z)/g(z) is seen to, be an entire function with no zeros. The result follows from Theorem 7.51 (see, also the discussion in the beginning)., Weierstrass’s theorem shows that for a preassigned sequence of points, we, can construct an infinite product that represents an entire function having, zeros at the preassigned sequence. What about the converse? Given an entire, function whose zeros are known, can we construct an infinite product representation for the function? In view of Theorem 12.22, this is always possible, up to a multiplicative exponential function. The explicit determination of the, exponential function is usually quite difficult. Before we solve this problem, for the function sin πz, it is appropriate to formulate Theorem 12.22 in the, following form., Theorem 12.23. Let {an }n≥1 be a sequence of nonzero complex numbers and, f (z) be an entire function that has zeros at an , listed with multiplicities. Suppose that f has a zero of order k ≥ 0 at zero. Then there exists an entire, function g(z) such that, , , ∞ , , z, z, f (z) = z k eg(z), 1−, En, ., an, an, n=1
Page 437 :
12.2 Weierstrass’ Product Theorem, , 429, , More generally, one can replace the product in the last equation by, , , ∞ , , z, z, 1−, Epn, a, a, n, n, n=1, where {pn }n≥1 is a sequence in N such that for each R > 0,, pn +1, ∞ , , R, < ∞., |an |, n=1, .n, Suppose that Pn (z) =, k=1 [1 + fk (z)] is a finite product of analytic, functions on a domain D. Then logarithmic differentiation gives, Pn (z) fk (z), =, ., Pn (z), 1 + fk (z), n, , k=1, , Pn (z)/Pn (z), , Note that, has poles at the zeros of Pn (z). This procedure continues to hold for uniformly convergent infinite products of analytic functions, and we state the following result whose proof is left as a simple exercise., Theorem 12.24. Suppose, .∞ {fn (z)}n≥1 is a sequence of analytic functions in, a domain D and let n=1 [1 + fn (z)] converge uniformly on D to f (z). Then, ∞, , f (z) fk (z), =, ,, f (z), 1 + fk (z), k=1, , where the sum converges uniformly on D when f (z) = 0., The following result is often useful for expanding entire or meromorphic, functions., Theorem 12.25. Let f be analytic except for simple poles at a1 , a2 , . . . and, be arranged so that, 0 < |a1 | ≤ |a2 | ≤ · · · ≤ |an | ≤ · · · ,, , with bn = Res [f (z); an ]., , Let {Cn } be a sequence of positively oriented simple closed contours such that, each Cn includes a1 , a2 , . . . , an but no other poles. Suppose that, Rn = dist (0, Cn ) → ∞ as n → ∞,, Ln = length of Cn = O(Rn ),, |f (z)| = o(Rn ) on Cn, (e.g., the last condition is satisfied if f (z) is bounded on all Cn ). Then,, , , ∞, , 1, 1, f (z) = f (0) +, bn, +, z − an, an, n=1, for all z except at these poles.
Page 439 :
12.2 Weierstrass’ Product Theorem, g(z), , sin πz = e, , , ∞ , , z2, z, 1− 2 ,, n, n=1, , 431, , (12.10), , where g(z) is some entire function. Suppose, for the moment, we could show, that g(z) is constant. Then (12.10) could be written as, , ∞ , , z2, sin πz = cz, 1− 2 ., n, n=1, Since limz→0 (sin πz)/z = π = limz→0 c, have, sin πz = πz, , .∞, , n=1 (1, , ∞ , , , 1−, , n=1, , − z 2 /n2 ) = c, we would then, , z2, n2, , , ., , (12.11), , The remainder of the proof will consist of showing that g(z) is indeed constant,, thus justifying (12.11). To this end, suppose that z is not an integer. Then we, form the logarithm derivative in (12.10) to obtain, π cot πz = g (z) +, = g (z) +, , ∞, , 1 , −2z, +, z n=1 n2 (1 − z 2 /n2 ), , (12.12), , ∞, , 1 2z, +, ,, z n=1 z 2 − n2, , ∞, where term-by-term differentiation is justified because n=1 Log (1 − z 2 /n2 ), converges uniformly on every compact subset of the open set that exclude the, integers. It suffices to show that g (z) ≡ 0. To do this, we use the method, developed in the proof of Theorem 12.25. Now we consider, %, 1, π cot πz −, for z = 0, f (z) =, z, 0 for z = 0, and Cn to be the square with vertices at z = (n + 1/2)(±1 ± i), n ∈ N. Then, f (z) is analytic at the origin having simple poles only at n, n ∈ Z \{0}, and, the contour Cn does not pass through the poles of f (z). Clearly, the function, 1/z is bounded on these squares and for each n ∈ Z \{0},, , , 1, π cos πz, cos πn, −, = 1., Res [f (z); n] = lim (z − n), =, z→n, sin πz, z, cos πn, Next we show that there exists a positive number M such that, | cot πz| ≤ M =, , 1 + e−3π, for z ∈ Cn ., 1 − e−3π, , To see this, we observe that for z = n +, , 1, 2, , + iy (y ∈ R),
Page 440 :
432, , 12 Entire and Meromorphic Functions, , | cot πz| = | tan(iπy)| =, , e−πy − eπy, ≤ 1 < M., e−πy + eπy, , Similarly, for z = x + i(n + 12 ) (x ∈ R), we can calculate, | cot πz| =, , 1 + e−(2n+1)π, 1 + e−3π, e2iπz + 1, ≤, ≤, = M., e2iπz − 1, 1 − e−3π, 1 − e−(2n+1)π, , Since cot(πz) = cot(−πz), the same bound is valid on the other two sides, of Cn , which confirms that cot(πz) is bounded on all contours Cn taken as a, whole. By Theorem 12.25, we get, k , , , f (z) = lim, , k→∞, , = lim, , k , , , k→∞, , =, , n=−k, n=0, , n=1, , ∞, , n=1, , z2, , 1, 1, +, z−n n, , , , 1, 1, +, z−n z+n, , , , 2z, ., − n2, , We conclude that the identity, ∞, , π cot πz =, , 1 2z, +, z n=1 z 2 − n2, , (12.13), , is valid for all nonintegral values of z. A comparison of (12.13) with (12.12), shows that g (z) ≡ 0. But this means that g(z) is constant, which verifies, (12.11)., Whenever an infinite product expansion for an entire function is found, it, is of interest to compare it with a power series expansion. This can often lead, to interesting relationships. To illustrate, we have, , ∞ , , (πz)5, z2, (πz)3, +, − ··· ., 1 − 2 = πz −, sin πz = πz, n, 3!, 5!, n=1, The z 3 term in the infinite product is, , , ∞, , 1, 1, 1, ., πz(−z 2 ) 1 + 2 + 2 + · · · = −πz 3, 2, 3, n2, n=1, By the uniqueness of the Maclaurin expansion, we must have, −π, , ∞, , π3, 1, ., =, −, n2, 3!, n=1, , (12.14)
Page 441 :
12.2 Weierstrass’ Product Theorem, , 433, , ∞, This gives the identity n=1 1/n2 = π 2 /6., As we have seen, the function, f (z) =, , ∞ , , z −z/n, 1+, e, n, n=1, , (12.15), , is an entire function having simple zeros at the negative integers and no other, zeros. The function f (z − 1) has the same zeros in addition to a zero at the, origin. Hence, f (z − 1) = eg(z) zf (z),, , (12.16), , where g(z) is some entire function. Next we show that g(z) is actually a, constant. Forming the logarithmic derivative in (12.16), we get, 1 f (z), f (z − 1), = g (z) + +, f (z − 1), z, f (z), which, by Theorem 12.24, gives, ∞ , , n=1, , 1, 1, −, n+z−1 n, , , , ∞, , 1 , = g (z) + +, z n=1, , , , , 1, 1, −, n+z, n, , , ., , (12.17), , The sum on the left-hand side of (12.17) can be expressed as, , , , ∞ , 1, 1, 1, −1 +, −, z, n+z−1 n, n=2, , , ∞ , , 1, 1, 1, 1, 1, −1+, −, −, +, z, n+z, n, n n+1, n=1, , , ∞, , 1, 1, 1, = −1+, −, +1, z, n, +, z, n, n=1, , ∞ , 1 , 1, 1, = +, −, ., z n=1 n + z, n, =, , Comparing this with the right side of (12.17), we find that g (z) ≡ 0. Thus, g(z) = γ, γ a constant. To determine γ, we set z = 1 in (12.16). This gives, 1 = f (0) = eγ f (1) = eγ, , ∞ , , n=1, , 1+, , 1, n, , , , e−1/n ,, , or, , e−γ = lim [(1 + 1)(1 + 1/2) · · · (1 + 1/n) exp (−(1 + 1/2 + · · · + 1/n))] ., n→∞, , Therefore, using the natural logarithm, we find
Page 442 :
434, , 12 Entire and Meromorphic Functions, , , , , 1, − ln 1 +, γ=, n, n, n=1, , , 1, 1 1, = lim 1 + + + · · · + − ln(n + 1) ., n→∞, 2 3, n, ∞ , , 1, , (12.18), , The constant γ in (12.18) is known as Euler’s constant; its numerical value, is approximately 0.577., The fact that this limit exists gives a “sophisticated”, ∞, way of showing that n=1 (1/n) diverges. The function, Γ (z) =, , ∞, e−γz , z −1 z/n, 1, :=, e, 1, +, eγz zf (z), z n=1, n, , (12.19), , is known as the gamma function. It represents an analytic function at all, points except the negative integers and zero, where it has simple poles. Since, g(z) = γ in (12.16), we have, f (z) = eγ (z + 1)f (z + 1)., This enables us to determine the most important property of the gamma, function; namely, the functional equation, Γ (z + 1) =, , 1, 1, = γz, = zΓ (z)., e f (z), eγ(z+1) (z + 1)f (z + 1), , (12.20), , When z = n, a positive integer, (12.20) shows that, Γ (n + 1) = nΓ (n) = n(n − 1)Γ (n − 1) = · · · = n(n − 1) · · · 2Γ (1)., But, Γ (1) =, , 1, 1, = γ −γ = 1,, eγ f (1), e e, , so that Γ (n + 1) = n!. We collect the above piece of information as, Theorem 12.26. The gamma function is analytic in C except at the simple, poles at 0, −1, −2, . . . . Also, Γ (z + 1) = Γ (z) and Γ (n + 1) = n! for n ∈ N., Moreover,, ∞ , , z −z/n, 1, = eγz z, 1+, e, Γ (z), n, n=1, is an entire function., An interesting relationship exists between the gamma function and the, sine function. Applying the relations, sin πz = πzf (z)f (−z) and f (z) = 1/zeγz Γ (z), in (12.19), we obtain
Page 443 :
12.2 Weierstrass’ Product Theorem, , sin πz =, , 435, , π, π, =, ., eγz Γ (z)(−z)e−γz Γ (−z), −zΓ (−z)Γ (z), , Since −zΓ (−z) = Γ (1 − z), it follows that, π, = Γ (z)Γ (1 − z), sin πz, for all! nonintegral values of z. In particular, setting z = 12 , we find that, Γ is obviously a positive function on (0, ∞), we deduce, Γ 2 12 = π., ! Because, √, 1, that Γ 2 = π. Also, it follows that Γ (z) is zero-free because for z =, 0, −1, −2, . . . , the gamma function is given by a convergent infinite product, of nonvanishing factors. Further, for n ∈ N, the functional equation gives, , , , , , 2n − 1, 2n − 1, 1, 1 · 3 · · · (2n − 1), 2n + 1, =, Γ, =, ,, Γ, Γ, 2, 2, 2, 2n, 2, whereas the above identity gives, , , π, 2n + 1, =, Γ −, 2, sin π(n + 32 ) Γ, , √, (−1)n+1 π, (−1)n+1 2n+1 π, !, ! =, ., 2n+3 =, 1 · 3 · · · (2n + 1), Γ 2n+3, 2, 2, 1, , Remark 12.27. In real analysis, the function, + ∞, tx−1 e−t dt (x > 0), Γ (x) =, 0, , is studied extensively. Interestingly enough, the gamma function defined by, (12.19) can be expressed in this integral form for all positive real values of z., In the next chapter, we will redefine the gamma function as a complex integral, and show that the two definitions represent the same function at all points, •, where the integral converges., Remark 12.28. We may also evaluate Γ ( 12 ) by this integral definition. We, have, + ∞, 1, =, t−1/2 e−t dt., Γ, 2, 0, A substitution of t = y 2 leads to, , √, + ∞, 2, 1, π √, Γ, = π,, e−y dy = 2, =2, 2, 2, 0, this last integral having been evaluated in Section 9.3., , •, , Questions 12.29., 1. Can an entire function be constructed that has “a” points at a preassigned sequence of points?, 2. If an → a ∈ C and if f is entire with zeros at an , then is f (z) ≡ 0 in C?
Page 444 :
436, , 12 Entire and Meromorphic Functions, , ∞, .∞, 3. If n=1 1/an diverges, can n=1 (1 − z/an ) be analytic anywhere?, 4. What is meant by a “convergence producing” factor?, 5. Is there an analog to Weierstrass’s theorem for functions analytic in an, arbitrary domain?, 6. What other choices for the sequence {Pn (z)} would have worked in the, proof, .∞, .∞, ∞of Weierstrass’s theorem?, 7. If .n=1 an converges, is n=1 (1+an z) entire? Is n=1 (1+an z 2 ) entire?, ∞, Is n=1 (1 + an p(z)) entire? Here p(z) is a polynomial., 8. What is the relationship between the infinite product and infinite series, expansion for entire functions?, 9. Knowing the infinite product expansion for sin πz, what other infinite, product expansions can we determine?, Exercises 12.30., 1. Construct an analytic function f in |z| < R such that f has zeros only, at z = −R + 1/n, n ∈ N., 2. Construct an entire function whose only zeros are at z = ln n (n =, 2, 3, 4, . . . )., 3. Construct an entire function f (z) with the following properties:, a) f (z) vanishes at z = 1, 2, 3, . . . and nowhere else., b) The zero of f (z) at z = n has multiplicity n., c) Construct an entire function f (z) such that f has zeros at z = n3/2, (n ∈ N) and nowhere else., d) Construct an entire function f (z) such that f has zeros at z = n3/4, (n ∈ N) and nowhere, .∞else., 2, 4. (a) Find the value, of, n=1 (1 + 1/n )., .∞, 4, (b) Show that n=2 (1 − 1/n ) = (eπ − e−π )/8π., 5. (a) Show that e2z − 1 and sin iz have simple zeros at the same points., (b) Set (e2z − 1)/ sin iz = eg(z) , and determine g(z)., 5, for the series and the product ex6. By comparing the term involving, ∞ z −4, pansion of sin πz, show that n=1 n = π 4 /90., ∞, −2, 7. Use , the summation formula for π cot πz to sum the series, n=1 n, ∞, −4, and n=1 n ., 8. Show that the convergence of (12.13) is uniform on all compact subsets, that contain no integers., 9. Show that, ∞ , , z z/n+(5z2 /n2 ), 1−, e, n, n=1, represents an entire function., , .∞ , 10. Suppose 0 < |a1 | ≤ |a2 | ≤ · · · → ∞. Show that n=1 1 − azn eQn (z), represents an entire function, where, 2, [ln n], z, z, 1 z, 1, Qn (z) =, +, + ··· +, ., an, 2 an, [ln n] an
Page 445 :
12.3 Mittag-Leffler Theorem, , 11. Prove that, cos z =, , ∞ , , , 1−, , n=1, , 12. Prove that, , 437, , , 4z 2, ., (2n − 1)2 π 2, ∞, , π coth πz =, , 1 2z, +, ., z n=1 z 2 + n2, , 13. If a is not an integer, show that, ∞ , , 1+, f (z) =, n=−∞, n=0, , z, a+n, , , , e−z/(n+a), , is an entire function, and that, f (z) =, , a sin π(z + a) −πz cot(πa)ez/a, e, ., (z + a) sin πa, , .∞, 14. Evaluate n=1 (1 + 1/n2 + 1/n4 )., 15. Use the product, , expansion, , for sin, πz to show that, 2·2, 4·4, 6·6, π, =, ··· ., (a), 2, 1 · 3 3 · 5 5 · 7, , √, 2·2, 6·6, 10 · 10, (b) 2 =, ··· ., 1 · 3 , 5 · 7 9· 11, , , √, 2·4, 8 · 10, 14 · 16, (c) 3 = 2, ··· ., 3·3, 9·9, 15 · 15, 16. Show that, , , ∞, 1, d Γ (z), =, ,, dz Γ (z), (z + n)2, n=0, where Γ (z) is defined by (12.19). Here the function Ψ = Γ /Γ is known, as the digamma function, or alternatively, as the Gauss psi-function., 17. Show that Γ (z) sin πz is an entire function., 18. Expand ez − 1 in an infinite product., 19. Construct an entire function having simple zeros at z = n2 (n ∈ N), and nowhere else., √ that one solution to this is given by the entire, √ Show, function (sin π z)/(π z)., , 12.3 Mittag-Leffler Theorem, A function is said to be meromorphic in a domain D if it has no singularities, other than possibly poles, in D. If no domain is specified, it will be, assumed that the function is meromorphic in the whole complex plane. Thus,, every entire function is meromorphic but the converse is not necessarily true., Sum and products of meromorphic functions are meromorphic. Quotients of
Page 446 :
438, , 12 Entire and Meromorphic Functions, , meromorphic functions are meromorphic, provided that the denominator is, not identically zero. For b ∈ C, the function 1/(z − b) is an example of the, simplest type of meromorphic function. As a consequence of the definition, we, see that the function f (z) which is meromorphic in C cannot have infinitely, many poles in a bounded region. For if it does, the sequence of poles must, have a limit point p. Since f (z) cannot be analytic in any neighborhood of p,, the point p is a singularity that is not a pole. Note that, 1, sin(1/(1 − z)), has infinitely many poles in the unit disk |z| < 1., Example 12.31. From Theorem 12.26, we observe that the gamma function defined by (12.19) is a meromorphic function in C with simple poles at, 0, −1, −2, . . . and the functional equation Γ (z + 1) = Γ (z) shows that, Γ (z + n) = (z + n − 1)Γ (z + n − 1) = (z + n − 1) · · · (z + 1)zΓ (z),, for z = 0, −1, −2, . . . . Using this, we see that, Γ (z + n + 1), (z + n − 1) · · · (z + 1)z, Γ (1), (−1)n, =, =, ,, (−1)(−2) · · · (−n + 1)(−n), n!, , lim (z + n)Γ (z) = lim, , z→−n, , z→−n, , and therefore Res [Γ (z); −n] = (−1)n /n!., , •, , The reciprocal of an entire function is meromorphic, its poles consisting, of the zeros of the entire function. More generally, we have the following, characterization of meromorphic functions., Theorem 12.32. A function is meromorphic if and only if it can be expressed, as the quotient of entire functions., Proof. First, suppose that f (z) = g(z)/h(z), where g(z) and h(z) are entire, functions with no common zeros (any common zero can be factored out). Then, the only singularities of f (z) are poles consisting of the zeros of h(z). Hence, f (z) is meromorphic., Conversely, suppose f (z) is meromorphic. Then by Weierstrass’s theorem,, there exists a function h(z) where zeros coincide in both position and order, with the poles of f (z). Therefore, g(z) = f (z)h(z) is an entire function because, the poles of f (z) are cancelled by the zeros of h(z). Thus f (z) = g(z)/h(z) is, the quotient of entire functions, and the proof is complete., In this section, we are going to prove a theorem for meromorphic functions, analogous to Weierstrass’s theorem for entire functions. Given any finite set, of points {b1 , b2 , . . . , bn }, the (meromorphic) rational function
Page 447 :
12.3 Mittag-Leffler Theorem, , 439, , 1, 1, 1, +, + ··· +, z − b1, z − b2, z − bn, has a simple pole at each point of the set. Therefore, any other meromorphic, function f (z) having a simple pole at bk with the principal part as above must, be of the form, n, , 1, f (z) =, + φ(z), z − bk, k=1, , where φ(z) is an arbitrary entire function. If the given set is infinite, the, problem of constructing a meromorphic function having a pole at each point, of the set is more complicated. Then we have to worry about the convergence., Consider the function, f (z) =, , ∞, , , 1, ,, 2 − n2, z, n=1, , which is defined and converges for all values of z except for the squares of, integers. For all z in some neighborhood of a given point not containing the, square of an integer, we have, 2, 1, ≤ 2, |z 2 − n2 |, n, for n sufficiently large. Hence by Theorem 6.31, the convergence of the series, for f (z) is uniform in some neighborhood of each such point. This shows, that f (z) is analytic at all points except z = n2 (n ∈ N) so that f (z) is a, meromorphic function having simple pole at z = n2 ., On the other hand, suppose we wish to construct a meromorphic function, having simple poles, the positive integers with residues equal to 1. The likely, at, ∞, candidate f (z) = n=1 1/(z − n) fails to converge anywhere. As in the case, of Weierstrass theorem, a convergence producing term is needed. Note that, the constant term in the power series expansion of 1/(z − n) about z = 0 is, −1/n. So we try with the function, ∞ , , n=1, , 1, 1, +, z−n n, , , =, , ∞, , , z, ., n(z, − n), n=1, , Note that for |z| ≤ R and N large enough so that N ≥ 2R, then, |z − n| ≥ n − |z| ≥ n − R ≥ n/2 for all n ≥ N ., Therefore,, 2, 1, ≤ 2 for n ≥ N, n(z − n), n, so that the series converges absolutely for all z excluding positive integers.
