Introduction to Complex Analysis (Translations of Mathematical Monographs)

Translations of

MATHEMATICAL

MONOGRAPHS Volume 168

Introduction to Complex Analysis Junjiro Noguchi

American Mathematical Society

Selected Titles in This Series 168 Junjiro Noguchi, Introduction to complex analysis, 1998 167 Masaya Yamaguti, Masayoshi Hate, and Jun Kigami, Mathematics of fractals, 1997 166 Kenji Ueno, An introduction to algebraic geometry, 1997

165 V. V. Iahkhanov, B. B. Lure, and D. K. Faddeev, The embedding problem in Galois theory, 1997

164 E. I. Gordon, Nonstandard methods in commutative harmonic analysis. 1997

163 A. Ya. Dorogovtsev, D. S. Siivestrov, A. V. Skorokhod, and M. I. Yadrenko, Probability theory: Collection of problems, 1997

162 M. V. Boldin, G. I. Slmonova, and Yu. N. Ty urin, Sign-based methods in linear statistical models, 1997 161 Michael Blank, Discreteness and continuity in problems of chaotic dynamics. 1997 160 V. G. OsmolovsklT, Linear and nonlinear perturbations of the operator div, 1997 159 S. Ya. Khavinson, Best approximation by linear superpositions (approodmate nomography), 1997

158 Hideki Omorl, Infinite-dimensional Lie groups, 1997 157 V. B. Kolmanovakil and L. E. Shalkhet, Control of systems with aftereffect, 1996 156 V. N. Shevchenko, Qualitative topics in integer linear programming, 1997 155 Yu. Safarov and D. Vassillev, The asymptotic distribution of eigenvalues of partial differential operators, 1997

154 V. V. Prasolov and A. B. Sossinsky, Knots, links, braids and 3-manifolds. An introduction to the new invariants in low-dimensional topology, 1997

153 S. Kh. Aranson, G. R. Belitsky, and E. V. Zhuzhoma, Introduction to the qualitative theory of dynamical systems on surfaces, 1996 152 R. S. Ismagilov, Representations of infinite-dimensional groups, 1996 151 S. Yu. Slavyanov, Asymptotic solutions of the one-dimensional Schri dinger equation, 1996

150 B. Ya. Levin, Lectures on entire functions, 1996 149 Takashi Sakai, Riemannian geometry, 1996 148 Vladimir I. Plterbarg, Asymptotic methods in the theory of Gaussian processes and fields, 1996

147 S. G. Gindikin and L. R. Volevich, Mixed problem for partial differential equations with quasihomogeneous principal part, 1996 146 L. Ya. Adrianova, Introduction to linear systems of differential equations, 1995 145 A. N. Andrianov and V. G. Zhuravlev, Modular forms and Hecke operators, 1995 144 O. V. Tl,oshidn, Nontraditional methods in mathematical hydrodynamics, 1995 143 V. A. Malyshev and R. A. Mintos, Linear infinite-particle operators, 1995 142 N. V. Krylov, Introduction to the theory of diffusion processes, 1995 141 A. A. Davydov, Qualitative theory of control systems, 1994

140 Aizik I. Volpert, Vitaly A. Volpert, and Vladimir A. Volpert, Traveling wave 139

solutions of parabolic systems, 1994 I. V. Skrypnik, Methods for analysis of nonlinear elliptic boundary value problems, 1994

138 Yu. P. Razmyslov, Identities of algebras and their representations, 1994 137 F. I. Karpelevich and A. Ya. Krelnin, Heavy traffic limits for multiphase queues, 1994 136 Masayoshl Mlyanlshi, Algebraic geometry, 1994 135 Masaru Takeuchi, Modern spherical functions, 1994 134 V. V. Prasolov, Problems and theorems in linear algebra, 1994 133 P. I. Naumkin and I. A. Shishmarev, Nonlinear nonlocal equations in the theory of waves, 1994

132 Hajime Urakawa, Calculus of variations and harmonic maps, 1993 131 V. V. Sharko, Functions on manifolds: Algebraic and topological aspects, 1993 130 V. V. Vershinin, Cobordisms and spectral sequences, 1993 (Continued in the back of this publication)

Introduction to Complex Analysis

Translations of

MATHEMATICAL MONOGRAPHS Volume 168

Introduction to Complex Analysis Junjiro Noguchi Translated by Junjiro Noguchi

American Mathematical Society Providence, Rhode Island

Editorial Board Shoshichi Kobayashi (Chair) Masamichi Takesaki

*f i

(Introduction to complex analysis) by Junjiro Noguchi Copyright © 1993 by Shokabo Publishing Company, Ltd. Originally published in Japanese by Shokabo Publishing Company, Ltd., Tokyo, 1993

Translated from the Japanese by Junjiro Noguchi 1991 Mathematics Subject Classification. Primary 30-01.

Library of Congress Cataloging-in-Publication Data Noguchi, Junjir6, 1948[Fltkuso kaiseki gairon. English[ Introduction to complex analysis / Junjiro Noguchi ; translated by Junjiro Noguchi. p. cm. - (Translations of mathematical monographs ; v. 168) Includes bibliographical references (p. - ) and index. ISBN 0-8218-0377-8 (alk. paper) 1. Functions of complex variables. 2. Mathematical analysis. I. Title. II. Series. QA331.7.N6413 1997 97-14392 515'.9-dc21 CIP

Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication (including abstracts) is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Assistant to the Publisher, American Mathematical Society, P. 0. Box 6248. Providence, Rhode Island 02940-6248. Requests can also

be made by e-mail to reprint-peraissioneaas.org.

1998 by the American Mathematical Society. All rights reserved. Translation authorized by Shokabo Publishing Co.. Ltd. The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America.

® The paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability.

Visit the AMS home page at URL: bttp://vw.aas.org/

10987654321

020100999897

TO AKIKO

Contents Preface

xi

Chapter 1. Complex Numbers 1.1. Complex Numbers 1.2. Plane Topology 1.3. Sequences and Limits Problems

1 1

3 9 15

Chapter 2. Complex Functions 2.1. Complex Functions 2.2. Sequences of Complex Functions 2.3. Series of Functions 2.4. Power Series 2.5. Exponential Functions and Trigonometric Functions 2.6. Infinite Products 2.7. Riemann Sphere 2.8. Linear Transformations Problems

Chapter 3. Holomorphic Functions 3.1. Complex Derivatives 3.2. Curvilinear Integrals 3.3. Homotopy of Curves 3.4. Cauchy's Integral Theorem 3.5. Cauchy's Integral Formula 3.6. Mean Value Theorem and Harmonic Functions 3.7. Holomorphic Functions on the Riemann Sphere Problems Chapter 4. Residue Theorem 4.1. Laurent Series 4.2. Meromorphic Functions and Residue Theorem 4.3. Argument Principle ix

91 91

94

99

CONTENTS

4.4. Residue Calculus Problems

106 114

Chapter 5. Analytic Continuation 5.1. Analytic Continuation 5.2. Monodromy Theorem 5.3. Universal Covering and Riemann Surface Problems

117

Chapter 6. Holomorphic Mappings 6.1. Linear Transformations 6.2. Poincare Metric 6.3. Contraction Principle 6.4. The Riemann Mapping Theorem 6.5. Boundary Correspondence 6.6. Universal Covering of C \ {0, 1} 6.7. The Little Picard Theorem

141

117 124

130 138

141

144 149 153 157 163

167

6.8. The Big Picard Theorem

170

Problems

174

Chapter 7. Meromorphic Functions 7.1. Approximation Theorem 7.2. Existence Theorems 7.3. Riemann-Stieltjes' Integral 7.4. Meromorphic Functions on C 7.5. Weierstrass' Product 7.6. Elliptic Functions Problems

179 179 185 192 194

202 210 226

Hints and Answers

229

References

241

Index

245

Symbols

249

Correction List

251

Preface

Complex analysis is an active research subject by itself, but, even more, it provides the foundations for broad areas of mathematics, and plays an important role in the applications of mathematics to engineering. This book is intended to describe a classical introductory part of complex analysis for university students in the sciences or engineering, and in particular to serve as a text or a reference book for juniors, seniors, or first-year graduate students in the sciences. The prerequisites are elementary calculus, from the real numbers through differentiation and integration, and linear algebra; set theory and some general topology are not required, but would be helpful. Historically, the contents of this volume had been discovered by the beginning of the twentieth

century, so they are not very new. Nevertheless, the author tried to arrange the presentation and choose terminology to agree with modern work in the field. Many books have been written on this subject, in many languages. Roughly speaking, they can be divided into two types. One tries to give an understanding

of the subject using intuitive arguments. The other puts more emphasis on rigorous proofs, presenting the subject as a fundamental theory of mathematics. This book falls in the latter group. It is well-known that there is some difficulty in dealing with curves, related to Cauchy's integral theorem, a point on which several published books are somewhat unsatisfactory. To deal with it rigorously, we give detailed descriptions of the homotopy of plane curves. This material on curves is something the reader may well have encountered in previous courses, so it is scattered through the text where needed. Readers whose mathematical background already includes this material may simply confirm the theorems and go on. Since residue theorem is important both in pure mathematics and in applications, we give a fairly detailed explanation of how to apply it to numerical calculations; this should be sufficient for those who are studying complex analysis for applications.

After this book, the student should be ready to take up value distribution theory, the theory of R.iemann surfaces, complex analysis in several variables, the theory of complex manifolds, and other subjects. Complex analysis will also xi

PREFACE

xii

provide fundamental methods for the theory of differential equations, algebraic geometry, number theory, and other fields. The author wrote Chapters 6 and 7 with these transitions in mind.

The author will be very pleased if this book helps students to understand the classical theory of complex analysis, to relish its beauty, and to master the rigorous treatment of mathematical demonstrations. This volume is based on lectures for third-year students given by the author at the Tokyo Institute of Technology. The class contained not just mathematics majors, but also physics and applied physics students. The author is very grateful to all of them. Finally, but not least, the author would like to express his deep thanks to Professor Mitsuru Ozawa, who led him to complex analysis, and to Professor Shingo Murakami, who recommended him to write this book. He is also obliged to Mr. Shuji Hosoki, of the Shokabo publishing company, for the proofreading. April, 1993

at Ohokayama Junjiro Noguchi

Added in English translation: The publication of this English translation was made possible by the sug-

gestion and the recommendation of Professor Katsumi Nomizu. The author expresses his sincere gratitude to him. February, 1997

at Ohokayama Junjiro Noguchi

CHAPTER 1

Complex Numbers

In this chapter we assume that the real numbers are known, and we explain the elementary facts of complex numbers.

1.1. Complex Numbers The set R of all real numbers cannot be enlarged as long as its order " 0; that is, of A. If a

A(a; r) n A = 0. This is absurd. Thus A = A.

To show the converse, we take an arbitrary z ¢ A. By assumption, z is not an accumulation point of A, either. Therefore, there is r > 0 such that A(z; r) n A = 0; that is, A(z; r) C A`. This shows that A' is open. ii) It suffices to show that A C A. Take an arbitrary accumulation point z of A. Since for any r > 0

A(z;r)nA96 0, we may take z1 E 0(z; r) n A. Set r1 = r - Iz1 - zj. Since z1 E A, we have

A(z1;r1)nA#0.

1.2. PLANE TOPOLOGY

5

Hence, A(z; r) n A 54 0, and then z E A.

.4

FIGURE 4

iii) Taking an arbitrary closed set E with E D A, we have to show that E D A. This is easy and is left to the reader.

For a subset A C C, a point z E A is called an interior point of A if there is a neighborhood U of z with U C A. Therefore, the set of all interior points of A, denoted by A, is open. An interior point of A` is called an exterior point of a A. The set 8A = A \ A is called the boundary of A, and a point of it a boundary point. In particular, 8A is a closed set. O

We set

C(a;r) = 80(a; r) = {Iz - al = r}, which is called the circle of radius r with center a; in particular, C(0;1) is called the unit circle. Let B be a subset of A. Then B is called an open set in A if there is an open set U C C with B = A n U. Similarly, B is called a closed set in A if there is a closed set E C C with B = A n E. In this way we get the relative topology of A. If B is an open (resp., closed) subset of A, then A \ B is a closed (resp., open) subset of A. If B has no accumulation point in A, B is said to be discrete in A. In general, a family {Ua}aEr of subsets Ua C C is called a covering of A if A C UaEr U, . In particular, if all Ua are open, the covering is called an open covering of A. In this case, {UQ n A}QEr is called an open covering of A (with respect to the relative topology), too. If there is an open covering {U1, U2} of A satisfying (1.2.3)

U1nA540,

U2nA340, UInU2nA=O,

we say that A is not connected. If there is no such covering, then A is said to be connected. A connected open subset of C is called a domain. (1.2.4) THEOREM. The following three conditions are mutually equivalent. i) A is connected.

1. COMPLEX NUMBERS

6

ii) Let {U1, U2} be any open covering of A with respect to the relative topol-

ogy, such that U1 fl U2 = 0. Then either U1 = A (U2 = 0), or U2 = A (U1 = 0).

iii) If any non-empty subset B of A is open and closed in A, then B = A. PROOF. i) -4:* ii) This is just the definition. ii) iii)

. iii) Put U1 = B and U2 = A \ B. . ii) Let the non-empty U; be B.

Let I be a closed interval of R. Then a continuous mapping 0 : 1 3 t

¢(t) _

x(t) + iy(t) E C (i.e., x(t) and y(t) are real-valued continuous functions on I) is called a curve or arc. In the case where 1 is a general interval (open interval, semi-closed interval, etc.), it is called a general curve. The image C = ¢(I) is also called a curve, but, strictly speaking, 0 should be called a curve. We say that "C is given by 0". In fact, the two curves 01

[0, 2a] 3 t - cost + i sin t E C,

dpi : [0,2w] 3 t -+ cos 2t + i sin 2t E C,

have the same image 01([0, 27r]) = 02((0,2r]). But 01 goes around the unit circle once, and 02 twice, so they should be distinguished. The variable t is called the parameter of 0.

FIGURE 5

A curve

: J -+ C is called a parameter change of 0 if there is a strictly mono-

tone continuous function r : J -F I from J onto I satisfying O(s) = 0(r(s)), s E

J. If I = [To, TI] and 0(To) = 0(T1), then 0 is called a closed curve. In this case, we consider also the following ip as a parameter change. Let J = [So, S1] (So < S1) be an arbitrary closed interval, and take -ro E (To, T1) and ao E (So, S1). We put (1.2.5)

tb(s)=0(ro+ s-So (T1-ro)), \\

VI (

s)

= 0 TO +

So:s

Qo,

ao - So s - ao

Si - ao

(ro

- To

)

,

Qo < s < S 1.

1.2. PLANE TOPOLOGY

7

Furthermore, a parameter change of this Tp is also considered as a parameter change of .0. 0(r.)=d(TJ

1

FIGURE 6

J

FIGURE 7

It is clear that parameter changes of curves give rise to an equivalence relation. In our arguments, it is inconvenient to consider a curve as a continuous mapping. Therefore, we call the equivalence class of 0 a curve, and denote it by C, C((b),

or C(O: I -- C). Here 0 stands for the representative. Still, we may call 0 and the image 0(1) a curve, but the meaning will be clear. ExERCISE 1. Let 0 : [To, T1 ] C be a curve, and [So, S1 ] (So < S1) an arbitrary closed interval. Then there is a parameter change : J -' C of 0 with

J=[So,S11. Let C(0 : [To,T1] -+ C) be a curve. Then O(To) is called the initial point, and 0(T1) the terminal point. The initial and terminal points of C are called the ends of C, and we say that C connects ¢(To) and 4)(T1). If 0 is one to one, C is called a simple curve (or a Jordan curve). If C is closed and 0 is one to one on [To, TI), C is called a simple (or Jordan) closed curve. When C is a closed curve, an arbitrary point of C can be the initial (terminal) point by a parameter change. Thus, in this case we consider that the initial (terminal) point of C is O(TO)), determined if a representative ¢ of C is fixed. If 0 is constant (0(t) then ¢ is called a constant curve. We say that 4) : [To, T1 ] - C is continuously differentiable if the real part x(t) of 4)(t) and the imaginary part y(t) of 0(t) are defined in an open interval containing [To, T1 ], are differentiable there, and their derivatives are continuous. We set T(t) _ 4)'(t) = dt (t) + i y (t). Here, in case 0(To) = O(T1), we also assume that do(To)/dt = d¢(T1)/dt. Moreover, 0 is said to be non-singular if dO(t)/dt # 0 for all t. If there is a partition To = to < t 1 < ... < t, = T1 such that the restrictions 01 [tt _ I, t, ] to [ti _ 1, t;], 1 < j < n, are continuously differentiable, ¢ is said to be piecewise continuously differentiable. A curve C is (piecewise) continuously differentiable if there is such a representative 0 of C. A non-singular curve and a piecewise non-singular curve are similarly defined.

1. COMPLEX NUMBERS

8

C) be a curve. On the image of C, 0(t) moves Now, let C(4 : [To,TI] from the initial point toward the terminal point as the parameter t E [To,T1) increases. Assume that C is not a constant curve. By the above Exercise 1 we may assume [To,T1] = [-1,1]. Put 0(t) = 0(-t). Then ip : [-1,1] -. C is not a parameter change of 0. If 0(-1) iO 0(1), then 0(t) moves from 0(-1) toward '(1) as t moves from -1 toward 1, and ?P(t) from 0(1) toward 0(-1) Therefore we may consider that the curve C(¢) already has an orientation in this sense. .

The curve given by >(i is called the curve with inverse orientation and is denoted

by-C.

o{-))=v())

IX 1)

FIGURE 8

Now, suppose that the terminal point of a curve C1 is the same as the initial point of another curve C2. Suppose that C1 is given by ¢I : [0,1] -. C and C2 by 02 : [l, 2} -. C. Then we define a curve Cs(03 :10,21 -. C) so that it is equal to 01 over [0,1] and to 02 over [1, 2]. The curve C3 is called the sum of C1 and C2, and we write C3 = C1 + C2.

For a subset A C C, C is called a curve in A if its image is contained in A. If two arbitrary points of A can be connected by a curve in A, A is said to be arcwise connected For instance, a disk i(a; r) and C itself are arcwise connected.

Ex itcISE 2. A closed interval [To, TI] (C R C C) is connected in the sense of Theorem (1.2.4). (1.2.6) THEOREM. An open subset of C is connected if and only if it is arcwise connected.

PROOF. The "if' part follows from Exercise 2. To show the "only if' part, we assume that A is a connected open set and A # 0. Fix a point zo E A. Let U1 be the totality of all points of A connected to z1 by curves in A, and let U2 be its complement. For any point a E Uj U = 1, 2), we take r > 0 with A(a; r) C A. Since 0(a; r) is arcwise connected, A(a; r) C Up Thus the U, are both open sets. Since A = U1 U U2 and U1 # 0, we have U2 = 0 by definition,

andsoA=UI. O

1.3. SEQUENCES AND LIMITS

9

A curve 0 : [To, T1 ] -+ C is called a piecewise linear curve (Streclcenzug) if < t = T1 of [To,T1] such that on each there is a partition To = to < t1 < [tj -I , tj ]

is written as

0(t) = 0(t3-1) + aj(t - tj-1),

ti-1 5 t 5 tj,

where a, E C. We immediately obtain from the proof of Theorem (1.2.6)

(1.2.7) COROLLARY. An open subset A of C is connected if and only if two arbitrary points of A can be connected by a piecewise linear curve. In general, let A be a subset of C. For two points z1, z2 E A we write z1 - z2

if there is a connected subset A' C A with z1, z2 E A'. We show that this is an equivalence relation. The reflexive law, "z1 - z2 . z2 - z1", is trivial. Assume that z1 - z2 and z2 - z3. Take connected subsets A', A" C A such that x1, z2 E A' and z2, z3 E A". It suffices to show that A' U A" is connected. Let {U1i U2} be an open covering of A' U A" with respect to the relative topology such that U1 fl U2 = 0. Since {U1, U2} is also an open covering of A', A' C U1 or A' C U2; we may assume A' C U1. In the same way, we see that A" C U1 or A" C U2. Since Z2 E A' fl A" C U1, A" C U1. Therefore, A' U A" C U1, and so its connectedness follows. For z E A we set A2 = {z' E A; z' - z}; this is called the connected component of A containing z. If A.., 34 Ax then A., f1 A., = 0. Taking representative elements za from all connected components of A, we have (1.2.8)

A = UA2,. Z.

This is called the decomposition of A into connected components. In particular, if A is open, then so are the connected components of A, so that their represen-

tatives za = xa + iya may be chosen as rational points. Since there are only countably many such points, we may enumerate them, and hence denote them by zi, i = I,- , N(< oo). Put A; = Az,, i = 1,... , N. Then they are disjoint domains and N

(1.2.9)

A=UAi. i=1

1.3. Sequences and Limits From now on, a complex number is simply called a number. { z } no of numbers (or points) is An infinite sequence (zo, z1, ... , z,, , ... called a sequence. A sequence {z} n o is said to converge to a E C or to have limit a if for an arbitrary e > 0 there is a number np E N such that

Iz - al<e

for all n>no.

1. COMPLEX NUMBERS

10

In this case we write

a = lim n--xzn or zn -+ a (n

(1.3.1)

oo).

Since (1.3.2)

max{IRezl,lImzl} < Izl S IRezl+IImzl

for z E C, (1.3.1) is equivalent to

a = lim n-=Zn,

(1.3.3)

0 = lim yn, n-+oc

where zn = xn + iyn and a = a + if3. A convergent sequence {zn }n o is always bounded (that is, for some M, IznI < M for all n). In particular, for a sequence (r,}"°-0 of real numbers we write

llm rn = +00 n-ao if for an arbitrary K > 0 there is a number no E N such that rn > K for n ? no. If lim zn and lim wn exist, and a, b E C, then we have lim(azn + bwn) = a lim z + b lim wn,

lim(znwn) = (lin1zn)(Iinwn), lnn zn = rim Zn ,

I IiMzn I = lim I zn l

EXERCISE 1. Show the above four equalities.

A sequence {zny}"_p formed by apart of (the order is not changed) is called a subsequence of {zn} 0. If {zn} o converges to a, then so does any of its subsequences. (1.3.4) THEOREM. Every bounded sequence has a convergent subsequence.

This is clear by the corresponding Weierstrass' theorem for real numbers, (1.3.2) and (1.3.3). A sequence {z,,}'o is called a Cauchy sequence if for an arbitrary e > 0 there

is an no ECsuch that

Izn-z,nl<e,

n,m>no.

In this case, {Rezn}

are both Cauchy sequences by (1.3.2). and {Imzn} Thus by Cauchy's theorem for sequences of real numbers we have (1.3.5) THEOREM. A sequence {zn}n o converges if and only if it is a Cauchy sequence.

Let A C C be an arbitrary subset. Note that A is closed if and only if any convergent sequence of points of A has a limit in A. We say that A is compact if any sequence of points of A has a convergent subsequence.

1.3. SEQUENCES AND LIMITS

11

(1.3.6) THEOREM. The following are equivalent for A C C: i) A is compact. ii) A is bounded and closed. iii) Let A C UQEr UQ be an arbitrary open covering. Then there are finitely many U , , , ... , UQ, such that I

AC UUQ,. =1

PROOF. The equivalence of i) and ii) follows immediately from Theorem (1.3.4).

Put S = {Re z; z E A}. Then S is a bounded closed subset of R. Let a be the minimum of S. Then we put

A[a,r]={zEA;a:Re z a. The corresponding statement for bounded closed subsets of R is known as the Heine-Borel theorem. Therefore A[a, a] is covered by finitely many

UQ,,1SjSk: k (1.3.7)

A[a, a] C U UQ, . j=1

FIGURE 9

It follows that there is a r1 > a such that k

A[a, r1] C U UQ, . j=1

(If there is no such rl, we can find z E A[a,a + 1/n] \ U;=1 UQ,, n = 1, 2, .... By "i) a ii)" shown above we may assume that converges to a. Then a E A[a, a], so that some UQ, 9 a. Thus z E UQ, for large n. This is absurd.) Now, let TO be the supremum of r such that A[a, r] is covered by finitely many UQ . Suppose ro < max S. It follows from the same reason as (1.3.7) that A[ro, rp]

1. COMPLEX NUMBERS

12

is covered by finitely many U. , and hence so is A[ro - 6, ro + 6] for some 6 > 0. Since A[o. ro-6] is covered by finitely many Up, so is A[u, ro+6]. This contradicts

the choice of r0. Thus r(j = maxS. Again by the same reasoning we see that A = A[o, max S] is covered by finitely many U0. Consider the open covering {,3(a;1);a E Al of A. By the assumption there are finitely many points a1, ... , a1 E A such that A C c7(a1; 1) U ... U 6(a,;1).

Therefore A is bounded. We next show that A is closed. Let a be an accumulation point of A. Suppose

a ¢ A. Put

Uo={zEC;Iz-a]>2 U = 1zEC;n+2 0 there is an no E N such that na

fl zn-1 <e

(1.3.12)

for aUn2>n1?no.

n=n,

PROOF. i) Take no as in (1.3.10), i), and put a = limrn_,, Hn no zn. Then a54 0 and

_

zm

lln=no Zn m_1 n=nr, zn

_ aa =1

(m-+oo).

ii) We first show the "only if" part. Take no as above. There is an M > 0 such that 1
0 there is an no (> n4) such that nz

11 n1

zn-1 <e

foralln2?n1_no.

It follows from this and (1.3.13) that n, Inriil n2

Zn - fl Zn n=no

n=no

I I

Zn Zn

In=no

I

I

I

I In°ni

Me.

PROBLEMS

15

gives rise to a Cauchy sequence, and so it converges by Hence { m ;b zn } M-o Theorem (1.3.5). By (1.3.13) the limit is not zero. 0

We will discuss infinite products again in the next chapter, after defining the exponential and logarithmic functions for complex numbers.

Problems 1. Write f, 'T1,

1 -

3i in the form of x + iy.

2. In general, put x + iy = u + iv. Express u and v as functions of x and y. 3. To a complex number z = x+iy we assign a real matrix 0(z) _ ( y V). Let O(z1)+¢(z2) and 0(z1)0(z2) stand for the addition and product as matrices. Then show the following. 0(zt ± z2) = O(zl) ± 0(z2), 0(z0O(z2) = O(zlz2),

04)- _ 0(z-1)

(z 34 0).

+ i sin 2, n E N, 0 < j < n - 1, on the 4. Take n points P3 = cos unit circle. Let POP, be the distance between Po and P3. Show that

.,-1-

j=1 POP; = n.

5. Let 0 < r < 1, and 0, E R, n = 0, 1, 2, .... Show that

a

E rn (COS On + i sin On) n=0

converges.

6. Define a sequence {zn}n`_o by zn+1 - zn = a(zn - z.-I) with 0 < 1al < 1. Write the limit lim zn in terms of zo and z1.

7. Let E. C C, a E t, be compact subsets such that EQ, n . . . n EQk # 0 for any finitely many EQ ... , EQk. Show that nQEr EQ 0 0. 8. Let B C A be a discrete subset in A (c C). Show that for any compact subset K C A, K n B is a finite set. 9. Show that the ring R(r1 i r2) = {z E C; r1 < Izj < r2} (0 < r1 < r2) is a domain.

10. Let D C C be a domain, and E C D be a discrete subset. Show that D\ E is also a domain.

CHAPTER 2

Complex Functions

In this chapter we use the properties of complex numbers described in the previous chapter to deal with complex functions, per series and analytic functions, and to study their properties. By making use of power series we define elementary functions such as exponential functions, trigonometric functions, etc. Then we define the Riemann sphere and introduce linear transformations. The material presented in this chapter may be said to form the foundation of complex functions theory.

2.1. Complex Functions Let A c C, and let f : A -. C be a mapping. We call f a complex function defined on A. We set

f (z) = f (z),

Ref=f+f 2

z E A,

'

Imf=f2i-f We call Ref (reap., Imf) the real (reap., imaginary) part of f . The function f is said to be bounded if there is a number M > 0 such that If (z) I < M for all z E A. Let zo E C be an accumulation point of A. Then we say that f has limit a E C at zo, if for an arbitrary e > 0 there is a b > 0 such

that

If(z) - al < 0,

zE(A(zo;6)\{zo}) f1 A,

and write

a = lim f (z). ExERC11E 1. A function f (z) has limit a at zo if and only if for an arbitrary f (z,,) = a. of A \ {zo} converging to zo,

sequence

17

2. COMPLEX FUNCTIONS

18

In particular, if zo E A and f (zo) = lim=_,, f (z), f is said to be continuous at zo. If f is continuous at all points of A, f is said to be continuous on A. Let g be another function on A. If f and g are continuous at zo (or on A) and a, b E C, then af(z) + bg(z),

f(z)g(z)

are continuous there; moreover, if g(zo) # 0, then f(z)/g(z) is continuous at zo, too.

EXERCISE 2. i) f (z) is continuous at zo if and only if Ref (z) and Imf (z) are continuous at zo. ii) If f (z) is continuous at zo, then so is if (z) 1.

We say that a complex function f : A -. C is uniformly continuous if for an

arbitrary or >0there is ab>0such that

Iz-w1 0 and a E A there is a b(a) > 0 such

that z E i (a,b(a)) n A

If(z) - f(a)I < e.

Since {A(a; 6(a) /2); a E A) is an open covering of A, there are finitely many ax, ... , ai E A by Theorem (1.3.6) such that I

AC

tJ A(aj,bj /2),

b; = b(aj).

Therefore, for an arbitrary z E A there is some j with &(a,,5J/2) 9 z, so that

If(z)l =1f(z) - f(ad) + f(a,)l < e + If(aJ)I <e+ max if(ai)Ic 15j 0 there is a number no such that for all n > no and z E A

Ifn(z) - f(Z)1 < f(2.2.1) THEOREM. i) A sequence {fn }n'=0 of complex functions on A converges uniformly if and only if for an arbitrary e > 0 there is a number no such

that for allm,n?no and zEA Ifm(z) - fn(z)1 < f. ii) If a sequence {fn}' o of continuous complex functions fn on A converges

uniformly, then the limit function f = limn, fn is continuous. PROOF. i) This is easy and is left to the reader. ii) Take an arbitrary e > 0. By assumption there is a number no such that

If (Z) - fb(z)I < f,

z E A.

Take a point a E A, and fix it. Since fro is continuous at a, there is a b > 0 such

that Ifno(z) - fno(a)I < f, z E A(a;b). It follows from the above two equations that for z E A(a; 6)

If(z) - f(a)I < if (z) - f,b(z)I + Ifno(z) - f, (a)I + Ifno(a) - f(a)I < 3f.

Thus f is continuous at a. 0

2. COMPLEX FUNCTIONS

20

A sequence (f,, } n o of complex functions on A is said to be uniformly bounded

if there is an M > 0 such that for all n and z E A

If"(z)I s M. A bounded sequence of complex numbers necessarily has a convergent subsequence (Theorem (1.3.4)). If a sequence of complex functions on A is uniformly bounded, then does it have a convergent subsequence? The answer is "No" in general (see problem 3 at the end of this chapter for an example). Furthermore, here we consider sequences of continuous complex functions, and require the limit functions to be continuous. With this in mind we introduce the following new concept: A family F of complex functions on A is said to be equicontinuous if for an

arbitrary c> 0 there is a 6> 0 such that for all f E F (2.2.2)

Z, W E A, Iz - wI < b

If (z) - f (w) I 0 such that IRezI 5 M, IImzI 5 M for all z E A. Set

F1={zEC;IRezISM, IImzISM}. Dividing each side of the closed square F1 into two equal parts, we have four closed squares F2,1, ... , F2,4. Repeating this process n times, we have 4" 1 squares F", J, 1 5 j 0, we choose a 6 > 0 so that (2.2.2) holds. By Theorem (1.3.6) A is covered by finitely many A(a3;b) with aj E A and 1 < j < 1. Since E 0(aj; 6) n E for every j. There is a number Ito E = A, there is a point. such that for all It, p' > Ito (2.2.4)

19v(z,.(J)) -go,(z,(,))I < C,

1Sj

Now, let a E A be an arbitrary point. Then t(a,; b) 3 a for some j. It follows from (2.2.2) and (2.2.4) that for the above p, p' > Ito 9,,'(a)f :519,,(a) - 9p(z,,(j))I +

9v'(z,.(,))l

+ Igo,(z,.(,)) -gv'(a)1 < 3e.

Therefore we infer from Theorem (2.2.1), i) that {g}N o converges uniformly on A.

Let D be an open subset of C and {f }n o be a sequence of complex functions on D. We say that o converges uniformly on compact subsets if it converges uniformly on every compact subset of D. o of complex functions on D converges uniformly on every compact subset if and only if for every a E D there is an r > 0 such that A(a; r) C D and If,, }n_0 converges uniformly on A(a; r). ii) The limit function of a sequence of continuous complex functions on D which converges uniformly on compact subsets is continuous.

(2.2.5) THEOREM. i) A sequence

PROOF. i) The "only if" part is clear. We show the "if" part. Let K be a compact subset of D. By assumption, there is an rd > 0 for every a E K such that A(a;r0) C D and o converges uniformly on 0(a;rQ). Since {I(a;rQ)}QEK is an open covering of K. there are finitely many al,... a, E K such that K C Vj-1 0(a,;rl) with r, = r0,. Since o converges uniformly on U.,=1 t(aj; r,), it also converges uniformly on K. ii) To show continuity we may restrict the complex functions to a disk neighborhood A(z; r) C= D of a given point z E D. The result then follows immediately from Theorem (2.2.1), ii).

We say that a sequence {f, }n o of complex functions on D is uniformly bounded (reap., equicontinuous) on compact subsets if it is uniformly bounded (resp., equicontinuous) on every fixed compact subset of D. (2.2.6) THEOREM. If a sequence {f,,} n o of complex functions on D is uniformly

bounded and equicontinuous on compact subsets of D, then it contains a subsequence converging uniformly on compact subsets of D. We prepare the following for the proof.

2. COMPLEX FUNCTIONS

22

(2.2.7) LEMMA. There are relatively compact subsets U,,, n = 1, 2, ... , of D satisfying i) U. C=

1, n = 1, 2, ... ; ii) for any compact subset K C- D, there is an n E N with K C U,,.

PROOF. If 8D = 0, then D = C, and we may put U = 0(n), n = 1, 2, ... . Assume that 8D 0. We define the boundary distance by d(z; 8D) = inf { Iz - wI; w E 8D},

(2.2.8)

z E D.

Take a sequence w E 8D, n = 0,1, 2, ... , so that

Iz -

-d(z; 8D) (n -oo).

o is bounded, and so it has a converging subsequence by Theorem (1.3.4). Hence we may assume from the beginning that limwn = wo. Since 8D is closed, wo E 8D. Therefore we have

Then {w },°,°

(2.2.9)

d(z; 8D) = Iz - woI = min{Iz - wI; w E 8D} > 0.

Forz'ED Iz' - woI - Iz - woI v }v o which converges uniformly on U2. Inductively, we take a subsequence which converges uniformly on U,, (ja = 1,2,... ). o of Then the subsequence { f n y 1 } i } o of {f,,} converges uniformly on every U. Hence, {fn(P+1),'}x o converges uniformly on compact subsets of D. 0

EXERCISE 1. Show that the functions f,, (z) = zn, z E A(1), n = 1, 2, ... , are not equicontinuous. EXERCISE 2. Show that if {fn(z)}°' 1 is uniformly bounded and equicontin-

uous on a subset A C C, then so are gn z =

n

2.3. Series of Functions In what follows, functions mean complex functions. Let {fn}no be a sequence 3o f,, is called a series of functions defined on A C C. Then the formal sum o f functions. For a number N (= 0, 1, - - -) the sum En=o f,, is called an Npartial sum or simply a partial sum. If the sequence {sv(z)}N-o converges at a point z E A, the series o fn of functions is said to converge at z, and we write

n

lira sN(x) _ fn(z). N--x n=0 If {sN}N=o converges at all points of A, we say that F_n ofn converges on A and write the limit function as

x A.

lim sN _ n=0

If two series, E o f,, and F_n o g, of functions on A converge at a point z E A (resp., on A), then for constants a, b E C

x

x

x

(afn(z) + bgn (z)) = a > f,, (z) + b> g. (z) n=0

n=0

n=0

at z (resp., on A). If ER`a Ifn(z)I converges at z (resp., on A), Ln o fn is said to converge absolutely at z (resp., on A). The following is clear. (2.3.1)

Theorem (1.3.8), iii) and iv) hold for a series Fx ofn of functions at z E A or on A.

We say that F,n o fn converges uniformly on A if {sN } N=o converges uniformly on A.

2. COMPLEX FUNCTIONS

24

(2.3.2) THEOREM. A series ER o f of functions on A converges uniformly on A if and only if for an arbitrary e > 0 there is a number no such that fj (Z) <e 1=m

for all n ? m >_ no and z E A. Moreover, if the series converges uniformly and all the functions fn are continuous, then the limit function Eno fn is continuous.

PROOF. Assume that F_R o f, = f converges uniformly. Then for an arbitrary e > 0 there is a number no such that n

IE fj(z) -f(z)

no. Then we have n

E f,(=)

- ('f(Z)_f(Z) j=o

fj(z) - f(Z)

j=m

)

j=o

m-I E fj(z) - f(z) < 2e.

fj(z) - f(z)

j0

)=0

This proves the "only if" part.

Take an arbitrary e > 0.

Now, we show the "if" part. Put f = E By assumption there is a number no such that 0

n

Ef3(z) - Efj(z) j=0

j=0

for all n ? m > no and z E A. Letting n

E f) (Z) < e

Pm+I

oo, we have

m

f(z) - Efj(z)

< C.

1=0

Thus Eno f, converges uniformly to f . Continuity is a direct consequence of Theorem (2,2.1), ii).

0

We say that a series Fn o M with M 0 is a majorant of a series F,' 0 fn of functions on A if

Ifn(z)I < M.,

z E A, n = 0, 1, ... .

The following theorem is called the majorant test or Weierstrass' M-test. This test is simple but quite useful.

2.4. POWER SERIES

25

o Mn be a majorant of a series En o fn of functions on A. If E' o Mn converges, then E'0 fn converges absolutely and uniformly

(2.3.3) THEOREM. Let

on A.

The proof is clear by definition and by Theorem (2.3.2). EXERCISE 1. Show the above Theorem (2.3.3).

Let T_n o fn be defined over an open set D C C. If the sequence {sN } of its partial sums converges uniformly on compact subsets of D, then =o fn is said to converge uniformly on compact subsets of D. By Theorem (2.2.5), ii) we have

(2.3.4) THEOREM. Let >n o f" be a series of functions on an open set D C C. If En o fn converges uniformly on compact sets of D. then the limit function is continuous.

2.4. Power Series

Let {an}n o be a sequence and c E C. Then the series >no an (z - c)' of functions is called a power series. By a translation we can reduce to the case c = 0, and so we henceforth mainly assume that c = 0. For a power series o anz" and zo E C we consider the following three conditions:

En

(2.4.1)

i) Eno anzo converges.

ii) limn_,, anzo = 0. iii) limn_,,,,Ianzp I < +oo.

Clearly, i) implies ii), and ii) implies iii).

(2.4.2) LEMMA. If any of (2.4.1) holds, then and uniformly on compact subsets of o(Izol).

oanz' converges absolutely

PROOF. It suffices to assume that zo 96 0 and (2.4.1), iii) holds. Then there is an M > 0 such that Janzo"I 0 such that Eno anz" converges. We call R the radius of convergence of E 0 anz", and {z E C; Iz1 = R} the circle of convergence. If R > 0, we call F,n o an z" a convergent power series. The limit n oanz" defines a continuous function on A(R).

2. COMPLEX FUNCTIONS

26

(2.4.3) THEOREM. The radius of convergence of > 0anzn is given by

R=

1

!im n- 3c

IanI

where 1/ + oc = 0 and 1/0 = +oo. PROOF. Suppose that R = 0. Then it follows from Lemma (2.4.2) that for any r > 0 there are infinitely many n with Ianlrn > 1. Hence

nlm "IanI>_T

Letting r \ 0. we have that tun

nx

IanI = +00.

Suppose that R > 0. Take arbitrarily 0 < r < R. By Lemma (2.4.2) there is a number no such that IanrnI < 1 for n no. Therefore

llm "

IanI < f.

Letting r / R, we have that lim " IanI < 1/R. Next we take r > 0 so that noo im " IanI < 1/r. Then there is a number no such that Ian Irn < 1 for all

n-x

n ? no. By Lemma (2.4.2), r no. Therefore ,r,n-no


0. Since _ 4 0' n > 2, (2n - 2)! (4n)! 1

letting x = 2 we have ;

? cost=l - 22 + 2 2! 4! 0 and 0 < x 0,

x sin x = = ((4m 1+I )! _ (4mx2+ 3)!) x m=o

1

> 0.

We take 05x 0 is called a disk neighborhood of oc. The subspace S\{N. S} is identified with C', which carries two complex coordinates z and i, and by (2.7.2) both coordinates give the same topology. The space S, whose subspace C\ {N} (resp., C \ {S}) is assigned the complex coordinate z (resp., i) is called the Riemann sphere, and denoted by C. The statements on the topology of C such as the convergence of sequences. accumulation points, closures, closed sets, open sets, connectedness, domains, continuous functions and mappings. etc.. continue to bold on C. For instance, let E2 be a mapping. Then f is said to E3 C C, j = 1, 2. be subsets, and let f : E1 be continuous if for an arbitrary point P E E1 and an arbitrary sequence o 0 converges to f (P), i.e., lim,, , f (P,) = f (P). In particular, if f is continuous, injective, and surjective. and if the inverse f -1 E2 - E1 is also continuous, then f is called a homeomorphism. For

in E1 converging to P, { f :

example, fo in (2.7.2) is a homeomorphism. Since the composition of analytic functions is analytic (see Remark (3.5.9) in

the next chapter), it follows from (2.7.5) that if a function in an open subset of C \ {0,o0} is analytic in z. it is analytic in i, and the converse holds, too. Therefore the notion of analyticity is well defined on an arbitrary open subset of C. A circle of C stands for a circle, or a line of C (plus the point at infinity oc to be precise). Here a line of C is considered as a circle passing through ao. EXERCISE 2. Let P (resp., P') be the point of S corresponding to z E C (resp.,

z' E C) by the stereographic projection from N. Express the distance d(P, P') between P and P' in R3 in terms of z and z'. Moreover, express the distance d(P, N) in terms of z. EXERCISE 3. Show that under the stereographic projection circles in C correspond to circles in S which are intersections of S and hyperplanes in R3, and vise versa. (This fact is referred to as the correspondence of circle to circle under stereographic projection.)

2. COMPLEX FUNCTIONS

40

2.8. Linear Transformations A mapping f from C onto itself written as (2.8.1)

f(z)

=

az + b

cz+d

is called a linear fractional transformation, a linear transformation, or a PvMobius transformation. Here the coefficients a, b, c, d E C must satisfy

ad-bc96 0. If c = 0, f (oo) = oo, and otherwise

f(oo) = a,

f

(_d)\ = 0,0_

The mapping f has the inverse f-1: f-1(W )

//

_ dw-b

-cur+a

which is a linear transformation. In particular, f : C C is a homeomorphism. The multiplication of a. b, c, and d by a common non-zero number does not change the linear transformation f (z), and hence we may assume that (2.8.2)

ad - be = 1.

The set of all 2 x 2 complex matrices ( d) satisfying (2.8.2) is denoted by SL(2,C). In general, a set G endowed with an operation a b (a, b E G) satisfying the following conditions is called a group:

i) Two arbitrary elements a, b E G determine a third element a b E G, and (a b) c = a (b c) holds for any three elements a, b, c E C. ii) There is an element e E G such that a e = e a = a for all a E G. (This e is unique and is called the unit element.) iii) For an arbitrary a E G there is an element b E G such that ab = ba = e. (This b is unique; it is called the inverse element of a and is denoted by EXERCISE 1. Show the uniqueness asserted in ii) and iii).

The set SL(2, C) forms a group under matrix multiplication, and is called the special linear group. By a direct computation we see that the composition of two linear transformations is also a linear transformation. It is easily seen that this composition satisfies the above conditions so that the set Aut(C) of all linear transformations is a group. The unit element of Aut(C) is the identity mapping.

2.8. LINEAR TRANSFORMATIONS

41

(2.8.3) THEOREM. 7he mapping

c

d) E SL(2, C) -+

cz + d

E Aut(C)

is a surjective group homomorphism (i.e., 4;(a p) = 44(a) Ker 0 (= $-1(the unit element of C)) is as follows:

The kernel

'1)}.

PROOF. The first half follows from a direct computation. For the latter half we use the identity az + b

cz+d

=

z

and (2.8.2). Then we have b = c = 0, ad = 1, and a/d = 1. Therefore a = d = I

ora=d=-l.

Identifying a E SL(2, C) and -a E SL(2, C), we have the following group:

SL(2, C)/ H0 1)}. By Theorem (2.8.3) the group Aut(C) is written as

Aut(C)

= SL(2, C)/ Hol 'I))-

The right-hand side of this equality is often denoted by PSL(2, C), and called the projective special linear group. (2.8.4) THEOREM. For two triples, (zl, z2, z3) and (wl, w2, w3), of distinct points

of a, there is a unique linear transformation f such that f (z;) = w i = 1, 2, 3. PROOF. We first show existence. If zl = oo, we set f, (z) = 1/z; otherwise, we

set fl(z) = z - zl. Then fl(z1) = 0. If fl(Z3) = oo, we set f2(z) = z; otherwise

we set f2(z) = z/(z - fi(z3)). Then f2 o fi(zl) = 0, and f2 o f,(z3) = oo. Moreover, we set f3(z) = z/f2 o h(22) and f = f3 o f2 o f1. Then f is a linear transformation satisfying f (zl) = 0, f (z2) = 1, and f (z3) = oo, Similarly, there is a linear transformation g satisfying g(wl) = 0, g(w2) = 1, and g(w3) = 00. Then g-1 o f is the required linear transformation. For uniqueness it suffices to show that if a linear transformation f satisfies f (0) = 0, f (1) = 1, and f (oo) = oo, then f (z) = z. This is easy. If c = 0, the linear transformation (2.8.1) gives rise to

f(z) = dz+

b ;

d

otherwise,

f (z)

a

=c

add be

z + d/c

2. COMPLEX FUNCTIONS

42

Therefore an arbitrary linear transformation is represented by a composition of the following linear transformations of three kinds-

(2.8.5)

i) ii)

f (z) = z + b,

bEC

f(z) = az,

a E C' (non-zero multiplication).

iii)

f (z) =

(translation). (inversion).

Z

A circle of C is written by

(2.8.6)

Iz-z1I =k IZ - Z2I

with distinct Z), z2 E C and k > 0, which is the so-called Appollonius' circle.

EXERCISE 2. Show that if k = 1, equation (2.8.6) presents a line passing through (z1 +z2)/2 and perpendicular to the line passing through z, and z2, and

that if k * 1 it presents a circle of C with center at. (z1 - k2z2)/(1 - k2) and with radius I(z, - z2)k/(1 - k2)I. Substituting a linear transformation z = f "1(w) of (2.8.5). i)-viii) into (2.8.6), we obtain

(2.8.7)

Iw- f(zl)I _k' (>0). Iw-f(z2)I

This defines a circle. Thus circles of C are mapped to circles of C by linear transformations. This property of linear transformations is called the correspondence of circle to circle. Take a circle C(a; r). If two points z1, z2 E C satisfy

(2.8.8)

(z1 - a)(z2 - a) = r2,

then z, is called the reflection of z2 with respect to the circle C(a; r), and vise versa. We also say that z1 and z2 are mutual reflections with respect to C(a; r). We define the center a and or, to be mutual reflections. In particular, a point of C(a; r) is the reflection of itself with respect to C(a; r). The points zl and z2 in (2.8.6) are mutual reflections with respect to the circle defined by (2.8.6). This

2.8. LINEAR TRANSFORMATIONS

43

is verified by Exercise 2 and a direct computation.

FIGURE 17

FIGURE 16

A line L of C was defined to be a circle of C. If two points z1,z2 E C are mirror symmetric with respect to L, then they are said to be mutual reflections with respect to L. The points on L, including oo, are reflections of themselves with respect to L. Therefore one deduces the following reflection principle. (2.8.9) THEOREM. Let f be a linear transformation, C a circle of C, and zl the reflection of z2 with respect to C. Then f(zj) and f(z2) are mutual reflections with respect to f (C).

Note that two points zl, z2 E C \ L are mutual reflections with respect to a line L of C if and only if L is defined by (2.8.6) with k = 1. The same holds for a circle C(a; r). That is, two points 21, z2 E C \ C(a; r) are mutual reflections with respect to a circle C(a; r) if and only if C(a; r) is defined by (2.8.6) with suitable a E C and r > 0. In general, a subset F of a group G is called a subgroup if F itself is a group under the operation of C. The unit element is always shared by all subgroups of G. For example, {± (o 0)j is a subgroup of SL(2, C). If a linear transformation f satisfies f(A(1)) = A(1), f is said to preserve s(1). We denote the set of all those f by Aut(A(l)), which is a subgroup of Aut(C). We determine the type of f E Aut(A(1)). For a E 0(1) we look for Oa E Aut(A(1)) such that Oa(C(0;1)) = C(0;1), 00(A(1)) = A(1), and 0.(a) = 0. The reflection of a with respect to C(0;1) is 1/a, and so 0a(1/a) = oo. Hence for a candidate we set

z-a (2.8.10)

oa(z) =

-az + 1'

2. COMPLEX FUNCTIONS

44

FIGURE 18

In fact, oa(a) = 0, and, if Ez) = 1,

z-a

_ Iz-aI

-az+1

la - z)

-- 1.

Furthermore, we see that al2

fzI < 1 b 1 -

=

1z -

Iaz - 112

(1 -

1a12)(1 - 1z12)

> 0.

Paz - 112

It follows that a(0(1)) _ i(1). Note that 0,-' = 0-

.