Page 448 :
440, , 12 Entire and Meromorphic Functions, , In one sense, the theorem we prove about meromorphic functions will be, more general than the corresponding theorem for entire functions. For we will, construct a meromorphic function not only with preassigned poles, but with, preassigned principal parts. Let {bn } be a sequence tending to ∞. For each, n, associate a rational function of the form, , , (n), (n), ak n(n), 1, a2, a, +, +, ·, ·, ·, +, ,, (12.21), = 1, Pn, z − bn, z − bn, (z − bn )2, (z − bn )kn, (n), , where the ai are complex constants, ak n(n) = 0. Note that kn is the order of the pole, and may vary with n. Our goal is to construct a meromorphic, ∞ function whose principal part for each n is Pn (1/(z − bn )). If the series, n=1 Pn (1/(z − bn )) does not converge, it will be shown that the convergence, producing factor for each n consists of a polynomial that is the partial sum, of the Maclaurin expansion for Pn (1/(z − bn ))., Before stating and proving our theorem, one final remark is in order. Suppose two meromorphic functions f (z) and g(z) have the same poles with the, same principal parts. Then f (z) − g(z) has no poles, and consequently must, be an entire function. Hence any two meromorphic functions with the same, principal parts can differ by at most an (additive) entire function. This entire function is the meromorphic analog to the (multiplicative) exponential, function of the corollary to Theorem 12.22. Now we formulate this discussion, as, Theorem 12.33. Let f (z) be a meromorphic function. If φ(z) is an arbitrary, entire function, then the most general meromorphic function g(z) which coincides with f (z) in its poles and the corresponding principal parts is given by, g(z) = f (z) + φ(z)., We now illustrate this result by an example. Consider, f (z) = cot z and g(z) =, , 2ie2iz, ., e2iz − 1, , Then both f (z) and g(z) are meromorphic functions in C having simple poles, at z = nπ, n ∈ Z. It is a simple exercise to see that, for both the functions,, residues at each of these poles are 1. Thus, the poles and the corresponding, principal parts of f and g are the same. Consequently, they differ by an, additive entire function. Again, it is easy to see that f (z) − g(z) = −i, an, entire function., The dominating result for meromorphic functions in C is due to MittagLeffler., Theorem 12.34. (Mittag-Leffler’s Theorem) Let {bn }n≥1 be a sequence of, points tending to ∞, and Pn (1/(z − bn )) be a polynomial in 1/(z − bn ) of the, form (12.21). Then there exists a meromorphic function f (z) that has poles, at the points bn (n ∈ N) with principal part Pn (1/(z − bn )), and is otherwise, analytic.
Page 449 :
12.3 Mittag-Leffler Theorem, , 441, , Proof. Without loss of generality, assume that none of the bn is zero, for we, can always add a rational function having a pole at the origin. It may also be, assumed that the sequence is so arranged that, 0 < |b1 | ≤ |b2 | ≤ |b3 | ≤ · · · ., For each n, the rational function Pn (1/(z − bn )) is analytic in the disk, |z| < |bn | and has a Maclaurin expansion, , , ∞, 1, (n), Pn, ak z k ., =, z − bn, k=0, , This series clearly converges absolutely in |z| < |bn | and uniformly in |z| ≤, |bn |/2. Thus, Pn (1/(z − bn )) can be approximated in |z| ≤ |bn |/2 by a partial, sum, nk, , (n), (n), (n), ak z k = a0 + a1 z + · · · + an(n), z nk, Qn (z) =, k, k=0, , as closely as we please. In particular, for a large value of n, we have, , , 1, 1, |bn |, Pn, ., (12.22), − Qn (z) < n for |z| ≤, z − bn, 2, 2, We will show that the function, , , ∞ , , f (z) =, Pn, n=1, , 1, z − bn, , , , , − Qn (z), , is the meromorphic function that we want. It suffices to show that the series, f (z) converges uniformly on an arbitrary compact subset |z| ≤ R of C that, excludes the points |bn | ≤ R. Choosing N so that |bN | ≥ 2R, we see from, (12.22) that, , , ∞, ∞, , , 1, 1, Pn, < ∞., − Qn (z) <, z − bn, 2n, n=N, , n=N, , The uniform convergence follows from the Weierstrass M -test (see Theorem, 6.31). Note that, , , , ∞ , , 1, Pn, − Qn (z), z − bn, n=N, , is analytic in |z| ≤ R because the poles of Pn (1/(z − bn )) lie outside |z| = R., For |z| ≤ R (≤ |bN |/2), , , , N, −1 , , 1, Pn, − Qn (z), z − bn, n=0, is an analytic function with no singularities except the prescribed poles. Since, R is arbitrary, f (z) is meromorphic in the plane.
Page 450 :
442, , 12 Entire and Meromorphic Functions, , Remark 12.35. Instead of choosing the convergence producing polynomials, {Qn (z)}, we, ∞could have chosen any other sequence of polynomials {Rn (z)}, for which n=1 [Pn (z) − Rn (z)] converges uniformly on compact subsets that, exclude the poles., •, Remark 12.36. In general, the, of the convergence producing polynodegree, ∞, mials {Rn (z)} varies with n. If n=1 Pn (1/(z −bn )) converges, we may choose, Rn (z) ≡ 0. If, , Pn, , 1, z − bn, , , , 1 z k, 1, =−, z−n, n, n, ∞, , =, , (|z| < n),, , k=0, , it was shown that we may choose Rn (z) = 1/n, a sequence of constant polynomials. The reader may also verify that if, , , ∞, √, 1, 1, 1 zk, √ = −√, (|z| < n),, =, Pn, k/2, z − bn, z− n, n, n, k=0, , √, we may choose Rn (z) = 1/ n + z/n., , •, , The function, ∞, , 1 , +, z n=1, , , , 1, 1, +, z−n z+n, , , , ∞, , =, , 1 2z, +, z n=1 z 2 − n2, , is seen to be a meromorphic function having a simple pole at each integer, with residue 1. Hence, ∞, , f (z) =, , 1 2z, +, + g(z),, z n=1 z 2 − n2, , (12.23), , where g(z) is entire, is the most general such meromorphic function. In the, special case that g(z) ≡ 0, a comparison of (12.23) and (12.13) shows that, f (z) = π cot πz., Throughout this section, we have seen similarities between Weierstrass’s, theorem and Mittag-Leffler’s theorem. As an application of Mittag-Leffler’s, theorem, we will now prove a generalization of Weierstrass’s theorem., Theorem 12.37. Let {an } be the sequence of distinct complex numbers approaching ∞. Then for any sequence {cn } of complex numbers, there exists, an entire function f (z) such that f (an ) = cn for every n., Proof. According to Weierstrass’s theorem we construct an entire function, g(z) that has simple zeros at z = an . Then g(an ) = 0 and g (an ) = 0 for, each n. According to Mittag-Leffler’s theorem, we can construct a meromorphic function h(z) that has simple poles at z = an with the principal part, cn /g (an )(z − an ) (if cn = 0, h(z) is taken to be analytic at z = an ).
Page 451 :
12.3 Mittag-Leffler Theorem, , 443, , Since the simple poles of h(z) are also the simple zeros of g(z), the singularities of f (z) = g(z)h(z) are removable. That is, f (z) is an entire function., For each n, we can expand g(z) in a Taylor series about the point z = an ., Then, g(z) = g (an )(z − an ) +, , g (an ), (z − an )2 · · · ., 2, , (12.24), , Also we may write, h(z) =, , cn, + h1 (z),, g (an )(z − an ), , (12.25), , where h1 (z) is analytic in some neighborhood of z = an . Combining (12.24), and (12.25), we find, f (an ) = lim f (z) = lim g(z)h(z) = cn ., z→an, , z→an, , This completes the proof., Remark 12.38. If cn ≡ 0, then Theorem 12.37 reduces to Weierstrass’s the•, orem., Questions 12.39., 1. What is the relationship between Weierstrass’s theorem and MittagLeffler’s theorem? Can one be derived from the other?, 2. What can be said about the sum of meromorphic functions? The product?, 3. How do the convergence producing factors of Weierstrass’s theorem and, Mittag-Leffler’s theorem compare?, 4. Why is the logarithmic derivative important in this chapter?, 5. In the proof of Mittag-Leffler’s theorem, why was it necessary to assume, that none of the bn were equal to zero?, 6. Is there a unique entire function that satisfies the conditions of Theorem, 12.37?, 7. Is Mittag-Leffler’s theorem still valid if we allow the principal part to, have essential singularities?, Exercises 12.40., 1. Construct a meromorphic function f (z) with the following two properties:, (i) f (z) has poles at z = 1, 2, 3, . . . and nowhere else., (ii) The pole at z = n has order n., 2. Show that (sin z)/(e2iz + 1) and 1/(2i cos z) differ by an entire function,, and determine, ∞, .∞it., 3. Show that n=1 (1 − z/an ) is entire if and only if n=1 1/(z − an ) is, meromorphic.
Page 452 :
444, , 12 Entire and Meromorphic Functions, , 4. Let (p ≥ 1) be an integer. If, ∞ , , n=1, , ∞, n=1, , 1/|an |p+1 converges, show that, , 1, 1, z, z2, z p−1, +, + 2 + 3 + ··· + p, z − an, an, an, an, an, , , , is meromorphic., 5. (a) Suppose that 0 < |b1 | ≤ |b2 | ≤ · · · (|bn | → ∞).Show that there, ∞, exists a positive sequence {an } such that f (z) = n=1 an /(z − bn ), is analytic except at z = bn ., (b) Consider the expansion f (z) = c0 + c1 z + c2 z 2 + · · · + cm z m + · · ·, and express cm, in terms of the quantities an and bn ., .∞, 6. Suppose g(z) = n=1 (1 − z/bn ) is entire. With the notation of the, previous exercise, show that h(z) = f (z)g(z) is entire and evaluate, , h(zn ) in terms, ∞ and αn 2and g (zn )., (|αn |/n ) converges. Show that, 7. Suppose n=1, ∞, (a) f (z) = 2z n=1 (−1)n αn /(z 2 − n2 ) is meromorphic., (b) g(z) = (sin πz/π)f (z) is entire, with g(n) = αn ., 8. Show that, ∞, , π2, 1, =, ., (z, −, n)2, sin2 (πz), n=−∞, n=0, , 9. Derive the Weierstrass’s theorem from Mittag-Leffler’s theorem., 10. Prove the identity, ∞, , 1 1, 1, 1, =, −, +, 2z, ., 2 + 4π 2 n2, ez − 1, z, 2, z, n=1, , 11. Show that the function Θ(z) defined by, Θ(z) =, , ∞, , , 1 + h2k−1 ez, , !, , 1 + h2k−1 e−z, , !, , k=1, , is entire and satisfies the functional equation, Θ(z + 2 log h) = h−1 e−z Θ(z)., , (0 < |h| < 1),
Page 453 :
13, Analytic Continuation, , We have previously seen that an analytic function is determined by its behavior at a sequence of points having a limit point. This was precisely the, content of the identity theorem (see Theorem 8.48) which is also referred to, as the principle of analytic continuation. For example, as a consequence, there, is precisely a unique entire function on C which agrees with sin x on the real, axis, namely sin z. But we have not yet explored the following question: If, f (z) is analytic in a domain D1 , is there a function analytic in a different, domain D2 that agrees with f (z) in D1 ∩ D2 ? Analytic continuation deals, with the problem of properly redefining an analytic function so as to extend, its domain of analyticity. In the process, we come across functions for which, no such extension exists. Finally, we apply our knowledge of analytic continuation to two of the most important functions in analysis, the gamma function, and the Riemann-zeta function, defined originally by a definite integral and, an infinite series, respectively., , 13.1 Basic Concepts, Consider the power series, f0 (z) =, , ∞, , , zn., , n=0, , This power series converges for |z| < 1, and hence, f0 (z) is analytic in the, disk |z| < 1 and represents there the function f (z) = 1/(1 − z). Although the, power series diverges at each point on |z| = 1, f (z) is analytic in C \{1}. For, any point z0 = 1, the Taylor series representation, f (z) =, , ∞, , f (n) (z0 ), (z − z0 )n, n!, n=0, , (13.1), , is valid when |z − z0 | < |1 − z0 | (see Figure 13.1). The disk in which (13.1)
Page 454 :
446, , 13 Analytic Continuation, , Figure 13.1., , converges may or may not have points in common with the disk |z| < 1. For, example,, ∞, , f (n) (eiα ), (z − eiα )n (0 < α < 2π), f1 (z) =, n!, n=0, converges in a disk that overlaps |z| < 1; but the disk, |z − 2| < 1, in which, f2 (z) =, , ∞, , f (n) (2), (z − 2)n, n!, n=0, , converges does not. In Figure 13.2, we show the domains in which f0 (z),, f1 (z), and f2 (z) converge. In their respective domains of convergence, they all, represent the same function f (z) = 1/(1 − z). In addition, the integral, + ∞, e−t(1−z) dt, 0, , Figure 13.2.
Page 455 :
13.1 Basic Concepts, , 447, , converges for Re z < 1 and it can be easily checked that the integral represents, ∞, f (z) = 1/(1 − z) in this half-plane. But they agree with f0 (z) = n=0 z n, (|z| < 1) for a certain value of z although they appear different. In fact, they agree with f (z) = 1/(1 − z) which is analytic for all z = 1. So we see, that apparently unrelated functions may actually represent the same analytic, function in different domains., Suppose f0 (z) is known to be analytic in a domain D0 . We wish to determine the largest domain D ⊃ D0 for which there exists an analytic function, f (z) such that f (z) ≡ f0 (z) in D0 . As we have just seen in the first example,, 1 in which an analytic function, C \{1} is the largest domain containing |z|, <, ∞, may be defined that agrees with f0 (z) = n=0 z n in |z| < 1. In our terminology, we say that f0 has an analytic continuation from the unit disk |z| < 1, into the punctured plane C \{1}. To see how one can carry out the process of, analytic continuations, we need to introduce several definitions., A function f (z), together with a domain D in which it is analytic, is said, to be a function element and is denoted by (f, D). Two function elements, (f1 , D1 ) and (f2 , D2 ) are called direct analytic continuations of each other iff, D1 ∩ D2 = ∅ and f1 = f2 on D1 ∩ D2 ., Whenever there exists a direct analytic continuation of (f1 , D1 ) into a domain, D2 , it must be uniquely determined, for any two direct analytic continuations, would have to agree on D1 ∩ D2 , and by the identity theorem (see Theorem, 8.48) would consequently have to agree throughout D2 . That is, given an, analytic function f1 on D1 , there is at most one way to extend f1 from D1, into D2 so that the extended function is analytic in D2 . Thus, one of the main, uses of this idea is to extend the functional relations, initially valid for a small, domain D1 , to a larger domain D2 . Sometimes such an extension may not be, possible. For instance, if D1 is the punctured unit disk 0 < |z| < 1 and D2 is, the unit disk, then the function f1 (z) = 1/z cannot be extendable analytically, from D1 into D2 . Similarly, if, D1 = C \{z : Re z ≤ 0, Im z = 0}, and D2 = C,, then, for f1 (z) = Log z, no extension from D1 to D2 is possible., Remark 13.1. Consider the series, f1 (z) =, , ∞, , zn, ., n2, n=1, , This series converges for |z| ≤ 1 and f1 (z) is analytic in the disk |z| < 1, and, represents the function, + z, Log (1 − t), dt., f (z) = −, t, 0
Page 456 :
448, , 13 Analytic Continuation, , However, f1 (z) cannot be continued analytically to a domain D with 1 ∈ D,, since, ∞, , n − 1 n−2, f1 (z) =, z, −→ ∞ as z → 1+ ., n, n=2, This observation shows that the convergence or divergence of power series at, a point on the circle of convergence does not determine whether the function, which defines the series can or cannot be continued along that point., •, The property of being a direct analytic continuation is not transitive. That, is, even if (f1 , D1 ) and (f2 , D2 ) are direct analytic continuations of each other,, and (f2 , D2 ) and (f3 , D3 ) are direct analytic continuations of each other, we, cannot conclude that (f1 , D1 ) and (f3 , D3 ) are direct analytic continuations, of each other. A simple example of this occurs whenever D1 and D3 have no, points in common. However, there is a relationship between f1 (z) and f3 (z), that is worth exploring., Suppose {(f1 , D1 ), (f2 , D2 ), . . . , (fn , Dn )} is a finite set of function elements with the property that (fk , Dk ) and (fk+1 , Dk+1 ) are direct analytic, continuations of each other for k = 1, 2, 3, . . . , n − 1. Then the set of function, elements are said to be analytic continuations of one another. Such a set of, function elements is then called a chain., Example 13.2. Define (see Figure 13.3), , Figure 13.3. Illustration for a chain with n = 3
Page 457 :
13.1 Basic Concepts, , 449, , f1 (z) = Log z for z ∈ D1, f2 (z) = Log z for z ∈ D2, f3 (z) = Log z + 2πi for z ∈ D3 ., Then {(f1 , D1 ), (f2 , D2 ), (f3 , D3 )} is a chain with n = 3. Note that 0 =, f1 (1) = f3 (1) = 2πi., •, Note that (fi , Di ) and (fj , Dj ) are analytic continuations of each other if, and only if they can be connected by finitely many direct analytic continuations. If γ : [0, 1] → C is a curve and if there exists a chain {(fi , Di )}1≤i≤n ,, of function elements such that, γ([0, 1]) ⊂ ∪ni=1 Di , z0 = γ(0) ∈ D1 , zn = γ(1) ∈ Dn ,, then we say that the function element (fn , Dn ) is an analytic continuation of, (f1 , D1 ) along the curve γ. That is a function element (f, D) can be analytically continued along a curve if there is a chain containing (f, D) such that, each point on the curve is contained in the domain of some function element, of the chain. As another example, the domains of a chain are also shown in, Figure 13.4. In some situations, analytic continuation of function element are, carried out easily by means of power series. In this case, a chain is a sequence, of overlapping disks., , Figure 13.4. Illustration for a chain, , Given a chain {(f1 , D1 ), (f2 , D2 ), . . . , (fn , Dn )}, can a function f (z) be, defined such that f (z) is analytic in the domain {D1 ∪ D2 ∪ · · · ∪ Dn }?, Certainly this can be done when n = 2. The function, , f1 (z) if z ∈ D1, f (z) =, f2 (z) if z ∈ D2 ,, is analytic in D1 ∪D2 . If D1 ∩D2 ∩ · · · ∩Dn = ∅, we can show by induction that, f defined by f (z) = fi (z) for z ∈ Di (i = 1, 2, . . . , n) is analytic. However,
Page 458 :
450, , 13 Analytic Continuation, , 0, , Figure 13.5., , the proof for the general case fails. Consider the four domains illustrated in, Figure 13.5. For a fixed branch of log z, set f1 (z) = log z in D1 . The function, element (f1 , D1 ) determines a unique direct analytic continuation (f2 , D2 ),, which determines (f3 , D3 ), which determines (f4 , D4 ). We thus have the chain, {(f1 , D1 ), (f2 , D2 ), (f3 , D3 ), (f4 , D4 )}. However, in the domain D1 ∩D4 it is not, true that f1 (z) = f4 (z). We actually have f4 (z) = f1 (z) + 2πi for all points in, D1 ∩ D4 . The difference in the two functions lies in the fact that the argument, of the multiple-valued logarithmic function has increased by 2π after making a, complete revolution around the origin. Note also that we can continue (f1 , D1 ), into the domain D3 by different chains and come up with different functions., For the chains {(f1 , D1 ), (f2 , D2 ), (f3 , D3 )} and {(f1 , D1 ), (g1 , D4 ), (g2 , D3 )},, we have the values of f3 and g2 differing by 2πi. Before we continue the, discussion, let us present our case by a concrete example., Example 13.3. Consider the function f (z), initially defined on the disk D =, {z : |z − 1| < 1} by the series expansion, 1, 1, f (z) = z 1/2 = 1 + (z − 1) − (z − 1)2 + · · · ., 2, 8, Here it is understood, that we start with the series representation of the prin√, cipal branch of z:, f (z) = e(1/2) Log z = (1 + (z − 1))1/2 ., Note also that f is analytic in D. Let γ : [0, 2π] → C be the closed contour, given by γ(t) = eit , starting from z0 = γ(0) = 1. Then f (z) actually has an, analytic continuation along γ. In fact, we have an explicit convergent power, series about eit (write z 1/2 = eit/2 [1 + (z − eit )/eit ]1/2 ):, , , 1 −3it/2, 1, 1 1, (z − eit )2 + · · · ,, ft (z) = eit/2 + e−it/2 (z − eit ) +, −1, e, 2, 2 2, 2!