Let a linear transfor-

mation f preserve A(1). Then g = f o 0 '(o) is a linear transformation, and g(0) = 0. Since the reflection point of 0 with respect to C(0;1) is oo, g(z) = orz with some a E C'. Since g(C(0;1)) = C(0;1), Ial = 1, so that a = eee (0 E R). Therefore ej& _az-a a = f -' (0). A z) = z + l'

(2.8.11) THEOREM. i) A linear transformation f E Aut(A(l)) is written as ff(z (z))= =

e`a

ii) For two arbitrary points

z-a E i(1), there is an f E Aut(A(1)) such

that f (a) =#. PROOF. The first half is already proved. The latter half follows from f =

0'o0.

Property ii) of the above theorem is referred to by saying Aut(A(1)) acts transitively on A(1). Set

H = {zEC;Imz>0}.

2.8. LINEAR TRANSFORMATIONS

45

H is called the upper half plane. Next we find a linear transformation ip E Aut(C)

such that 1(H) = 0(1).

FIGURE 19

We set a condition, V5(i) = 0. Since ?(i(R U {oo}) = C(0;1), it follows from Theorem (2.8.9) that -0(-i) = oo. Hence for a candidate we set z

a

Cz) = z + i

(2.8.12)

By easy computations I7/i(z)I = 1 for z E R, and IO(z)I < 1 if and only if z E H. Let Aut(H) denote the set of all linear transformations preserving H. Then

Aut(H) is a subgroup of Aut(C). For an element f E Aut(H) there is a g E Aut(A(1)) such that f = tP -1 ogoz/i. From this and Theorem (2.8.11) we deduce

that Aut(H) acts transitively on H. Since If (0), f (1), f (oo)} C R U fool, we see that az + bd' ad - be = 1, f (z) - cz + where a, b, c, d E R. Hence f (R U {oo}) = R U {oo} and

Im z > O b Im f (z)

1 faz+b az+b _ Imz = 2i

cz + d

cz + d

Icz + d12

> 0.

It follows that f (H) = H. Let SL(2; R) denote the set of all real matrices of SL(2, C). We hence obtain the following theorem. (2.8.13) THEOREM. i) The mapping

(a b) E SL(2, R) - f (z) =

az + b E Aut(H) cz + d

is a surjective group homomorphism, and Ker D = {± o ° )} . ii) Aut(H) acts transitively on H. We have the following expression:

Aut(H) = SL(2,R)/ { f ( 0

0)1.

2. COMPLEX FUNCTIONS

46

The right-hand side of the above equation is denoted by PSL(2, R). The group SL(2, R) carries subgroups such as

d) a, b, c, d E Z, ad - be = 1

SL(2, Z) _

;

)ESL(2.Z); Cc a

r (n) (o

a)

d)

-0

0) (mod n) }

(n ? 2).

(mod

n) means that a-1, b, c, and d- I are integral multiples of n. The subgroup f (n) or C(n)/ {f (p ° )} is called the principal congruence subgroup, and in particular C(1) = SL(2, Z) or PSL(2, Z) the modular group. These groups play important roles in the theory of elliptic functions and the theory of automorphic forms. Here (ac

i)

EXERCISE 3. Show that I'(n) is a subgroup of SL(2, R).

Problems log(1 + z)

ez - 1

sin z

, and lim 1. Compute the limits lim z , lim :-0 z-.0 2 Z 2. Let f (z) = (z" - 1)/(z - 1). Find the supremum and the infimum of If (z)l on A(1). 3. Let fn (r) = sin nx, n = 1, 2, ... , be a sequence of functions of the real variable x E [0,2ir]. Then {fn(x)}n , is uniformly bounded. Prove that any subsequence of { f"(x)},°,°_1 does not converges on (0, 2ir). (This requires

a knowledge of Lebesgue integration. If the reader finds it too difficult, see the answer in the back of the book.) 4. Find the radii of convergence of the following power series:

x

i) E npz"

00

ii)

(p > 0).

>q"2z"

n=0 iii)

00

iv) 1+Ez"

j(n!)""z".

(tql < 1). n-1

(a+Y)(0+v)

(Gauss' hypergeometric function)

5. (Stolz' domain and Abel's continuity the-

orem)Let (EC(0;1),and let G((;r)denote a subdomain of 0 (1) hedged by two line segments passing through ( and form-

ing an angle 0 < r < 7r/2 with the line passing through 0 and C. We call G((; r) Stolz' domain with vertex (. (a) Show that G((; r) n A((; cos r) C 2

cosy} nA((;cosr). {,E.12-(I1 - Izj
1, take r = cos' -21 E (0, n/2). Thr(en, show that

l

(zE0(1);1z(1 1,

fi (1 - a)n converges uniformly on compact subsets

n=1

of C.

00

9. Show that II (1 - zn) converges absolutely and uniformly on compact subn=1

sets of 0(1).

10. Show that

II 00 (1 + z2n) =

1

1-z n=o 11. The cross ratio of four points zl, .... Z4 of C is defined by /

lZ1,Z2,Z3,Z4) =

(ZI - Z3)(Z2 - Z4) (Z1 - Z4)(z2 - Z3)

Here, if some zJ = oo, the above right hand side is supposed to represent

the limit as z., - oo. Let f E Aut(C) be a linear transformation and ZI, Z2, Z3 E C such that P Z2) = 1, f (z3) = 0, and f (z4) = oo. Show that f (Z) = (Z, Z2, Z3, Z4)

12. Show that for any linear transformation f and four points zr, ... '24 of C (z1, Z2, z3, z4) = (f (zi ), A Z2), f (Z3), A Z4))

(the invariance of the cross ratio).

13. Let f (z) = (az + b)/(cz + d) j9 z, ad - be = 1, be a linear transformation. Prove that if a + d = ±2, then f has one fixed point z (a point z such that f (z) = z); otherwise, f has two fixed points. 14. Let a and 6 be the above fixed points of f . Show that if a 0 Q, w = f (z) is given by

w-a _KQisZ - a z-f3 w-f3

K>0, 9ER.

2. COMPLEX FUNCTIONS

48

The linear transformation f is said to be hyperbolic if e`B = 1, elliptic if K = 1, and loxodromic otherwise. 15. Let or and Q be as above. Suppose that a = t3. Show that w = f (z) is given by 1

1

w-a

z-a

o0),

(a

00).

In this case, f is said to be parabolic. 16. Let f (z) be as in problem 13. Show that if a+d is real, then f is hyperbolic, elliptic, or parabolic according to whether Is + dl > 2, < 2, or = 2; show that if a + d is not real, f is loxodromic. 17. Find a general form for linear transformations preserving 0(R), 0 < R < oo. 18. i) Find a linear transformation which maps 0, 1, oo to i,1 + i, 2 + i. ii) Find a linear transformation which maps -I, i,1 to -2, i, 2.

CHAPTER 3

Holomorphic Functions

In this chapter we first define holomorphic functions, and then prove Cauchy's integral formula, which is the most fundamental and important theorem in complex analysis. One sees the close relationship between the topological structure of a domain and complex analysis on it.

3.1. Complex Derivatives Let D be an open set of C, and let f be a function on D. We say that f is complex differentiable at a point a E D if there is a number a E C such that for an arbitrary > 0 there is ab >0 with f(a + z) - f(z) al

(3.1.1)

h

<e

for all h E A(b) \ {0}. We call a the complex derivative of f at a, and denote it by f'(a) = JL (a). The expression (3.1.1) is equivalent to (3.1.2)

f(a + h) = f(a) + ah + o(h),

a = f'(a),

where o(h) is a term such that limh.o o(h)/h = 0. If f is complex differentiable at a, then f is continuous at a. If f is complex differentiable at every point of D, f is said to be holomorphic. In this case the function f' : z E D f'(z) E C is called the derived function or the derivative of f , and is also denoted by df /dz. If f is holomorphic, so is 1 If outside the set (f = 0}, and

f

( f),

When fl and f2 are holomorphic functions, and a, and a2 are constants, the following linearity is clear:

(aifi + a2f2)' = aifi + a2f2 We have the Leibniz formula:

(fl f2)'=fi f2+fi f2 49

3. HOLOMORPHIC FUNCTIONS

50

EXERCISE 1. Show that for n E Z, (z")' = nz"-I

We set z = x + iy, and consider f (z) as a function f (x, y) in x and y. Set f (x, y) = u(x, y) + iv(x, y),

where u(x, y) (resp., v(x, y)) is the real (resp., imaginary) part of f (x, y). Assume that f is holomorphic. Then by (3.1.2) u(x, y) and v(x, y) are totally differentiable. Letting h --+ 0 with real h and with purely imaginary h, we have

au ax

av .au a5 8x - 8y av

Therefore

su (3.1.3)

5;

=

au = - ax av .

av

Si

8y ,

These are called the Cauchy-Riemann equations. EXERCISE 2. Let u(x, y) and v(x, y) be real valued, totally differentiable functions satisfying the Cauchy-Riemann equations (3.1.3). Show that the complex valued function f (x, y) = u(x, y) + iv(x, y) is holomorphic as a function of

z=x+iy. Partial differential operators 8: = a/az and a= = 0/ft are defined by

a=f-az-2(8x+tay) of aZf-a8-

1

Of _ 1 of

ax

i0y)'

Then (3.1.3) is equivalent to a=f = 0.

(3.1.4)

This is called the a-equation (d-bar-equation). For a holomorphic function f,

f =0Zf.

(3.1.5)

Let +/ : I -. D be a continuously differentiable curve. For t, t' E I with t # t'

f o b(t') - f o t'(t) = , a fi(001) 001) - +'(t) + n

t'-t

f

t'-t

(

(t') - v (t) t,

t

Since limt.-t(o(t') - P(t))/(t` - t) = v(t), (3.1.6)

.f c

00(t) 00-

(3.1.7) THEOREM. If a holomorphic function f on a domain D satisfies f' = 0, then f is constant.

3.1. COMPLEX DERIVATIVES

51

PROOF. If we set f(z) = f(x,y) = u(x,y) + iv(x,y), we have by (3.1.5) or (3.1.3)

8u

8u

8v

8v =

ax=8y=8x8ySince D is connected, u and v are constant.

(3.1.8) THEOREM. Let f be a holomorphic function on a domain D. Then, if Ref or Imf is constant, so is f . PROOF. It is sufficient to deal with the case where Ref is constant. Setting

f = u + iv, we have 8u/8x = 8u/8y - 0. It follows from (3.1.3) that 8v/8x = 8v/8y - 0. Hence Theorem (3.1.7) implies the constancy of f .

Let f be a holomorphic function on a domain D. Let D' be a domain of C containing the image f (C). Let g be a holomorphic function on Y. Fix zo E D arbitrarily, and set wo = f (zo). Then for a small h 9(f (zo + h)) - 9(f (zo)) = g(wo + hf'(zo) + o(h)) - g(wo) = g'(wo)(hf'(zo) + o(h)) + o(hf'(zo) + o(h))

= g'(wo)f'(zo)h + o(h). Therefore

l hm

9(f(zo + h)) - 9(.f(zo)) _ 9V (zo))f (zo) h

Thus we have the following:

(3.1.9) THEOREM. The composite g o f of two holomorphic functions f and g is again holomorphic, and (go f)'(z) = 9 (f(z))f'(z)

We take a convergent power series about a E C:

f (z) = 00 E a, (Z - )n. n=o

Let R > 0 be its radius of convergence. (3.1.10) THEOREM. The above f (z) is a holomorphic function on 0(a; R), and

nan(z - a)'-

f'(z) _ n=1

Here the radius of convergence of the right side is R, too.

3. HOLOMORPHIC FUNCTIONS

52

Since limn

PROOF. Without loss of generality we may assume a = 0.

"n+1=1,

lim " Ian I = Um nx n-oo

Ian I = lim n-oo

`

Ian+1

= lim n-ocV (n + 1)IanI. Thus the radius of convergence of

x

g(z) = >

nanzn-1

n=1

is R. Take an arbitrary point z E t(R). Fix r with IzI < r < R. By Lemma (2.4.2) there is an M > 0 such that (3.1.11)

n = 1,2,... .

Ianrn1 < M,

ForhEi(r-IzI)\{0}we have If (z + hh - f(z)

_

9(z)I

(z + h)n - zn

_

an

h

n=1 n

'C

a E nF, n=2 v=2

_ E Ian n=1

< M x

(n) Zvhn-v-1

n

0o

Ia.I n=2

(IZI + Ihj)n - Zn l

Ihi IZI + IhI

1

ihI(

)

r

IZlvlhln-v-1

v=2

- nlzln-1

n

(v)

I n

n

(r }-r (Lzl)n-'] IZI

rHere we used (3.1.11) in the last estimate. It follows from (2.4.8) and (2.4.9)

that If (z + h) - f(z)

_

h

< M

=

_

1

9(z)I

IzI + IhI

IhI `r - (IzI + IhI) MrIhI

(r - IzI)2(r - (IzI + IhI))

-

IzI

l

r - IzI)

-0

(h

r (r - IzI)2 0)

O

(3.1.12) COROLLARY. An analytic function is holomorphic.

We mention that the converse of the above statement holds, and will be proved in Theorem (3.5.7). It follows from Theorem (3.1.10) that a function f (z) expressed by a convergent power series about a is complex differentiable arbitrarily many times, and the coefficients an are determined by the n-th derivatives f(n) (a) of f(z): (3.1.13)

an = nifi"1(a).

3.1. COMPLEX DERIVATIVES

53

Hence we see that if a function is expanded to a power series, the coefficients of the power series are uniquely determined.

Let f (z) = u(x, y) + iv(x, y) be a holomorphic function of z = x + iy E D. The Jacobian J f (x, y) of f : (x, y) E D -+ (u(x, y), v(x, y)) E R is given by

Jf (x, y) _

(x, y) k (x, y) (x,y)

I7Z(x,y)

The Cauchy-Riemann equations (3.1.3) imply

(z=x+iy).

Jf(x,y)=If'(z)I2

(3.1.14)

Therefore J f (x, y) > 0 everywhere. If f'(z) 4 0, then J f (x, y) > 0, so that f gives rise to an orientation preserving mapping from (x, y)-plane to (u, v)-plane.

That is, to vectors Xj = (aj1,aj2), j = 1,2, at z we assign vectors

Uj =lad,

8u

+aj2

3u

Ot,

M

,ajIT +a, j2ft

7 = 1, 2'

at f (z). If {X1, X2} is a right-hand system (det ( x , ) > 0), then so is (U1, Uz).

X7

X1

f(z)

FIGURE 20

Now let zo E D, and assume f'(zo) 0 0. For the sake of simplicity, we assume that zo = 0 and f (zo) = 0. Let Cj (0) : [0, tj] D), j = 1, 2, be two curves such

that the limit lime....+oOj(t)/t = aj exists. Take 0, = arga,, j = 1,2, so that 0 5 8z - 81 < 27r. Then we call 0 = 8z - 81 the angle pinched by C1 and C2. (3.1.15) THEOREM. Let the notation be as above. Then the angle pinched by f (C1) and f (C2) is equal to that pinched by C1 and C2.

PROOF. Since f is holomorphic, it follows from (3.1.6) that lim

+o

f(0j(t)) = f'(0)aj = If'(0) I t

Iajle'(e;+"9 f'(o))t

Thus the angle pinched by f (C1) and f (C2) is 8z - 81. 0

3. HOLOMORPHIC FUNCTIONS

54

FIGURE 21

The property of f in the above theorem is called conformality. Hence, an injective holomorphic function f with nowhere vanishing f' is called a conformal mapping.

EXERCISE 3. Let u(x, y) and v(x, y) be totally differentiable real functions. Set f (x, y) = u(x, y) + iv(x, y). Show that if J f (z, y) # 0 and f is conformal, then f is holomorphic. EXERCISE 4. Show the following:

dze2=e',

d Iog(1 + z)

+z

zEC. z E O(1).

EXERCISE 5. Let f (z) = f (x, y) be a continuously differentiable function on a domain D C C. Let K be a compact subset of D. Fbr z E K, set

f(z+h) = f(z)+ Lz(z)h+

Lz(z)h+o(h).

Show that o(h)/h -. 0 uniformly in z E K as h -+ 0; i.e., for an arbitrary e > 0,

there is a >0 such that lo(h)/hl 0 there is a number n such that

It - t'I
0sothat

{It)_t'I< e,

It - t,I < 60

If a m(t) - f o (t ')I < E.

Letting 60 > 0 be smaller if necessary, we have for I(d)I < bo r

S + E > E if 0 j=1

10(tj) - m(tj-1)l

3.2. CURVILINEAR INTEGRALS

59

On the other hand, it follows from the definition that there are (0) : To = To < < Tn = T1 and (k E [Tk_1i Tk] with I(is)I < bo such that 7-1 < S-E
i If 0 0(ti)I - 10(t,) -O(t;-1)1 1=1

> S-e{L(C)+3nmax{If o0I}+1}. Hence we have shown (3.2.9). We write

S = f If(z)IIdzI C

It follows that (3.2.10)

ifc

f(z)dzl s f

c

If(z)IIdz1

m(Tk-1)I

3. HOLOMORPHIC FUNCTIONS

60

Let fn, n = 0, 1,... , be continuous functions on D. If {fn}n 0

or F " fn converges uniformly on compact subsets of D, then by (3.2.10) we have respectively fn(z)dz, lim fn(z)dz = Jim n-oo JC J n--oo 00

(3.2.11)

/

J

fn(z)dz =

J fn(z)dx.

n=0 C Let 0(t) = 01(t)+i02(t) be piecewise continuously differentiable. As in (3.2.2) it follows from (3.2.7) that n=O

+e1i(ti - ti-1))

S((d);

=1

+i¢z(ti-1 +02i(ti - ti-1))}(ti - ti-0

(0 0 is arbitrary, we have a contradiction. 0 REMARK. A good point of the above proof is that it is not necessary to assume

the continuity nor the integrability of the derivative f'(z) of f(z). (3.4.9) LEMMA. On an arbitrary disk 0(a; r) C D, f has a primitive function. PROOF. For a point z E A(a; r) we define a curve CZ by

0:tE(0,11-+a+t(z-a)EO(a;r), and set

F(z) = fc* f ( ()d(. We show that F(z) is a primitive function of f (z).

FIGURE 30

Fix an arbitrary point zo E 0(a; r). For an arbitrary e > 0 there is a 0 < 6 < r - Ial such that (3.4.10)

If (zo + h) - f (zo) I < e,

IhI < 6.

Let rh denote the line segment from zo to zo + h. It follows from Lemma (3.4.1) and (3.4.10) that

F(zo + h) - F(zo) h

=

I

_ f (zo)

f f(z)dz-f(zo)I =lh,h(f(z)-f(zo))dz h

< IhI f If(z) - f(zo)IIdzl < I' L(I'h) = C ti

Therefore F is holomorphic, and F' = f .

0

3. HOLOMORPHIC FUNCTIONS

70

Let C(V : [0; 1] -' D) be a curve. As in the proof of Lemma (3.3.4), there is < t,, = I of [0,1] fine enough so that (3.3.5) holds. a partition (d) : 0 = to < Then the piecewise linear curve C(d) connecting W(to), ..., and p(t,,) in order by line segments is homotopic to C. (3.4.11) LEMMA. The curvilinear integral fC(d) f(z)dz is independent of (d), provided only that (3.3.5) is satisfied. PROOF. It is sufficient to prove that for a refinement (d) of (d) (3.4.12)

Jc{d

f (z)dz =

f (z)dz.

For we see by (3.4.12) that for two partitions (dl) and (d2)

J

G'(f(z)dz = (

fd3)

f(z)dz = f(dz) f(z)dz,

d1)

where (d3) is the partition formed by all partition points of (dl) and (d2). To show (3.4.12) it is sufficient to deal with the cage where one partition point tJ is

added to (t,,t,fl). at,)

r

Pu")

FIGURE 31

In this case we denote by E the triangle with vertices cp(t,), :p(t;), fp(tj+1), and by r the perimeter of E. Lemma (3.4.1) implies

Id.) f (z)dz - fe(d) f (z)dz = / f (z)dz = 0.

0

By Lemma (3.4.11) we can define the curvilinear integral off along C by (3.4.13)

f f(z)dz = J (d) f(z)dz,

where (d) satisfies (3.3.5). The next theorem is called Cauchy's integral theorem: (3.4.14) THEOREM. Let f be a holomorphic function on D, and let C be a closed curve in D which is homotopic to a point. Then

jf(z)d.z=0.

3.4. CAUCHY'S INTEGRAL THEOREM

71

PROOF. Let C be given by p : [0,1[ - . D, and set zo = ap(0) = v(1). There is a homctopy

45:[0,1]x[0,1]

D

connecting W and zo. Then ap(t) = 45(t, 0), and 45(t,1) _- zo. Denoting by C. the curve given by 45,(t) = 45(t, s), we define

A=sup{8E[0,1]; J f(z)dz=J f(z)dz, 0Sa'5s} c

llC.,

ll

As in Lemma (3.3.7) we take 6 > 0 and 0 = to
A+6', which is a contradiction. We have A = 1. Taker > 0 so that

A(zo;r) CD. There is a6">0such that C,CO(zo;r)forall 1-6" 0. Show that fjo Rj f(z)dz and f[-Ro f(z)dz converge as R -. +oo, and that fR f (z) = Rlim

it- R,R]

f (z)dz = 0.

EXERCISE 2. On A(1), find a primitive function of f (z) = (z2 -1)-2 in power series.

3.5. Cauchy's Integral Formula Let C(a; r) be the perimeter circle of the disk 0(a; r) given by 0: 9 E [0, 2a] -+ a + te'B E C(a; r).

(3.5.1)

f

(3.5.2) THEOREM. J

C(a:r) z - a

dz = 2iri.

PROOF. Set z = 0(9) = a + re'°. Since qV(O) = riese, we have by (3.2.12) r 1 r2* rie; e O J dz= 1 d6=2rri. C(a:r) z - a 0 77C.6

3.5. CAUCHY'S INTEGRAL FORMULA

73

Let D C C be a domain, and a E D. We define d(a; 8D) = inf { Ia - wI; w E 8D} < +oo.

If d(a; 8D) < +oo, there is a w E 8D with d(a; 8D) = is -wl. The next theorem is called Cauchy's integral formula (3.5.3) THEOREM. For an arbitrary a E D and 0 < r < d(a; OD)

f (z) =

1

f (S) dC,

.

C(a;r) l; - z

27ft

z E 0(a; r).

PROOF. For the sake of simplicity we may let a = 0. Let z = Izle'BO E A(r). Take a 6 > 0 with 0(z; 6) C i(r). Let Cl = C(0; r), let C2 be the line segment from z+6e`BO to rei00, and let C3 = C(z;6). We may assume that the initial and terminal point of C, is reie0, and that of C3 is z+6e'e0. Set C = Cl -C2 -C3+C2. Then C is homotopic to the curve C given in Figure 33.

FIGURE 33

Let Cbe given by ¢: 10, 1] - D, and let 0: (t, s) E [0, 11 x [0,1[ - (1 - s).3(t) - sreieO E D \ {z}

be a homotopy connecting C and the point -re'B° in D \ {z). Hence C is homotopic to a point in D \ {z}. Since the function f in D \ {z}, it follows from Theorem (3.4.14) that

z) is holomorphic

Jid(=0. C - z

f C)=f f.

One obtains

J

,

C-z

C, ( - z

Since

f

f2x

f(C)

3C-z

o

f (z + bee)

(z+beie)-z

f2w

=

f (z

+ 6eie)id8

2xif(z)

(6

0),

3. HOLOMORPHIC FUNCTIONS

74

one gets

J

27ri

Fix z E 0(a; r). Ash

J h

, C

("

0

d(= f (z).

0, the following convergence is uniform in ( E C(a; r):

_ f(C) I

f(C)

(-(z+h) (-z

f(()

_4

Thus we deduce the integral formula for the derivative off from Theorem (3.5.3):

f,(z) =

1 f(C) 2;i- 1C(o;.) (C - Z)2 4,

z E 0(a;r).

Repeating this n times, we find that the n-th derivative f (") (z) is expressed by f(n)(z)

(3.5.4)

f(() r = n) 17r, z)"+, dC C(a;r) ((-

zE0(a;r), n=0,1,.... E C(a; r) one has

For z E A(a; r) and 1

_

1

1

_

00

1

fz-a

(-any

a

Here, note that

{ z-a l
0 such that if (z) I < M for z E C. For an arbitrary R > 0 one has by Theorem (3.5.20), i)

la"I R

where M > 0 and R > 0 are constants. Show that f (z) is a polynomial of degree at most n.

3.6. MEAN VALUE THEOREM AND HARMONIC FUNCTIONS

81

EXERCISE 5. Let f be a continuous function on A(1) which is holomorphic in A(1). Assume that f (eie) is constant on an arc {eie; 0 < 81 < 0 < 02 5 27r}. Then, show that f (z) is constant on A(1). EXERCISE 6. Let f be a holomorphic function on D. For an arbitrary point w E f (D), show that the inverse image f -1(w) is either the whole of D, or a discrete subset of D.