Page 459 :
13.1 Basic Concepts, , 451, , where z ∈ Dt = {z : |z − eit | < 1}. Thus, after one complete round along the, unit circle, we end up at z = 2π by, , , 1, 1, 2, f2π (z) = − 1 + (z − 1) − (z − 1) + · · ·, 2, 8, √, which is just the other branch of z. The initial and final function elements, in this case are (e(1/2) Log z , D) and (−e(1/2) Log z , D), respectively. Also, we, observe that the domain formed by the union of all the domains Dt (which, can be clearly covered by finitely many such disks), 0 ≤ t ≤ 2π, surrounding, the origin is not simply connected. In the case of a simply connected domain,, the result of the continuation will be unique, no matter what chain is used., •, This is the substance of the Monodromy Theorem., The difference between single-valued and multiple-valued functions may, be viewed from another point of view. Suppose f (z) is analytic in a domain, D. A point z1 is said to be a regular point of f (z) if the function element (f, D), can be analytically continued along some curve from a point in D to the point, z1 . The set of all regular points of f (z) is called the domain of regularity for, f (z)., ∞, As we have seen, the function f0 (z) = n=0 z n has domain of regularity, {z : z = 1}. Note that the function f (z) = 1/(1 − z) is analytic in the domain, of regularity for f0 (z) and agrees with f0 (z) at all points where they are both, analytic., Consider now the function, , + z, + z , ∞, ∞, , z n+1, n, (|z| < 1),, f0 (ζ) dζ =, ζ, dζ =, F0 (z) =, n+1, 0, 0, n=0, n=0, where the path of integration lies in the unit disk. The function, + z, dζ, = − Log (1 − z), F (z) =, 1, −ζ, 0, agrees with F0 (z) in the disk |z| < 1, and is analytic everywhere in the plane, except z = 1 and the ray Arg (1 − z) = π (i.e., the ray along the positive real, axis beginning at z = 1). The function, F1 (z) = − log(1 − z), , (0 < arg(1 − z) < 2π), , is a continuation of F (z) from the half-plane 0 < Arg (1 − z) < π to the whole, plane, excluding the point z = 1 and the ray Arg (1 − z) = 0., Thus the domain of regularity for, F0 (z) =, , ∞, , z n+1, n+1, n=0
Page 460 :
452, , 13 Analytic Continuation, , is {z : z = 1}. Note, however, that there does not exist a function that is, both analytic in the domain of regularity for F0 (z) and agrees with F0 (z) in, the disk |z| < 1. As we shall see by the next theorem, this phenomenon occurs, only because {z : z = 1} is a multiply connected domain., Remark 13.4. We say that the multiple-valued function log(1 − z) is regular, in the domain {z : z = 1} because each such point is a regular point. Some, authors allow multiple-valued functions to be analytic. Their definition of, analytic then corresponds to our definition of regular. This next theorem shows, us that a regular function is always single-valued (hence analytic) in a simply, connected domain., •, Theorem 13.5. (Monodromy Theorem) Let D be a simply connected domain, and suppose f0 (z) is analytic in a domain D0 ⊂ D. If the function, element (f0 , D0 ) can be analytically continued along every curve in D, then, there exists a single-valued function f (z) that is analytic throughout D with, f (z) ≡ f0 (z) in D0 ., Proof. We outline the proof, leaving some details for the interested reader., Suppose the conclusion is false. Then there exist points z0 ∈ D0 , z1 ∈ D,, and curves C1 , C2 both having initial point z0 and terminal point z1 such, that (f0 , D0 ) leads to a different function element in a neighborhood of z1, when analytically continued along C1 than when analytically continued along, C2 (see Figure 13.6). This means that (f0 , D0 ) does not return to the same, function element when analytically continued along the closed curve C1 − C2 ., , Figure 13.6., , To prove the theorem, it thus suffices to show that the function element, (f0 , D0 ), D0 ⊂ D, can be continued along any closed curve lying in D and, return to the same value. In the special case that the closed curve C is a, rectangle, the proof will resemble that of Theorem 7.39., Divide the rectangle C into four congruent rectangles, as illustrated in, Figure 7.16. Continuation along C produces the same effect as continuation, along these four rectangles taken together. If the conclusion is false for C,
Page 461 :
13.1 Basic Concepts, , 453, , then it must be false for one of the four sub-rectangles, which we denote by, C1 . We then divide C1 into four congruent rectangles, for one of which the, conclusion is false. Continuing the process, we obtain a nested sequence of, rectangles for which the conclusion is false. According to Lemma 2.25, there, is exactly one point, call it z ∗ , belonging to all the rectangles in the nest., Since z ∗ ∈ D, there exists a function element (f ∗ , D∗ ) with z ∗ ∈ D∗ ⊂ D., For n sufficiently large, the rectangle Cn of the nested sequence is contained, in D∗ . But this means that f ∗ (z) is analytic in a domain containing Cn ,, contrary to the way Cn was defined. This contradiction concludes the proof, in the special case in which the curve is a rectangle. For the general proof, see, Hille [Hi]., Suppose f (z) is analytic in a domain D and z0 is a boundary point of D., The point z0 will be a regular point of f (z) if, for some disk D0 centered at, z0 , there is a function element (f0 , D0 ) such that f0 (z) ≡ f (z) in the domain, D0 ∩ D. Any boundary point of D that is not a regular point of f (z) is said, to be a singular point of f (z)., ∞, For the function f (z) = n=0 z n (|z| < 1), we have seen that each point, on the circle |z| = 1 is a regular point except for the point z = 1. That, all points on the circle cannot be regular is a consequence of the following, theorem., ∞, Theorem 13.6. If the radius of convergence of the series f (z) = n=0 an z n, is R, then f (z) has at least one singular point on the circle |z| = R., Proof. Denote the disk |z| < R by D, and suppose that all points on |z| = R, are regular points. Then, for each point zα on the circle, we can find a function, fα defined in a disk Dα centered at zα such that the function element (fα , Dα ), is a direct analytic continuation of (f, D). Since ∪α Dα covers the compact set, |z| = R, a finite subcover (D1 , D2 , . . . , Dn ) may be found. The function g, defined by, , f (z) if z ∈ D, g(z) =, fi (z) if z ∈ Di ,, is analytic in the domain D = D ∪ D1 ∪ D2 ∪ · · · ∪ Dn . Since D contains, the disk |z| ≤ R, the domain must also contain the disk |z|, R + for some, ≤, ∞, positive . Hence the power series representation g(z) = n=0 an z n is valid, in the disk |z| < R + , contradicting the fact that the Maclaurin series for, f (z) has radius of convergence R., Corollary 13.7. If f (z) is analytic in the disk |z − z0 | < R and the Taylor, series expansion about z = z0 has radius of convergence R, then f (z) has at, least one singular point on the circle |z − z0 | = R., Proof. Set ζ = z − z0 , and apply the theorem to f (ζ)., Although we are guaranteed that a power series must have singular points, on its circle of convergence, determining their location is, in general, a difficult
Page 462 :
454, , 13 Analytic Continuation, , problem. By placing a restriction on the coefficients, we can locate a particular, singular point. Here is one of the results that we have in this direction., ∞, Theorem 13.8. Suppose f (z) = n=0 an z n has radius of convergence R <, ∞. If an ≥ 0 for every n, then z = R is a singular point of f ., Proof. If z = R is not a singular point, then f (z) is analytic in some disk, D0 : |z − R| < . For a positive number ρ (< R) sufficiently close to R, we can, find an open disk D1 centered at z = ρ that contains the point z = R and is, contained in D0 . Then the Taylor series, ∞, , f (n) (ρ), (z − ρ)n, n!, n=0, , (13.2), , converges at a point z = R + δ (δ > 0) (see Figure 13.7)., , Figure 13.7., , ∞, n, According to Theorem 13.6, the series, n=0 an z has a singular point, iθ0, somewhere on the circle |z| = R, say Re . Hence the Taylor series, ∞, , f (n) (ρeiθ0 ), (z − ρeiθ0 )n, n!, n=0, , has radius of convergence R − ρ (if the radius of convergence were larger, then, Reiθ0 would not be a singular point). Note that for each n we have, f (n) (ρeiθ0 ) =, , ∞, , , k(k − 1) · · · (k − n + 1)ak (ρeiθ0 )k−n ., , k=n, , Since an ≥ 0, we obtain from (13.3) the inequality, f (n) (ρeiθ0 ) ≤, , ∞, , k=n, , k(k − 1) · · · (k − n + 1)ak ρk−n = f (n) (ρ)., , (13.3)
Page 463 :
13.1 Basic Concepts, , Thus, 1, f (n) (ρeiθ0 ), = lim sup, R−ρ, n!, n→∞, , 1/n, , ≤ lim sup, n→∞, , f (n) (ρ), n!, , 455, , 1/n, , ,, , which means that the radius of convergence of (13.2) is at most R − ρ. This, contradicts the fact that the series converges at z = R + δ. Therefore, z = R, is a singular point of f (z)., We have shown that a power series must have at least one singular point on, its circle of convergence. The question arises as to whether there is an upper, bound on the number of singular points on the circle. We will show that it is, possible for every such point to be singular. If f (z) is analytic in a domain, whose boundary is C, and every point on C is a singular point of f (z), then, C is said to be the natural boundary of f (z). In such a case, the domain of, regularity is the same as the domain of analyticity., We will make use of the following lemma in constructing a power series, with a natural boundary., ∞, Lemma 13.9. Suppose that f (z) = n=0 an z n has a radius of convergence, R. If f (reiθ0 ) → ∞ as r → R, then the point Reiθ0 is a singular point of f (z)., Proof. If Reiθ0 is a regular point, then there is a function g(z) that is analytic, in a disk centered at Reiθ0 and agrees with f (z) for |z| < R. But then, lim f (reiθ0 ) = lim− g(reiθ0 ) = g(Reiθ0 ),, , r→R−, , r→R, , contradicting the fact that the limit on the left side is infinite., Consider now the function, f (z) =, , ∞, , , n, , z2 = z + z2 + z4 + z8 + · · · ,, , n=0, , which converges (and so is analytic) in the disk |z| < 1. We will show that the, circle |z| = 1 is a natural boundary for the function f (z). First observe that, f (z) → ∞ as z → 1 along the real axis, so that z = 1 is a singular point (this, is also a consequence of Theorem 13.8). Note that f (z) satisfies the relation, f (z) = z + f (z 2 ). Hence f (z) and f (z 2 ) simultaneously approach ∞. But then, f (z 2 ) → ∞ when z 2 → 1 through real values, thereby making −1 a singular, point. This gives insight into the general method. The function f (z) satisfies, the recursive relationship, f (z) = z + z 2 + z 4 + · · · + z 2, , n−1, , For each fixed n, we have, n, , |f (z)| ≥ f (z 2 ) − n, , n, , + f (z 2 )., , (|z| < 1).
Page 464 :
456, , 13 Analytic Continuation, n, , Since f (z 2 ) → ∞ along each ray tending to a 2n -th root of unity, it follows, that each 2n -th root of unity is a singular point. That is, all points of the form, n, e(2kπ/2 )i , where k and n are positive integers, are singular points. Now every, neighborhood of any other point on the unit circle must contain one of these, 2n -th roots of unity. Hence no point on the unit circle is a regular point. That, is, |z| = 1 is a natural boundary for f (z)., A similar argument may be used for, f (z) =, , ∞, , , z n! ,, , n=0, , which is analytic in the disk |z| < 1. If z = re2π(p/q)i , where p and q are, positive integers and 0 < r < 1, then (since e2π(p/q)n!i = 1 for all n ≥ q) it, follows that, |f (z)| =, , q−1, , , rn! e2π(p/q)n!i +, , ∞, , , rn! ≥, , n=q, , n=0, , ∞, , , rn! − q., , (13.4), , n=q, , Since the right-hand side of (13.4) tends to ∞ as r tends to 1, all points of, the form e2π(p/q)i are singular points. But these points are dense on |z| = 1,, so that the unit circle is a natural boundary for f (z)., Since a power series converges in a disk, its boundary must be a circle. But, we have defined natural boundary to include a function for which the domain, of analyticity need not be a disk. Consider the function, f (z) =, , ∞, , , e−n!z ., , n=0, , Since the series converges uniformly for Re z ≥ δ > 0, the function f (z) is, analytic for Re z > 0. We now show that the imaginary axis is a natural, boundary for f (z)., Suppose z = x + 2π(p/q)i, where p is an integer, q is a positive integer,, and x is a positive real number. Then, |f (z)| =, , q−1, , n=0, , e−n!(x+2π(p/q)i) +, , ∞, , n=q, , e−n!x ≥, , ∞, , , e−n!x − q., , (13.5), , n=q, , Because the right side of (13.5) tends to ∞ as x tends to 0, it follows that, all points of the form 2π(p/q)i are singular points. But these points are dense, on the imaginary axis so that the imaginary axis furnishes us with a natural, boundary for f (z)., Remark 13.10. Let Δ be the unit disk |z| < 1 and let γ : [0, 1] → Δ be a, curve with γ(0) = 0 and D , be such that 0 ∈ D ⊆ Δ. Then there is always, ∞, an analytic continuation of ( n=0 z n! , Δ) along γ. However, if γ1 : [0, 1] → C
Page 465 :
13.1 Basic Concepts, , 457, , ∞, is given by γ1 (t) = 2it, there is no analytic continuation of ( n=0 z n! , Δ), along γ1 ., ∞, n, •, Similar comments apply for the function element ( n=0 z 2 , Δ)., Questions 13.11., 1. If f (z) = z in a domain D0 , can f (z) be analytic in a domain D1 even, though f (z) = z in D1 ?, 2. Can two functions, analytic in the disk |z| < 1, agree at infinitely many, points there and not agree everywhere in the disk?, 3. Can an analytic continuation always be transformed into a direct analytic continuation?, 4. Is it possible that the function elements (f, D1 ) and (g, D2 ) can be, connected by an infinite chain of function elements, but by no finite, subchain?, 5. Why is the domain of regularity a domain?, 6. What is the difference between a singular point and a singularity? A, regular point and a point of analyticity?, 7. Can infinitely many points on the boundary C of a domain be singular, without C being a natural boundary?, 8. If D1 , D2 , . . . , Dn are domains, when is their union a domain?, 9. Is the converse of Lemma 13.9 true?, 10. Is there a relationship between gaps in the coefficients of the Maclaurin, series for f (z) and the circle of convergence being a natural boundary?, 11. Is there a relationship between the Cauchy Theorem and the Monodromy Theorem?, √, 12. What does the Monodromy theorem tell us about log z? About z?, Exercises 13.12., 1. Given a set of real numbers 0 ≤ θ1 < θ2 < · · · < θn < 2π, construct a, function f (z) such that, (i) f (z) is analytic in |z| < 1;, (ii) the only singular points of f (z) on the unit circle are at, eiθ1 , eiθ2 , . . . , eiθn ., ∞, 2. Given (f1 , D1 ), where f1 (z) = n=0 z n and D1 = {|z| < 1}, construct, a chain {(f1 , D1 ), (f2 , D2 ), . . . , (fn , Dn )}., 3. Show that the set of regular points of an analytic function is open, and, the set of singular points, ∞is closed., n+1, n+1, 4. (a) Show that f (z) = n=0 [z 2 /(1−z 2 )] is analytic in the domain, |z| < 1 and the domain |z| > 1, and that |z| = 1 is a natural, boundary for the function in each domain., (b) Determine f (z) in each of these domainsin closed form., n, ∞, 5. Show that, = 1 is a natural boundary for n=0 z 3 ., |z|, ∞, 6. Suppose n=0 an z n! (an > 0) has radius of convergence R. Show that, |z| = R is a natural boundary.