3.6. Mean Value Theorem and Harmonic )Functions Let D C C be a domain, let f be a holomorphic function on D, and let a E D. By Theorem (3.5.3) we have f2w (3.6.1)

f(a) =

0 < r < d(a;BD).

f(a + reie)dO,

2w

This is called the mean value theorem for holomorphic functions. We set a = 0 and r = 1 for the sake of simplicity in what follows. The value f (O) of f at the center 0 of the disk A(1) is expressed by the integration of f over the boundary circle C(0;1) of A(1). We consider a similar expression for a point of A(1) other than the center. Let z E A(1) and take O_(() =

< E A(1).

z(+1 '

and 0j 1 are injective and surjective mappings As seen in Theorem (2.8.11), between A(1), and holomorphic in A(1/IzI), a neighborhood of A(1). Therefore f o 0.1(() is holomorphic in a neighborhood of A(1), and fo.=1(w)dw.

fo0=1(0)= 1f 2ari

W

C(o;l)

Changing the parameter by w = z((), C = eie, 0 < 0 5 tar, one gets from (3.2.18)

j2* 1

(3.6.2)

f(e,e),O:(e'e)ieed9 4,/ (e1e)

2ari Jo Tar

f

2x

f(e'e) We

IzZI2 dB.

(3.6.3) THEOREM. Let f be a continuous function on A(a; R) which is holomor-

phic in A(a; R). Then

f2,r P z) = tar

f (a + Re`s) I

2 _I(z

aa)12

'

z E A(a; R).

3. HOLOMORPHIC FUNCTIONS

82

PROOF. By the change of variable w = (z - a) /R, it is reduced to the case where R = 1 and a = 0. Let z E A(1) be an arbitrary point, and take 0 < r < 1. Applying (3.6.2) to g(C) = f (r(), one obtains r2A f(reis)I

f(rz) = 2_

IZ12

d$.

Letting r -+ 1, we have the desired formula. 0 The function of ( and z

K((' z) -

2_ z2

I

IC

- ZI2

C # z,

> 0'

is called the Poisson kernel. The formula of Theorem (3.6.3) is written as (3.6.4)

f(z) =

1f 2R

f(C)K(C - a,z - a)dO. (a;R)

This is called the Poisson integral.

Let a = 0 and set f (z) = u(z) + iv(z). Then it follows from (3.6.4) that (3.6.5)

1 u(C)K(C,z)d8. Za f (a;R)

u(z) =

The same holds for v(z). Since

K((,z)=ReC+i,

(3.6.6)

the function u(Z)

2,r C(O;R)

U(()

C+z

C-z is holomorphic on A(R), and Reu(z) = u(z). The above integral is called the complex Poisson integral. Thus we have (3.6.7)

f (z) = u(z) + i Im f (0).

The second order differential operator (3.6.8)

A=(+) =4&&=4&8:

is called the Laplacian. Since 3.f = 0, we have the so-called Laplace equation (3.6.9)

4u=0.

A real-valued C2-function u(z) = u(x, y) which satisfies the partial defferential equation (3.6.9) is called a harmonic function. The real and imaginary parts of a holomorphic function are harmonic. Conversely, for a given harmonic function u(z, y) we will construct a holomorphic function f with real part u.

3.6. MEAN VALUE THEOREM AND HARMONIC FUNCTIONS

83

Let D be a simply connected domain. Let u(x, y) be a harmonic function on D. Then 8=u is holomorphic by (3.6.8) and (3.6.9). Fixing a point a E D, by Theorem (3.4.15) we have a holomorphic function

f(z)=J

=28=udz+u(a), a

which is a primitive function of 8,u. Let 0 < r < d(a; 8D). Then f (a + reie) = / r 28=u(a + te'9)e'9dt + u(a)

=/ (cos8+sm9)dt 0

+

Jr( sine-

clu

cose)dt+u(a).

Setting a = al + ia2, one gets

Ref (a + rei8) = f

u(aI + t cos 0, a2 + t sin e)dt + u(a)

= u(al + r cos e, a2 + r sin 9) - u(al, a2) + u(a)

= u(a + reie)

Set D' = {z E D; Ref (z) = u(z)}. Since D' a, D' # 0. Clearly, D' is closed. To conclude that D' = D, it is sufficient to show that D' is open. Let b E D' be an arbitrary point. For z E 0(b; d(b; 8D)) f (z)

n

28,udz + u(a) z

rb

a

28,udz +

20,u(z)dz + u(a). b

The above computation and the assumption b E D' yield

Ref (z) = Re f.b 28,

+ u(a) + Re

rz

28,u(z)dz

= u(b) + u(z) - u(b) = u(z). Hence A(b; 8D) C D', and so D' is open. (3.6.10) THEOREM. i) For a harmonic function u on a simply connected domain

D, there exists a holomorphic function f = u + iu' on D with real part u, which is unique up to a purely imaginary constant. ii) Let u be a harmonic function on a domain D. If ulU _ 0 on a non-empty open subset U C D, then u = 0 on D.

3. HOLOMORPHIC FUNCTIONS

84

iii) Let u and D be as in ii). Let a E D, and 0 < r < d(a; 8D). Then for z E A (a; r)

u(z) =2ir1 J

u(()K(( - a, z - a)dO,

(= a + reiB

( ax)

In particular,

u(a) = 2n f,(a;r) u(a + reiB)d6

(the mean value theorem).

iv) Let u be as in ii). If u is not constant, then u does not attain the maximum nor the minimum in D.

PROOF. i) It remains to show the uniqueness; this follows from Theorem (3.1.8).

ii) Let D' denote the set of points z E D such that there is a neighborhood V C D of z with ujV - 0. By definition, D' is open. Since D' D U, D' # 0. Let a E D' n D. There is a point b E D' such that a E 0(b; d(b; 8D)).

FIGURE 37

Let f be a holomorphic function on i(b;d(b;BD)) such that Ref = u. Since b E D', f is constant, so that uIA(b; d(b; 8D)) = 0. Thus a E D'. Since D is connected, D' = D. iii) This follows from i) and (3.6.5).

iv) Assume that a attains the maximum value at a E D. (For the minimum, it suffices to take -u.) For 0 < r < d(a; 8D) 1

2r.

2n

f(u(a) - u(a + reio))dO 2'

1

= u(a)

-j

u(a + reiB)dO = 0.

0

Since u(a) - u(a + rei°) is continuous in 0 and non-negative, u(a + reiB) = u(a),

0 < 0 < 27r.

Therefore uI0(a; d(a; 8D)) _- u(a), and it follows from ii) that u - u(a) on D. This is a contradiction.

3.6. MEAN VALUE THEOREM AND HARMONIC FUNCTIONS

85

The function u' in i) is called the adjoint harmonic function of u, and it is unique up to a constant term. EXAMPLE. a) The adjoint harmonic function of x is y. b) The adjoint harmonic function of log jzj is arg z.

EXERCISE I. Show that x2 - y2 is harmonic. What is its adjoint harmonic function?

The importance of harmonic functions for the study of holomorphic functions is made clear by the existence of adjoint harmonic functions. Generally speaking,

it is easier to construct a harmonic function than to construct a holomorphic function. The problem of finding a harmonic function on a domain D with prescribed boundary values on 8D is called the Dirichlet problem. We solve it for a disk:

(3.6.11) THEOREM. Let h be a continuous real valued function on C(0;1). Then the real valued function defined by u(z) = j27r h(eae)K(e:B, z)dO

2rr

is harmonic on A(1), and for every point e" E C(0;1)

lim, u(z) = h(e't). PROOF. It follows from (3.6.6) that u(z) is the real part of the holomorphic function

f(z) =

1 27r

f

h(eie)e

+zdd

e

(o:1)

ie,

z

and hence harmonic in A(1). Applying (3.6.5) to u = 1, one gets

f2r 2>r

h(e`O)K(e`$,z)d6.

Noting that K(eie, z) > 0, we have for a small 6 > 0 Ju(z) - h(e't)1

If 2L

s,+

{h(eie)

o

0 such that Ju(z) - h(e")I < 2e, so that limx_e,i u(z) =

h(e").

z E a(1) fl A(e,'; b'),

0

REMARK. By the above proof one sees that if h is bounded, h(e';) holds at a point e" where h is continuous. EXERCISE 2.

u(z) _

Let D and D' be domains of C, and let f : D -+ D' be a

holomorphic function with values in D' (which is called a holomorphic mapping from D to D'). Show that if u' is a harmonic function on D', so is the composite

u of on D. EXERCISE 3 (Harnack inequality). Let u(z) be a harmonic function on A(R) such that u(z) >-- 0. Show that

R - IzI u(0) < u(z) R + IzI S u(0) R- IzI R+ IzI EXERCISE 4. Show that if u is a harmonic function on C, and u ? 0, then u is constant.

3.7. HOLOMORPHIC FUNCTIONS ON THE RIEMANN SPHERE

87

EXERCISE 5. Express a harmonic function u on A(1) by the Poisson integral which has boundary values 1 on C(0;1)n{Imz > 0} and -1 on C(0;1)n{Imz < 0}.

3.7. Holomorphic Functions on the Riemann Sphere Let D be a connected open subset of the Riemann sphere C. Then D is called a domain of C, as in the case of C. On the Riemann sphere, there are complex coordinates z on 6\ {oo} and i on a\ {0}. They are related on C' = 6\ {0, oo} by

On C', z = 1/i is a holomorphic function of i, and the converse holds, too.

Let f be a function on D. On C', there are two coordinates, z and i. If f is holomorphic as a function of z on D n C', then by Theorem (3.1.9) it is holomorphic with respect to i; the converse holds, too. We define f to be complex differentiable at oc, provided that oo E D, if f is complex differentiable

as a function of i at i = 0. We say that f is holomorphic on D if f is complex differentiable at every point of D. We know that the complex coordinates z and i are related by zi = 1, and represent the same point of the Riemann sphere C. When f is considered as a function of z (resp., i), we write f (z) (reap., f o i). For example, we have

foi=f

(z),

f0i(0)=f(oo)

The various properties of holomorphic functions on domains of C remain valid for those on domains of C. We need, however, the following notion of differential to deal with curvilinear integrals. We set

Uo=DnC\{oc},

U1 =Dn{0}.

A pair w = {rlo(z),171(i)} of holomorphic functions, rlo(z) on Uo and rl1(i) on U1 is called a differential on D if on Uoo n( U1 the following holds: (3.7.1)

r10(z) = r11

(I) `-z2) _ III

To represent this change of variables, we write (3.7.2)

w = rlo(z)dz.

Here dz is considered to be transformed to dz = (-1/i2 )di. Then, on Uo n U1, w = r1o(z)dz =171(i)di.

Let C(¢ : I - D) be a curve in D. Assume that C C D n C \ {0} and it has a finite length. Then frio(z)dz =

jt7i(i)d2

3. HOLOMORPHIC FUNCTIONS

88

where C in the right side is a curve given by i o 4(t) = 1/0(t). Thus we may define the curvilinear integral of w along C by

/w=

(3.7.3)

j

go(z)dz =

j m(i)di.

c c For a general curve C(q5 : [To, T1 J -e D), if there is a partition (d) : To = to < < t = T1 such that 0(It;-1it)]) C Uo or ¢([tj-1,tj]) C U1, and such t1 < that (3.7.4)

JCI[tj-.f)

IdzI < +oo,

or J Iits-i til Idi[ < +oo,

we define the curvilinear integral of w along C by (3.7.5)

f=>f

w.

If it exists, it is independent of the choice of the partition (d). The curvilinear integral fC f (z)dz with C C C is considered as a curvilinear integral of the differential f (z)dz along C. A differential w = {rlo(z),r)1(i)} on D is called a holomorphic differential if rb(z) (reap., r,1(i)) is holomorphic in z (reap., z). A holomorphic function F on D is called a primitive function of w if

dzF(z) = qo(z),

z E Uo,

d Foz=rtl(z),

zEU1.

We write dF = w for this fact, and call it the differential of F. If C is another primitive function of w, then the difference F - G is constant. All properties of holomorphic functions and curvilinear integrals of holomorphic functions on domains of C which have been proved up to the last section also hold for holomorphic functions and for holomorphic differentials on domains of d which may contain the infinite point oo. For example, let w be a holomorphic differential on a domain D C C and let C be a curve in D. Take a piecewise linear curve C for C as in Lemma (3.3.4). Since a piecewise linear curve always satisfies (3.7.4), one may define the curvilinear integral of w along C by Jcw=

jew.

As in the proof of Lemma (3.4.11), the above right side is independent of the choice of C so long as C satisfies (3.3.5), and Theorem (3.2.19) extends to (3.7.6) THEOREM. Let w and D be as above. Let F be a primitive function of w, and let C be a curve in D with initial point zo and terminal point z1. Then

jw = F(z1) - F(zo).

PROBLEMS

89

Cauchy's integral Theorem (3.4.14) extends to (3.7.7) THEOREM. Let w and D be as above. Let C be a closed curve in D which is homotopic to a point. Then w = 0.

IC (3.7.8) THEOREM. Any function holomorphic on the whole C is constant.

PROOF. Let f be a holomorphic function on C. Since a is compact, if I attains the maximum at some point a E C. If a 0 oo, Theorem (3.5.21) implies that f is constant; if a = oo, then the function f o z, holomorphic in z, gives rise

to a constant. Thus f is constant. 0 If we deal with non-constant holomorphic functions on D, by this theorem we may exclude the case of D = C. If D 0 C, we take b E C \ D. If b = oo, then D C C; if b # oo, we may reduce the case to D C C through the transformation z 1/(z - b).

Problems 1. Show that f (z) = x2 + iy2 satisfies the Cauchy-Riemann equations at the origin 0, but is not holomorphic in any neighborhood of 0. 2. Let f (z) be a holomorphic function on a domain D. Prove that f (z) is a holomorphic function on D' _ {z; z E D}. 3. Let a E C` and define f (z) _ (1 + z)° = e° °s(1+;i. Take a branch of f (z) on 0(1), and expand it to a Taylor series about 0. 4. Let C(O : [a, b] -+ C) be a curve. Show that L(C) = 1cdlim0 L(C; (d))

(5 +oo),

where (d) is a partition of [a, b].

5. Let C be the perimeter of the square {(x, y); -r 5 x 5 2r, -r 5 y 5 2r} (r > 0) with anti-clockwise orientation. Compute the curvilinear integral of f (z) = x2 + iy2 along C. 6. i) Let f (z) = (z - a)/(z -)8) with Q i3 (E C). Show that f is holomorphic in a neighborhood of oo.

ii) Let L denote the line segment from cr to J3. Show that D = C \ L is simply connected.

7. Let f be a holomorphic function on 0(R) such that If (z) 15 M for all z E A(R). Prove that MRn! = (R - IzI)n+1

I f(n)(z)I
0, then f is constant. 10. (Hadamard's three circles theorem) Let f be a holomorphic function on a

domain {z E Q RI < Izl < R2} with 0 < Rl < R2, which is called an annulus. Set M(r) = max{I f (z)I; IzI = r} for r E (R,, ,R2). Show that logM(r) is a convex function with respect to logr: that is, for Rl < r, < r2 < r3 < R2

logM(r2)
1).

(lal < 1),

log la - e`sld9 = 10

log lal

(lal > 1).

13. Find the adjoint harmonic function of the harmonic function x+y+x2 -y2. 14. Show that u (x, y) = ez (x cos y - y sin y) is harmonic, and find a holomorphic function f (z) with real part u(x, y).

15. Let {un(x, y)}n o be a sequence of harmonic functions on a domain D, converging to u(x, y) uniformly on compact subsets. Prove that u(x, y) is harmonic.

16. Let f (z) be a bounded function, holomorphic on H, and continuous up to R. Prove that f (z) = r J_ 00

(t z) - t2

I 1

Ref (t)dt + Imf (i).

17. Let h(t) be a bounded continuous function on R. Show that the function u(x, y) =

!J

y h(t)dt z (t-x)2+N2

is harmonic on H. Moreover, as x -+ to E R and y -+ +0, u(x, y) - h(to).

CHAPTER 4

Residue Theorem

We begin in this chapter with Laurent series, and then explain meromorphic functions and the residue theorem. By making use of it, we will prove the Prinzip von der Gebietstreue (open mapping theorem), and the inverse function theorem. In the last section we explain methods for applying the residue theorem to calculate various kinds of definite integrals. These are direct consequences derived from Cauchy's integral theorem proved in Chapter 3; nevertheless, the reader may sense the depth of content of the theorem.

4.1. Laurent Series

For aECand O 0, a is called a pole of order m of f , and one writes f (a) = oo. If an = 0 for all n < 0, then f is holomorphic in A(a; r2), and f(z)=a,n(z-a)m+a,n+,(z-a)m+l+...,

If m > 0, a is called a zero of order m of f .

am #0,

m>_0.

4. RESIDUE THEOREM

94

For a = oo, we define the notion of zeros and poles in the same way as above by making use of the complex variable z. Note that

{i;0 < lil < R} = S z; IzI >

l .

T1 The relation between the Laurent series of f with respect to i and that with respect to z is given by o0

f(z) = >00 anzn = E a_nZ = f o i.

(4.1.8)

n=-oo

n=-oo

Therefore we have the following: (4.1.9)

f (z) = E'n=-. anzn is holomorphic at oo if and only if an = 0 for all n ? 1;

(4.1.10)

the differential

_

(n

f

°°

1 anzn ] dz = oc

a_nzn

/

)

(_.) dz

is holomorphic at oo if and only if an = 0 for all n >_ -1. EXERCISE 1. Obtain the Laurent series of f (z) = el/= about z = 0. EXERCISE 2. Obtain the Laurent series of f (z) = 25(11+ z) about z = 0.

4.2. Meromorphic Functions and Residue Theorem Let D be a domain in C. A function on D which has at most a pole at every point of D is called a meromorphc function. Let f be a meromorphic function on D. If a E D is a pole of f, the function

fuo f(z)=

1

f (z)

is holomorphic in a neighborhood of a. In this sense f uniquely defines a continuous mapping (4.2.1)

f:D-.C.

For example, a rational function P(z)/Q(z), defined as a ratio of two polynomials P(z) and Q(z) with Q(z) $ 0, is meromorphic on C. By definition, the set E of all poles off is a discrete subset of D. If E = 0, then, of course, f is holomorphic

in D. Two meromorphic functions f and g on D are said to be equal if there is a discrete subset F C D such that f and g are holomorphic on D \ F, and if f = g on D \ F. This property is independent of the choice of F by the identity Theorem (2.4.14). If f 0, l/f is meromorphic, and the sum and product of two meromorphic functions are meromorphic. Thus the set of all meromorphic

4.2. MEROMORPHIC FUNCTIONS AND RESIDUE THEOREM

95

functions on D forms a field. For meromorphic functions we have the following identity theorem.

(4.2.2) THEOREM. Let fl and f2 be meromorphic functions on D. Let A C D be a subset with an accumulation point in D. If fl = f2 on A, then f, = f2 on D. PROOF. We may assume that f2 = 0. Let zo E D be an accumulation point of A, and E be the set of all poles of f . If zo ¢ E, Theorem (2.4.14) implies that f1 - 0 on D \ E, so that f1 = 0. Assume that zo E E. If zo 36 oo, we take

0 0 such that A(a; b) C D, and (A (a; b) \ {a}) n {f = o} = 0.

Here, {f = 0} stands for (z E D; f (z) = 0}. For C(a; b) with anti-clockwise orientation 1

f

) dz >= 1.

i JC(a;b) f (Z)

Since {f,, } converges uniformly on A(a; b), there is an no such that t

f (z)I < min{I f (w)j; w E C(a;b)},

z E C(a; b), n ? no.

It follows from Theorem (4.3.7) that the number of zeros of f in A(a; 6) equals

that off + (f - f) = f (n ? no). Thus fn has a zero in i(a; b); this is a contradiction.

Let D C C be a domain. A meromorphic function f in D_is said to be univalent if f is injective (one-to-one) as a mapping from D into C. (4.3.10) THEOREM. Let {fn}n o be a sequence of holomorphic functions in a domain D C C such that it converges uniformly on compact subsets of D to f. If all fn are univalent, then f is either univalent or constant.

PROOF. Assume that f is not constant, nor univalent. Then there are two points z1 i z2 E D with f (z1) = f (Z2) = wo. Take neighborhoods U; of z i = 1, 2, with U1 n U2 = 0. Since {f,, - wo} 0 converges to f - wo uniformly on compact subsets of D, it follows from Theorem (4.3.9) that there are some n = wo. That is, f,, (z') = fn(z2), and this and zi E U;, i = 1, 2, such that f,. is a contradiction.

a point z E D For a meromorphic function f in D and for a point w E such that f (z) = w is called a w-point of f; an oo-point of f is nothing but a pole of f . When w # oo and the order of zero off - w at z is m, m is called the order of the w-point z; the order of the oo-point z of f is defined as that of the pole off at z. (4.3.11) THEOREM. Let f be a meromorphie function in a domain D C C. Let zo E D be a wo-point of order m of f with wo = f (zo). Then there are neighborhoods V of zo and W of wo such that for an arbitrary w E W \ {wo} there are

exactly m distinct w-points, z,... , z,,,, of order 1 off in V.

4.3. ARGUMENT PRINCIPLE

103

PROOF. We assume that zo and w are not oo; otherwise, we argue similarly by using z or tD about oo. By assumption we may expand f to a Taylor series about zo: f(z)=wo+a,,(z-zo)'"+...,

a,,,#0.

There is 0 < b < d(zo; OD) such that A(zo; b) n {f = wo} = {z0}, (4.3.12)

(4.3.13)

0(zo;6)n{f'=o}\{zo}=0, f '(Z)

tai ./C(w;a) f (z) - wo

dz = m.

There is0<e e,

z E C(zo;6).

Applying Theorem (4.3.7) to f (z) - to = (f (z) - wo) + (wo - w) with to

E

A(wo; e) \ {wo}, we infer that the number of zeros of f (z) - wo in A(zo; b) is equal

to that of f (z) - to for to E A(wo; e). If z E A(zo; 6) and f (z) = to E A(wo; e), then z is a u)-point of order 1 of f by (4.3.12). Therefore there must be exactly m such points z. The following theorem is called the Prinzip von der Gebietatreue (open mapping theorem). (4.3.14) COROLLARY. Let f be a non-constant meromorphic function in a domain D C C. Then, as f is considered to be a continuous mapping from D into C, the image f (D) is a domain of C.

PROOF. By Theorem (4.3.11) f (D) is an open subset. Take two arbitrary points f (z1), f (z2) E f (D). There is a curve C(45) connecting z1 and z2 in D. Then f o 0 is a curve connecting f (z1) and f (z2). Thus, f (D) is connected. We give one more application of Theorem (4.3.11), which is called the inverse function theorem.

(4.3.15) THEOREM. Let f be a holomorphic function in a neighborhood of zp E C, and set wo = f (zo). If f'(zo) 0, there are neighborhoods U of zo and W

of" such that f (U) =W and the restriction f I U : U- W off to U has a holomorphic inverse function g : W -, U of f JU with 9(wo) = zo.

PROOF. Take neighborhoods U of zo and W of wo as in Theorem (4.3.11). Since f'(zo) 96 0, m = 1, so that for an arbitrary to E W there is a unique point

4. RESIDUE THEOREM

104

z E U with f (z) = ur. Write g(w) = z. We first check the continuity of g. Take an arbitrary w' E W, and set z' = g(w'). In a neighborhood V' of z', we have (4.3.16)

f (Z) - f (Z') = al (z - z') + a2(z - z')2 + ...

(a, # 0)

= (z - z')(al + a2(z - z') + -) = (z - z')h(z).

Since h(z') = a, 34 0, after taking V' smaller if necessary, there is a positive number b such that (4.3.17)

z E V.

lh(z)I >_ 6,

Then f(V') = W' is an open neighborhood of w'. It follows from (4.3.16) and (4.3.17) that

619(w)-9(0I 5 1w-ti1. Hence g(w) - g(w) -' 0 as w -+ w'; that is, g is continuous. Again by (4.3.16) we have for wE W' with w76 w'

g(w) - g(w') w - w'

_

I h(g(w))

The continuity of g implies that lim,,,_,, h(g(w)) = h(g(w')) = h(z) = f'(z), and so lira g(w) - 9(&) w-.w' w - w'

= g'(w') =

1

P(z')

Thus, g is complex differentiable in W'; i.e., g is holomorphic. Then we may set

U=9(W)=f-'(W)- o (4.3.18) THEOREM. Let f be a univalent meromorphic function in a domain D C C. Then there is a meromorphic function g in the domain W = f (D) C C such that the mapping g : W -. C is the inverse mapping off : D - W.

PRooF. By Theorem (4.3.14), the image W = f (D) C C is a domain, and we have the inverse mapping g : W Doff since f is univalent. Take an arbitrary point zo E D, and set wo = f(zo). It follows from Theorem (4.3.11) and the univalence off that zo is always a we-point of order 1. Therefore, in the case of zo 0 oc and wo 96 oo, f'(zo) 96 0. By Theorem (4.3.15) g is holomorphic in a neighborhood of wo. In the case of zo = oo and wo 0 oo, we have that f o i/di(0) 0. Thus z o g(w) = 1/g(w) is meromorphic in a neighborhood of ", and hence g(w) is meromorphic there. In the case of zo = oo and " = oo, we see that riw o f = 1/f is holomorphic in a neighborhood of i = 0. Thus, for the same reason as above, g is meromorphic in a neighborhood of oo. 0

4.3. ARGUMENT PRINCIPLE

105

It should be noted that Theorem (4.3.18) does not hold for real differentiable functions, nor for real analytic functions. For example, take f (t) = t3 for t E R. Then f : R R is injective, but the inverse function ' t E R is not differentiable

at t = 0. Let f be a holomorphic function in a neighborhood of zo E C. If f'(zo) # 0, then by Theorem (4.3.15) f has an inverse function g defined in a neighborhood of wo = f (zo). In what follows, we consider the case of f'(zo) = 0. For the sake of simplicity, we assume that zo = wo = 0. Let m be the order of zero of f at 0. Then m > 1, and we have

f (z) =amzm +a,.+ l = amzm

zm+l + .. .

+ ...1 \(1 +'-+1 am

= a,,, z'h l (z)

(a,,, # 0, h l (0) = 1).

Taking 6 > 0 small, we have

hi(z) E A(1;1),

z E 0(6).

Fixing one branch of log in 0(1;1), we get h1(z) = elg hi (s), and set

h2(z)=expl mlogh1(z)). we set

Taking one value of

h3(z) =/zh2(z). Then f (z) = (h3(z))m, and h3(z) _ /z + ... # 0.

h3(0) =

It follows that v(z) = h3(z) has the inverse function z = gi(v) with gl (0) = 0. Let the Taylor series expansion of g1 be

91(v) = clv + c2v2 + ..., Since w (4.3.19)

f(z)=vm,wewrite v= i,andset 92 (W)

= C1

VW-

+ c2 ("vf_)2 + .. .

Then f(g2(w)) = w, and so one may consider g2(w) = f-1(w). But, 92(w) is a holomorphic function of v = and for the variable w, 92 (w) = g1( w-) is a multi-valued function with m values if w # 0. In general a series such as (4.3.19) in the powers of Vw- is called a Puiseux series, and in the present

4. RESIDUE THEOREM

106

case, in particular, (4.3.19) is called the Puiseux series expansion of the inverse

function f -'(w). 4

0

0001.7.

V"

FIGURE 41

EXERCISE 2. Show that the numbers of zeros and poles of a rational function counting multiplicities over the Riemann sphere d are equal. EXERCISE 3. Set f (z) = z2/(1 + z). Obtain the Puiseux series expansion of f -1(w) about w = 0.

4.4. Residue Calculus In this section we explain bow to calculate the values of some definite integrals

as applications of residue Theorem (4.2.7) or (4.2.14), classifying the definite integrals into several types. i) Let R(x, y) be a rational function of two variables x and y such that it has no pole on the circle C(0;1). We want to calculate the value of !2w

1= /

R(cos t, sin t)dt.

Jo

For z=e'twehave Cost

=Z(z+ZI

slat=

///

dt =

2ilz-z \\ //

1 dz. sz

It follows from Theorem (/4.2.7) that

I=J

C(0;1)

=2w

R(1 Iz+lZ 1, 1

zW ldz

` Res(a;R(2 (+)2i

nE0(1)

(z-

zl)l

4.4. RESIDUE CALCULUS

107

102 a

EXAMPLE. We calculate I = 1

C(o;1) a+ 2 (z + _) 2

_

tzdz

a + cost

, a > 1. Then

_2

1 z2+2az+1dz

C(o:i) 1

C(o;I) {z - (-a - a -1)}{z-(-a+ a -1)}

? 27ri Res

(-a + `

dz

1

a2 - 1;

{z - (-a - a 1)}{z-(-a+ a -1)} {z - (-a + a --1)j = 47r lim z-.-a+ a -1 {z - (-a - /a2 - 1)} {z - (-a+ a --I)) _

27r

e

a -1*

In the latter half we used (4.2.10) to obtain the value of the residue.

FIGURE 42

EXERCISE 1. Calculate the following definite integrals. dB

)1"

i

ii) f iii)

(0 1).

1i) Let R(z) be a rational function of z. We want to get (4.4.1)

I =1

00

R(x)dx = oc

10

o

R(z)dz.

1-10

Express R(z) = P(z)/Q(z) with coprime polynomials P(z) and Q(z). For the convergence of the integral (4.4.1) at ±oo the following condition is necessary and sufficient: (4.4.2)

degP-degQ 0, 1 < i S k, be all the roots of Q(z) = 0 in the upper half-plane. Let b, E R, 1 < j 51, be all the real roots of Q(z) = 0. Then we are going to show that k

(4.4.4) p.v.

for,

R(x)dx = 2xi

1

Res(a:; Rz)) + Sri E

R(z)).

j_1

FIGURE 43

Take r > ro > maxi; {la, I, Ib; 1), and small e > 0 so that A(a e) do not intersect It, and 0(ai; e) and 0(bj; e) are mutually disjoint. Let C(,.) denote the half circle from r to -r in the upper half plane, and let C(r,,) denote the curve from -r to r on the real axis where about every bJ the curve goes around the half circle C- (b,; e) of C(b,; e) in the lower half plane (see Figure 43). Then (4.4.5)

/

r

R(z)d.- = 2ri

Res(ai; R(z)) +

R,es(b;; R(z))}

.

By condition (4.4.2) there are an M > 0 and a ro > 0 such that for all r > ro IR(Te`s)I S

ra

IN

4.4. RESIDUE CALCULUS

and hence

M -rdz = < = Icf, r2 r

R(z)dz

JC()

y0

(r --+ oo).

Expand R(z) about every b., to a Laurent series:

R(z) = C-I + co + cl (z - b,) + -

z-bj C-I

z-b3 We get

f

C-I z

bj

+ h(z) dz -+ rric_I

(e - 0).

Since c_1 = Res(b,; R(z)), we obtain (4.4.4) by letting r - oc and e -b 0 in (4.4.5).

EXAMPLE. We calculate 1 = p.v. r x41

1

dx = 2 P.V.

j x41

1 dx.

Since

Q(z) = z4 - 1 = (z - 1)(z + 1)(z - i)(z + i), conditions (4.4.2) and (4.4.3) are satisfied. We have

_

1

Res (1' Q(z)) Therefore

4

Res (-1; Q(z))

1 = 2 [2viRes (i; =

__1

1

1

7ri

{Res

Res

i

4

(1;

1

Q(z))

_ i 4

Res (-1; Q(z) )}]

22ai 4) _ 4

EXERCISE 2. Calculate the following integrals. °°

i)

p.v.1. 2 P.V. r J ao

1

X2

x

0o

1

ii) 1

dx.

+

x4-+-X21 2 dx.

iv) 1o

x4 + 1 dx.

x + 1 )°

dx (n = 1, 2,... ).

( 2

Hi) We let R(z) = P(z)/Q(z) be as in ii), but relax condition (4.4.2) to

deg P - deg Q < -1.

(4.4.6)

We keep condition (4.4.3). Then the following holds: (4.4.7)

p.v.

foo

R(x)e'=dx = tai E R s(a; R(z)e':) + si E Res(b; R(z)e'=). Ima>O

Imb=O

4. RESIDUE THEOREM

110

The proof is the same as in ii); the only difference is to show that

fC)

R(z)e"dz - 0

(r -. oo)

under condition (4.4.6). Set z = re' (0 5 0:5 ir). Then Ie'r there is some M > 0 such that JR(re'BW< Mr. Therefore

sI

= e-r°in9, and

a-redo.

R(z)e'= < M 0 To

If one knows Lebesgue integrals, it is clear that the above right side converges to 0 as r - oo. Relying only on Riemann integrals, one deduces the same from the following estimate:

r

a-r sin Bd0

=

a-r sin e W

2

0


0, we have a Taylor expansion of 0 about to: For t E (to - b, to + b) 30

c E C.

fi(t) = zo + E n=1

(5.1.6)

Introduce a new complex coordinate

1;=t+is,

t,sER.

It follows from (5.1.6) that oc

OW = zo +

cn(t - to)"

if - tol < b,

n=1

is a convergent power series, and defines a holomorphic function. Since 0'(to) = c1 96 0, by the inverse function Theorem (4.3.15) 0(1:) may be assumed to be injective on A (to; 6). Set

u+ = A(to; b) n {Iml; >o}, u- = L V

= (t0-6,t0+5), = 00(to; b)),

A(to; b) n {Imp 0 if necessary, that

OD n v = Co. It does not lose generality to assume that V+ C D (if V- C D, one may replace s by -a). Then C E U+ and Z E U- are the reflection points of

5. ANALYTIC CONTINUATION

172

each other with respect to L. We say that O(C) and O(Z) are the reflection points of each other with respect to Co.

FIGURE 50

We show that this concept of reflection does no depend on the choice of OD be a change of parameter of 0, which parameters of y. Let 4 : [To,T1I gives C and is non-singular and analytic. Take 4 with 0(io) = 0(to), and define A(bo; b), Ut, L, V. ..., etc., as above. Let b > 0 be sufficiently small so that the values of the holomorphic function 0' o t on A(io; 6) are contained in A(to; 6). Set h = 0-1 o I,1(to; 6) : A(to; 6) --' A(to; 6).

The holomorphic function h takes real values on L: h(L) C L. Therefore it follows from Theorem (5.1.4) and the uniqueness of the analytic continuation that for f E A(to; 6), h(t) = h(1). Setting t = h(t), we have 0w _ k)T

0(7 =0(0-

Hence, two points which are reflection points with respect to Co are also reflection

points with respect to Co. In the case of zo = oo, we use the coordinate i to define the notion of reflection points as well.

EXERCISE 1. Show that if C is a circle (the case of a line is included), the notion of reflection points defined in Chapter 2, §8 coincides with one defined here.

(5.1.7) THEOREM. Let D; C C, i = 1, 2, be domains such that a part C, of 0 8D, is a simple non-singular analytic curve and 8D, \ C, does not accumulate 0

at any point of C.,._Let f be a meromorphic function on D1 with values in D2 as a mapping into C. Assume that f is extended to a continuous mapping over 0 0 0 D1 U C1 so that f (Cl) C C2. Then, if z E DI is sufficiently close to C1, to its 0 reflection point z' with respect to Ci there corresponds the reflection point f (z)' 0 of f (z) with respect to C2, so that f is analytically continued, crossing C1, to a 0 domain containing C1 as a meromorphic function.

5.1. ANALYTIC CONTINUATION

123

FIGURE 51

0

0

PROOF. Take arbitrarily a point z1 E C1, and set z2 = f(z1) E C2. We suppose that in a neighborhood of z,, C, is represented by a simple non-singular analytic curve O.: (-b;, b;) - C; with b; > 0. As described above, we extend 4;

over a complex variable {, E 0(b;). We may assume that O; : 0(b;) - MA00) is injective. Let U, denote the part of A(b;) in the upper half plane. Take b1 > 0 sufficiently small. Then the composite function 021 o f o 01 : Ui - U2 is holomorphic, extends continuously over the real axis (-bl, b1), and takes real values on it. By Schwarz' reflection principle, Theorem (5.1.4), 021 o f o 01 extends to a holomorphic function g on 0(b1) with values in i (b2). Set h1 = 02 o g o 0,' on V1 = 01 (A(61)). Then f is analytically continued to h1 on V1. e V1(zj ). Then W is a domain, We write V1(zl) for this V1, and set W = U 0

z,EC,

C1, and there is a meromorphic function h such that h = f on W n D1. Thus f is analytically continued to D1 U W. 0 W

REMARK. When C1(0 : [To, T1] - 8D) is a closed curve in Theorem (5.1.7), there are two possible cases:

i) There is a change m : [To,TI] - OD of parameter of

such that

is

non-singular analytic and for some io E (To, T1), ¢(to) _ O(To) ii) There is no such change of parameter of 0. If C2 is not of case i) at f oO(To), f cannot be analytically continued to a domain containing i(To) no matter whether C1 is of case ii) or i) (see Figure 52, below). If C1 (reap., C2) is of case i) at 0(To) (reap., f o O(To)), then f is analytically

124

5. ANALYTIC CONTINUATION

continued to a domain containing C1.

f(Z) =

FIGURE 52

EXERCISE 2. Let f (z) be a continuous function on 0(R) that is holomorphic

in 0(R), and let If (z)I - K (constant) on the boundary C(0; R). Show that f (z) is a polynomial. EXERCISE 3. Show that if an entire function f (z) takes real values on the

real axis, and purely imaginary values on the imaginary axis, then f (z) is an odd function.

5.2. Monodromy Theorem A meromorphic (resp., holomorphic) function element about a point a E C is defined as a pair (Fa, Ua) of a meromorphic (resp., holomorphic) function in a disk neighborhood Ua of a, and U. is called the disk of definition. We sometimes abbreviate (Fa, Ua) by Fa, calling it a meromorphic or holomorphic function element about a. Let C(O : [0, 1] -+ C) be a curve. Assume that a meromorphic function element (F,(t), Ut) is given for every t E [0, 11. We write Ft = Fo(t). We say that {(Ft, Ut)}tE(o,11 is an analytic continuation (of (Fo, Uo)) along C if for an arbitrary to E [0, 11 Fto - F t

on Uta fl Ut,

t E (t1, t2),

where t1 = inf{t E [0, to]

q5([t,to]) C Uto},

t2 = sup{t E [to, 11 : O([to, t]) C U10 }.

FIGURE 53

FIGURE 54

5.2. MONODROMY THEOREM

125

In this case, the disk Ut of definition of every Ft is written as 0(t) j4 oo,

Ut = {(t);rt),

{z E C; IzI < rt},

4(t) = oo,

and by increasing rt if necessary we may assume (cf. Figures 53 and 54) that (5.2.1)

inf{rt; t E [0, 1]} > 0.

(5.2.2) THEOREM. Let {(Ft, Ut)}tE[o.1! and {(Gt, Vt)}tE[o,1] be two analytic con-

tinuations along a curve C(t¢ : [0,1] C). If Fo = Go in a neighborhood of ¢(0), then F1 = G1 in a neighborhood of 0(1).

PROOF. Set E _ {t E [0, 1]; Ft = Gt in a neighborhood of 0(t)}. By the assumption, E # 0, and E is clearly open in [0,1]. Take an arbitrary to E E. Take a connected neighborhood Ut, of q(to) so that it is contained in the disks of definition of Ft,) and Gto. If t' E E is sufficiently close to to, O(t') E Ut,. Since Fta and Gt,, coincide on a neighborhood of O(t'), Ft. = Gto on Ut0 by Theorem

(4.2.2). Thus to E E, and so E is closed in [0, 1]. We have that E = [0,1]. 0 Let C} (O, : [0,1] x [0, 1] --+ C), j = 0, 1, be two curves which are homotopic to each other. Let the homotopy connecting them be

4':[0,1]x[0,1]-4C, 4'(t, 0) = 00 (t),

4'(t,1)

(t).

Then we have the following monodromy theorem

(5.2.3) THEOREM. Let CJ, j = 1, 2, and 4 be as above. Suppose that for every (t, s) E [0, 1] x [0,1] a meromorphic function element F(t,,) about 4i(t, a) is given so that for each fixed a E [0, 1] {F(t.,)}t gives rise to an analytic continuation along C,(0. = If F(o,o) = F(o,8) in a neighborhood of the initial point P = t>5o(0) (0 < s < 1), then F(1.o) = F(1,1) in a neighborhood of the terminal point Q = 0o(1)

PROOF. Set S = {s E (0, 1]; F(1,.) = F(1,o) in a neighborhood of Q). Since 0 E S, S 0 0. We write U(t,,) for the disk of definition of F(t.9). Note that U(t,3) is connected. Fix arbitrarily so E S. We may assume (5.2.1) for {F(t,,0) }t. By the uniform continuity of 4 there is 6 > 0 such that for every a E (so -b, so+b)fi [0,11 (5.2.4)

0 (t, s) E Ut.,,,

0 < t < 1.

126

5. ANALYTIC CONTINUATION

FIGURE 55 We

FIGURE 56

fixsE(so-b,so+6)fl[0,1],andset T = it E (0, 1); F(t,,) = F(t,,,) on U(t.,) fl U(e.,o) }.

Since 0 E T, T # 0. In the same way as in the proof of Theorem (5.2.2) one sees that T is open and closed in [0, 1]. Hence T = 10, 11, and so in a neighborhood of Q, F(1,,) = F(,.,o) = F(i,o). It follows that S is open. In the above arguments, without assuming so E S, we infer that if there is an s which satisfies (5.2.4) and belongs to S, then so E S. Hence S is also closed, and then S = [0, 1].

Let f be a meromorphic function on a domain D C C. Let P E D, and consider the meromorphic function element obtained by the restriction of f to a disk neighborhood of P in D. We consider meromorphic function elements (FQ, UQ) produced by all possible analytic continuations of f along curves in C with initial point P, and set b = U UQ. Then b is a domain containing D, and we have a multi-valued meromorphic function 1(Q) = Fq(Q). It follows from Theorems (3.3.6) and (5.2.3) that (5.2.5) THEOREM. Let f and b be as above. i) Let Q E D. Identify the meromorphic function elements (FQ, UQ) and (Gq, Vq) if Fq and Gq are identical in a neighborhood of Q contained in Uq fl VQ. Then the number of meromorphic function elements of f defined about Q is at most countable. ii) Let Do c D be a simply connected domain such that for a point Q E Do a meromorphic function element (Fq, Uq) about Q can be analytically continued along any curve in Do with initial point Q. Then f defines a one-valued meromorphic function fo on Do.

The above fo in ii) is called a branch of f. The identification of meromorphic function elements used in the above i) induces an equivalence relation in {(FQ, Uq)}. Let [Fq, Uq] denote the equivalence class of (FQ, Uq). We denote by X the set of all [FQ, UQJ. We introduce a topology on X as follows. It is sufficient to define neighborhoods of [FQ, Uq] E X. For every P E Uq, we take a

5.2. MONODROMY THEOREM

127

disk neighborhood Up C UQ and the restriction Fp = FQJUp of FQ to Up. We define the system of neighborhoods of (Fq, Uq] by {[Fp, Up]; P E V},

where V runs over all neighborhoods of Q in UQ. Thus X is a topological space, which is called a Riemann surface defined by f. Defining f ([FQ, Uq]) = Fq(Q), f gives rise to a one-valued function on X. (5.2.6) EXAMPLE. The Riemann surface defined by the multi-valued meromorphic function

f(z)=logz, zED=C\{0}, is obtained as follows. Take countably many complex planes Cj, -oo < j < co. Let L,1 (resp., L,) denote the real pos-

=C

itive axis in CJ \ {0} as accumulation points of the upper (resp., lower) halfplane, and distinguish between Lt and L7. Then we identify L with L,_1 of

CJ_1 \ {0} for all -oo < j < oo (see Figure 57). Let X denote the resulted

topological space. Defining f(i) = 0 at 1 E Co\ {0}, we obtain a one-valued func-

tion f on X.

FIGURE 57

Making use of g(w) = ew, we identify injectively G; = {z E C; 2ja < Im z < 2(j + 1)ir} with (C3 \ {0}) U L+:

g1G.' :Gj-+(C.\{0})UL?. G; = C, X is identified with C. The logarithmic function w = log z is the inverse function of the holomorphic function z = z(w) = e', and so the above Riemann surface X is also called the Riemann surface of the inverse function of z(w) = e'°. Since trJ,°

FIGURE 58

5. ANALYTIC CONTINUATION

128

(5.2.7) EXAMPLE. The Riemann surface X of the inverse function w = f of z(w) = w2 is obtained by gluing (identifying) two copies of C along the nonnegative real axis (see Figure 58). Let Do be a simply connected domain in

C \ {0}, and take a branch fo(z) of f on Do. The other branch of f is f, (z) = - fo(z). Set G; = f,(Do), j = 0,1. Then G; is a domain of C \ {0}, the mapping

fi : Do -. Gi is injective, and Go n G, = 0.

FIGURE 59

In general, if the function fin Theorem (5.2.5) is expanded to a Puiseux series with an integer p > 2,

f(z)

_

an(

z - a)n

(a

oo,0 < N < oo),

n>-N

foz

=

an(v z)n

(a = oo, 0 < N < oo),

n>-N and if p cannot be reduced to a smaller one by another Puiseux series expansion,

then a is called a brunch point, and p - 1 is called the order of the branch. If in a neighborhood of a f is expressed as a meromorphic function of log(z - a) or log z, and is an infinitely-many-valued function, then a is called a logarithmic branch point. For instance, if a E R is irrational, then

f(z) = z° has logarithmic branch points at z = 0 and coo.

5.2. MONODROMY THEOREM

129

(5.2.8) EXAMPLE. We consider the Riemann surface X of the inverse cosine function w = cos-1 z. 00

00

w = C08-1 z

FIGURE 60

Let L 1 denote the half line Re w = -ir, Im w >_ 0 in the w-plane from oo to -7r. Let L2 denote the line segment of the real axis from -7r to 0, and let L3 denote the half line Re w = 0, Im w >_ 0 from 0 to oo. Let Go be the domain surrounded by them. Let Cj be the image of Lj mapped by z = cos w with orientation for j = 1, 2, 3. Then C1 runs from oo to -1 on the negative real axis, C2 from -1 to 1 on the real axis, and C3 from 1 to oo on the positive real axis. Since a holomorphic function preserves the orientation except at points where the derivative vanishes, Go is injectively mapped to the upper half plane Do. This mapping is analytically continued by the reflection principle, Theorem (5.1.7) or Theorem (5.1.4). The reflection of Do with respect to C1 is the lower half plane D1, and its image by cos-1 z is the reflection G1 of Go with respect to L1. We carry out this continuation process with respect to Cj, j = 1, 2, 3, as far as possible. The branch points of cos' -z are -1 and 1, and their values are 2nir and (2n + 1)ir (n E C), respectively. The orders of branch are 1. Thus the Riemann surface X is identified with C. EXERCISE 1. Construct the Riemann surface of sin-1 Z. EXERCISE 2. What is the Riemann surface of 1 - cos z? Show that in fact it defines two 1-valued entire function on C.

Let F(z1i... , zn) be a complex function of n variables, z1, ... , z,,. Set zj = xj + iyj, 1 5 j 5 n. If F is continuously differentiable in the variables x., and 1 5 j 5 n, and if for every j the Cauchy-Riemann equations (5.2.9)

1

8

2 \ 8X,

1 8 )F=O, i 8yj

1<j:n,

hold, then F is said to be holomorphic in (z1, ... ,

Let fi (z),1 5 j 5 n, be n holomorphic functions of z, and assume that the composite F(f1(z),... , fn(z)) is defined. Then, if F is holomorphic, so is Let C(O : (0,1] -+ C) be a curve and let {fjt)1i 1 5 j 5 n, F(f1(z),... , be analytic continuations of holomorphic functions along C. Let F(w1, ... , w)

5. ANALYTIC CONTINUATION

130

be a holomorphic function such that all composites F(flt(z),... , f,,t(z)) are defined. Then it follows from the identity Theorem (2.4.14) that (5.2.10)

F(f11(z),... ,f",(z)) ° 0.

F(f1o(z),... ,f,,o(z)) =-O

This is referred as the principle of the permanence of the functional relation of analytic continuation.

5.3. Universal Covering and Riemann Surface Let D C C be a domain and fix a point P E D. Let 7r1(D)p denote the set of all homotopy classes (Cl of closed curves C in D with the initial and terminal point P. A multiplication in 7r1(D)p is defined by

{Cl} {C2} _ {C1 +C2}. Here it is sometimes written as {C1 } + {C2 }, while the multiplication is not commutative in general. Let e denote the homotopy class {C} E 7r1(D)p of C which are homotopic to the point P. We set {C}-1 = {-C) for {C} E 7r1(D)p. Then 7r1(D)p forms a group with unit element e, and is called the fundamental group of D with base point P. For another point Q E D we have a group isomorphism (5.3.1)

tQ : 7r1(D)p 9 {C}

{CQ + C - CQ} E 7r1(D)Q,

where CQ is a curve from Q to P in D.

FIGURE 61

Thus the groups irl (D)p and 7r1(D)Q are mutually isomorphic as abstract groups.

We simply denote it by 7r, (D), and call it the fundamental group of D. Let D' be another domain, and f : D D' be a continuous mapping. Then we have a group homomorphism (5.3.2)

f.: 7r1(D)p E) {C} -+ {f(C)} E 7r1(D')fiQi.

Here f (C) is defined by f o 0 with C = C(O).

The continuous mapping f : D -* D is called a covering mapping, and D is called the covering of D' if (5.3.3) i) f (D) = D';

5.3. UNIVERSAL COVERING AND RIEMANN SURFACE

131

ii) for every point w E D' there is a neighborhood V C D' of w such that for any connected component U of f -'(V), the restriction off to U,

flu: U- V, is a homeomorphism (a surjective and injective continuous mapping with continuous inverse mapping).

EXAMPLE. 1) f : C 3 z - ez E C' is a covering mapping. 2) f : C' 3 z z2 E C' is a covering mapping. EXERCISE 1. Let f : D -+ D' be a covering mapping. Show that f (w) has no accumulation point for every w E D'. EXERCISE 2. Show that the fundamental group of C' is isomorphic to Z.