Page 466 :
458, , 13 Analytic Continuation, , ∞, 7. Suppose f (z) = n=0 an z n is analytic for |z| < 1 and that an is real, k, for each n. If n=1 an → ∞ as k → ∞, show that z = 1 is a singular, point for f (z). , ∞, 8. Suppose f (z) = n=0 an z n has radius of convergence 1 and that the, only singularities on the circle |z| = 1 are simple poles. Show that the, sequence {an } is bounded., *1, 9. , Show that f (z) = 0 (1 − tz)−1 dt is an analytic continuation of f0 (z) =, ∞, n−1, /n from the unit disk |z| < 1 into the whole complex plane, n=1 z, minus the interval, [1, ∞)., ∞, 10. Suppose f (z) = n=0 (−1)n an z n has radius of convergence R and an ≥, 0 for every n. Show that z = −R is a singular point., , 13.2 Special Functions, There are functions which arise so frequently in complex analysis that they, have intrinsic interest. The gamma function of Euler and the zeta function, of Riemann are two such “special functions” which require special attention., As we have seen in the previous chapter, the gamma function is meromorphic, with simple poles at 0, −1, −2, . . . , and it is free of zeros. Its reciprocal is an, entire function, with a simple zero at each nonpositive integers and with no, other zeros. This may be expressed as, ∞ , , z −z/k, 1, = zeγz, e, ,, 1+, Γ (z), k, , (13.6), , k=1, , , , where, γ = lim, , n→∞, , , n, , 1, − ln n ., k, , k=1, , Thus we may rewrite (13.6) as, , n , , , , z+k, 1, = lim ze[1+(1/2)+(1/3)+ ··· +(1/n)]z−z ln n lim, e−z/k, n→∞, n→∞, Γ (z), k, k=1, ', &, n , , , z, = lim ze−z ln n, 1+, n→∞, k, k=1, , z(z + 1)(z + 2) · · · (z + n), = lim, ., n→∞, nz n!, This leads to an alternate expression for the gamma function, namely, n!nz, ,, n→∞ z(z + 1) · · · (z + n), , Γ (z) = lim, , (13.7)
Page 467 :
13.2 Special Functions, , 459, , which is defined for all values except zero and the negative integers. Equation, (13.7) is referred to as “Gauss’s formula”. Therefore, for all values of z with, z = 0, −1, −2, . . . , we get that, , , n!nz, nz, = zΓ (z)., Γ (z + 1) = lim, n→∞ z + n + 1, z(z + 1) · · · (z + n), In this way, we obtain an alternate proof of the functional equation of the, gamma function shown in the previous chapter. There is still one more method, to obtain this equation as we shall see soon., In real analysis, the gamma function is defined in terms of the improper, integral, + ∞, tx−1 e−t dt (x > 0)., (13.8), Γ (x) =, 0, , Note that the integral (13.8) makes no sense when x ≤ 0. Indeed, as e−t > e−1, for all t ∈ (0, 1), and for 0 < δ < 1, , , + 1, +, 1 1 x−1, 1 1 − δx, x−1 −t, t, e dt ≥, t, dt =, e δ, e, x, δ, which approaches ∞ as δ → 0+ for x < 0. Thus, the improper integral (13.8), diverges for x < 0. It is easy to see that it also diverges at x = 0., To see that the integral (13.8) converges for all positive x, we write, + 1, + ∞, Γ (x) =, tx−1 e−t dt +, tx−1 e−t dt = I1 + I2 ., 0, , 1, , −t, , Since e ≤ 1 for t ≥ 0, it follows that the integral (13.8) converges at t = 0, because for each δ > 0,, + 1, + 1, 1, 1 − δx, x−1 −t, <, t, e dt ≤, tx−1 dt =, x, x, δ, δ, so that I1 ≤ 1/x. For large t,, tx−1 e−t ≤ et/2 e−t = e−t/2, so that the integral converges at ∞. In fact, since limt→∞ (tx−1 /et ) = 0, the, integrand of I2 is also bounded so that, + ∞ x−1, + ∞, 1, t, (N ≥ N (x))., tx−1 e−t dt ≤, dt =, x+1, N, N, N t, Hence Γ (x) is defined for all x > 0. An integration by parts gives, + ∞, + ∞, ∞, tx, Γ (x + 1) =, tx e−t dt = − t, +x, tx−1 e−t dt = xΓ (x)., e, 0, 0, 0, , (13.9)
Page 468 :
460, , 13 Analytic Continuation, , Note that (13.6) has been shown to satisfy Γ (x + 1) = xΓ (x) for complex, values of x. From (13.9) and the fact, + ∞, e−t dt = 1,, Γ (1) =, 0, , it follows that Γ (n + 1) = n! for all positive integers n., Consider now the complex-valued function, + ∞, tz−1 e−t dt., Γ (z) =, , (13.10), , 0, , For z = x + iy, x > 0, we have, tz−1 = e(x−1) Log t+iy Log t = e(x−1) Log t = tx−1 ., Hence the integral (13.10) converges absolutely for x > 0, with, + ∞, tz−1 e−t dt = Γ (x), |Γ (z)| ≤, 0, , so that (13.10) is well defined in the half-plane Re z > 0. We wish to show, that (13.10) has two important properties: first, it is analytic for Re z > 0;, second, it agrees with (13.7) for Re z > 0. This will justify the apparently, inexcusable notation in which the same letter is used for (13.10) and (13.7)., Let K be a compact subset of the half-plane Re z > 0. For z = x + iy ∈ K,, choose x0 , x1 so that 0 < x0 ≤ x ≤ x1 < ∞. Then, we have, + 1, + ∞, |Γ (z)| ≤ Γ (x) ≤, tx0 −1 e−t dt +, tx1 −1 e−t dt < Γ (x0 ) + Γ (x1 )., 0, , 1, , Thus Γ (z) is bounded in the infinite strip, x0 ≤ Re z ≤ x1 ., For n ≥ 1, we set, , +, Γn (z) =, , n, , (13.11), , tz−1 e−t dt., , 1/n, , We will show that Γn (z) is analytic for Re z > 0, with, + n, , tz−1 e−t ln t dt., Γn (z) =, 1/n, , To this end, we show that, on any strip of the form (13.11), the expression, , h, + n, + n, t −1, Γn (z + h) − Γn (z), z−1 −t, z−1 −t, −, − ln t dt, t e ln t dt =, t e, h, h, 1/n, 1/n, + n, th − 1, − ln t dt, ≤, tx−1 e−t, h, 1/n
Page 469 :
13.2 Special Functions, , 461, , can be made arbitrarily small for |h| sufficiently small. Using the mean-value, theorem and the uniform continuity of ln t on the interval [1/n, n], we can, show that (th − 1)/h converges uniformly to ln t for 1/n ≤ t ≤ n. It thus, follows when |h| < δ( ) that the last integral above is bounded above by, + n, tx−1 e−t dt < Γ (x) < (Γ (x0 ) + Γ (x1 ))., 1/n, , Hence Γn (z) is analytic (for x0 < Re z < x1 ), with, + n, , tz−1 e−t ln t dt., Γn (z) =, 1/n, , But, lim Γn (z) = Γ (z), , n→∞, , for x0 ≤ Re z ≤ x1 . Since Γn (z) is locally uniformly bounded in the right, half-plane, Montel’s theorem (Theorem 11.14) may be applied to show that, Γ (z) is analytic for Re z > 0., We now show that the integral definition (13.10) agrees with (13.7) for, x = Re z > 0. Set, n, , + n, t, tx−1 1 −, dt (x > 0, n ≥ 1)., Γn∗ (x) =, n, 0, Integrating by parts, we obtain, , , n n, n−1, +, t, t, tx, 1 n x, 1−, +, t 1−, dt, Γn∗ (x) =, x, n, x 0, n, 0, n−1, , +, t, 1 n x, t 1−, dt., =, x 0, n, Integrating by parts n − 1 more times, we get, 1, 1 n−1, n−2, ···, ×, x n(x + 1) n(x + 2), n(x + n − 1), (n − 1)!nx+n, = n−1, n, x(x + 1) · · · (x + n), n!nx, ., =, x(x + 1) · · · (x + n), , Γn∗ (x) =, , +, , n, , tx+n−1 dt, 0, , Thus for x > 0,, n!nx, ., n→∞ x(x + 1) · · · (x + n), , lim Γn∗ (x) = lim, , n→∞, , If we can now show that, , (13.12)
Page 470 :
462, , 13 Analytic Continuation, , lim Γn∗ (x) =, , +, , n→∞, , ∞, , tx−1 e−t dt, , 0, , on the interval [1, 2], it will then follow from the identity theorem that, + ∞, n!nz, =, tz−1 e−t dt, lim, n→∞ z(z + 1) · · · (z + n), 0, in the largest domain containing the interval [1, 2] in which both functions are, analytic; that is, the representations (13.7) and (13.10) will have been shown, to be equal in the right half-plane., For n > N , we have, n, , + N, t, ∗, x−1, Γn (x) >, 1−, t, dt (1 ≤ x ≤ 2)., (13.13), n, 0, The sequence of polynomials fn (t) = (1 − t/n)n converges uniformly to e−t, on any finite interval [a, b]. Furthermore,, fn (t) ≤ fn+1 (t) ≤ e−t, for n sufficiently large. Hence for each fixed x, the integrand of (13.13) (as a, function of t) converges uniformly to tx−1 e−t on the interval [0, N ]. Therefore,, , n, + N, + N, t, lim Γn∗ (x) ≥ lim, tx−1 1 −, dt =, tx−1 e−t dt., n→∞, n→∞ 0, n, 0, Since N is arbitrary, it follows that, lim Γn∗ (x) ≥, , *n, 0, , ∞, , tx−1 e−t dt., , (13.14), , tx−1 e−t dt, so that, + ∞, lim Γn∗ (x) ≤, tx−1 e−t dt., , (13.15), , n→∞, , But Γn∗ (x) ≤, , +, , tx−1 e−t dt ≤, , *∞, , 0, , 0, , n→∞, , 0, , Combining (13.14) and (13.15), we see that (13.10) agrees with (13.7) for, 1 ≤ x ≤ 2, and consequently they must agree in the right half-plane. Hence, (13.7) (or (13.6)) may be viewed as a direct analytic continuation of the, function, + ∞, tz−1 e−t dt, , 0, , from the domain Re z > 0 to C \{0, −1, −2, . . . }., Our next discussion concerns the function, ζ(s) =, , ∞, , 1, ,, ns, n=1, , (13.16)
Page 471 :
13.2 Special Functions, , 463, , known as the Riemann-zeta function. (Here we use the traditional notation, denoting the complex variable s = σ + it rather than z = x + iy.) This is, one of the most challenging and fascinating functions which has a natural link, connecting the set of prime numbers with analytic number theory. We have, already met this series at s = 2 and s = 4 with (p. 433 and Exercise 12.30(6)), ζ(2) = π 2 /6 and ζ(4) = π 4 /90., Since fn (s) = n−s = e−s Log n is an entire function and for s = σ + it,, |n−s | = e−σ Log n = n−σ ,, we see that the series (13.16) converges absolutely for Re s > 1 and uniformly, for Re s ≥ σ0 > 1. Hence ζ(s) represents an analytic function in the half-plane, Re s > 1. Consequently,, ζ (s) =, , ∞, , , fn (s) = −, , n=1, , ∞, , , (ln n)n−s for Re s > 1,, , n=2, , and more generally,, ζ (k) (s) = (−1)k, , ∞, , , (ln n)k n−s for Re s > 1., , n=2, , Now, to see its link with the collection of prime numbers, we prove the following, Theorem 13.13. (Euler’s Product Formula) For σ > 1, the infinite product, ., −s, ) converges and, p (1 − p, , , 1, 1, =, 1− s ,, ζ(s), p, p, , (13.17), , where the product is taken over the set P = {2, 3, 5, 7, 11, . . . } of all prime, numbers p., −s, Proof. Since the series, p converges absolutely for all Re s > 1, and it, converges uniformly on every compact subset of the half-plane Re s > 1, the, infinite product (13.17) converges. Next we note that for σ > 1, ζ(s), so that, , 1, 1, 1, 1, = s + s + s + ···, 2s, 2, 4, 6, , , , 1, 1, 1, ζ(s) 1 − s = 1 + s + s + · · · ., 2, 3, 5, , Similarly, one can find that
Page 472 :
464, , 13 Analytic Continuation, , , , , 1, 1, 1, 1, 1, ζ(s) 1 − s, 1 − s = 1 + s + s + s + ··· ., 2, 3, 5, 7, 11, More generally,, ζ(s) 1 − 2−s, , !, , !, ! −s, 1 − 3−s · · · 1 − p−s, =, m = 1 + p−s, N, N +1 + · · ·, , and because of the unique factorization of integers, we can continue the procedure to obtain in the limiting case, , !, ζ(s), 1 − p−s = 1,, p, prime, , as desired., Next, we wish to find an analytic continuation of the function element, (ζ(s), Re s > 1)., To do this, we will first establish a connection between the Riemann-zeta, function and the gamma function. Recall the integral representation, + ∞, xs−1 e−x dx (Re s > 0)., Γ (s) =, 0, , The substitution x = nt gives, +, Γ (s) = ns, , ∞, , ts−1 e−nt dt., , (13.18), , 0, , Applying the identity, k, , n=1, , e−nt = e−t, , , , 1 − e−kt, 1 − e−t, , , =, , 1 − e−kt, et − 1, , to (13.18), we get, + ∞, k, , 1, 1, 1 − e−kt s−1, t, =, dt., s, n, Γ (s) 0, et − 1, n=1, Thus for Re s > 1 and k a positive integer, we have, + ∞ s−1, + ∞ −kt, k, , 1, 1, 1, t, e, dt −, ts−1 dt, =, s, t−1, t−1, n, Γ, (s), e, Γ, (s), e, 0, 0, n=1, , (13.19), , because both integrals converge. It will now be shown that the last integral, tends to 0 as k → ∞., Since |ts−1 | = tσ−1 (σ = Re s), it follows that
Page 473 :
13.2 Special Functions, , +, , ∞, , 0, , Given, , e−kt s−1, t, dt ≤, et − 1, , +, , ∞, , 0, , e−kt σ−1, t, dt., et − 1, , 465, , (13.20), , > 0, choose δ small enough so that, +, 0, , δ, , e−kt σ−1, t, dt ≤, et − 1, , +, 0, , δ, , tσ−1, dt <, et − 1, , for all k. Next choose k large enough so that, + ∞ σ−1, + ∞ −kt, e, t, σ−1, −kδ, t, dt < ., dt, ≤, e, t, e −1, et − 1, δ, δ, , (13.21), , (13.22), , Combining (13.21) and (13.22), we see that the integral in (13.20) becomes, arbitrarily small for k sufficiently large. Upon letting k approach ∞ in (13.19),, we obtain the following result which relates the zeta function with the gamma, function., Theorem 13.14. For Re s > 1,, ζ(s) =, , + ∞ s−1, ∞, , 1, 1, t, dt., =, s, t−1, n, Γ, (s), e, 0, n=1, , (13.23), , The problem of extending the domain of definition for the Riemann-zeta, function is that the integral in (13.23) diverges for Re s ≤ 1. Now, we wish, to extend ζ(s) analytically to be a meromorphic function on C with a simple, pole at s = 1. To do this, we represent ζ(s) as a contour integral with the help, of (13.23) that avoids the origin, so that the resultant function will be shown, to be entire. The continuation will then be accomplished by relating this new, function to the integral in (13.23)., Let C consist of the part of the positive real axis from ∞ to (0 < < 2π),, the circle centered at the origin of radius traversed in the counterclockwise, direction, and the positive real axis from to ∞ (see Figure 13.8). Notice, that the contour is chosen with the usual positive orientation. Compare the, contour in Figure 13.8 with the contour in Figure 9.6. Write, , Figure 13.8.
Page 474 :
466, , 13 Analytic Continuation, , Figure 13.9., , z s−1 = e(s−1) log z ,, where log z = ln |z| + i arg z with arg z = 0 when z lies in top edge of the, branch cut of the real axis from ∞ to , whereas arg z = 2π when z lies on, the bottom edge of branch cut from to ∞. Now, consider the function, +, z s−1, dz (0 < < 2π)., (13.24), f (s) =, z, C e − 1, The integral converges, and it represents an entire function of s. Note that, the value of the integral in (13.24) is independent of . To see this, suppose, that 0 < 1 < 2 < 2π. The region C 2 − C 1 , illustrated in Figure 13.9, is seen, to be simply connected. Cauchy’s theorem may thus be applied to show that, +, +, +, z s−1, z s−1, z s−1, dz, =, dz, −, dz = 0., z, z, z, C2 −C1 e − 1, C 2 e − 1, C1 e − 1, Therefore,, , +, C2, , z s−1, dz =, ez − 1, , +, C 1, , z s−1, dz., ez − 1, , To evaluate the integral in (13.24), we first assume that Re s > 1, and we, express it in the form, +, + ∞ (s−1)(ln t+2πi), +, ts−1, z s−1, e, dt, +, dz, +, dt (13.25), f (s) =, t−1, z −1, e, e, et − 1, |z|=, ∞, +, + ∞ s−1, t, z s−1, 2πis, dt, +, dz., − 1), = (e, z, et − 1, |z|= e − 1, Suppose that σ = Re s > 1. From the identity, ez − 1 = z +, , z3, z2, +, + ···, 2!, 3!, , we see that for |z| sufficiently small, |ez − 1| ≥ |z|/2. Hence
Page 475 :
13.2 Special Functions, , +, |z|=, , z s−1, dz ≤ 2, ez − 1, , +, , σ−1, , |z|=, , |dz| = 4π, , σ−1, , 467, , ,, , which approaches 0 as → 0. Hence, for Re s > 1, f (s) tends to a limit as, → 0. Since f (s) is independent of , we may evaluate f (s) by letting → 0, in (13.25). This yields, + ∞ s−1, t, f (s) = (e2πis − 1), dt (Re s > 1)., (13.26), t−1, e, 0, We can compare (13.26) with (13.23) to get, Theorem 13.15. For the branch of z s−1 and the contour C indicated above,, we have, ζ(s) =, , f (s), − 1)Γ (s), , (e2πis, , (Re s > 1)., , (13.27), , That is, we have an identity valid for all Re s > 1:, +, z s−1, 1, ζ(s) = 2πis, dz., (e, − 1)Γ (s) C ez − 1, Although (13.27) was proved only for Re s > 1, as f (s) is an entire function, the identity theorem may be used to extend this to a larger domain., Each simple pole of Γ (s) is cancelled by a simple zero of e2πis − 1. Hence, (e2πis − 1)Γ (s) is an entire function. We have thus expressed ζ(s) in (13.27), as the quotient of entire functions, that is, as a meromorphic function. The, poles of ζ(s) must occur at points where, (e2πis − 1)Γ (s) = 0., Now Γ (s) = 0, and e2πis − 1 = 0 at the integers. But for zero and the, negative integers, we have the zeros of e2πis − 1 being cancelled by the poles, of Γ (s). Hence the only zeros of (e2πis − 1)Γ (s) occur at the positive integers., However, ζ(s) was already shown to be analytic for Re s > 1 (thus, f (s) = 0, for s = 2, 3, 4, . . . ). In conclusion, the ζ function is analytic for all values of s, except s = 1 and hence, it continues analytically to C \{1}., Therefore, the only possible pole for ζ(s) occurs at s = 1. To prove that, s = 1 actually is a pole, we must show that f (1) = 0. From (13.24), we see, that, +, 1, dz., f (1) =, z −1, e, C, Since the only singularity of 1/(ez − 1) inside C (0 < < 2π) is a simple pole, at z = 0, an application of the residue theorem shows that, f (1) = 2πi lim, , z→0, , z, = 2πi = 0., ez − 1
Page 476 :
468, , 13 Analytic Continuation, , Hence ζ(s) has a simple pole at s = 1 with residue, lim, , s→1, , (s − 1)f (s), s−1, = 2πi lim 2πis, = 1., s→1 e, (e2πis − 1)Γ (s), −1, , It follows that, , 1, as s → 1 ., s−1, Thus the, ∞equation (13.27) represents a direct analytic continuation of the, series n=1 1/ns (Re s > 1) to a function analytic in C, except for a simple, pole at s = 1. We have established the following, ζ(s) ∼, , Theorem 13.16. The zeta function is meromorphic in C with only simple, pole at s = 1 with residue 1., It follows that the complete Riemann-zeta function may be expressed as, ζ(s) =, , 1, + g(s),, s−1, , where g(s) is some entire function. Of course, for Re s > 1,, ∞, , 1, 1, ., −, s, n, s, −, 1, n=1, , g(s) =, In view of the identities, , e2πis − 1 = 2ieπis sin πs and Γ (s)Γ (1 − s) =, , π, ,, sin πs, , (13.27) also takes the form, ζ(s) =, , f (s)Γ (1 − s), Γ (1 − s) −πi(s−1), e, =−, f (s)., 2πieπis, 2πi, , (13.28), , We may rewrite this as, Γ (1 − s), ζ(s) = −, 2πi, where, , +, C, , (−z)s−1, dz, ez − 1, , (0 < < 2π),, , (−z)s−1 = e(s−1) Log (−z) for z ∈ C \[0, ∞)., , The representations (13.27) and (13.28), though valid in C, give no insight, into the location of the zeros for the Riemann-zeta function. To aid us in, this endeavor, we shall develop a recursive relationship for the Riemann-zeta, function, providing explicit information, namely, Theorem 13.17. (Functional Equation of Zeta Function) For all s ∈ C,, the ζ-function satisfies the functional equation, πs , ζ(s) = 2s π s−1 sin, Γ (1 − s)ζ(1 − s)., 2
Page 477 :
13.2 Special Functions, , 469, , For the proof of this theorem, the following lemma will be helpful., Lemma 13.18. Let D be the domain consisting of the whole plane, excluding, disks of the form |z − 2kπi| < 12 , k an integer. Then there exists a real number, δ > 0 such that |ez − 1| ≥ δ for z in D., Proof. Since ez − 1 is a periodic function of period 2πi, it suffices to prove the, inequality for the region R consisting of the strip −π ≤ Im z ≤ π, excluding, the disk |z| < 12 . Observe that ez − 1 tends to ∞ as z approaches ∞ in the, right half-plane on R, and approaches −1 as z tends to ∞ in the left half-plane, of R. Choose δ, 0 < δ < 1, such that |ez − 1| ≥ δ on the circle |z| = 12 . Since, ez − 1 never vanishes in R, the minimum modulus theorem may be applied to, show that |ez − 1| ≥ δ for all z in R., Now we proceed to prove Theorem 13.17. For the proof, we modify the, contour C used to define f (s) in (13.24). Fix s with s < 0. Let 0 < < 2π, and k be a positive integer. Let Ck differ from the contour in Figure 13.8 only, in that the circle has radius (2k + 1)π instead of . Define, +, 1, z s−1, fn (s) =, dz., 2πi Ck ez − 1, The idea is to relate the integral (13.24) defined for C with a new integral, defined for Ck but with a factor 1/2πi, introduced for convenience. Then the, function, z s−1, ez − 1, has simple poles inside the contour Ck − C at the points, z = ±2nπi, , (n = 1, 2, . . . , k)., , The residue at ±2nπi is, , , z ± 2nπi, lim, z s−1 = (±2nπi)s−1, z→±2nπi, ez − 1, = e(s−1)[ln(2nπ)+i arg(±πi)], = (2π)s−1 ns−1 ei(s−1) arg(±πi) ., As arg(iπ) = iπ/2 and arg(−iπ) = i3π/2, making use of the residue theorem,, we have, 1, 2πi, , +, Ck −C, , We substitute, , , , z s−1, s−1, (s−1)iπ/2, (s−1)i3π/2, dz, =, (2π), +, e, ns−1 ., e, ez − 1, n=1, k
Page 478 :
470, , 13 Analytic Continuation, , , , e(s−1)iπ/2 + e(s−1)i3π/2 = e(s−1)iπ e−(s−1)iπ/2 + e(s−1)iπ/2, = −2eiπs sin(πs/2)., Hence it follows that, 1, 2πi, , +, Ck −C, , k, , z s−1, 1, s−1 iπs, dz, =, −2(2π), e, sin(πs/2), ., z, 1−s, e −1, n, n=1, , (13.29), , By Lemma 13.18, we have, 1, 2πi, , +, |z|=(2k+1)π, , 1, z s−1, dz ≤, z, e −1, 2πδ, , +, |z|=(2k+1)π, , |z s−1 | |dz|, , (13.30), , 1, {(2k + 1)π}s, δ, → 0 as k → ∞ (since s < 0)., ≤, , In view of (13.30), we let k → ∞ in (13.29) to obtain, +, 1, z s−1, dz = −2(2π)s−1 eiπs sin(πs/2)ζ(1 − s), 2πi −C ez − 1, , (s < 0)., , That is the function f (s) in (13.24) may be expressed as, f (s), = 2(2π)s−1 eiπs sin(πs/2)ζ(1 − s), 2πi, , (s < 0)., , A substitution of (13.31) into (13.28) yields the identity, πs , Γ (1 − s)ζ(1 − s) (s < 0)., ζ(s) = 2s π s−1 sin, 2, , (13.31), , (13.32), , Since both sides of this identity are meromorphic functions of s, this hold for, all s, by the identity theorem. The proof of Theorem 13.17 is now complete., Much of the analytic interest in the zeta function follows from the functional equation (13.32). For instance, the expression (13.32) enables us to, locate some of the zeros of ζ(s). Note first that, as a consequence of Theorem, 13.13, ζ(s) has no zeros in the half-plane Re s > 1. Since Γ (1 − s) and ζ(1 − s), are both analytic and nonzero for Re s < 0, the only zeros of ζ(s) there are, due to the zeros of sin(πs/2), that is, at the points s = −2, −4, . . . . These, zeros are called the trivial zeros of the Riemann-zeta function., The only zeros unaccounted for must lie in the strip 0 ≤ Re s ≤ 1, which, is called the critical strip. Now, we formulate, Theorem 13.19. The only zeros of the zeta function not in the critical strip, are at −2n, n ∈ N.