Now, let D be as above, and fix a point Po E D. Let X be the set of all homotopy classes ap of curves from Po to P in D, where P moves over D. Take an arbitrary point P E D and a disk neighborhood Up. For a point Q E Up we take a curve L from P to Q in Up. Then the homotopy class ap + {Cp } is independent of the choice of Cp , since Up is simply connected.

FIGURE 63

FIGURE 62

Set

V(ap) = {aP + {C }; Q E UP}.

If ap 0,3p, then (5.3.4)

V(ap) fl V(13p) = 0.

Let co E X be the homotopy class of the constant curve P0, and let (5.3.5)

7r:X 3ap-+PED

be the natural projection. We define a topology of X by taking neighborhoods V(ap) for each point ap E X; that is, a point sequence {an}°°_0 of X converges to ap E X if and only if for an arbitrary neighborhood V(ap) of ap there is a number no such that an E V(ap) for all n > no. The restriction mapping (5.3.6)

7rIV(ap) : V(ap) -+ Up

5. ANALYTIC CONTINUATION

132

is a homeomorphism. Take two distinct points ap 31 fig E X. If P 36 Q, then, taking Up and UQ with Up n UQ = 0, we define V(ap) and V(13q) so that

V(ap)nV(fQ)=0; if P = Q, the same holds by (5.3.4). Therefore, X is a so-called Hausdorff topological space. For ap = {C(Q+ : [0,1] -+ D)} we have a continuous mapping (5.3.7)

0 : [0,1] 30 t

am(p) _ {C(0J[0, t])) E X,

and 0(0) = co and O(1) = ap. Thus X is arcwise connected. While we do not know if X is a domain of C, the mapping a : X - D defined by (5.3.5) satisfies the properties of (5.3.3), i), ii). We also call this a covering. If we choose another

Po E D, we have another X, which is homeomorphic to the original one, and so we may identify them as a topological space. We call this X the universal covering of D, and it : X --; D the universal covering mapping. EXERCISE 3. Let Po E D and X be as above. We take another point FO E D, and then obtain another X' as X. Let CP° be a curve from Po to Po. Show that (5.3.8)

: X' E) apo

{CPo } + apo E X

is a homeomorphism. An element -y E 1r1 (D) = art (D)p,, of the fundamental group defines a homeomorphism (5.3.9)

ry:X3a-+ry+aEX.

For two elements -r1. y2 E a1(D) (5.3.10)

'Y1 C ry2 = 71

'72 (= 71 +'Y2),

where the left side is the composition of the mappings and the right side is the multiplication in the group ir1(D). In this case, we say that the group a1(D) acts on X. The set {ry(a)-,1y E ir1(D)} with an element a E X is called the ir1(D)-orbit of a. It follows that {ry(ap);'y E ir1(D)} = {y(aQ);7 E 1r1(D)}

FIGURE 64

P = Q.

5.3. UNIVERSAL COVERING AND RIEMANN SURFACE

133

Considering a r1 (D)-orbit as one point, we get a quotient space X/r1 (D) of X by the action of rl(D). We have

X/r1(D) = D.

(5.3.11)

When an element 7 of r1 (D) is considered as a homeomorphism defined by (5.3.9), r1(D) is called the deck transformation group or the covering transformation group. For distinct ap, /3p E X with P E D, set 7 = Op - ap E r1 (D). It follows from (5.3.9) that -y: X

X satisfies

7(V(ap)) = V0p)

(5.3.12)

In general, let Y be a closed interval of R, an open interval of R, a domain of C, or some finite product of them. Let f : Y --. D be a continuous mapping.

A continuous mapping f: Y - X is called a lifting of f if f= r o f:

D

Y f

For a curve C(O : [0,11

D) with the initial point P E D and ap E X we set

1 : [0,11 9 t - ap + {CI[0, t1} E X.

(5.3.13)

Then r o fi(t) = '(t), so that tai is a lifting of ' and gives a curve CQP, called a lifting of C. We see by (5.3.6) that a lifting of C is unique if the initial point ap is given. Therefore, for a lifting Cpp of C with the initial point /3p 0 ap we have

CQP nOp. =0.

(5.3.14) ca,

x j

p

D C

FIGURE 65

ar

D

C p

FIGURE 66

If C is closed, then -y = ap + {C} - ap E r1 (D) and it follows from (5.3.9) that (5.3.15) i) -y(ap) = the terminal point of CQP.

5. ANALYTIC CONTINUATION

134

ii) {C} is homotopic to a point if and only if y(ap) = op. Since X is arcwise connected, the notion of curves, their homotopy, and simple connectedness is defined for X in the same way as in the case of domains of C. (5.3.16) LEMMA. Let it : X - D be as above. Let 4) : [0,1] x [0,1] - D be a

continuous mapping, and set P = 0(0,0). For an arbitrary op E X there exists a unique lifting 4: [0,1] x (0,11 - X with 4)(0, 0) = op. PROOF. Take a lifting 4(., 0) of 0) : [0,1] 3 t --+ 4(t, 0) E D with 4)(0,0) = op. By making use of (5.3.6) we see that there exist a a > 0 and a continuous mapping 4: [0, 1] x [0, or]

X such that 4)(0, 0) = op and tr o 4 = 0.

Let S be the set of all such o E [0,11. In the same way as in the proof of the monodromy Theorem (5.2.3) we infer that S is open and closed, so that S= [0, 1]. To show the uniqueness, we take a continuous mapping 'P : [0,1] x [0,11 - X such that W(0, 0) = op and it o is = 4). Set E = {(t, s) E [0,11 x [0,1J; 'Y(t, 8) _ +$(t, s)}.

Then E is clearly closed, and E 3 (0, 0). It follows from (5.3.6) that E is open in [0,1] x [0, 1]. Thus E = [0,11 x [0, 1], and 1i = 4. Let 4' be another lifting of 4) in Lemma (5.3.16). Then, by (5.3.12) there is a y E 7r, (D) such that (5.3.17)

6'=yo4.

(5.3.18) THEOREM. Let D and X be as above. i) X is simply connected. ii) Let Do C C be an arbitrary simply connected domain, and let f : Do -. D be a continuous mapping. Then, for an arbitrary of(.,) E X with

zo E Do there is a unique lifting f : Do - X such that f (zo) = of(...) and n o f = f. If j' is another lifting of f, then there is a y E Ai(D) such that j' = y o f . PROOF. i) Take a closed curve e in X with the initial point op E X. Then C = ir(C) is a curve in D, and C is a lifting of C with the initial point ap. It follows from (5.3.15) that C must be homotopic to a point. Let fi : [0,11 x [0,11 ---

D be a homotopy connecting C and P. By Lemma (5.3.16) there is a lifting 4) : [0, 1] x [0, 1] - X with 4)(0, 0) = op. It follows that $ is a homotopy connecting C and op. ii) Take an arbitrary point z E Do and a curve C from zo to z. Let C be

the lifting of f (C) with the initial point afand denote the terminal point by j(z). We see by Lemma (5.3.16) that %(z) is independent of the choice of C. For every point z' E Do we take a sufficiently small neighborhood W C Do so

5.3. UNIVERSAL COVERING AND RIEMANN SURFACE

135

that for neighborhoods Uf(z,) of f (z') and V(f (z')) of j (z') satisfying (5.3.6), f (W) C Uf(z,), and zEW. f(z)EV(f(z')), Since it o j (z) = f (z) is continuous and 7rlV (f (z')) is a homeomorphism, f is continuous. The latter half follows from (5.3.17). EXERCISE 4. Let ire : X3 -. D3, j = 1, 2, be universal covering mappings,

X2 be a lifting of and let f : D1 - D2 be a continuous mapping. Let f : X1 f o 7r1i and let f.: n1(Dl) -+ Tri (D2) be as in (5.3.2). Then, show that

Jo. =f.(7)°f,

7Enl(D1)

EXERCISE 5. Set Dj = C', j = 1, 2, and A : D1 3 z -- z" E D2. 1 + t E D2. i) Obtain all liftings of the curve C, given by 01 [0,1] 3 t e2'"'t E D2. ii) Obtain all liftings of the curve C2 given by 02 : [0, 1] 3 t EXERCISE 6. The mapping it : C (= D1) 3 z -+ e2' E C' (= D2) is a universal covering mapping. In this case, answer the questions similar to the above i) and ii).

Let X be as above. Then every point ap E X has a neighborhood V(ap) which is homeomorphic to a disk neighborhood Up (c D) of P. We call V(ap) a disk neighborhood of ap. There are a complex coordinate z in Up for P 34 oo, and z in U,,,. Through the restriction irIV(ap), the points of V(ap) are identified with the points of Up, so that we introduce a complex coordinate zv(Qp) in V(cep) by making use of z or z. The mapping (5.3.19)

zv(ap) : V(ap)

zv(ap)(V((P))(C C)

is homeomorphic, and if V (ap) n V (,OQ) 0 0, (5.3.20)

ozvI P) : zv(ap)(V(ap) nV(AQ)) -' zv(RQ)(V(aP) nV(fQ)) is a homeomorphism, and a holomorphic function as well. We call zv(ap) a ZV(RQ)

holomorphic local coordinate in V(ap) of X. In general, if every point of a connected Hausdorff topological space has a neighborhood endowed with a complex function satisfying (5.3.19) and (5.3.20), then the space is called a Riemann surface, and the complex function is called a holomorphic local coordinate. The Riemann sphere C is a Riemann surface with holomorphic local coordinates z and z, which is compact and simply connected. The Riemann surfaces of a multi-valued meromorphic function and of the inverse function dealt with in the last section are Riemann surfaces in this sense. The above X gives rise to a simply connected Riemann surface.

EXAMPLE. The mapping it : C (= D1) 3 z - e2x`z E C- (= D2) is a universal covering mapping, and C is the Riemann surface of the inverse function

136

5. ANALYTIC CONTINUATION

(I/2wri) log w of the holomorphic function w = e2*'z. The fundamental group of

C' is Z, and acts on C by

C3z-.z+nEC,

nEZ.

REMARK. We have a "global' holomorphic Coordinate z in C or in its domain D. For multi-valued meromorphic functions on D or for the universal covering of D, the coordinate z is no longer "global". Here it becomes necessary to introduce the notion of Riemann surfaces. For Riemann surfaces the notion of holomorphic functions and holomorphic mappings is defined in what follows, and all theory which has been described can be developed with suitable generalizations. It can

be shown that a simply connected Riemann surface must be one of C, C, or A(1). This fact is known as the unifortnization of Riemann surfaces, and the proof requires considerable preparations of function theory on Riemann surfaces. We will prove this fact for domains of C in Chapter 6.

In general, let X1 and X2 be Riemann surfaces. A continuous mapping f : X2 is defined to be holomorphic if for every point P E X1 the function w o f (z) is holomorphic in z, where z (resp., w) is a holomorphic local coordinate in a disk neighborhood V (reap., W) of P (resp., f (P)) with f (V) C W. By (5.3.20) this definition is independent of the choice of holomorphic local coordinates z and w. Setting X2 = C, we call a holomorphic mapping f : X1 C a holomorphic function on X1. The mapping f : D C (cf. (4.2.1)) defined by a meromorphic function on D is a holomorphic mapping. Conversely, a holomorphic mapping f : D C with f ; oo is defined by a meromorphic function. Hence, a meromorphic function on D is a holomorphic mapping from D into C which is not constantly oo. X1

EXERCISE 7. Let X be a Riemann surface. Show that the followings are

defined independently by the choice of holomorphic local coordinates.

i) A function u : X -+ R (or C) is defined to be of class Ck (k 5 oo) if in a disk neighborhood V of an arbitrary point P E X with a holomorphic local coordinate zv(Q) = z = x + iy (Q E V), u o zV_ I is of class Ck as a function in (x, y). [To, TI] -, X is said to be of class Ck if ii) A continuous mapping for every point to E [To, Tl ] the function zv o O(t) is of class Ck in a :

neighborhood of to, where zv is a holomorphic local coordinate in a disk neighborhood V of 0(to). The notion of piecewise class Ck is defined in the same way as above. Let 7r : X -+ D be the universal covering defined in (5.3.5), and consider X and D as Riemann surfaces. Of course, the mapping n is holomorphic. Let Do C C be a simply connected domain, and let f : Do - D be a holomorphic

mapping. The following is clear by definition:

5.3. UNIVERSAL COVERING AND RIEMANN SURFACE

137

(5.3.21) A lifting f : Do -. X of f is holomorphic. For a holomorphic mapping between Riemann surfaces, the Prinzip von der Gebietstreue holds as in Corollary (4.3.14). As in Theorem (4.3.18), the inverse mapping of a univalent (injective) holomorphic mapping f : Xl - X2 between Riemann surfaces X1 and X2 is called a biholomorphic mapping. In this case, by Theorem (4.3.18) the inverse f -1 is holomorphic, too, and X 1 and X2 are said to be mutually biholomorphic. In the special case of X1 = X2, a biholomorphic mapping f : X1 - X1 is called a holomorphic transformation, and the set of all of them is denoted by Aut(Xi ), which forms a group under the law of composition. We call Aut(X1) the holomorphic transformation group of X1. Assume that for every point P E X2 a disk neighborhood Wp with a holomorphic local coordinate wp and a function hp(wp) in wp are assigned. We call the family h = {hp(wp)}pErz an hermitian (reap., pseudo-) metric or a conformal (reap., pseudo-) metric if the following conditions are satisfied:

(5.3.22) i) hp(wp) > 0 (reap., > 0), and hp(wp) is of class C°°. ii) If Wp n Wq 0, -2

hp(wp(a))

'W (wp(a))

= hg(wQ(a)),

a E Wp n WQ.

Because of the above equation we write h = hpdwp dwp = hp(dwp(2. We consider dwp dip = idwp 12 to be transformed as dwp dwQ

2

dwQ dwQ.

X2, the pull-back f'h of h by f is

For a holomorphic mapping f : X1

defined as follows. Take a disk neighborhood V (reap., W) of X1 (reap., X2) with holomorphic local coordinate z (reap., w) so that f (V) C W, and h = h(w)(dw(2

in W. Set (5.3.23)

f'h = h o f

dw o f dz

(dz(2

It is clear that f'h is independent of the choice of holomorphic local coordinates. If dw o f/dz 54 0, then f'h defines an hermitian metric there. If f is biholomorphic, then f'h is an hermitian metric. Let 0 : (To, T1 ] -, X2 be a piecewise continuously differentiable curve. For a t E (To, T1] at which 0 is differentiable, take a disk neighborhood W containing 0(t) and a holomorphic local coordinate to in W as in (5.3.23). Then the function dw

h(O(t)) I

t(t) I

d

(>

0)

5. ANALYTIC CONTINUATION

138

is independent of the choice of w by (5.3.22), ii). We define the length Lh(C) of the curve C = C(0) with respect to h by T

(5.3.24)

Lh(C) =

h(O(t))

! To

Idw

d0(t) I

dt.

For a subset E of X2 we define the area Ah(E) with respect to h by (5.3.25)

Ah(E) = J h(w)dudv,

w = u + iv.

Here it is easily checked in the same way as in the case of length that the right side of the above equation is locally well-defined independently from holomorphic

local coordinates and hence globally, provided that the integration exists. Idzj2. On a\ 101 EXERCISE 8. i) On C\ {oo} = C, we set h = 4(l +1212)-2 C, we set h = 4(1 + IzI2)-2Jdz"I2. Show that these define an hermitian metric on the Riemann sphere C, which is called the Fe+bini-Study metric. ii) Obtain the length of the real axis R with respect to the above Fubini-Study metric. iii) Obtain the area of C with respect to the Flibini-Study metric. EXERCISE 9. Assume that the composition 9o f of two holomorphic mappings f and g is defined. Let h be an hermitian metric defined on a domain containing the image of g. Show that (g o f'(g'h).

Problems 1. Construct the Riemann surface of the multi-valued function z(z -1). What are the branch points and the branch orders? 2. Construct. the Riemann surface of the inverse function of f (z) = z2 - 2z (z E C). What are the branch points and the branch orders? 3. Let X be a Riemann surface and let u be a real valued function of class C2. We define u to be harmonic if for an arbitrary point P E X and for a disk neighborhood V of P with holomorphic local coordinate zy = z = x + iy, the function u o zy 1(x, y) is harmonic in (x, y). Show that this definition is independent of the choice of a holomorphic local coordinate.

4. Let D C C be a simply connected Riemann surface and let f be a nonvanishing holomorphic function in D. Show that there is a holomorphic function g in D such that f (z) = es(z).. 5. Let f be a holomorphic function on a Riemann surface X. Show that if III takes a maximum at a point of X, then f is constant. 6. Let u be a harmonic function on a Riemann surface X. Show that if u takes the maximum or the minimum at a point of X, then u is constant. 7. On A(R)(0 < R < oo) we define an hermitian metric 2 4RzI2)2 Jdz12,

h= (R2

139

PROBLEMS

which is called the Poincar# metric. Obtain the length of the curve Ct,,el(0 :

[0,r] 3 t te'B E .(R)) (0 < r < R), and the area of the disk A(r) with respect to h. 8. Let h = 2a(z)ldzI2 (local expression) be an hermitian metric on a Riemann surface X. Show that the function Kh(z) _

44 loga(z) _

.9. 5. log a(z)

is defined independently of the choice of the local holomorphic coordinate z. We call Kh the Gaussian curvature of h. 9. i) Let hi be the Fabini-Study metric on C (see §3, Exercise 8). Show that

Kh, =1. ii) Show that for the hermitian metric h2 = jdzj2 (the Euclidean metric) on

C, Kh, = 0. iii) Show that for the Poincari metric h3 (cf. problem 6) on 0(R) with

0 0 so that C A(0;R). Let 0C

9(z) = Ea. Zn n=O

be the Taylor expansion of g. The Laurent expansion of h(z) = goi = g(1/z), z E c.(1/R) \ {0}, is given by 0

00

h(i) _ E anz-n = E a_nzn.

-a

n=0

Note that h restricted to A(1/R) \ {0} does not take any value in i(b;1). By Casorati-Weierstrass' Theorem (5.1.2) i = 0 is not an isolated essential singularity of h; that is, z = 0 is at most a pole of h. Therefore, there is a number no such that ano 0 0 and an = 0 for all n > no, and so g(z) is a polynomial no

0"00.

g(z) = b + E anzn, n=1

Since g is biholomorphic, g(z) has no zero. The fundamental theorem of algebra,

Theorem (3.5.23), implies that no = 1, and hence g(z) = a1z+b,g'(z) = a1 96 0.

The next result is called Schwarz' lemma:

(6.1.2) LEMMA. Let f (z) be a holomorphic function on e(1) such that f (0) = 0

and If(z)I < 1. Then

lf(z)I s Iz

If'(0)l 0. Let C((: (0,1] -y A(1)) be a piecewise continuously differentiable curve with initial point 0 and terminal point 4. Set

fi(t) _ 01 (t) + i#s(t),

3(t) _ Since

(t),

j = 1,2.

1 - Io1(t)12,

1

6 2 12

(1 - J0(t)I2)2 = (I - 1(t)2)2 We have a curve C1 = C1(01) given by 431. Then the initial (resp., terminal) point of C1 is 0 (resp., 4), and by (6.2.12) Lo(C1)-

Ln(1)(C)

0. Furthermore, we have

The equality holds if and only if 432(t) L o 1) (C ) = 1

f

21d11(t)J

1

o

(

1

- 01(t)2

dt

> f1

f

o

f

=

1

2

01(t)2

\\1+x 1+ 4 log

+

1 - x2

\ 1 -x111

dz

dx = [l og(l+ x )-log (1- x ) 1o

1-4. It is to be noted that the equality in the above inequality holds if and only if o 1(t) > 0; for such an example, we may take 431(t) = t4. Thus we see that L4%(1)(C) >- log{(1 + 4)/(1 - 4)}, and the equality holds if and only if C is a line segment C(o,,zz) connecting 0 and 4. It follows that (6.2.13)

do(1)(z1, z2) = do(1)(01 zi) = log

l-

_1112+1

Rom this, (6.2.10), iii) follows. In general, a curve C connecting two points 21i z2 E 0(1) is called a hyperbolic geodesic if

do(1)(z1,z2) = LA(1)(C)

The above C(o,Za) is the unique geodesic connecting 0 and 4, and C(,,,,,,) (C(asz)) is the unique geodesic connecting z1 and z2. The mapping f is analytically continued to a neighborhood of 0(1). By the conformality of f (Theorem (3.1.15)) C(s1,s2) is a part of the circle (including the case of a line), passing through z1 and z2 and orthogonally crossing the boundary circle C(0;1) of A(1). We denote by L(z1iz2) the part of that circle contained in A(1). For

6.2. POINCARE METRIC

147

any two points w1, w2 E L(z1, z2) the segment of L(zl, z2) connecting w1 and w2 is the geodesic C(2 ,,,,W2) connecting wl and w2. We also call L(zl, z2) a

hyperbolic geodesic. Take the third point z3 E 0(1) outside L(zl,z2). Then there are infinitely many hyperbolic geodesics passing through z3 and having no

intersection with L(zl, z2) (see Figure 67). To confirm this, it is easier to set z3 = 0 by a linear transformation of Aut(A(1)). If we associate a hyperbolic geodesic to a line in Euclidean geometry, the above property contradicts the axiom of parallels. The geometry in which hyperbolic geodesics in A(l) are considered as lines is called non-Euclidean or hyperbolic.

FIGURE 67

The metric do(1) defines the so-called metric-topology on A(1). That is, a sequence {z} n° o of points zn E A (l) is said to converge to a E A(1) if

slim do(I)(zn,a) = 0. We say that do(1) is complete if any Cauchy sequence {zn}n o with respect to d,%(1) (i.e., for an arbitrary e > 0 there is a number no such that dp(1)(zn, z,n) < e for all n,m > no) converges to a point of A(1). In fact we have the following theorem. (6.2.14) THEOREM. i) A sequence {zn}no of points zn of i(1) converges to a point a E 0(1) with respect to do(1) if and only if it converges to a as a sequence of points of C. ii) dj(1) is a complete metric. PROOF. i) By (6.2.13) d j(1) (z, w) is continuous in the two variables z and w. Therefore llm Izn - al = 0 lira da(1)(z,,,a) = 0.

n-x

nac

Conversely, we assume limn-ao d&(1) (zn, a) = 0. By the transformation 0a E Aut(A(1)) in (6.2.13) we may assume a = 0 without loss of generality. It follows from (6.2.13) that (6.2.15)

1+

dA(1)(zn,0) = log 1 - Iznl

-' 0

(n -. oo).

6. HOLOMORPHIC MAPPINGS

148

Hence, limn Iznl = 0. ii) Let {zn}°°_o be a Cauchy sequence in A(1) with respect to dA(1); that is, for an arbitrary e > 0 there is a number no such that (6.2.16)

da(1)(zn, zm) < E

for all n, m. > no. In particular, da(1)(zn,zn(,) < e, n > no, and so

da(1)(0,zn) 0 there is a number vo such that da(1)(zn,,,a) < E,

(6.2.17)

v ? ve.

We may assume that n,,,, >_ no. Thus, by (6.2.16) and (6.2.17)

do(1)(zn,a) dD, (f (a), f (b)) for all a, b E D1.

This property of the hyperbolic metrics is called the contraction principle. For the proof we first deal with the case where D1 = D2 = 0(1). The next is called Schwarz-Pick's lemma:

(6.3.11) LEMMA. For an arbitrary holomorphic mapping f : 0(1) - A(1) we have

i) f'g1 5 g1, and if the equality holds at a point, then f E Aut(0(1)); ii) d4,(1)(a,b) ? da(1)(f(a), f(b)) for all a,b E 0(1).

6. HOLOMORPHIC MAPPINGS

152

PROOF. i) We want to show that (6.3.12)

f'91(a)

91(a)

at an arbitrary point a E A(1). By making use of

z+a ox+1 we get 0' ag1(0) = g1 (a) from Theorem (6.2.3). Thus (6.3.12) is equivalent to

f'0-a91(0) < 01a91(0)i

that is, 91(0) > 0-a °f'°0' a91(0) = (0-aofo0a)'g1(0). Therefore it suffices to prove (6.3.12) at a = 0 for an arbitrary holomorphic mapping f : A(1) - A(1): (6.3.13)

f*91(0) < 91(0)

Since f'91(0),

(xf(0))'91(0) =

we may assume that f (0) = 0. By the definition of g1i (6.3.13) with f (0) = 0 is equivalent to (f'(0)I < 1. This was proved by Lemma (6.1.2).

Suppose that there is a point a E 0(1) such that f'g1(a) = g1(a). By the above arguments, we may assume that a = 0 = f (0). Then it follows from Lemma (6.2.1) that f E Aut(A(1)). ii) Take a piecewise continuously differentiable curve C from a to b in 0(1). It follows from i) that Lo(1)(C) ? Lo(1)(f(C)) > do(1)(f(a),f(b))

Thus do(1)(a,b) > do(1)(f(a), f(b)). 0 PROOF OF THEOREM (6.3.10). i) Let irk : A(1) - D,, j = 1, 2, be the universal coverings. Let F : A(1) --+ A (l) be a lifting of f o ir1 : A(1) -s D2:

i(1)

P

1 ir1

D1

A(1) 1 W2

-L

D2

It follows from Lemma (6.3.11) that F'g1 < g1. Hence, the definition (6.3.1) implies that f' hp, < hD,. ii) The proof is similar to that of Lemma (6.3.11), ii). 0 The notion of hyperbolic distance was generalized for general complex manifolds by S. Kobayashi (181, and it has played an important role in complex analysis.