Page 479 :
13.2 Special Functions, , 471, , The problem of classifying the zeros of the zeta function is a formidable, (and unsolved) task. Thus far, infinitely many zeros have been found in this, strip; remarkably, all are situated on the line Re s = 1/2, which is called the, critical line. Also, it is proved that the zeta function has zeros neither on the, line Re s = 1 nor on the line Re s = 0. A theorem of Hardy proves that there, are infinitely many zeros inside the critical strip. See the books by Edwards, [E] and Ivic [I] for further information., Encouraging the student to further pursue mathematics, we end this book, not with a theorem but with a famous conjecture known as the, Riemann Hypothesis. All the nontrivial zeros of ζ(s) lie on the line Re s =, 1/2., Questions 13.20., 1., 2., 3., 4., , Is Γ, * (0+) = ∞?, Is |z|=1/3 Γ (z) dz = 2πi?, *, For n ∈ N, what is the value of |z|=n+1/3 Γ (z) dz?, For n ∈ N, is the function, Γ (z) −, , (−1)n, n!(z + n), , analytic in the disk |z + n| < 1?, 5. Is there a function f (z) = Γ (z), analytic in the right half-plane, that, satisfies the relationship f (z + 1) = zf (z)?, 6. What properties of the gamma function can most easily be proved by, (13.7)?, 7. What identities can be found by comparing (13.6), (13.7), and (13.10)?, 8. In showing the equivalence of (13.7) and (13.10) in the right half-plane,, why was it necessary to first show that they agreed on a finite interval?, 9. What properties do the gamma function and the Riemann-zeta function, have in common?, ∞, ∞, 10. How do the properties of n=1 an z n and n=1 (an /z n ) compare?, 11. What kind of function is (1 − s)ζ(s)/Γ (s)?, 12. What information about the Riemann-zeta function, other than the location of some zeros, can we obtain from (13.32)?, 13. Is the set of all zeros of the zeta function symmetric with respect to, both the critical line and the real axis?, Exercises 13.21., 1. Prove Legendre’s duplication formula, √, πΓ (2z) = 22z−1 Γ (z)Γ (z + 1/2).
Page 480 :
472, , 13 Analytic Continuation, , 2. Show that the gamma function may be expressed as, +, , ∞, , Γ (z) =, , e−t tz−1 dt +, , 1, , ∞, , , (−1)n, ., n!(z + n), n=0, , 3. Show that Re ζ(s) > 0 when Re s ≥ 2., s−1, )ζ(s) is an entire function and may be represented, 4. Show, ∞that (1−1/2, n+1, /ns for Re s > 1. Where else does this series converge?, as n=1 (−1), 5. For 0 < Re s < 1, show that, 1, ζ(s) =, Γ (s), , +, , , , ∞, , t, , s−1, , 0, , 1, 1, −, t, e −1, t, , , dt., , 6. Show that ζ(1 − s) = (1/2s−1 π s ) cos(πs/2)Γ, ∞ (s)ζ(s)., 7. Determine an analytic continuation of n=1 z n /n1/4 ., 8. Consider the analytic function, f (z) =, , ∞, , 1+c n, z, n, +c, n=1, , (c > −1)., , Determine the largest domain to which f can be analytically continued?, Determine an analytic continuation of f from the unit disk to a larger, domain?
Page 481 :
References and Further Reading, , [A], [Ap], [CG], [C], [DO], [E], [G], [Ge], [GK], [H], [Hi], [I], [LR], [M], [N], [Ne], [P], [P1], [P2], [R1], , Ahlfors, L., Complex Analysis, 2nd ed., McGraw-Hill, New York, 1966., Apostol, T., Mathematical Analysis, Addison-Wesley, Reading, MA, 1967., Carlesson, L., Gamelin, T.W., Complex Dynamics, Springer, Berlin, Heidelberg, New York, 1993., Conway, J.B., Functions of One Complex Variable, 2nd edition, Springer,, Berlin, Heidelberg, New York, 1978., Depree, J., Oehring, C., Elements of Complex Analysis, Addison-Wesley, Publishing Company, Reading, MA, 1969., Edwards, H., Riemann Zeta Function, Academic Press, New York, 1974., Gamelin, T.W.: Complex Analysis, Springer, Berlin, Heidelberg, New York,, 2001., Gelbaum, B.R.: Modern Real and Complex Analysis, John Wiley & Sons,, Inc., 1995., Greene, R.E., Krantz, S.G., Function Theory of One Complex Variable, John, Wiley & Sons, Inc., 1997., Henrici, P., Applied and Computational Complex Analysis, Vol. I, II, III, John Wiley & Sons, Inc., 1974, 1977, and 1986., Hille, E., Analytic Function Theory, Vol. 2., Ginn, Lexington, MA, 1962., Ivic, A., The Zeta Function, John Wiley & Sons, Inc., 1987., Levinson, N., Redheffer, R., Complex Variables, Holden-Day, 1970., Markushevich, A.I., Theory of Functions of a Complex Variable, Chelsea,, 1977., Nehari, Z., Introduction to Complex Analysis, Allyn and Bacon, Boston,, 1968., Newman, M., Elements of the Topology of Plane Sets of Points, 4th ed.,, Cambridge University Press, 1961., Palka, B.P., An Introduction to Complex Function Theory, Springer, Berlin,, Heidelberg, New York, 1991., Ponnusamy, S., Foundations of Complex Analysis, 2nd ed., Narosa Publishing House, India, 2005., Ponnusamy, S., Foundations of Functional Analysis, Narosa Publishing, House, India, 2003., Rudin, W., Principles of Mathematical Analysis, 2nd ed., McGraw-Hill, New, York, 1964.
Page 482 :
474, , References and Further Reading, , [R2], , Rudin, W., Real and Complex Analysis, 3rd ed., McGraw-Hill, New York,, 1987., Sprecher, D., Elements of Real Analysis, Academic Press, New York, 1970., Taylor, A., Advanced Calculus, Blaisdell, New York, 1955., Titchmarch, E.C., Theory of Functions, Oxford University Press, 1960., Veech, W.A., A Second Course in Complex Analysis, W.A. Benjamin, Inc.,, New York, 1967., , [S], [T], [Ti], [V]
Page 483 :
Index of Special Notations, , Symbol, , Meaning for, , ∅, , empty set, , a∈S, , a is an element of the set S, , a ∈ S, , a is not an element of S, , {x : . . . }, , set of all elements with the property . . ., , X ∪Y, , set of all elements in X or Y ;, i.e., union of the sets X and Y, , X ∩Y, , set of all elements in X as well as in Y ;, i.e., intersection of the sets X and Y, , X⊆Y, , set X is contained in the set Y ; i.e., X is a subset of Y, , X ⊂ Y or X Y, , X ⊆ Y and X = Y ;, i.e., set X is a proper subset of Y, , X ×Y, , Cartesian product of sets X and Y ,, {(x, y) : x ∈ X, y ∈ Y }, , X \Y or X − Y, , set of all elements that live in X but not in Y, , X, , c, , complement of X, , =⇒, , implies (gives), , ⇐⇒, , if and only if, or ‘iff’, , −→ or →, , converges (approaches) to; into, , −→, , or →, , does not converge, , =⇒, , , does not imply, , N, , set of all natural numbers, {1, 2, · · · }, , N0, , N ∪ {0} = {0, 1, 2, · · · }, , Z, , set of all integers (positive, negative and zero), , Q, , set of all rational numbers, {p/q : p, q ∈ Z, q = 0}, , R, , set of all real numbers, real line, , R∞, , R ∪ {−∞, ∞}, extended real line, , C, , set of all complex numbers, complex plane
Page 484 :
476, , Index of Special Notations, , C∞, , extended complex plane, C ∪ {∞}, , R, , n-dimensional real Euclidean space,, the set of all n-tuples x = (x1 , x2 , . . . , xn ),, xj ∈ R, j = 1, 2, . . . , n, , n, , iR, , set of all purely imaginary numbers, imaginary axis, , z, |z|, , z := x − iy, complex conjugate of z = x + iy, , x2 + y 2 , modulus of z = x + iy, x, y ∈ R, , Re z, , real part x of z = x + iy, , Im z, , imaginary part y of z = x + iy, , arg z, , set of real values of θ such that z = |z|eiθ, , Arg z, , argument θ ∈ arg z such that −π < θ ≤ π;, the principal value of arg z, , lim sup |zn |, , upper limit of the real sequence {|zn |}, , lim inf |zn |, , lower limit of the real sequence {|zn |}, , lim |zn |, , limit of the real sequence {|zn |}, , sup S, , least upper bound, or the supremum,, of the set S ⊂ R∞, , inf S, , greatest lower bound, or the infimum,, of the set S ⊂ R∞, , inf x∈D f (x), , infimum of f in D, , max S, , the maximum of the set S ⊂ R;, the largest element in S, , min S, , the minimum of the set S ⊂ R;, the smallest element in S, , f : D −→ D1, , f is a function from D into D1, , f (z), , the value of the function at z, , f (D), , set of all values f (z) with z ∈ D;, i.e., w ∈ f (D) ⇐⇒ ∃ z ∈ D such that f (z) = w, , f −1 (D), , {z : f (z) ∈ D}, the preimage of D w.r.t f, , f, , −1, , (w), , the preimage of one element {z}, , f ◦g, , composition mapping of f and g, , dist (z, A), , distance from the point z to the set A, i.e., inf{|z − a| : a ∈ A}, , dist (A, B), , distance between two sets A and B, i.e., inf{|a − b| : a ∈ A, b ∈ B}
Page 485 :
Index of Special Notations, , [z1 , z2 ], , closed line segment connecting z1 and z2 ;, {z = (1 − t)z1 + tz2 : 0 ≤ t ≤ 1}, , (z1 , z2 ), , open line segment connecting z1 and z2 ;, {z = (1 − t)z1 + tz2 : 0 < t < 1}, , Δ(a; r), , open disk {z ∈ C : |z − a| < r} (a ∈ C, r > 0), , Δ(a; r), , closed disk {z ∈ C : |z − a| ≤ r} (a ∈ C, r > 0), , ∂Δ(a; r), , the circle {z ∈ C : |z − a| = r}, , Δr, , Δ(0; r), , Δ, , Δ(0; 1), unit disk {z ∈ C : |z| < 1}, , ∂Δ, e, , unit circle {z ∈ C : |z| = 1}, , n, exp(z) = n≥0 zn! , an exponential function, , Log z, , ln |z| + iArg z, − π < Arg z ≤ π, , log z, , ln |z| + i arg z := Log z + 2kπi, k ∈ Z, , , ∂, 1 ∂, −i, , Cauchy–Riemann operator, 2 ∂x, ∂y, , , ∂, 1 ∂, +i, 2 ∂x, ∂y, , z, , ∂, ∂z, ∂, ∂z, fz, , ∂f, , partial derivative w.r.t z, ∂z, , fz, , ∂f, , partial derivative w.r.t z, ∂z, , Int (γ), , interior of γ, , Ext (γ), , exterior of γ, , γ1 + γ2, , sum of two curves γ1 , γ2, , L(γ), , length of the curve γ, , f, , (n), , (a), , n-th derivative of f evaluated at a, , f (z) = O(g(z)), as z → a, , there exists a constant K such that |f (z)| ≤ K|g(z)|, for all values of z near a, , f (z) = o(g(z)), as z → a, lim zn = z,, , n→∞, , lim, , ⎫, ⎬, , z→a, , f (z), =0, g(z), , or zn → z, or ⎭, d(zn , z) → 0, , sequence {zn } converges to z with a metric d, , Res [f (z); a], , residue of f at a, , 477
Page 486 :
Index, , nth roots of unity, 19, arg z, 15, cos z, 96, zeros of, 96, neighborhood, 26, sinh z, 98, sin z, 96, zeros of, 96, Aut (D), 398, absolute convergence of a series, 155, absolute convergence of power series,, 174, absolute value, 7, absolutely convergent, of infinite product, 415, additive inverse, 3, algebraic number, 13, analytic automorphisms, 397, analytic continuation, chain, 448, direct, 447, of gamma function, 462, of Riemann-zeta function, 463, analytic function, 131, analytic functions, Poisson integral formula, 360, analytic logarithm, 239, analytic part, 289, angle between curves, 381, antiderivative(s), 195, 217, 218, 222, 232, arc, 197, length of, 209, arc length, 209, , argument, 15, Argument principle, 333, associative law, 2, automorphism group, 398, bianalytic, 384, Bieberbach Conjecture, 409, bilinear transformations, 68, disk onto disk, 76, 77, half-plane onto disk, 74, 75, half-plane onto half-plane, 76, Bolzano–Weierstrass, 36, Borel–Carathéodory, 355, boundary of a set, 28, boundary point, 28, bounded sequence, 34, 159, bounded set, 26, branch, 110, of logarithm, 110, of square root function, 116, branch cut, 111, 116, branch of z 1/2 , 116, Casorati–Weierstrass theorem, 298, Cauchy Criterion, 37, Cauchy criterion, 154, for sequences, 154, Cauchy sequence, 36, Cauchy’s “weak” theorem, 220, Cauchy’s inequality, 263, Cauchy’s integral formula, 244, generalized, 247, Cauchy’s theorem, 234, 237, for a disk, 233
Page 487 :
480, , Index, , Cauchy’s theorem (Continued), for a multiply connected region, 237, for a rectangle, 226, weak form of, 220, Cauchy–Goursat, 226, Cauchy–Hadamard formula, 177, Cauchy–Riemann equations, 125, complex form, 127, polar form of, 135, sufficient condition for analyticity,, 132, chain, 448, rule, 135, chordal distance, 57, chordal metric, 57, circle in C, 9, circle of convergence, 175, closed curve, 197, closed set, 27, closure, 27, commutative law, 2, commutative property, 6, compact normal family, 394, compact set, 40, complement, 28, complete, 35, 37, complex exponents, 91, complex line integral, 202, complex logarithm, 109, complex number, 2, nth roots of, 18, absolute value of, 7, conjugate of, 7, modulus of, 7, polar form of, 15, vector representation, 6, complex number system, 45, extended, 45, complex numbers, 1, complex plane, 6, complex Poisson integral, 362, complex-valued function, 29, conformal mapping, 381, bilinear transformations, 386, conformal self-mapping, 398, conformally equivalent, 399, conjugate, 7, connected set, 29, continuous curve, 197, , continuous function, 50, continuous piecewise smooth, 200, continuously differentiable, 200, contour, 208, contour integral, 202, convergence, circle of, 175, disk of, 176, of sequences, 32, of series, 153, pointwise, 164, radius of, 175, uniform, 164, convergence of a series, 153, convergence producing factor, 426, countable set, 40, critical line, 471, critical point, 386, critical strip, 470, cross ratio, 73, invariance, 73, curve, 197, continuous piecewise smooth, 200, piecewise smooth, 208, De Moivre’s law, 92, De Moivre’s theorem, 18, Dedekind property, 35, 159, deleted neighborhood, 27, 57, of infinity, 57, dense set, 392, differentiable function, 123, differentiation of series, 180, digamma function, 437, direct analytic continuations, 447, Dirichlet problem, 364, for a disk, 364, for a half-plane, 370, disk of convergence, 176, divergence of series, 153, domain, 29, multiply connected, 198, simply connected, 197, domain of regularity, 451, domain set of a function, 48, dominated convergence test, 170, entire function, 131, equation of circle, 9
Page 488 :
Index, equation of line, 8, equicontinuous, 391, essential singularity, 298, Euclidean distance, 57, Euler’s constant, 434, exponential function, 92, 187, addtion formula, 93, extended complex plane, 45, extended real number system, 44, Fibonacci sequence, 192, 194, field, 3, ordered, 4, finite complex plane, 45, fixed point, 71, Formula, Cauchy’s integral, 244, Cauchy–Hadamard, 177, function, 48, cos z, 96, sin z, 96, analytic, 131, bilinear, 68, continuous, 121, at z0 , 50, on D, 50, differentiable, 123, domain set, 48, entire, 131, exponential, 92, 187, harmonic, 142, inverse, 49, meromorphic, 437, one-to-one, 48, onto, 48, preimage, 48, sectionally continuous, 208, univalent, 396, function element, 447, Fundamental Theorem of Algebra, 268,, 341, proof by argument principle, 346, proof by Liouville’s theorem, 268, proof by minimum modulus theorem,, 278, proof by residue concept, 330, proof by Rouché’s theorem, 341, fundamental theorem of calculus, 217, , 481, , fundamental theorem of integration,, 230, gamma function, 434, analytic continuation of, 462, integral definition of, 460, limit definition of, 459, product definition of, 434, 458, Gauss psi-function, 437, Gauss’s formula, 459, Gauss’s mean-value theorem, 275, geometric series, 158, glb, 34, greatest lower bound, 34, Green’s theorem, 218, Growth Lemma, 267, harmonic conjugate, 143, harmonic function, 142, congugate, 143, mean value property, 352, Harnack’s inequality, 371, Harnack’s principle, 373, Heine–Borel theorem, 41, Hurwitz’s theorem, 341, identity principle, 270, identity theorem, 271, image of a set, 48, imaginary axis, 6, imaginary part, 4, imaginary unit, 2, Inequality, Cauchy’s, 263, Harnack’s, 371, Schwarz, 14, triangle, 7, infinite product, 411, absolutely convergent, 415, converges, 411, diverges, 411, interior point, 26, inverse function, 49, theorem, 347, inverse of a function, 49, inverse points, 77, w.r.t circle, 78, inversion, 63, isogonal, 382, isolated point, 52
Page 489 :
482, , Index, , isolated singularity, 293, at infinity, 300, Jordan curve, 197, Jordan Curve Theorem, 197, Koebe function, 409, Laplace’s equation, 142, polar form of, 143, Laurent series, 286, analytic part of, 289, principal part of, 289, Laurent’s theorem, 286, least upper bound, 34, Legendre’s formula, 471, Lemma, Growth, 267, Schwarz’s, 280, length of a vector, 7, limit inferior, 159, limit of a sequence, 32, limit point, 27, limit superior, 159, line in C, 8, line integral, 203, complex, 202, real, 203, linear fractional transformations, 66, linear function, 62, Liouville’s theorem, 264, generalised versions, 265, harmonic analog, 352, proof by Schwarz’s lemma, 281, locally bianalytic, 384, locally constant, 54, logarithm, 109, branch, 110, natural, 92, principal branch of, 111, logarithmic derivative, 331, logarithmic spiral, 103, lub, 34, M-L Inequality, 211, Maclaurin series (expansion), 183, magnification, 62, magnitude (length), 7, Maximum modulus theorem, 275, 276, , proof by open mapping theorem, 344, Maximum principle, 354, for harmonic functions, 354, mean value property, 352, meromorphic function, 437, Minimum modulus theorem, 279, Minimum principle, for harmonic functions, 354, Mittag-Leffler theorem, 440, modulus, 7, Monodromy theorem, 199, Monodromy Theorem, 452, Montel’s theorem, 393, Morera’s theorem, 255, Mousetrap principle, 157, multi-valued functions, example of, 110, multiply connected domain, 198, natural boundary, 455, 456, natural logarithm, 92, neighborhood, 25, 26, deleted, 27, neighborhood of infinity, 45, 57, non-isolated singularity, 293, normal family, 392, compact, 394, north pole, 46, 57, one-point compactification, 45, 46, one-to-one function, 48, onto function, 48, open connected set, 29, open cover, 40, Open mapping theorem, 343, open set, 26, order relation, 4, ordered field, 4, orientation, negative, 199, positive, 199, orientation of a curve, 199, orthogonal circles, 10, Parallelogram identity, 10, parallelogram rule, 6, parameterized curve, 197, partial product, 412, partial sum, 153, period, 94
Page 490 :
Index, periodic, 94, Picard’s great theorem, 299, Picard’s theorem, 266, piecewise smooth, 208, curve, 208, point at infinity, 25, 57, pointwise convergent, 164, Poisson integral formula, for analytic functions, 359, for harmonic functions, 361, Poisson kernel, 360, Polar form of complex numbers, 15, pole, 297, power series, 174, absolute convergence, 174, radius of convergence of, 175, preimage, 48, principal branch, 111, of z 1/2 , 116, of log z, 111, principal part, 289, principle, Argument, 333, Mousetrap, 157, uniqueness/identity, 270, quotient of entire functions, 438, radius of convergence of series, 175, range of a set, 48, ratio test, 162, rational functions, 304, real axis, 6, real part, 4, rectifiable curve, 211, reflection, 63, region, 29, regular, 199, regular point, 451, 453, removable, 297, removable singularity, 297, residue, 308, at ∞, 313, at a finite point z0 , 308, at a pole, 311, Residue at a pole of order k, 311, Residue at a simple pole, 311, Residue theorem, 310, 314, Riemann mapping theorem, 400, , 483, , Riemann sphere, 46, Riemann’s theorem on removable, singularity, 295, Riemann-zeta function, 463, analytic continuation, 464, 468, Root test, 162, roots of complex numbers, 18, rotation, 62, Rouché’s theorem, 338, Schwarz inequality for complex, numbers, 14, Schwarz’s Lemma, 280, Schwarz’s theorem, 364, Schwarz–Christoffel transformation, 403, sectionally continuous, 208, sequence, bounded, 34, of partial sums, 153, series, 153, absolutely convergent, 155, Laurent, 286, Maclaurin, 183, power, 174, product of, 189, root test, 162, Taylor, 184, 188, set, bounded, 26, closed, 27, compact, 40, connected, 29, countable, 40, open, 26, open connected, 29, simple closed curve, 197, simple curve, 197, simply connected, 197, singular point, 453, singularity, 293, at infinity, 300, essential, 298, isolated, 293, non-isolated, 293, pole of order k, 297, removable, 297, smooth, 200, curve, 200, piecewise, 200
Page 491 :
484, , Index, , special functions, 458, stereographic projection, 46, subsequence, 34, symmetric points, 77, Taylor series, 184, Taylor series (expansion), 188, Taylor’s theorem, 249, Theorem, Open mapping, 343, Bolzano–Weierstrass, 36, Borel–Carathéodory, 355, Casorati–Weierstrass, 298, Cauchy Criterion, 37, Cauchy’s, 234, 237, Cauchy’s “Weak”, 220, Cauchy–Goursat, 226, Cauchy–Hadamard, 177, De Moivre’s, 18, Gauss’s mean-value, 275, Green’s, 218, Harnack’s principle, 373, Heine–Borel, 41, Hurwitz’s, 341, Identity, 271, inverse function, 347, Jordan Curve, 197, Laurent’s, 286, Liouville’s, 264, M-L Inequality, 211, Maximum modulus, 275, 276, Maximum Principle (harmonic), 354, Mean Value Property), 352, Minimum modulus, 279, Mittag-Leffler, 440, Monodromy, 199, 452, Montel’s, 393, Morera’s, 255, , Picard’s, 266, Picard’s great, 299, Residue, 310, 314, Riemann mapping, 400, Riemann’s, 295, Rouché’s, 338, Schwarz’s, 364, Taylor’s, 249, Uniqueness, 270, Weierstrass product, 427, transcendental number, 13, transformation, Schwarz–Christoffel, 403, translation, 61, triangle inequality, 7, trichotomy, 4, trivial zeros, 470, uniform continuity, 54, uniform convergence, 164, of products of functions, 418, uniformly bounded, 390, locally, 390, uniformly Cauchy, 168, uniformly continuous, 54, uniqueness principle, 270, Uniqueness Theorem, 270, univalent function, 396, variation of the argument, 337, vector, 6, Weierstrass M-test, 170, Weierstrass product theorem, 427, winding number, 222, zeros, of an entire function, 423, of Riemann-zeta function, 471
Page 492 :
Hints for Selected Questions and Exercises, , Questions 1.1:, 1. The set {0, 1, 2, . . . , p}, p a prime, is a field under the operations of, addition and multiplication modulo p., 2. Between any two elements in an ordered field there is another element., 8. We can see clearly the relationship between a complex number and a, point in the plane., 10. Closure. Even though (1, 1) > (0, 0), we have (1, 1)(1, 1) = (0, 2)., Exercises 1.2:, 4. (a) (−5, 14), , (b) 18−9i, , (c) −2+2i, , (d) −4, , (e) 2(n/2)+1 i sin nπ/4., , Questions 1.7:, 4. |z1 +z2 | = |z1 |+|z2 | if and only if z1 and z2 lie on the same ray emanating, from the origin., 7. Because their product is rational., Exercises 1.8:, , √, √, 1. (b) 13 − 6i (d) 2 (f) 2., 3. (b) (x + 5)2 + y 2 > 42 (c) −1 ≤ x ≤ 1, y = 0, 12. We require |z1 | = |z2 | = |z1 − z2 | so that, , (d) y 2 = −20(x − 5)., , |z1 |2 = a2 + 1 = 1 + b2 = (a − 1)2 + (b − 1)2 ., This gives a = ±b, a2 − 2a + 1 − 2b = 0, b2 − 2b + 1 − 2a = 0. Note that, a = ±b ⇒ b2 ∓ 2b + 1 − 2b = 0, ⇒ b2 − 4b + 1 = 0, √, ⇒ b=2± 3, √, ⇒ b = 2 − 3 ( as 0 < b < 1), √, ⇒ a = 2 − 3 ( as 0 < a < 1)., Thus the given points form an equilateral triangle if a = b = 2 −, , √, 3.
Page 493 :
486, , Hints for Selected Questions and Exercises, , 13. As |zj | = 1 (j = 1, 2, 3), we have |z3 − z1 | = |z2 − z3 | iff z1 z3 = z2 z3 . The, latter equation follows if we multiply z3 = −(z1 + z2 ) by z1 and z2 and, compare the resulting equations. Similarly, we can get |z3 −z1 | = |z2 −z1 |, and the assertion follows., Questions 1.13:, 2. The most important such function is f (x) = ln x., 9. Note that arg(1/z) = arg z., 13. Follow from text. A careful discussion may also be found in DePree and, Oehring [DO]., 14. Expanding (1.15) gives some useful identities., 18. The roots of unity form a group under multiplication., 20. This is discussed in Chapter 4., Exercises 1.14:, , √, √, √, 10. (a) ±(1 + i)/ 2 (d) ± 4 2(cos(π/8) + i sin(π/8)) (g) ±(1 − i)/ 2, 11. We have ω 3n+1 = ω and ω 3n+2 = ω 2 and so it is easy to see that, ⎧, −2ω if n = 3m, ⎪, ⎪, ⎪, ⎨1−ω, if n = 3m + 1,, for m = 1, 3, 5, 7, . . . ,, Sn = 1 + ω, ⎪, ⎪, ⎪, 1, ⎩, if n = 3m + 2, ω, and, , ⎧, ⎨, Sn =, , 0 if n = 3m, 1 if n = 3m + 1,, ⎩, 1 − ω if n = 3m + 2, , for m = 0, 2, 4, . . . ., , 12. There is nothing to prove if ω = 1. Therefore, we assume that ω is, different from 1. Since ω is a cube root of unity, we have ω 2 = −1 − ω, and therefore, (a + bω + cω 2 )3, = (a + bω − c(1 + ω))3, = (a − c)3 + 3(a − c)[(b − c)ω][a − c + ω(b − c)] + (b − c)3, = (a − c)3 + (b − c)3 + 3(a − c)(b − c)[(a − c)ω + (b − c)ω 2 ]., As (a − c)ω + (b − c)ω 2 = (a − c)ω − (1 + ω)(b − c) = (a − b)ω − (b − c),, the right-hand side of the above expression is real iff, 0 = Im [(a − b)ω − (b − c)], and this holds if a = b. In this way we see that the required condition, is that a, b, c are not all different., 15. As n → ∞ the sum approaches 2π, the circumference of the unit circle.
Page 494 :
Hints for Selected Questions and Exercises, , 487, , 5, , 16. Rewrite the given equation as ((1 + z)/(1 − z)) = 1, since z = 1 is not, possible. So, with z = 1, the solution is given by, 1 − ei2kπ/5, 1+z, = ei2kπ/5 ; i.e., z =, , k = 0, 1, 2, 3, 4., 1−z, 1 + ei2kπ/5, Writing the above equation as, z=, , −2i sin(kπ/5), e−ikπ/5 − eikπ/5, , k = 0, 1, 2, 3, 4,, =, 2 cos(kπ/5), e−ikπ/5 + eikπ/5, , we see that all the roots of the given equation lie in the imaginary axis, at zk = −i tan(kπ/5), k = 0, 1, 2, 3, 4., 17. The roots of equation (z−1)5 = −1 are the vertices of a regular pentagon, having center at 1 and vertex at the origin respectively. Comparing the, above equation with the given equation we obtain α = −5, β = 10,, γ = −10, δ = 5 and η = 0., n, 18. Since |(1 + ix)/(ix − 1)| = 1, the equation ((ix + 1)/(ix − 1)) = ζ, iθ, (ζ = e ) becomes, ix + 1, = ei(θ+2kπ)/n ,, ix − 1, that is, 1 + ei(θ+2kπ)/n, 2 cos[(θ + 2kπ)/2n], ix =, ,, =, i(θ+2kπ)/n, 2i sin[(θ + 2kπ)/2n], −1 + e, where k = 0, 1, . . . , n − 1. So, , , θ + 2kπ, , i.e., θ = −2kπ − 2n cot−1 (x)., x = cot −, 2n, Since ζ = eiθ , we have, n, , −1, −1, ix − 1, = eiθ = ei(−2kπ−2n cot x) = e−2in cot (x), ix − 1, proving the assertion., Questions 2.9:, 4. Any set of real numbers that is closed is also a closed subset of the, plane. The empty set is the only set that is open in both the real line, and the plane., 6. The integers have no limit point., 7. A boundary point of a connected set with more than one point must be, a limit point., 8. A ∩ B ⊂ A ∩ B. To see that the containment is proper, let A denote the, set of irrational numbers and B denote the set of rational numbers., 10. This is known as a convex set.
Page 495 :
488, , Hints for Selected Questions and Exercises, , 11. This is known as a starlike set. The plane minus the negative real axis, is starlike with respect to the origin, but is not convex., Exercises 2.10:, 4. (a) Bounded open set., (b) Not open, not closed, not connected, not bounded., (e) Open on the real line, connected, unbounded., Questions 2.22:, 2. Let xn = (−1)n and yn = (−1)n+1 ., 4. The sequence {xn }, where, , 1/n if n is odd, ,, xn =, n if n is even, 5., 6., 7., 10., , is an unbounded sequence with a limit point., When the set is closed., 2, 0, 1, 1, 1, . . . ., The set of rational numbers may be expressed as a sequence., The sequence {bn } is increasing., , Exercises 2.23:, 2. Construct disjoint neighborhoods about two distinct limit points., 4. (b), (c), (d)., 5. Suppose that zn → z0 . Then, given > 0 there exists an N (assume, N ≥ 2) such that |zn − z0 | < /2 for all n ≥ N . Now for each n ≥ N,, we have, 1, 1, 1, zk − z0 =, (zk − z0 ) ≤, |zk − z0 |, n, n, n, n, , n, , k=1, , n, , k=1, , =, , 1, n, 1, n, , N, −1, , k=1, N, −1, , , k=1, , n, 1 , |zk − z0 | +, |zk − z0 |, n, k=N, , 1, (n − (N − 1)), n, 2, k=1, %, ), N −1, 2 , < , whenever n ≥ max N,, |zk − z0 | ., , <, , |zk − z0 | +, , k=1, , 6. (a) 1, 2, 12 , 2, 13 , 2, . . ., 2n + 1, k + 1 2k + 1, nk + 1, 3 5, , ... ,, ,, , ... ,, , ..., (b) 1, 2, . . . , n, , , . . . ,, 2 2, 2, k, k, k, (c) A sequence consisting of all the rational numbers., 8. 1, 2, 2 12 , 3, 3 13 , 3 23 , 4, 4 14 , 4 24 , 4 34 , 5, 5 15 , . . . .
Page 496 :
Hints for Selected Questions and Exercises, , 489, , Questions 2.28:, 2. The infinite union may not be compact. Each integer is compact,, but, ∞, their union is not. Also, [0, 1−1/n] is compact for each n, but n=2 [0, 1−, 1/n] is not compact., 3. It is open and unbounded., 5. If there are no limit points., 6. Infinitely many., Exercises 2.29:, 6. For p and q relatively prime positive integers, f (p, q) = 2p 3q is a oneto-one mapping of the positive rationals into the positive integers., 7. For each open set in the cover, choose a point in the set both of whose, coordinates are rational., Questions 2.30:, 2. A set containing a neighborhood of ∞ is unbounded, but the converse, does not hold. For instance, {z : Re z > 0} does not contain a neighborhood of ∞., 6. They are identified with themselves., 8. The n + 1 sphere., 11. The image of the line x + y = 1 is given by the plane x1 + y1 + u1 = 1,, where x21 +y12 +u21 = 1. This is the intersection of the plane x1 +y1 +u1 =, 1 and the Riemann sphere, which is a circle passing through the north, pole (0, 0, 1)., Exercises 2.31:, 1. If z0 is a limit point, then N (∞; |z0 | + 1) does not contain infinitely, many, points of the, sequence., , y1, x1, ., ,, 4., 1 − u1 1 − u1, 6. Review Exercises 5 and 6 after reading Chapter 3., 7. See the book by S. Ponnusamy [P1]., Questions 2.45:, 1. It might be confused with our definition of a domain as an open connected set., 4. If points are always closer in the w plane, the function is uniformly, continuous. The converse is not true. Consider f (z) = 2z on a bounded, set., 5. All sequences are uniformly continuous. Just choose δ = 1., 7. It need not be a limit point; for instance, a constant function., 9. No such thing. The function is uniformly continuous on |z| ≥ for all, > 0., 10. A mapping from the unit disk onto two points.
Page 497 :
490, , Hints for Selected Questions and Exercises, , Exercises 2.46:, 1. (a) 6i (b) −(8 + 6i)/5 (c) 0 (d) 1/2., 2. (a) is continuous whereas (b) is uniformly continuous; (c) and (d) are, discontinuous at the origin., 8. f (z) = sin πz., 18. Use Exercise 12., Questions 3.2:, 3. A half-plane., 1, 1, 5., = + b., z+b, z, 6. Rotation and magnification., Exercises 3.3:, (a) v = −3(u − 2) (d) (u − 3)2 + (v − 1)2 = 8., (a) Im w > −1 (b) Im w > 0., The strip between the lines v = u − 3 and v = u − 7., (c) The triangle with vertices −1 + 11i, −13 + 5i, and 2 − 10i., (b) (u − 12 )2 + (v + 1)2 = 54 ., (a) (u − 23 )2 + v 2 = ( 13 )2, (d) (u − 12 )2 + v 2 < 14 ., 9. 0 ≤ Re z ≤ 2 maps onto the right half-plane minus the disk |w − 1/4| <, 1/4., 1., 2., 3., 6., 7., 8., , Questions 3.27:, 3. A linear transformation., 13. Use the fact that lines and circles map onto lines and circles. Review, this question, after reading Chapter 11., ⎧, ⎨ 0 for z = ∞, 15. f (z) = ∞ for z = 0, ⎩, z otherwise., If we require continuity, the function must be bilinear., Exercises 3.28:, z + 2(1 − i), iz − 1, (d) w = −, ., 2. (a) w =, −3z + i, z−2, 1, 1 2, 1 2, 2, 4. (a) Im v ≤ 2 (b) u + (v − 2 ) ≥ ( 2 ) ., 5. (c) (u − 1)2 + (v − 1)2 < 1 (d) u2 + (v + 1)2 > 2., 9. What is the image of the real line under a bilinear map?, 11. Choose z2 = ∞ in the previous exercise., 12. Suppose w = (az + b)/(cz + d). If a = d, b = c = 0, then there are, infinitely many fixed points. If (a − d)2 = −4bc, then there is one fixed, point.
Page 498 :
Hints for Selected Questions and Exercises, , 491, , 18. For instance, set, T (z) = z + 1, S(z) = z/(z + 1) and U (z) = (z + 1)/z., Then, T (S(z)) =, , z+1, 2z + 1, , S(T (z)) =, z+1, z+2, , T (U (z)) =, , z+2, 2z + 1, , U (T (z)) =, ., z, z+1, , and, , 19. w = A(iz − z0 )/(iz − z 0 ), |A| = 1., 24. We present a direct proof. Clearly the equation of the line L is y = x + 1, and the equation of the line passing through 3i and 2 + i is y = −x + 3., Note that these two lines are perpendicular. Solving these two equations, give 1 + 2i as its point of intersection. Note that, √, |(1 + 2i) − 3i| = |1 + 2i − (2 + i)| = 2., Thus z1 and z2 are inverses with respect to the given line., 26. (3 + 6i)/5., Questions 3.29:, 1. It doesn’t., 2. Any half-plane whose boundary passes through the origin., 4. When the ray (extended) passes through the origin., Exercises 3.30:, 2. We have f (z) = x2 − y 2 + 2ixy = u + iv, where u = x2 − y 2 and v = 2xy., Note that
Page 499 :
492, , Hints for Selected Questions and Exercises, , x = 1 ⇒ u = 1 − y 2 , v = 2y, i.e., u = 1 − (v 2 /4), y = 1 ⇒ u = x2 − 1, v = 2x, i.e., u = (v 2 /4) − 1, , u = x2 − (1 − x)2 = 2x − 1, x+y =1 ⇒, , i.e., v = (1 − u2 )/2., v = 2x(1 − x), 4. Rewrite the given function as, 1, z, =, 2, (1 − z), 4, , , , 1+z, 1−z, , 2, , 1, − ., 4, , 7. Set w1 = T1 (z) = z n and w2 = T2 (z) = eiα (z − z0 )/(z − z 0 ), where, α ∈ R and z0 with Im z0 > 0 are fixed (choose for example, α = 0 and, z = i). Then the composed mapping w = (T2 ◦ T1 )(z) gives a mapping, with the desired property., 8. (a) The upper half-plane. (b) The plane minus the positive real axis., 9. The unit disk n times., Questions 4.7:, 2., 4., 8., 11., , Infinite strips of width 2π., Unbounded along any ray other than one along the real axis., No, as will be shown in Chapter 8., ez is unbounded along every ray in the right half-plane, while ez + z is, unbounded along every ray., 14. tan z = i ⇐⇒ eiz − e−iz = i2 (eiz + e−iz ) ⇐⇒ eiz = 0., Exercises 4.8:, , √, 1. (a) 2kπ/3i (b) ±(1 + i) kπ (k ≥ 0), (c) ln |2kπ| + i(π/2, + nπ)., , , y, y, 2, 2, x/(x +y ), 6. (a) e, − i sin 2, cos 2, x + y2, x + y2, 2, 2, 8. (c) | sin z| + | cos z| ≥ | sin2 z + cos2 z| = 1., , (b) |e1/z | ≤ e1/ ., , Questions 4.10:, 1., 2., 4., 9., 11., , The inverse of the exponential function., Yes., The further in the right half-plane, the larger is the area of its image., exp(f (z))., At all points except z = π/2 + 2kπ. This will be better understood after, Chapter 10., , Exercises 4.11:, , √, √, 1. (a) The line segment from [(1 + i)/ 2]e−5 to [(1 + i)/ 2]e5 ., (c) The part of the annulus in the upper half-plane bounded by the, semicircles |w| = 1/e2 and |w| = e.