EXERCISE 1. Let D be a hyperbolic domain, and let hD be the hyperbolic metric. Show that f'hD = hD for f E Aut(D).

6.4. THE RIEMANN MAPPING THEOREM

EXERCISE 2.

153

Let D be as above, and let r : A(1) -p D be the universal

covering. Show that for two points wl, w2 E D dD(wl, w2) = min {do(1) (zl, z2); zz E 0(1), r(xi) = w3, 9 = 1, 21.

6.4. The Riemann Mapping Theorem Let D C C be a domain. Let { f } - 0 be a sequence of holomorphic functions on D which converges uniformly on compact subsets to f . Then it follows from Theorem (3.5.19) that f is holomorphic, and that

(6.4.1)

d n

d

dz

dz

4`foi

df of

di

di

(n-oo) onDflC, (n

oo)

on D n C \ {0},

where the convergence is uniform on compact subsets. Now let F be a family of holomorphic functions on D. We say that F is a normal family if any sequence of holomorphic functions in F contains a subsequence which converges uniformly on compact subsets. For normal families the following Montel's theorem is fundamental.

(6.4.2) THEOREM. If a family F of holomorphic functions on D is uniformly bounded, then Y is normal.

PROOF. There is an M > 0 such that for every f E .F If(a)I 0. PROOF. As in (2.8.10), we set

Z+a

f(z) = 0-a(z) = dz+ 1 Then f E Aut(A(1)), and f (0) = a. A simple computation yields

f'(0)=1-1a12>0. Thus we have shown the existence of such f E Aut(.(1)). Let g E Aut(A(l)) satisfy the conditions. Then,

g-1

o f E Aut(©(1)) satisfies

9-1 o f (0) = 0,

(g-1 o f)'(0) > 0.

It follows from Theorems (6.1.4) and (2.8.11) that g-' o f(z) = e'Bz with 0 E R.

Since (g-' o f)'(0) = e'e > 0, e'9 = 1, and henceg-' o f(z) = z; i.e., g = f. 0 Our next theorem is called the Riemann mapping theorem.

(6.4.4) THEOREM. Let D C C be a simply connected domain such that the boundary 3D contains at least two distinct points. Then, for an arbitrarily given a E D there exists a unique bih olomorphic mapping f : D -+ 0(1) such that f (a) = o, CV&

f'(a)>0,

a

z (0) > 0,

a = oo.

oo,

PROOF. We are going to reduce D to a bounded domain of C. Take distinct points bl, bZ E 3D. We may assume without loss of generality that b1 E C. We take a linear transformation

z - bl Here, if b2 = oc, we set z/b2 = 0. Then D is biholomorphic to 01(D) C C \ {0}. We may assume that D C C \ {0}. Since D is simply connected, one may take by Theorem (5.2.5), ii) a branch 02(z) of f on D. As shown in Example (5.2.7), 02 : D 02(D) is biholomorphic, and 02(D) n (-$2(D)) = 0.

6.4. THE RIEMANN MAPPING THEOREM

155

Thus, ¢z(D) C C has an exterior point b3 E C. Set

&(z) =

1

b3

Then, 03o02 : D 03o02 (D) C C is biholomorphic, and 03o02(D) is bounded. Therefore one may assume that D is a bounded domain of C. Let F be the family of all univalent holomorphic functions f on D satisfying

f (a) = 0,

f(a) =1.

Since (z - a) E F, F # 0. Set IIf1100 =sup{If(z)1;zED},

f EF,

p=inf{IlfII ;f E F}.

Since (z-a)EF, p no, 11 fn II00 < p + E. Letting n -' oo, we have Ilfo II < p, and hence II fo ll = p. It suffices to show that

fo(D) = o(p).

(6.4.5)

By the maximum principle (Theorem (3.5.21)) it is clear that fo(D) C A(p). Assume that fo(D) 96 0(p). Taking c E 0(p) n 8(fo(D)), we set P(w - c) 04(w) - -ft + p2 The mapping 04 : E(p) - 0(1) is biholomorphic, and 04 (c) = 0. Taking a branch

O4 o fo(z) on D, we set

Os=P 04ofo(z)ED(p),

zED.

It follows that 0s is univalent, and 05(a) = v1--p-c. Moreover, set

06 =

PZ(05(z) - 05(a)) E A(P), -05(a)i5(z) + p2

z E D.

156

6. HOLOMORPHIC MAPPINGS

The mapping 4s is univalent, too, and 06(a) = 0,

110600

P-

A simple computation yields

06(a) = 2 Since 0 < jcl < p, I¢6(a)I > 1. Setting

07(z) = -e(a)46(z),

zED,

we have 07 E F, and 11071130 1

a)I < A

This contradicts the choice of p. Thus, (6.4.5) is proved. To show the uniqueness, we take a biholomorphic mapping g : D

0(1) such

that g(a) = 0 and g'(0) > 0. Then fo9' E Aut(A(1)), fog-1(0) = 0, and (fog-1)'(0) > 0. Lemma (6.4.3) implies that fog-1(z) = z, and so f = g. 0 (6.4.6) THEOREM. An arbitrary simply connected domain D of C is biholomor-

phic to C itself, C, or A(1); furthermore, these three domains are not biholomorphic to each other.

PROOF. If 8D = 0, D = C. If 8D consists of only one point a E C, we may assume a = oc by a linear transformation; hence, D = C. If 8D contains more than one point, it follows from Theorem (6.4.4) that D is biholomorphic to A(1). The latter assertion easily follows from Theorems (3.5.22) and (3.7.8). 0 It is known that the above Theorem (6.4.6) holds in fact for a general simply connected Riemann surface. It is called the xniformization theorem, and was proved by Koebe (1907), and independently by Poincares (1908). The Riemann mapping theorem was first claimed in his doctoral dissertation (1851), and a compact simply connected Riemann surface with boundary was dealt with. Riemann's proof depended on the existence of a solution of a variational problem which had not been established, and was criticized by Weierstrass. Riemann did not answer this criticism, but Hilbert did later (1904). Theorem (6.4.4) was first completely proved in its current form by Osgood (1900). An arbitrary domain or a Riemann surface, even if it is not simply connected, carries the universal covering which is simply connected. It is biholomorphic to C, C or 0(1) by the uniformization theorem. Thus the domain or the Riemann surface is C itself, or is represented as a quotient space of C or z(1) by the deck transformation group (cf. (5.3.11)).

EXERCISE 2. Let f : D - A(1) be a biholomorphic mapping with f(zo) = 0 (zo E D). Let g be a holomorphic function on D such that Ig(z)I < I on D, and g(zo) = 0. Show that I9'(z)1 < 1f'(zo)1.

6.5. BOUNDARY CORRESPONDENCE

157

EXERCISE 3. Show in Theorem (6.4.4) that if a E R and D is symmetric with respect to the real axis, then f (z) = f (z). EXERCISE 4. Obtain a biholomorphic mapping from D = {z = x + iy; y2 > 4c2(x + c2)} (c > 0) to the upper half plane H.

6.5. Boundary Correspondence The boundary of the upper half plane H in a consists of the real axis R and the infinity oo. The two ends, -oo and +oo, of R coincide with oo.

FIGURE 70

We endow 8H with the orientation from -oo to +oo via 0. We take an ordered triple (PI, P2, P3) of distinct points of OR Suppose that it is one of the following:

(6.5.1) i) P3 = oo, P1 < P2;

ii) P2 = co, P3 < P1; iii) P1 = 00, P2 < P3; iv) oo # P1 < P2 < P3 0 00In this case we say that the triple (P1, P2, P3) has the positive orientation. By Theorem (6.1.4), ii) and Theorem (2.8.13) an arbitrary f E Aut(H) is represented by (6.5.2)

f (z) = cz + d'

(c d)

E

SL(2, R).

The restriction

fIOH:OH -iOH is injective and homeomorphic in the sense of the topology induced from C. Since P(Z) -

ad - be (cz + d)2

f'(z) > 0,

1

=

(cz + 1)2

z E R,

it follows that f IR is monotone increasing. Therefore, a triple (P1, P2, P3) of positive orientation is mapped to a triple (f (PI), f (P2), f (P3)) of positive orientation. Hence we say that f preserves the orientation of M.

6. HOLOMORPHIC MAPPINGS

158

Now, cases ii)-iv) of (6.5.1) are reduced to case i) by a linear transformation

f (z) = x - P3 E Aut(H). Moreover, case i) is reduced to (PI1P2, P3) = (0, 1, oo)

(6.5.3)

by a linear transformation

=

f(z)

1

P2 - Pt

(z - P1) E Aut(H).

That is, every triple (P1, P2, P3) of positive orientation is mapped to (0, 1, oc) by the boundary correspondence of some f E Aut(H). Conversely, assume that f E Aut(H) fixes the triple (0, 1, oo):

f(0)=0, f(1)=1, f(oo)=00. As in (6.5.2), write f (z)

= cz + d'

(c

d)EsL(2R).

Since f (oo) = oc, c = 0; f (0) = 0 implies b = 0. Since f (1) = 1. a/d = 1. On the other hand, ad = 1. Thus, a = d = ±1, and so f (z) = z. That is, if f E Aut(H) fixes the triple (0, 1, oo) of positive orientation, then f must be the identity. Summarizing the above, we see that for two arbitrary triples (P1, P2, P3) and (P'1, P2, P3) of positive orientation, there exists a unique f E Aut(H) such that

f(P1)=1'j, f(P2)=P'2,

f(P3)=P.3-

The disk A(1) is biholomorphic to H by (z)

z + i' ty : H - A(1),

0(i) = 0.

By ip the boundary 8H = RU {oo} corresponds to 8A(1) = C(0;1) as follows: ij(oc) = 1,

P(R) = C(0;1) \ {1}.

6.5. BOUNDARY CORRESPONDENCE

159

7

1=v(1) W(P

v(R) 1=v(00)=u'(I',)

FIGURE 71

A triple (P,, P2, P3) of positive orientation with P; E 8H,1 S j < 3, is mapped by ?+/' to a triple (+J'(P, ), z/i(P2), 7/'(P3)) of points of C(0; 1) in anti-

clockwise order. Let (Q,, Q2, Q3) be a triple of points of C(0;1). We say that the triple (Q,, Q2, Q3) of C(0;1) is of positive orientation if they are in anti-clockwise order. A triple (Q1, Q2, Q3) of positive orientation of C(0; 1) is mapped by z[i-1 to a triple (V)- I (Q I ),1' I (Q2), V) - I (Q3)) of positive orientation of 8H. Thus we have proved the following theorem.

(6.5.4) THEOREM. Let D be H or A(1). Then every f E Aut(D) preserves the orientation of OD, and for arbitrary triples (P, i P2, P3) and (Pi, P2, P'1) of positive orientation, there exists a unique f E Aut(D) such that

f(P,)=Pi, f(P2)=P2, f(P3)=P2. Lemma (6.4.3) gives the uniqueness of holomorphic transformations at an interior point, and Theorem (6.5.4) does it by the boundary correspondence. EXERCISE 1. Obtain f E Aut(H), mapping (0, 1, oo) to (-1, 0, 1). EXERCISE 2. Obtain a biholomorphic mapping f : H - A(1), mapping (0, 1,00) to (1,i, -1).

(6.5.5) LEMMA. Let E3 C C, j = 1, 2, be subsets, and let F : E1 , E2 be an injective continuous mapping. If E, is closed, then F is a homeomorphism. PROOF. It suffices to show the continuity of F-I : E2 E,. Let Q E E2 be an arbitrary point, and let be an arbitrary sequence converging to Q. 0 Set P = f `(Q), and P = f - I n = 1, 2,.... Note that El is compact. Let P' be an accumulation point of i°_0. It is sufficient to show that P' = P. Let {P,,,, }v o be a subsequence of {P,,}n o which converges to P. Since F is continuous,

F(P') = Iim F(P,,,,) = lim "-.00Q,,,, = Q = F(P). V-00 The injectivity of F implies P' = P.

6. HOLOMORPHIC MAPPINGS

160

Let D C C be a simply connected domain, and assume that D : C; that is, D has an exterior point. If D is not a bounded domain of C, we take a point c E C\D, and set f (z) = 1/(z-c). Then D is biholomorphic to D1 = f (D) C= C, and f is extended continuously to a homeomorphism between the boundaries:

f:D-,D1.

(6.5.6)

(6.5.7) THEOREM. Let D C C be a simply connected domain such that D # 1b. Assume that the boundary 8D is given by a closed Jordan curve C(O : [0, 1] - C)

(that is, cD = O((0,1])). Then every biholomorphic mapping f : 0(1) -, D is extended to a homeomorphism from E(1) to D.

PROOF. By (6.5.6) we may assume that D is a bounded domain of C. Take an arbitrary point (E C(0; 1). For p > 0 we set OP = a(1) n 0((; p), W, = f(1v), A(p) = the area of we,,

r,

rp = C(() A yP = f(r), L(p) = the length of yp.

f

FIGURE 72

It is sufficient to prove that w, accumulates to one point of C as p 0. Furthermore, it is sufficient to prove it for a sequence {pn},, 1 (p > 0), converging

to 0. Since 1imp-o A(p) = 0, for an arbitrary f > 0 there exists 5(E) > 0 such that A(p) < E,

0 < p < b(E).

By Schwarz' inequality we have

f

e

0L(p)dp = fin

I f'(( + te'B)jtdtdg 6

\ 1/2

(ff

f'(( + te')tdtd9 I

/

i6

< n1 we have (6.5.11)

C _ {a,,, },

&p. = yp U {a.,).

6. HOLOMORPHIC MAPPINGS

162

Therefore, the diameter of wpn is less than L(pn), and by (6.5.8) n .-- oo.

WP- -{ anl,

This means that for an arbitrary e > 0 there is a number no such that for every n Vi wp.. C A(an,;E). Assume that an bn for any n. We write 91 for C(0;1). There is an injective and surjective continuous mapping

0: S'

8D = C.

It follows from Lemma (6.5.5) that the inverse mapping 0-' : C -. S' is continuous. Hence, 0-' (Cn) = In is a connected closed subset of S1. The points an, bn E C divide C into two curves C;,, C,,. One of them is Cn. Let sn, to be the end points of In. Then it follows from (6.5.9) and (6.5.10) that

Isn - t,,-'0,

InDIn+iJ....

Therefore, f,°°_i In = {r} with r E S', and so

Cn-'c=O(r),

n -'oo.

Since limn- x a. = limn_.n, bn = c, and limn-.x L(-yo,) = 0, we have by the above and (6.5.10) that wp P. -' C,

n - oo.

Thus, f is extended to a continuous mapping f : Y(1) -- D, and f(A(1)) = D. In the above proof, the case of (6.5.11) does not occur. For, if an, = bn,, f takes the value an, constantly on the arc S' f1 A((; pn, ). It follows from Chapter 3, §5, Exercise 5 that f (z) = an z E z(1). This is a contradiction. Therefore we see that f is injective on the boundary, and so by Lemma (6.5.5) f : A(1) -+ D

is a homeomorphism. 0 Let D be as in the above theorem, and let fo : 0(1) mapping. Extend it to a homeomorphism fo A(1)

D be a biholomorphic

D. We say that a

triple (Qi, Q2, Q3) of points of 8D is of positive orientation if (f0 1(Q1), fo 1(Q2), fo 1(Q3)) is of positive orientation. This is a property independent of the choice of fo (Theorem (6.5.4)). We immediately have the following by Theorems (6.4.4), (6.5.4), and (6.5.7)

(6.5.12) THEOREM. Let D C C be a simply connected domain such that D 54

C and the boundary 8D is given by a Jordan closed curve. Let (P1, P2, P3) be a triple of positive orientation of 8A(1), and let (Q1, Q2, Q3) be a triple of positive orientation of 8D. Then there ezists a unique biholomorphic mapping

f :0(1)-yD such that f(Pi)=Q;,1 0),

0' = A'(1).

The universal covering of 0' is given by w:H3z-4e2"'sE0'.

(6.8.1)

FIGURE 79

A fundamental domain is

{z=x+iy;0 0). Suppose that there are three distinct points a, b, c E C such that f ({a, b, c}) is a finite set. Then there is a 0 < Ro < R such that

f'i({a,b,c}) n0'(Ro) = 0. We show that in this case f will be meromorphically extended to 0' (R0). By the change of variable z/Ro, f may be assumed to be defined on 0', and f (0')n {a, b, c} = 0 as well. We regard f as a holomorphic mapping (6.8.6)

f :0' -aC\{0,1}.

172

6. HOLOMORPHIC MAPPINGS

By Theorem (6.6.7) C \ {0, 1} carries the complete hyperbolic metric hc\(o,1} Take a sequence {zn}°O=1 of 0' so that (6.8.7)

lim zn = 0,

Iznl > Izn+1I,

n-.oo

an = d (zn),

n-.oo

lim an = a E C.

Take a disk neighborhood U(a) of a. If for a positive number r < 1 (6.8.8)

f(A*(r)) C U(a),

by Riemann's extension Theorem (5.1.1) f extends to a holomorphic mapping from 0(r) into C. Assume that (6.8.8) does not hold. Then, there is a sequence {wn}10°_1 of 0' such that

lim wn noo

= 0,

On = f(wn) E C \ U(a).

Since a \ U(a) is compact, by taking a subsequence one may assume that

Q = lim f (wn) E C \ U(a). Moreover, taking subsequences of {zn} and {wn} and renumbering them, one may assume that Iznl>Iwnl>Izn+1I,

n=1,2,....

FIGURE 80 Set

Rn={zEC,IwnI-no.

Take a curve Cn in An from zn to wn. Then f (Cn) intersects C(a; ro/2). By the Prinzip von der Gebietstreue, Corollary (4.3.14), f (Rn) is a domain, and its boundary is a subset of f (1'n) U f (An). Thus, it follows from (6.8.11) that (6.8.12)

f (Rn) D E.

6. HOLOMORPHIC MAPPINGS

174

We consider the area of E with respect to hC\{o,l} It is clear that Ac\{o.l) (E) > 0.

(6.8.13)

On the other hand, by the contraction principle, Theorem (6.3.10), and (6.8.4)

Acs(o.l)(E) < f f`hcN{o.l}

R

< AA. (A*(IzmD)) =

in.

hA .

I log IznI I

.0

(n - co).

This contradicts (6.8.13).

If a meromorphic function f on C has an isolated essential singularity at oo, f is called a transcendental meromorphic function; that is, if f is not transcendental, f is rational. The following is a direct consequence of Theorem (6.8.5). (6.8.14) COROLLARY. A transcendental meromorphic function on C takes infinitely many times all values of C except for at most two values.

For example, f (z) = e' misses 0 and oo, and in fact takes all other values infinitely many times. Hence the above "two" cannot be made smaller. The proof of Theorem (6.8.5) essentially depended on the fact that the universal covering of C \ {0, 1, oo} is the upper half plane. This fact was proved in §6 by making use of the Riemann mapping Theorem (6.6.7). Hence it may be said that the Riemann mapping Theorem (6.6.7) played a key role in the proof of the big Picard Theorem (6.8.14). The Riemann mapping theorem is a deep result, but is a speciality of the one variable case. It is, however, sufficient for the proof to construct a complete hermitian metric h on C \ {0, 1, oo} for which the contraction principle holds. There is a quantity K(h), the so-called Gaussian curvature of h. and the contraction principle for h holds if there is a constant Co > 0 such that

K(h) < -Co. From this viewpoint, the big Picard theorem can be generalized to the higher dimensional case. For this, cf. [181 and X91.

Problems (2,-2), and dotty (2, 2. Show in Theorem (6.3.10) that if f' h p, (a) = hp, (a) holds at a point a E D1, then the equality holds at all points of D1. 1. Calculate

z).

3. Let D be a hyperbolic domain. Show that for arbitrary points z; E D, i = 1.2, there exists a continuously differentiable curve C from zl to z2 such that dD(zl, 22) = LD(C).

PROBLEMS

175

4. Let f be an entire function, and let N be a natural number. Prove that if the cardinality of f -i (w) is not more than N for all w E C, then f is a polynomial of degree at most n.

5. Obtain a biholomorphic mapping from D = {z E A(1);Imz > 0} to the upper half plane H. 6. Obtain a biholomorphic mapping from D = {z = re'5; 0 < r < 1, 0 < 0 < a}

(0 < a < 2ir) to 0(1). 7. Show that by w = z+1/z, 0(1) is univalently mapped onto D = C\ [-2,21. Where is C \ 0(1) mapped to? 8. Show that. by w = z(1 + ez)2 with e = -e-'°, a E R (this is called Koebe's function), 0(1) is univalently mapped onto (w E C; w 54 re, r > 1/4). 9. For a meromorphic function f on a domain D C C, set f.,,(z) 3 z) s

f

This is called the Schwarzian derivative of f. Show the following.

(a) if; z} = 0 if and only if f (z) = (az + b)/(cz + d). (b) For (c e) E SL(2,C), { -+b ;z} = {f;z}. (c) Set w = f (z), and assume that f'(z) iA 0. Then, {w; z}

(dw\J {z; w}. dz

(d) Set w = f (z), and let g(w) be a meromorphic function in w. Then, {go

f;z}={g;w}

(d)2 +{w;z}.

10. (Vitali) Let E be a subset of a domain D which has an accumulation point in D. Let (f,,) be a sequence of holomorphic functions which is uniformly bounded. Prove that if {fn } converges on E, then { f } converges uniformly on compact subsets in D. 11. Let f f } be a sequence of holomorphic functions on a domain D that converges at a point zo E D. Prove that if {Refs(z)} converges uniformly on compact subsets in D, then so does {f,) itself.

12. Let D be a hyperbolic domain, and let f : D - D be a holomorphic mapping. Show that if there is a point zo E D such that f(zo) = zo and f'hD(za) = hD(zo), then f E Aut(D). 13. Let D i = 1, 2, be rectangle domains of C, and let f : Dt D2 be a biholomorphic mapping. Show that f extends continuously up to the boundary of DI. Moreover, show that if each vertex of D1 is mapped by f to a vertex of D2, then D1 is similar to D2. 14. Let D be the interior of an n-gon in C, and let f : H

D be a biholomorphic

mapping. Suppose that for -oo < at < .. < a < +oo, b; = f (a,) are

6. HOLOMORPHIC MAPPINGS

176

vertices of D. Let ajx (0 < aj < 2) be the inner angles of the vertices b,. Prove that f can be expressed as follows: (z-aj)a1-1dz+C2

f(Z) = C1 J 11

(an

+oo),

j=1 Z n-1

1 (z) = C1

JJ (z - aj)°'-1dz + C2 J j=1

(an = +oo),

where C1 (# 0) and C2 are constants. (The above formula is called SchwarzChristoffel's formula) Hint 1. Suppose that an 0 +oo. Then, by Schwarz' reflection principle the function f (z) is analytically continued to a multi-valued holomorphic function on C \ {a1};=1. Hint 2. By making use of a branch oft = (f (z) - bj )1 /01j, one sees that t is analytically continued to the lower half plane in a neighborhood of aj, and has a zero of order 1. Thus, one gets an expansion f(z) - bj = co(z - aj)°3(1 + c1(z - aj) + )°3,

and sees that the principal part of f"(z)/f'(z) is (aj - 1)/(z - aj). Hint 3. Note that the multi-valuedness of f (z) is caused by the even number of repetitions of reflections with respect to line segments. Hence, letting f (z) denote another local branch of f (z), one gets

f' (z) = Af (z) + B

(A (0- 0) and B are constants),

and f " (z) / f(z) is invariant, so one-valued, and at oo holomorphic with value 0. It follows that f" (z)

_-

f'(z)

aj

4z _1 - aj

.

Now, integrate this. Hint 4. In the case when an = +oo, take a linear transformation 0 E Aut(H) such that 0(an) # +oo. Transform the formula obtained above by ¢-1.

15. Replace H by 0(1) in problem 14. Then show that f(z) can be expressed as r2 n

fl(z - a1)°'-1dz + C2.

f(z) = C1

J j=1

Here, Ia.. I = 1 and bj = f (aj).

16. Set

f(z)=

j

1

v'z(l _-Z 2 )

dz.

Prove that f (z) maps H biholomorphically to the interior of a square in the first quadrant with one edge a real interval from 0 to t(1/4)2/1 2x.

PROBLEMS

177

17. Let P(X) be a polynomial of degree > 2 without double root, and let f (z) be a holomorphic function on A'(1) (resp., C). Show that if P(f (z)) has only finitely many zeros (reap., no zero), f (z) has at most a pole at 0 (resp., f is a constant). 18. Show that the number of exceptional values of a non-constant rational or meromorphic function on C' is at most 2. 19. Prove the big Picard Theorem (6.8.5) by making use of Montel's Theorem (6.7.6).

CHAPTER 7

Meromorphic Functions

In this chapter we investigate more deeply the properties of meromorphic functions. Beginning with an approximation theorem, we

prove two kinds of existence theorems, the introductory part of value distribution of meromorphic functions on C, the infinite product expression of meromorphic functions, and the basic part of elliptic functions. These lead to function theory of several complex variables, the value distribution theory of one and several variables, the theory of Riemann surfaces, and the theory of elliptic curves and automorphic forms.