Page 500 :
Hints for Selected Questions and Exercises, , 493, , 2. (a) The region |w| ≤ 1, Im w ≥ 0., (b) The region |w| ≤ 1, Re w ≤ 0., (c) The region |w| ≥ 1, Re w ≥ 0., Questions 4.14:, 1. The imaginary part of the logarithm is the argument., 2. This is perfectly consistent, although sometimes inconvenient., 7. Only if −π < Arg z1 − Arg z2 ≤ π., Exercises 4.15:, 1. (c), , 1, 2, , ln(x2 + y 2 ) + i tan−1 (y/x)., , Questions 4.22:, No, because 2kπ = 0., (a + bi)c+di is real if d ln |a + bi| + c arg(b/a) = kπ, k an integer., It assumes at most mn distinct values., Because the function is not single-valued in any neighborhood of the, origin. The function is also discontinuous at the origin., 8. One is an n-valued function, and the other is an n-to-one mapping., 9. Only when m and n are relatively prime. For instance, (z 2 )1/2 has two, vales, whereas (z 1/2 )2 has only one., , 1., 3., 4., 6., , Exercises 4.23:, 1. (b) π e ei(π/2+2kπ)e (c) 12 ., 8. (c) 12 tan−1 (2/ − 1) + (i/4) ln 5., Questions 5.13:, 2. Because only one “bad” path need be found., 7. The derivative of, 2, x sin 1/x if x = 0, f (z) =, 0, if x = 0,, exists but is not continuous at the origin., 8. Usually when we are involved with expressions like x2 + y 2 ., 11. If f (z) exists, then f (z) is constant., 12. Nowhere when f (z) = |z|. Nowhere except at z = 0 when f (z) = |z|2 ., Exercises 5.14:, 1. (c) and (d)., 2. (a), (b), (c), (e), (f) differentiable at the origin, (d) differentiable everywhere., √, (x2 + y 2 )n/2, 7. (a), (b) y/ x., y
Page 501 :
494, , Hints for Selected Questions and Exercises, , Questions 5.28:, 2., 4., 11., 12., 14., 20., , Local versus global behavior., A function that is differentiable everywhere in the plane., If f (z)/g (z) is continuous at z0 ., No. 1/z, 0 < |z| < 1., Because the argument is not defined., (Arg z)2 is continuous on C \{0} and (Arg z)3 is discontinuous on the, negative real axis., 21. No., , Exercises 5.29:, 2. (a) a = b, c = −1 (b) a = b = c/2 (c) a = 1, b = 2kπ, (d) a = b = 0., 8. Let f = u + iv be entire. Then (see also Example 5.25), as, f (z) = ux + ivx ≡ vy − iuy ,, it follows that ux = 0 = vy and so, u = φ(y) and v = ψ(x). But then,, f (z) = φ(y) + iψ(x) and f (z) = iψ (x) = −iφ (y), which shows that Re f (z) = 0 and therefore, f (z) is a constant. Hence,, f (z) = az + b for some constants a and b with Re a = 0., 16. (a) 1/3 (b) 0 (c) 2 (d) Does not exist., Questions 5.40:, 1. If a property holds for analytic functions whenever it holds for its real, and imaginary parts., 4. If f (z) is analytic, then |f (z)| is continuous and ln |f (z)| is harmonic, when f (z) = 0., 5. No. See Chapter 10 for details., Exercises 5.41:, x, + c (c) v = 3x2 y − y 3 + c, x2 + y 2, 2, 2, (d) v = − ln |z| + c (e) v = ex −y sin 2xy + c., Follow the idea of Example 5.35., We have v(x, y) = (y 2 − x2 )/2 + k and u + iv = −iz 2 /2 + ik, where k is, some real constant., y2, x2, a = 3, v = 3xy 2 +, − x3 −, + c., 2, 2, 2, 2, Note that ux = 3ax + y + 1, uy = 2xy and so, , 2. (a) v = ay − bx + c, 4., 6., 7., 9., , (b) v =, , uxx + uyy = 6ax + 2x = 2x(3a + 1) = 0, gives a = −1/3 to make u harmonic in C. As a derivative formula for, f = u + iv is given by f (z) = ux (x, y) − iuy (x, y), it follows that, f (z) = −x2 + y 2 + 1 − 2ixy = −z 2 + 1. This gives
Page 502 :
Hints for Selected Questions and Exercises, , 495, , f (z) = −(z 3 /3) + z + c, where c is a constant. Note that harmonic conjugates v are given by, v(x, y) = Im (−z 3 /3 + z + c) = −(1/3)(3x2 y − y 3 ) + y + Im c., 13. ux vx = −uy vy!., 14. As Im f 2 (z) = 2uv and f 2 (z) is analytic on D, uv is harmonic., 1, 1, 1, ., 19. (a) − √, (b), (c) , 2, 2, 2, 1+z, 1−z, z (z 2 − 1), Questions 6.18:, , ∞, 5. If an > 0 and , n=1 an converges, then there exists a positive sequence, ∞, {bn } such that n=1 bn converges with an /bn → 0., 8. See Exercises 6.19 (8) and 6.19 (9)., 10. All sequences have a limit superior, but we have to avoid expressions, like ∞ − ∞., 12. The conclusion is valid as long as one sequence does not approach ∞, while the other approaches −∞., 13. No. (1/2n)1/n < 1 for every n., , Exercises 6.19:, 4. (a) Set an = rn − rn+1 , and apply the Cauchy criterion., √, √, √, (b) Show that an / rn < 2( rn − rn+1 )., Questions 6.36:, 2. The sequence {z + 1/n} is unbounded in the plane and converges uniformly., 5. A point is a compact, set., , 1/n if z is real,, 7. Define fn (z) =, 0 otherwise., The sequence {fn } converges uniformly to zero in the plane., 8. fn (z) = (−1)n does not converge, although |fn (z) − 1| = 0 for infinitely, many n. , n, k, 2, 10. fn (z) =, k=1 (z /k ) converges uniformly in |z| ≤ 1, but the limit, function is not differentiable at z = 1., Exercises 6.37:, , ∞, 8. The sequence {z/n} converges uniformly to 0 on |z| ≤ 1, but n=1 (z/n), diverges for z = 0., 10. (a) Converges uniformly to 0, where defined., (b) Converges uniformly for Re z ≥ > 0 and pointwise for Re z > 0., (c) Converges uniformly for Re z ≤ 0., (d) Converges uniformly to 1 for |z| ≤ r < 1 and pointwise to 1 for, |z| < 1; converges uniformly to 0 for |z| ≥ R > 1 and pointwise to 0 for, |z| > 1; converges to 12 when z = 1.
Page 503 :
496, , Hints for Selected Questions and Exercises, , 11. (a) Absolutely for |z| < 1 and uniformly for |z| ≤ r < 1., (b) Absolutely for |z 2 + 1| > 1 and uniformly for |z 2 + 1| ≥ R > 1., (c) Absolutely where defined (z = n), and uniformly on bounded subsets of the plane that exclude disks centered at the integers., (d) Absolutely for |z| > 1 and uniformly for |z| ≥ R > 1., Questions 6.57:, 1. |z0 | ≤ |z1 |., 2. Anentire function., ∞, ∞, 6. If n=0 nan z n converges, then n=0 an z n converges. The converse is, false., 7. It is at least 1., 9. fn (z) = z + n does not converge, although fn (z) = 1 converges., 11. In Chapter 8, it will be shown that all analytic functions have power, series representations., Exercises 6.58:, n, 3. |an z n | ≤, 0 | |z| for all n., |a, ∞, 5. Either n=0 an diverges or the series is a polynomial., 10. (a) |a| (b) 1/|a|, |a| > 1; 1, |a| ≥ 1 (c) 1 (d) 1/e (e) 1; For (f),, because of the presence of factorials, it is more convenient to use ratio, test. Now, , 11. (a) 2, , (n + 1)2, 1, an+1, = lim, = lim, =, n→∞ (2n + 2)(2n + 1), R n→∞ an, √, (b) 1/ 3 (c) ∞ (d) 1 (e) 5/3; For (f ), we, ⎧, ⎨ 0 if |z| < 1, 2, 2 if |z| = 1, |2n z n |1/n = 2|z|n →, ⎩, ∞ if |z| > 1, , 1, ., 4, note that, , and therefore,, n the series converges, nfor |z| < 1.n, 12. Let sn = i=1 ai , and consider i=m ai bi = i=m (si − si−1 )bi ., 14. (b) 9 − 2(z + 2) − 3(z + 2)2 + (z + 2)3 ., 1/n, 16. Given R−1 = lim supn→∞ |an |1/n = lim supn→∞ |1/an | . It is easy to, see that, 1, 1, =, =R, lim sup, 1/n, lim sup |an |1/n, n→∞ |an |, 2, and so the last, ∞equation givesn R = 1., 17. Let f (z) = n=0 cos(nπ/3)z . We first, ⎧, (−1)k, ⎨, (−1)k /2, an = cos(nπ/3) =, ⎩, (−1)k+1 /2, , compute, if n = 3k, if n = 3k + 1 , k ∈ N0 ,, if n = 3k + 2
Page 504 :
Hints for Selected Questions and Exercises, , so that, , 497, , , , 4, , z, z, z2, z5, −, − z3 −, −, − z6 + · · ·, 2, 2, 2, 2, , , 2, z, z, −, − z 3 (1 − z 3 + z 6 − · · · ), = 1+, 2, 2, , , z2, 1, z, −, − z3, = 1+, 2, 2, 1 + z3, (1 + z)(1 − z/2), =, 1 + z3, 1 − z/2, =, ., 1 − z + z2, , f (z) = 1 +, , Questions 6.66:, 2. At all points where the denominator is nonzero. This follows from the, fact (proved in Chapter 8) that an analytic function admits a power, series expansion., 4. 1/(1−z) is analytic for z = 2. A function cannot be analytic everywhere, on |z| = R, which is proved in Chapter 13., 8. The radius of convergence of the Taylor series about a point is the, distance between that point and the nearest zero of the denominator., Exercises 6.67:, 1. (a) Inequality, holds when, , , 1, if n is odd, 1/2n if n is odd, and bn =, an =, n, 1/2 if n is even ,, 1, if n is even ., 2. (a) R (b) R (c) ∞ (d) 0., n, −1/k, 4. , If an ≡ 1, then R = 1 for both, for, ∞ series.n2If an = 2 , then R = 2, ∞, kn, a, z, and, R, =, 1, for, a, z, ., n=0 n, n=0 n, 5. (b) lim sup nk /n!, , 1/n, , = 0., , n→∞, , 6. (a) 52 (b) 13, (c) 5 − 5i., 12, ∞, ∞ n, 7. Note that radius of convergence of n=1 zn2 and n=1 (−3)n z n are 1, and 1/3, respectively. According to Theorem 6.62, the radius of convergence R of the sum of these two series must be at least 1/3. Is R = 1/3?, 11. Substituting z = 0 in the functional equation gives, f (0)[1 − f (0)] = 0, i.e., either f (0) = 0 or f (0) = 1., If f (0) = 0, then, by differentiating the functional equation, we find that, f (2z) = f (z)f (z) which gives f (0) = 0. Continuing this process, we, get f (k) (0) = 0 for all k ∈ N. Thus, f (z) = 0 on Δr . If f (0) = 1, then, by differentiating the last equation we have, 2f (2z) = (f (z))2 + f (z)f (z)
Page 505 :
498, , Hints for Selected Questions and Exercises, , so that f (0) = (f (0))2 . In this way, we conclude that, f (z) =, , ∞, , (f (0))n n, z = exp(f (0)z)., n!, n=0, , Questions 7.12:, 2., 4., 7., 11., , No, because the initial and terminal points coincide., The complement of a simply connected domain is connected., The plane minus the integers., It is the continuous image of a compact and connected set., , Exercises 7.13:, 1. A circle if a = b and an ellipse if a = b., 3. As R → ∞, z(t) becomes a circle centered at the origin with radius 1., 5. (a) z(t) = t + i(1 − 2t) (0 ≤ t ≤ 1), (d) z(t) = 1 + 2 cos t + i2 sin t (−2π/3 ≤ t ≤ 2π/3)., 6. z(t) = t + i(2t2 − 3) (−1 ≤ t ≤ 2)., 7. (b) 4r2 cos2 θ + r2 sin2 θ = 1., 8. (a) 2πi (b) 4πi (e) −30 + 25πi., 9. (b) 2πi (d) 2π + 4πi., Questions 7.28:, 3. A finite number of discontinuities will not prove significant., 4. Yes, because a contour is compact., 5. As f is continuous at the origin, given > 0 there exists a δ > 0 such, that |f (δeiθ ) − f (0)| = |f (δeiθ )| < for δ < min{δ , r}, θ ∈ [0, 2π]. So,, +, , +, , 2π, , f (δe ) dθ ≤, , 2π, , f (δeiθ ) dθ < 2π → 0, , iθ, , 0, , 0, , as, , → 0., , Similarly,, +, |z|=δ, , f (z), dz = i, z, , +, , 2π, 0, , f (δeiθ ) dθ → 0, , as δ → 0., , 7. The parametrization is easier to deal with., Exercises 7.29:, , √, 4. (a) 12π, (b) 2(eπ − +e−π )., +, +, 1−i, 1−i, 5. (b), ,, ,, x dz =, y dz = −, z dz = 1,, 2 + C, C, +C, +2, (c), x dz = πi,, y dz = −π,, z dz = 2πi., C, , C, , C
Page 506 :
Hints for Selected Questions and Exercises, , 499, , 6. Along the line segment from the origin to 1 + i,, +, , +, , +C, , +, +, 1+i, 1−i, |z| dz = √ ,, z|dz| = √ ,, |z| |dz| = 1;, 2 + C, C +, C, +2, z dz = 0 =, |z| dz =, z|dz|,, |z| |dz| = 2π., , z dz = i,, , |z|=1, , 9. 2πi., 10. (a) e2+2i − 1, , |z|=1, , |z|=1, , (b) e1−i − e−(1−i) + 2(1 − i), , |z|=1, , (c), , i 1−i, (e, − e−1+i ), 2, , (e) (−7 + 5i)/3., 11. 43 ., 12. We have, +, +, + R, + π, iθ, iθ, 3, z|z| dz =, x|x| dx +, (Re )(R)iRe dθ = iR, C, , −R, , 0, , π, , e2iθ dθ = 0., , 0, , Questions 7.37:, 2., 5., 7., 9., , In the use of the Fundamental Theorem of Calculus., See Question 7.37 (2). *, It need not be analytic: |z|=1 (1/z 2 ) dz = 0., See Section 9.3., , Exercises 7.38:, 1. (a) 16/3 (b) Traversed in the positive sense −5/3. (c) 0, (e) 128/5., 6. Regardless of the contour chosen: (a) 0 (b) i(1 + 1/e)., , (d) πr2 /4, , Questions 7.55:, 1. Not, + necessarily, because |f (z)| is not analytic., dz, = 0., 3., |z|=1 z, 7. In order to apply the Cauchy-Riemann equations., 10. Any integral multiple of 2πi., 11. If g0 (z) is a solution, then so is g0 (z) + 2kπi., Exercises 7.56:, 3. f (z) = 1/z 2 ., 5. (a) π (b) −π (c) 0 (d) 0., 6. (a) 2πi (b) 0 (c) 2πi. These solutions will be easy to verify after, reading Section 8.1., 9. As Re z = (z + 1/z)/2 for |z| = 1, we have, +, f (z), I=, (z 2 + 1 + 2αz) 2 dz = 2πi(f (0) + 2αf (0))., z, |z|=1
Page 507 :
500, , Hints for Selected Questions and Exercises, , 10. As f (z) = az = ez Log a is an entire function, its primitive is given by, F (z) =, , az, ez Log a, =, ., Log a, Log a, , Questions 8.23:, 4., 6., 8., 10., , If they are analytic at the point, they are identical., {z n } does not converge uniformly on |z| < 1., not be analytic., *They need, 2, |z|, dz, = 0, but |z|2 is not analytic. Morera’s theorem is not ap|z|=1, plicable because the integral is not zero along every contour., ∞, ∞, , , 1, (|z| < 1), f (2) = 1 =, zn =, nz n−1 . Here, 1/(1 −, 12. f (z) =, 1, −, z, n=0, n=1, ∞, z) is not defined by n=0 z n at z = 2. See Chapter 13., Exercises 8.24:, 3. (a) 2πie2 (b) 2πie4 (c) 8πie4 (d) 2πie2 (sin 2+cos 2) (e) 2πe−2 sin 2, (f) 144πi., 4. (a) 0 (b) 0 (c) 0 (d) 12πi., 6. As |z| = 1, for the first integral, we may rewrite Re z = (z + 1/z)/2. For, the second and third integrals, we write z − 1 = eiθ so that, z = 1 + e−iθ = 1 +, and, Im z = Im (z − 1) =, , 7., 8., 9., 11., 16., , z, 1, =, z−1, z−1, , 1, z−z, =, 2i, 2i, , , , z 2 − 2z, z−1, , , ., , Now, use the Cauchy integral formula., π, 4πi, 2π 8πi, 2π, (c), −, (e), −, (a), 17, 17, 17, 17, 17, Use the Cauchy theorem for multiply connected domains and the, Cauchy integral formula., 1 5, (a) z + z 2 + 13 z 3 − 30, z, 3 2, 7 3, 27 5, 4, (c) 1 + z + 2 z + 6 z + 25, 24 z + 40 z ., 1, (a) 2 (b) 1 (c) 1 (d) 0., Express as eα log(1−z) and expand., , Questions 8.55:, 2. No. Indeed if f (z) =, , ∞, n=0, , an z n with |an | ≥ n!, then, , 1, = lim sup |an |1/n ≥ lim sup(n!)1/n = ∞, i.e., R = 0., R, n→∞, n→∞, 5. ez is bounded for Re z ≤ 0., 6. No, as will be shown in Chapter 11.
Page 508 :
Hints for Selected Questions and Exercises, , 501, , 12. If f (z) = 0 in C, then φ(z) = 1/f (z) is entire and |φ(z)| → 0 as, |z| → ∞. So, φ is entire and bounded on C. Consequently, φ and hence, f is constant, a contradiction. Thus, f has a zero in C., 13. No, because sin πz and 0 agree at the integers., 17. Consider φ(z) = e−(a−ib)f (z) . Then |φ(z)| = e−(au+bv) ≤ e−c = M ., Exercises 8.56:, 11. Set h(z) = f (z)/g(z) and show that h (z) = 0., 12. For instance, f (z) = sin(1/(z − 1)), sin((1 + z)/(1 − z))., 15. Define g(z) = f (z)−f (z). Then g is entire and g(an ) = 0 for all n ≥ 1. As, every bounded sequence of real numbers has a convergent subsequence,, it follows that g(z) = 0 in C, by the identity theorem. So, f (z) is real, on the real axis. As f (x) is real for x ∈ R, we can apply the mean value, theorem of calculus on the interval [a2n+1 , a2n ]. Thus, for each n, there, exists a cn such that, a2n+1 ≤ cn ≤ a2n and f (cn ) = 0., As an → 0, we see that cn → 0. Consequently, f (z) = 0 in C by the, identity theorem., Questions 8.72:, 2. f (z) = z n on |z| = r., iθ1, 3. Not necessarily. If f (z) = ez , then |ere | < |ez | for some |z| < r, whenever θ1 = 2kπ., 4. ez on {z : Re z < 0} ∪ {0} attains a maximum at z = 0., Exercises 8.73:, 3. For each n ∈ N, let fn (z) = z n f (z), and Fn (z) = fn (z)fn (z). Then, Fn, is analytic and for > 0 there exists an n such that |Fn (ζ)| < for all, ζ ∈ C. By the Maximum modulus principle, this inequality yields that, |x2n f (x)| ≤ for x ∈ (0, 1)., 4. (a) Max at z = r, min z = −r., (b) Max and min everywhere., (c) Max at z = r and min z = ir., (d) Max at z = −r, min z = r., 2, 2, 6. Observe that |ez −iz | = eRe (z −iz) , and then maximize the quantity, 2, Re {z − iz}., 9. Use Theorem 8.38 and the fact that |eP (z) | is continuous., 12. Use Schwarz’s lemma., 13. Suppose that such an f exists. Define, F (z) = φ3/4 ◦ f ◦ φ1/2 (z),, , φa (z) =, , a−z, ., 1 − az, , Then F satisfies the hypothesis of the Schwarz lemma. But a computation gives that F (0) = 32/21, which is a contradiction. Thus, we see
Page 509 :
502, , Hints for Selected Questions and Exercises, , that no such f can exist. See also Example 8.70. Define F (z) = f (z/3)., Then, F is analytic for |z| ≤ 1 and |F (z)| ≤ 1 for |z| ≤ 1 so that F (z), has zeros at ak = wk /3, k = 0, 1, . . . , n − 1. Therefore, F (z) is Blaschke, product with zeros at ak and so,, |F (0)| =, , n−1, , k=0, , 1, wk, = n., 3, 3, , Questions 9.8:, ∞, 2. No. n=1 (z n /n2 ) converges uniformly on the annulus 12 ≤ |z| ≤ 1, but, is, analytic on |z| = 1., not, ∞, z, (1/n, ) is analytic in a half-plane., 4., n=1, 5. Only for constant functions., 7. If f (z) is analytic in an annulus, then the identity is valid in that annulus., Exercises 9.9:, 1. Use partial fractions., ∞, ∞, , z 2n 1, +, ., 4., n!, z 2n n!, n=0, n=0, ∞, ∞, , 1 an − bn, (z − a)n−1, 5. (iii), (iv), (−1)n, ., n+1, a − b n=0 z, (a − b)n+1, n=0, 6. We note that, , , 1, 1, 1, 1, f (z) =, −, z z−b z−a b−a, , , , , 1 1, 1 1, 1, 1, 1, −, −, −, =, z−b z b, z−a z a b−a, , , b−a, b, a, 1, −, +, =, ab(b − a), z, z−a z−b, and the rest of the calculation is routine as in Example 9.7., 7. The calculation is routine once we write f (z) as, f (z) = 1 +, , 3, 1, +, ., z+1 4+z, , i, 1, 1, (d) 3 − ., 4(z − i), z, 6z, , ∞, ∞, n , , , , zn, an, n−1 1, 9. (a), ., (b), , where an =, k − 1 k!, k!z k, zn, n=0, k=0, k=1, 4, 8, 4, 4, 11. 4 + 3 + 2 + ., z, z, 3z, 3z, 8. (a) −
Page 510 :
Hints for Selected Questions and Exercises, , 503, , Questions 9.21:, 1. 1/ sin(1/z) has infinitely many singularities in the compact set |z| ≤ 1., All but one, z = 0, are isolated., 2. A counterexample will be constructed in Chapter 12., 5. ez − (7 − 2i)., 7. Only constant functions., 9. No, by definition., Exercises 9.22:, 4. (b) Simple poles at z = 1/(2k + 1)πi, nonisolated essential singularity, at z = 0., (e) Isolated essential singularities at z = 0 and z = ∞., (g) Branch point at z = 1 and simple pole at z = −1., 6. (a) Removable (b) Simple pole., (c) (d), (f) are all isolated essential singularities., (e) is a nonisolated essential singularity., z − z0, e1/(z−z2 ) ., 7. f (z) =, (z − z0 )(z − z1 )k, 13. Set f (z) = A/(1 − z/z0 ) + F (z), where A is a constant and F (z) is, analytic for |z| < R., 14. Set (1 + z)1/z = e(1/z) log(1+z) = e(1/z)[ Log (1+z)+2kπi] ., Questions 9.36:, 3. Morera’s theorem cannot be applied because sin(1/z 2 ) is not continuous, at z = 0., 5. None. The residue theorem is just a convenient form of Cauchy’s theorem., 8. Because 1 + x2n+1 has a singularity on the real axis., Exercises 9.37:, 1. (a) At z = π/2 + kπ, the residue is (−1)k+1 ., (b) At z = 1, the residue is −2; at z = 2, the residue is 2., ⎧, if n is even,, ⎪0, ⎪, ⎪, 1, ⎨, −, if n = 4k + 1,, (c) At z = 0, the residue is, (n + 1)!, ⎪, ⎪, 1, ⎪, ⎩, if n = 4k + 3., (n + 1)!, 3. As f (a) = 0, f (z) − f (a) = 0 in a deleted neighborhood of a. Therefore,, lim (z − a), , z→a, , 1, 1, = , = 0, f (z) − f (a), f (a), , and the conclusion is a consequence of the Residue theorem., 5. (a) 0 (b) 2πi(1 + 2e + 2e4 ).
Page 511 :
504, , Hints for Selected Questions and Exercises, , 6. We may rewrite the given integral as, +, zn, 1, ,, f (z) dz, f (z) =, I=, 2πi |z|=2, (z − eiφ )(z − e−iφ ), The poles of f (z) are at z = eiφ , e−iφ and both lie inside |z| = 2. It, follows easily that, I = Res [f (z); eiφ ] + Res [f (z); e−iφ ] =, , sin nφ, ., sin φ, , 7. (a) (2 − 4n)i (b) (2 + 4n)i., √, √, 9. Note that f (z) = 0 implies that z ∈ {∗ −3 + 2i, ∗ −3 − 2i}, where, √, √, ∗ −3 + 2i = {±a} and ∗ −3 − 2i = {±a},, , , where, a=, , , √, √, −3 + 13, 3 + 13, +i, ., 2, 2, , The poles lying in the upper half-plane are a and b := −a. They are, simple poles. Therefore we find that, , 2, a2, a, ia, z, ;a = , =, =−, Res, f (z), f (a), 4(a2 + 3), 8, as a2 = −3 + 2i. Similarly we have, , 2, b, b, ib, z, ;b =, =, = ., Res, 2, f (z), 4(b + 3), 4[(−3 − 2i) + 3], 8, 10. Consider the polynomial equation f (z) = a0 + a1 z + · · · + z n . Since, |f (z)| → ∞ as z → ∞, for sufficiently large R, we have |f (z)| > 0 for, |z| ≥ R. If we let F (z) = f (z)/f (z) and C = {z : |z| = R}, described, in the positive direction, then, +, +, 1, 1, f (z), dz =, F (z) dz = −Res [F (z); ∞], 2πi C f (z), 2πi C, , , F (1/z), = Res, ;, 0, z2, , , f (1/z), = lim z 2, z→0, z f (1/z), = n., 12. Keep c fixed and let R → ∞.
Page 512 :
Hints for Selected Questions and Exercises, , 505, , Questions 9.58:, 1., 6., 7., 8., , e1/z , z = 0., |z| maps the plane onto the ray Re u ≥ 0, v = 0, which is not open., ez maps the plane onto the punctured plane, which is not closed., Consider the exponential function f (z) = ez ., , Exercises 9.59:, 3. The sum of the roots of P (z)., 4. If f (z) = a0 + a1 z + · · · + an−1 z n−1 + z n , n ≥ 1 and R is chosen large, enough so that |f (z)| ≥ 1 for all |z| ≥ R then, for |z| ≥ R, we have, nz n−1 + · · ·, n, 1, f (z), =, = + terms in k , k ≥ 2., f (z), zn + · · ·, z, z, Thus,, 1, 2πi, , +, |z|=R, , n, f (z), dz =, f (z), 2πi, , +, |z|=R, , dz, = n., z, , Since f has no poles in C, the fundamental theorem of algebra follows., 10. For |z| = 1, z = x + iy, | − az n | = a > e > ex = |ez |., 12. If p(z) = z 3 + iz + 1, then, p(x) = 0 =⇒ x3 + 1 = 0 and x = 0, which is not possible. Similarly,, p(iy) = 0 =⇒ 1 − y = 0 and y 3 = 0, which is again not possible., 13. For |z| = 1,, |z 4 + 1| ≤ |z|4 + 1 = 2 < | − 6z| = 6, and for |z| = 2,, | − 6z + 1| ≤ 6|z| + 1 = 13 < |z|4 = 24 ., 16. For |z| = 3/2,, |z 3 + 1| ≤ |z|3 + 1 = 27/8 + 1 < |z|4 = 81/16, and for |z| = 3/4,, |z|4 = 81/256 ≤ −|z|3 + 1 = −27/64 + 1 ≤ |z 3 + 1|., 18. Use Corollary 9.47., 19. Use Hurwitz’s theorem.
Page 513 :
506, , Hints for Selected Questions and Exercises, , Chapter 10:, Questions 10.15:, 5. Not according, to Picard’s theorem., , x + y if |z| < 1,, 8. u(z) =, 1, if |z| = 1., 11. Theorem 10.14 generalizes Theorem 8.35 because Re f (z) ≤ |f (z)|, and, reduces to Theorem 10.4 when λ = 0., 14. By the mean value theorem, u(2, −1) = −2., 15. u(z) = ay + b for some constants a and b., 18. As f = u + iv is analytic in D, Im (f 2 ) = 2uv is harmonic. Also u2 is, harmonic on D iff u is constant., 19. No., Exercises 10.16:, 10. u = 1., 11. u = y., 13. Use Cauchy’s integral formula in conjunction with Theorem 10.13., Questions 10.34:, 2. Yes, by finding an upper bound on the entire function using (10.14),, and then applying Theorem 8.35., 3. The proof of uniqueness is easy in the case when F is continuous on, |z| = R. Then u would be harmonic for |z| < R, continuous on |z| = R,, and equal to F on |z| = R. Let u1 be another such function. Then, u − u1 = 0 on |z| = R showing that u = u1 for |z| < R (by Corollary, 10.10)., 4. In Chapter 11, we shall discuss mappings from the disk to other domains., This will enable us to solve the Dirichlet problem for other domains., 5. Theorem 10.6 gives the value of the harmonic function at the center of, the circle, whereas Theorem 10.18 gives the value for all points inside., Exercises 10.35:, 1, 1 − r2, 6. u(reiθ ) = tan−1, π, 2r sin θ, 8. Show that, tan−1, , (0 ≤ tan−1 t ≤ π)., , r sin θ, = Im log(1 + z), 1 + r cos θ, , (z = reiθ )., , 9. With C = [−R, R] ∪ ΓR , where ΓR is the upper semi-circular contour, from R to −R, we write, , , +, 1, 1, 1, f (z) =, −, dζ, f (ζ), 2πi C, ζ −z, ζ −z, +, +, y R, y, f (t) dt, f (ζ) dζ, =, +, ., π −R (t − z)(t − z) π ΓR (ζ − z)(ζ − z)
Page 515 :
508, , Hints for Selected Questions and Exercises, , Choosing the determination of the inverse tangent whose values lie in, the interval [0, π], we have, , , 1, 1 − r2, u(reiθ ) = tan−1 −, ., π, 2r sin θ, With this determination,, , , 1 − r2, π if 0 < θ < π, −1, −, =, lim tan, 0 if π < θ < 2π, r→1, 2r sin θ, the desired boundary conditions are satisfied., 15. Use for example, (10.14)., Questions 10.45:, 2. In view of Theorem 10.38, the mean-value property does not hold, whenever the product of two harmonic functions is not harmonic., 4. Yes, just consider {−un (z)}., 5. Consider |z| < R and Re f (z) > α., * 2π, * 2π, 7. We used the fact that 0 |u(reiθ )| dθ = 0 u(reiθ ) dθ., Exercises 10.46:, 5. Set g(z) = (1 − α)f (z) + α, where Re f (z) > 0. Then apply Theorem, 10.42. Why can’t α exceed 1?, 7. Set f (z) = (g(z) − α)/(1 − α), and apply Theorem 10.44., Questions 11.7:, 1., 2., 3., 9., , By convention the line itself is the tangent line., Only at the point of intersection., Not if the partial derivatives are continuous. See Nehari [N]., A one-to-one map is conformal if it is analytic; a conformal map is, locally one-to-one., 12. Only the composition., Exercises 11.8:, 1. f (z) = e3πi(z−z0 )/ ., Questions 11.16:, 2. No. Even the family of constant polynomials is not., 3. {z n } is uniformly bounded on |z| < 1, but {nz n−1 } is not., 4. No. Let F consist of one function f (z) defined by, , 1 if z = 1/n,, f (z) =, n if z = 1/n., This is unbounded in every neighborhood of the origin.
Page 516 :
Hints for Selected Questions and Exercises, , 509, , 5. We have assumed that the line segment between any two points lies in, the domain. Thus the proof is valid for any convex domain., 6. A sequence must be countable., Questions 11.30:, 2. f (z) = az, a > 0, maps the annulus r < |z| < R conformally onto the, annulus ar < |w| < aR., 5. No, because the punctured disk is not simply connected., 13. In the construction of the analytic square-root function., 14. In order to ensure that the normal family constructed is nonempty., 19. A desired map is given by φ(z) = eπz/α ., 20. A desired map is given by f (z) = (eiz − 1)/(eiz + 1)., Exercises 11.31:, 2. Suppose the plane were conformally equivalent to a simply connected, domain D other than itself. Since D is conformally equivalent to a, bounded domain, there would have to be an entire function mapping, onto a bounded domain., r2, 6. f (z) = z., r1, Questions 11.43:, 2. Theorem 11.32 enables us to prove theorems about S from theorems, about T ., 4. Functions of the form z/(1 − eiα z)2 are unbounded., 6. This follows from the fact that if J(f ) is a continuous functional defined, on a compact family F, then the problem |J(F )| = max has a solution, for some f ∈ F., Exercises 11.44:, 1., 4., 7., 8., , 1/z., Its derivative is 0 at z = − 12 e−iα ., ∞, n, n, Consider, ∞ f (z1 ) − f (z0 ) = (z1 − z0 ) + n=2 an (z1 − z0 )., If n=2 n|an | > 1, show that f (r0 ) = 0 for some 0 < r0 < 1., , Questions 12.16:, 3. See Example 12.11 and Exercise 12.17 (6)., 4. It can approach ∞, 0, or oscillate. , ∞, 10. The sequence {an } → 0. The series n=0 an does not converge absolutely. It may or may not converge., 13. It is an entire function., Exercises 12.17:, 3. Set − Log (1 − an ) = an + a2n (1/2 + an /3 + · · · ), and apply Theorem, 12.5.
Page 517 :
510, , Hints for Selected Questions and Exercises, , 5. Apply Exercise 3., 8. (a) |z| < 1 (b) |z| < 1, , (c) Re z > 1., , Questions 12.29:, 1. This will be established in the next section., 7. It is necessary that the convergence be absolute. For instance,, , ∞ , , (−1)n+1 z, √, 1+, diverges at z = 1., n, n=1, 8. The series expansion can be determined from the product expansion,, but the converse is not true., Exercises 12.30:, 1. For example,, , .∞ , n=1 1 −, , 1, n(R+z), , , , e1/(R+z)n ., , ∞ , , z (z/ ln n)+(1/2)(z/ ln n)2 + ··· +(1/n)(z/ ln n)n, 2., + ···., 1−, e, ln n, n=2, ∞, , 2, 3. (b) f (z) =, (1 − z/n)n ez /2n ., n=1, , 4. (a) Use the product expansion for sin πz. and note that the value is, (eπ − e−π )/(2π)., 5. (b) g(z) = z Log (−2i)., 11. Use the identity cos z = (sin 2z)/(2 sin z)., 13. Using the series expansion of e−z/n , it follows that, , , ∞, , n(1 − k) + a k, z, 1+, e−z/n =, (−1)k, z := 1 + an (z)., a+n, (n + a)k!nk, We observe that, , ∞, k=0, , k=0, , |an (z)| converges for all z, because, lim, , n→∞, , an (z), = |az|., 1/n2, , Thus, the given product represents an entire function., 2, 2, 2, 2, 9 eπ + e−π − (eπ /3 + e−π /3 ), ., 14., 8, π4, 1, 15. (b) First set z = 2 and then z = 14 . Now divide the latter expression by, the former., (c) Same as above, with z = 13 and z = 16 ., 16. Logarithmic differentiation of Γ (z) defined by (12.19) gives, , ∞, ∞ , , , 1, 1, 1, 1, 1, Γ (z), =− −γ+z, =− −γ−, −, ,, Γ (z), z, n(z + n), z, z+n n, n=1, n=1, for z ∈ C \{0, −1, −2, . . . }. Another differentiation yields the formula.
Page 518 :
Hints for Selected Questions and Exercises, , 511, , 18. Use the identity ez − 1 = 2iez/2 sin(z/2i)., Questions 12.39:, 2. They are meromorphic., 5. In order to obtain a Maclaurin expansion., 6. No, since the entire function g(z) constructed in the proof of Theorem, 13.8 is not unique., Exercises 12.40:, ∞, , 1, 1. f (z) =, ., (z, −, n)n, n=1, i, 2. e−iz ., 2, 8. Differentiation of the partial fraction decomposition of π cot(πz) gives, the desired result., 10. Take logarithmic derivatives of both sides in the identity proved in Exercise 12.30 (18).., !, !, n, 11. Define Pn (z) = k=1 1 + h2k−1 ez 1 + h2k−1 e−z and, Qn (z) =, , n, , , 1 + h2k−1 ez+2 log h, , , ! , 1 + h2k−1 e−(z+2 log h) ., , k=1, , Then, , !, !, .n, 2k+1 z, 1 + h2k−3 e−z, e, Qn (z), k=1 1 + h, ., = n, 2k−1 ez ) (1 + h2k−1 e−z ), Pn (z), k=1 (1 + h, =, , 1 + h−1 e−z, 1, 1 + h2n+1 ez 1 + h−1 e−z, →, = z, 1 + hez 1 + h2n−1 e−z, 1 + hez, he, , as n → ∞ which confirms the truth of the functional equation., Questions 13.11:, 1. Only if D0 ∩ D1 = ∅., 1, 1, 2. sin, ., = 1 when zn = 1 −, 1 − zn, nπ, 8. When their intersection is nonempty., 10. It can be shown that if, f (z) =, , ∞, , , ank z nk, , with, , nk+1 > (1 + )nk, , ( > 0),, , n=1, , then the circle of convergence of the series is a natural boundary of the, function., Exercises 13.12:, 1, ., (z, − eiθk ), k=1, , 1. f (z) = .n
Page 519 :
512, , Hints for Selected Questions and Exercises, , 2. Write, ∞, , 1, (z + p)n, 1, =, =, ,, 1−z, 1 + p − (z + p) n=0 (1 + p)n+1, , |z + p| < 1 + p,, , and set, for example,, fp+1 (z) =, , ∞, , (z + p)n, with Dp+1 = {z : |z + p| < 1}, (1 + p)n+1, n=0, , where p = 0, 1, 2, . . . ., 7. Expand g(z) = f (z)/(1 − z) in a series, and show that (1 − z)g(z) → ∞, as z → 1 through real values., ∞, , α2, αk, α1, +, +, ·, ·, ·, +, +, bn z n , where, 8. Set f (z) =, z − eiθ1, z − eiθ2, z − eiθk, n=0, ∞, n, n=0 bn z is analytic for |z| ≤ 1., 10. Apply Theorem 13.8 for f (−z)., Questions 13.20:, 1. As e−t > e−1 for all 0 < δ ≤ t ≤ 1,, , , + 1, +, 1 1 x−1, 1 1 − δx, x−1 −t, t, e dt >, t, dt =, e δ, e, x, δ, which approaches ∞ as x → 0+ , for each δ > 0., 8. We need the uniform convergence of the sequence {(1 − t/n)n }. This, sequence does, uniformly on the line., ∞not converge, n, is analytic in a disk and converges, 10. The series, n=1 an z, ∞ uniformly, on compact subsets of the disk; it can be shown that n=1 (an /nz ) is, analytic in a half-plane and converges uniformly on compact subsets of, the half-plane., 11. Entire., Exercises 13.21:, 1. First separate the product for Γ (2z) into even and odd terms:, , ∞ , ∞ , , z, z −z/k , 1, 2γz, = 2ze, e, 1+, 1+, e−z/(k+1/2) ., Γ (2z), k, k + 1/2, k=1, , k=0, , Deduce that Γ (2z)/[Γ (z)Γ (z +1/2)] has the form aebz . Finally, evaluate, a and b by setting z = 1/2 and z = 1. One can also use (13.7) to prove, this formula., ∞, 3. On the line Re s = 2, show that Re ζ(s) > 1 − n=2 1/n2 ., 6. Use the identities Γ (z)Γ (1 − z) = π/ sin(πz) and sin 2θ = 2 sin θ cos θ in, (13.32).
Page 520 :
Hints for Selected Questions and Exercises, , 513, , 7. By (13.18), we have, 1, n1/4, , 1, =, Γ (1/4), , +, , ∞, , t(1/4)−1 e−nt dt, , 0, , so that, + ∞, + ∞, ∞, ∞, , , 1, 1, zn, z, −3/4, −t n, dt., =, t, (e z) dt =, 1/4, 3/4 (et − z), Γ, (1/4), Γ, (1/4), n, t, 0, 0, n=1, n=1, The integral on the right defines an analytic function outside of the, interval [1, ∞).