7.1. Approximation Theorem Let U C C_ be an open set. In the present chapter we mainly deal with the case of U # C. By a linear transformation we may assume U C C without loss of generality.

(7.1.1) LEMMA. Assume that U C C is a bounded open subset. Let K C U be a compact subset, and let f be a holomorphic function on U. Then, for an arbitrary e > 0 there is a rational function F such that all poles are contained in 9U and If (z) - F(z)I < e,

z E K.

That is, f can be approximated uniformly on K by such F.

PROOF. Let n be a natural number, and let k and j be integers. Set (7.1.2) k)=z=x+iy,2n<x'2n1,2

En (7>

En =

I


_ no arbitrarily. If z is not a boundary point of En (j, k)'s, then there is a unique (jo,ko) with z E En(jo,ko). By Cauchy's integral formula, Theorem (3.5.16), and Cauchy's integral Theorem (3.4.14) we see that for E,, (.j, k) C U 1 f (z) d( 27ri JOE, (3,k) (- Z

_ f (z), (j, k) = (jo, ko),

otherwise. 10, Here, the orientation of 8En(j,k) is anti-clockwise. If En(j,k) and E,(j',k')

share an edge, then the sum of the curvilinear integrals over the edge with respect to 8En (j, k) and 8E,, (j', k') is zero. Therefore we obtain (7.1.4)

f(z) = 2'ri

I

S

(zz

Both sides of the above equation are continuous in z E K, and so it holds for all z E K. Since K is compact, d(K; 8U,,) = inf {d(z; 8Un); z E K}

= min{d(z; (9Un); z E K} > 0, d(K; 8U,,) > d(K; 8Un_ 1).

Set 60 = d(K; BU,,). Choose and fix n

no so that

(7.1.5)

10

1

2n

2

Let ry1,... , ryi denote the line segments of length 2-n that form the boundary 8Un. The rectangle with an edge -y which lies on the opposite side of En (j, k) C

7.1. APPROXIMATION THEOREM

181

U with the edge y intersects 8U. Let . be a point of the intersection. It follows from (7.1.5) that for z E K and ( E y

IS - I 2 < vF 20 < 1 Iz - .J.

(7.1.6)

Figure 83

For these z and C we have 1

(-z

(7.1.7)

_

1

1

z__

r

1

(-G

,

=o

By (7.1.6) this power series converges uniformly and absolutely in z E K and (E -y,,. Therefore, for an arbitrary E > 0 there is a number N. such that

+

I'll

1

z

j_o

(S - G)' (z - G)J+t

G E.

We set

F (z)

I 27ri

(z(

-)i+ f (()d(

1 F(z)

L F,.(z), =1

M = max{If(()I;( E 8U}.

7. MEROMORPHIC FUNCTIONS

182

Then, we have

If (z) - F(z)I y=1

27ri

7r

< 2"M

f-dC + F1,(z)I C3

(7.1.8) LEMMA. Let K C C be a compact subset, and let a, b E C \ K be two points in the same connected component of C \ K. Let F be a rational function which possibly has a pole only at a. Then F is uniformly approximated on K by national functions which possibly have a pole only at b.

PROOF. By the assumption we have that F(z) = Ek _,v Ck(Z - a)k with 0 S N < oo and ck E C. It is sufficient to prove the assertion for every 1/(z-a)k with k > 0. Take a curve from a to b which does not intersect K.

Figure 84

Set bo = min(Iz - wI; z E K, w E C) > 0, and take points of C:

a=ao,ai,...,a, =b,

jai+1-ail
0) is uniformly approximated on K by rational functions with a pole only at ai+1. It follows that for z E K

ai - a,+1

z-a,+1


0, the function

0(t) =

C21vIt (y2 + C, t2)2

takes the maximum value O(to) = 9c2/16

c,y2 at to = IyI//. Therefore,

we have m

E

c21bin

= n 1 (?l2 +clt2)2

_ 0 such that Ism(z)I < C3

lyl,

S.(z)I 5 c3/IyI, we get

Since lg(z)I = I (7.2.8)

Ih(z)I 5

I

z I+

lg(z)I < '+1C3

-

1Y

z=x+iy, 05x527r,

Iyj>41r.

It follows from (7.2.7) and (7.2.8) that F(z) is bounded on {z = x + iy; 0 5 x 5 21r}. By the periodicity, F(z) is a bounded holomorphic function on C, and so by Liouville's Theorem (3.5.22) F(z) is constant. It follows from (7.2.7) and (7.2.8) that liml,,_.«, F(iy) = 0, and hence F = 0. Thus, one gets the following identity:

sinz

z

_

+ n=-oo (-1) ao

1

z

+

G +n7r

na

2z

E(-I)n

z2

- n272'

where >' stands for the sum with the term for n = 0 omitted. EXERCISE 1. Show that 7r cot 7rz = 1

EXERCISE 2. Show that

>r sing 7rz

+ L/

r

z

I

n=-oo (-_----

n=00-oo (z - n)2

+ 1) . n

7.2. EXISTENCE THEOREMS

189

We are next going to show the existence of a holomorphic function possessing prescribed orders of zeros at prescribed discrete points in a domain D. This is called Weierstrass' theorem:

(7.2.9) THEOREM. Let D C C be a domain, and let {an}n 1 (N S oc) be a discrete set of points of D. Let vn, 1 < n < N, be integers. Then there exists a holomorphic function f (z) which has zeros of order vn at an, n = 1, ... , N, and no zeros elsewhere.

PROOF. By Theorem (7.1.14) there is a Runge increasing covering {Un}n 1 of D. Because of the construction of Un (cf. (7.1.13)), the following holds: (7.2.10) Any connected component of D \ U,, is not relatively compact in D. For every n we set

Pn(z) = fJ (z - aj)"'. ajEU

For a, E Un+1 \U,, we take a point bj E D\Un+1 which belongs to the connected

component V of D\Un containing a,. Then, the function log(z-a,)/(z-bj) has a one-valued branch in a neighborhood of Un. To show this, we take a piecewise linear curve C from a, to b, in V. The curve C consists of line segments L. from zk to zk+1, where

ai = zo,z1,... ,z1 = b,.

The function (z-zk)/(z-zk+1) is holomorphic and non-vanishing in C\Lk. Since C \ Lk is simply connected, 1og(z - zk)/(z - zk+1) has a branch in C \ Lk D U,,. Therefore, the function

- a, EOlogz-zk+1 z - zk log z -b; 1-1

z

(7.2.11)

k=

has a branch in a neighborhood of U,, . We take n = 1 and set f1(z) = P1(z). For a, E U2 \ U1, we take b, E D \ U2 as above. Set 92(z) =

P2(z)

1

-

n(z - b,)"' fl (z)

a,EUIul

z - a, z - bi /

Then log g2(z) has a branch in a neighborhood of U1. For an arbitrary E2 > 0, there is a holomorphic function h2(z) in D such that (7.2.12)

1 log92(z) + h2(z)i < e2,

z E U1.

Set

P2(z) f2(z) =

II(z - bj)"j

e112(2).

7. MEROMORPHIC FUNCTIONS

190

Then, in a neighborhood of U1 the function log

L2 (-z)

f1 (z)

= logg2(z) + h2(z)

has a branch, and by (7.2.12) it satisfies (7.2.13)

log

f2 (z)

I < C2-

h(z)

1

Here, we take E2 = 1/2. Repeating this process, we construct holomorphic functions f,(z) in neighborhoods of U such that (7.2.14) i)

fn(Z) Pn(4)

ii)

log f `+ (}) has a branch in a neighborhood of TI., and

Z E U,,, n = 1,2,...

0, 00.

z

,

zEU,.

logf fn(Z))1 0). Then

f

I

b(t)do(t) _ 0(0),

0 < 6 < bo.

a

To see this, we take a partition, -6 = to < t1 < do such that

< tI = b. Then, there is a

t" -I 0. It follows from (7.4.4) that (7.4.5)

27r

z' 0

log lf(re'B)1d8 = m(r,f) -

m

(r, fll/

Set

(7.4.6)

n(r, f) = the number of b,,'s in 0(r), n(0, f) = the number of by's such that bµ = 0,

N(r,f) = jr n(t,f) -n(O,f)dt+n(0,f)logr. 0

We call n(r, f) and N(r, f) the counting functions (of poles) of f . (7.4.7) LEMMA. Let the notation be as above. Then we have

o

ro/2.

Therefore, M(r, f) 0 such that n(t) < tµ' for t > r1. Hence, we have n(r)

1

0

rµ
0,

and that their sum is not greater than two. The above inequality is called Nevanlinna's defect relation, and immediately implies that the number of Picard exceptional values is not greater than two. EXERCISE 2. Let f3, j = 1, 2, ... , n, be meromorphic functions on C. Show the following:

i) T r, En f! '5 E" E" I T (r, f,) + log n,

ii) T (r,fl ..i f,) :5 E T(r,f,) EXERCISE 3. Show the following for f (z) = eZ.

i) N(r, f) = 0 and m(r, f) = r/r. (Thus, T(r, f) = r/ir.) ii) For a E C', we have

N(r, fla)=a+O(logr),

m(r, f1a)=O(logr).

EXERCISE 4. What is the exponent of convergence of an

/=

n°, n = 1, 2, ... ,

with a>0? EXERCISE S. i) For f (z) = sin z, cos z, show that 2rr

+O(logr) 5T(r, f) 5 72 rr +0(1).

ii) For f(z) = tanz, show that T(r, f) =

2r

+O(1).

7.5. Weierstrass' Product Let f be a meromorphic function on C. By Theorem (7.4.23) the exponent of convergence of zeros of f does not exceed the order of f. Conversely, for a discrete sequence of C, whose exponent of convergence is p, we may ask whether there exists a meromorphic function of order p whose zeros are exactly (a,). In the present section we describe Weierstrass' product, which answers this question.

7.5. WEIERSTRASS' PRODUCT

203

Let p E Z+ and define Weierstrass' irreducible factor E(z; p) by

E(z;O) = 1 - z,

p>1.

E(z;P)_(1-z)eZ+sF+...+°,

(7.5.1) LEMMA. i) For IzI < 1/2, I log E(z; p)I < ii) For an arbitrnry z E C we have

2IzIp+'.

log IE(2; 0)I < 1og(1 + IzI),

log JE(z; p) I < A(p) min{IzI". Izlp+i }.

p ? 1,

where A(p) = 2(2 + logp).

PROOF. i) For IzI < 1 - 1/2(p+ 1)(> 1/2) we have by (2.5.3) p

2

(7.5.2)

log(1-x)+z+ 2 + +zP

I logE(z;p)I =

zp+2

zp4-

ff

p+1

Ip+

1

-1p+1p+2+...I

(1+IzI+IzI2+...) zIp}1

< 2IzIp+

(p+ 1)(1 - IzI) =

ii) The case of p = 0 is clear. Assume that p > 0. For IzI > 1 we have 12. (7.5.3)

log IE(z;p)I no so that no

ny

Ls yY-1

(1)P

n>n1.

2

Y=2

Then, for n > n1 and t E [0, k]

0 < log a-t - log l l -

tn)" < 2e.

Thus, the uniform convergence on compact subsets is proved. Now, we prove the convergence of (7.5.13). Let I C (0, oo) be an arbitrary bounded and closed interval. Take an arbitrary e > 0 and no E N so that

J * e-ctz-ldt < e,

x E I.

a

This and (7.5.14) imply that for n > no and x E I

0 no such that for all nf> nl to

0
e-tt=-ldt - f

0

0

(1- t )n tx_ldt < E. \

n

Therefore,

n > nl,

0 < 1`(x) - 'n(z) < 3e, By repeating the integration by parts for

we get

nzn! On (-T) =

z E t.

x(x + 1)...(x + n).

7.5. WEIERSTRASS' PRODUCT

209

By making use of the expression of the right side, we extend the defining domain of qn (x) to Re z > 0, and define

0n(z) - z(z +

nX n!

Rez>0.

n)'

It follows from the above definition of 0n(z) that 1

6, (W l

TT

=

v= II

(1+ L) e

logn converges to a positive constant C, called It is well known that Euler's number. It follows from Lemma (7.5.4) that the convergence 00

(7.5.15)

1

On(z)

- zec' [I (1 + t/z) e'=1

is absolute and uniform on compact subsets of C; thus the limit is an entire function. It follows from (7.5.13) and the identity theorem (sharing the same

=-

values on the real positive line) that

00

zec,

(7.5.16) r(z)

z) e_tL

fi (1 + V

B=t

Therefore, we deduce that r(z) is analytically continued to a meromorphic function on C which has poles n E Z, n < 0, of order one, and has no pole nor zeros elsewhere. It also follows from (7.5.15) that (7.5.17)

r(z) = hm

n--0o z(z +

nZn! n)*

Equation (7.5.16) is called Weierstrass' infinite product expression, and (7.5.17) is called Gauss' infinite product expression.

The gamma function r(z) plays an important role in studying a number of other special and important functions, e.g., Riemann's zeta function ((z). Here, ((z) is defined by

((z)=Es °O

Rez>1.

n=1

It is known that {(z) can be analytically continued to a meromorphic function on C which has an pole of order one only at z = 1, and has zeros of order one at the negative even integers (called trivial zeros). It is the famous olden Riemann this has been hypothesis that all the other zeros of ((z) lie on the line Re z = z; an open problem since 1859. EXERCISE 1. Obtain the infinite product expression of cos z. EXERCISE 2. What are the exponents of convergence of the following sequences, and their Weierstrass' products?

i) an = n(log n)2,

n = 2,3,... .

ii) a = n!, n = 1, 2, ... .

7. MEROMORPHIC FUNCTIONS

210

EXERCISE 3. Show that for an arbitrary p = 0 there is an entire function of order p. EXERCISE 4 (H. Bohr and J. Mollerup). Show that if a function ,(x) in x > 0 satisfies the following three conditions, then o(x) = r(x): i) 0(1) = 1,

ii) O(x + 1) = x¢(x),

iii) log O(x) is a convex function.

7.6. Elliptic Functions Let f be a meromorphic function on C. If a number w E C satisfies f (z + w) = f (z),

z E C,

w is called a period of f . For example, f (z) = e= has a period 2iri, and f (z) _ sin z, cos z have a period 27r. Let 11 denote the set of all periods of f . Then 11 forms an additive group; that is, 0 E 11, and if wi, w2 E fl, then wl - wz E fl. We call fl the period group of f. (7.6.1) LEMMA. Let f and 12 be as above, and assume that f is not constant. Then ft is one of the following three cases: i) f2 = {0}. ii) There is an wi E C \ {0} such that t1= {nwi; n E Z}. iii) There are R-linearly independent wi,w2 E 0 such that fl = {nlwi + n2(02; nl, n2 E Z}.

PROOF. Assume that n 96 {0}. Since f is not constant, by the identity Theorem (2.4.14) fl is discrete. Take wi E fl \ {0} such that (7.6.2)

Iw1I = min{iwl;w E R).

Take an arbitrary element w E iZ \ {0}, and set r = w/w1. Assume that r E R. By making use of Gauss' symbol [r], we have that w = [r]wi + (r - (r])wi. Hence, (r - [r])w1 E 0, and J (r - [r])wi I < Iwi J. It follows from (7.6.2) that r = [r] E Z. Therefore, if there is no element of fl which is linearly independent of w1 over R, fl is of case ii).

Figure 89

7.6. ELLIPTIC FUNCTIONS

211

Assume that IT = {w E 11;w/w1 ¢ R} 0 0. Let w E Sl', and denote by , w] the parallelogram with vertices, 0, wl, w, w1 + w. Let IQ[wl, w] I be the area of Q[wl, i). Take w2 E St' such that

Q[W l

1Q[W1, W2]I = min IQ[wi,w]l

Let w E fl be an arbitrary element. Then there are unique r1, r2 E R such that W = r1Wi +r2W2

= [r1] i1 + [r21W2 + (rrl - [ri1)wl + (r2 - [r2])W2-

We want to show that r1, r2 E Z. By the above equation, it suffices to show that if 0 S rj < 1, j = 1, 2, then rj = 0. Since IQ[Wi, w]I = r21Q[wl, w2)I, the definition of w2 implies that r2 = 0. Thus, w = rlWl. The choice of w1 implies rl = 0. In this case, iii) holds. 0 The above Sl is a discrete subgroup of C, and acts on C by wES2.

As described in Chapter 5, §3, the set C/U of all !2-orbits [z] = {z + w; w E U} gives rise to a Riemann surface. If Sl = {0}, then C/tl = C, and this is the most likely case for f . For instance, if f (z) is a non-constant polynomial, or rational

function, then fl = {0}. For, if there exists an w E 12 \ {0}, then for zo E C f (zo + nw) = f (zo), n E Z, and hence f (z) = f (zo). Next we consider the case of ii). Take the universal covering (cf. Figure 90)

a: C9 z-+e2"fz/4'3 EC'.

Thus, C/il = C'. Then, f(z) is expressed by a meromorphic function g(w) in w = e2w'x/"' . We take a ring domain R(rl, r2) _ {rl < IzI < r2} so that it does not contain any pole of g(w). Then, g(w) is expanded there to a Laurent series 00

cnwn

g(w) n=-oo

Therefore, we have the expression, 00 cne2wiz/W,

f(z) _

n=-oo

212

7. MEROMORPHIC FUNCTIONS

If f (z) is entire, the above expansion converges uniform on compact subsets in C.

FIGURE 90

We next deal with the remaining case of iii), which is the the main theme of the present section. In what follows, we assume that 0 is of case iii). Then, f is called a doubly periodic function or an elliptic function; w1 and W2 are called the fundamental periods. (The origin of the name, "elliptic function" will be explained at the end of the section.) The set Q[wi,W2] is called the period parallelogram of f , or of Q. In the sequel, we assume that Q[wl, w2] contains the edges twI and tw2i 0 0, are biholomorphic to each other if and only if

rl

cr2

+ d'

(c d)

E

SL(2, Z).

7. MEROMORPHIC FUNCTIONS

214

PROOF. The "if" part follows from Lemma (7.6.5). Conversely, suppose that there is a biholomorphic mapping

f : C/f1[1,r1] - C/fl[1,r2]By (7.6.3) we may assume without loss of generality that f ([0]) = [0]. Let xj : C C/n[1, rj] be the universal covering mappings. As in (7.6.7) there

is a lifting F : C -. C of f o a, such that F is biholomorphic, F(0) = 0, and f o z, = n2 o F. It follows from Theorem (6.1.1) that F(z) = az with a E Co. Then, F(f1[l,r,]) = fl[1,r2J; in particular, F(1) = a E (1[1,T2], and F(r1) = ar, E f2[l, r2]. Thus, a and art generate 11(1, r2]. By Lemma (7.6.5) there is a matrix (a e ) E SL(2, Z) such that

(a1

(a

I

d)(1)'

Therefore, r, = (are + b)/(cr2 + d). 0 We investigate holomorphic and meromorpbic functions on 1-dimensional complex tori. (7.6.9) THEOREM. A doubly periodic holomorphic function f is necessarily constant.

PROOF. Let il[wi,w2] be the period group of f. Then, if(z)I < max{If(w)I;w E Qw1,4,-2 },

z E C.

Therefore, f is bounded on C, and by Liouville's Theorem (3.5.22) f is a con-

stant. D By this theorem we see that for a given lattice f2[w1, w2] there is no nonconstant holomorphic function with period group f2[w1,w2]. One may next naturally ask about the existence of non-constant doubly periodic meromorphic functions. Later, we will construct such functions by means of Weierstrass, but for now we deal with the general properties of doubly periodic meromorphic functions. Let f) = il[w1, w2] be a lattice, and let f be a meromorphic function on C such that every w E ft is a period of f. In this case, f is said to be 11-invariant. We define the degree deg f off as the number of poke off in the period parallelogram Q[w1, w2], counting multiplicities.

(7.6.10) THEOREM. Let f and fZ = Q[w1iw2] be as above. Then we have the following.

i) EaEQ(,,,,21 Res(a; f) = 0. ii) If f is not constant, then deg f > 2. iii) For an arbitrary a E C, the number of a-points of f, counting multiplicities, is deg f .

7.6. ELLIPTIC FUNCTIONS

215

PROOF. i) Let a;, i = 1, 2.... n, be poles of f contained in Q[w1, w2] (n deg f ).

FIGURE 93

Take a small c E C so that the interior of the shifted parallelogram c+Q[w1,w2]

contains all a;. Let Cj,1 < j < 4, be its perimeter, so that C2 = w1 + C4 and C3 = w2 + C1, and the orientation of Ej4=1 Cj is anti-clockwise. Then it follows from the residue Theorem (4.2.7) (cf. (4.2.14)) that Res(a;; f) =

. f (z)dz

tai J tai

E j=1

IfE c'

tai{f&z ,

21ri

EJ 7=fr' 1

f (z)dz

f(z) dz +

j

CM+ s

- f(z+w2))dz+

c.

(f(z)-f(z+w1))dz} 11

=0. In the last equality, the periodicity of f was used. ii) By Theorem (7.6.9) it suffices to show that there is no f with deg f = 1. If so, f has only one pole a1 E Q]w1, w21 with order 1. Then,

E Res(a; f) = Res(a1; f) 54 0. aEQ[w,,W2J

This contradicts i). iii) Let b1, 1 2, the sum EWEn IwI-Q is convergent. PROOF. Set 1 k = {Awl + mw2; Inl - 2 run over all prime numbers, and set 1

((z) = f 1 -

Rez > I.

Prove that this infinite product converges absolutely and uniformly on com-

pact subsets, and moreover, show that it coincides with Riemann's zeta function.

8. Let f (z) be a meromorphic function, and let (- bd) E SL(2, Z). Show that

of +b

T (r),cf+d

T(r,f)+O(1).

9. Let 0 = S2[w1 i W2] be a lattice of C, and let f be an fl-invariant nonconstant meromorphic function. Let a1, ... , ad (resp., b1,... , bd) be zeros (reap., poles) of f in Q[w1,w2], counting multiplicities (d = deg f ). Prove

that 10. Let e; E C,1 < i < 4, be distinct, and set P(w) = n; 1(w - e;). Show that by a linear transformation of the variable, the elliptic integral

is reduced to an elliptic integral

11

dz ,

Q(z)

where Q(z) is a polynomial of degree 3. 11. Let p(z) be Weierstrass' pe-function associated with a lattice 9 of C. Show

that if u+v+wEfi, p(u)

p'(u)

p(v) 00

1

1 = 0.

p'(w) 1 12. Observe that the solution of the dynamics of the pendulum shown below is described by an elliptic integral. p(w)

9 (gravity)

FIGURE 100

Hints and Answers

Here we give hints and answers to some of the exercises and prob-

lems presented in the text. The reader is urged to think over the exercises and the problems for at least a week before looking at these hints and answers.

Chapter 1 §2. Ex. 2. Assume that A = [To, Tl J is covered by two open subsets, Uo and U1, such that Uo n UI n A = 0. We may assume that To E Uo. It suffices to show that A C Uo. Set a = sup{t E (To,T1]; [To,t] C Uo}. V U , which leads to a contradiction (why?). Therefore, a E Uo. In the same way, if a < Ti, a contradiction follows; hence, a = Tl , and (To, Tl ] C Uo. §3. Ex. 2. For z E UQ take i(z;r(z)) C UQ,, r(z) > 0. Moving z

and or,, we obtain an open covering {A(z; r(z)/2)} of A. There are finitely many {A(z,,;r(za)/2)} which cover A. For UQ,, let V be the union of all {0(z,%;r(za)/2)} such that A(za;r(za)) C UQ,.

Problems. 1. E.g., set = x + iy. Then, x2 - y2 = 0, and 2xy = 1. The first implies x = ±y, but by the second, x =may. Hence, x = y = ±1/v"2-.

2. u=f

x2+y2+x/V`,v=± y/

x2+y2+x. (z-1)(z-Pj)zn-1=(z-1)(z"-I+

4. Use the identity 5. Use the absolute convergence.

+1).

6. lim z = (z1 - azo)/(1 - a).

7. Take an arbitrary ao E r. If n E, = 0, EQO C UQEr(C \ E,). Thus, there are finitely many EQ, , 1 < i 0, there is an irrational number 9 > 0 such that `e2"O - 11 < b. Since {e2n"'9} is dense in C(0; I), there is a subsequence e2n°"'8 -' -1, L - 00. Thus, I(e2"O)n - lI -' 2, and so 229

HINTS AND ANSWERS

230

j f" } is not equicontinuous.

§3. Ex. 1. IF,-=. fk(z)I :5 E'=,,, Ifk(z)I < Ex=m Mk. Then, use Theorem (2.3.2).

§4. Ex. 1. Set rn = (nlogn)I/n 10grn =

n(logn)2 , 0. Hence, r --+ 1, and

the radius of convergence is 1.

Ex. 2. Use Theorem (2.4.4). n!n-"/(n + 1)!(n + 1)-(n+1) = (1 + 1/n)n -+ e. The radius of convergence is e.

§5. Ex. 2. z' = 1/z. §6. Ex. 1. It follows from (2.6.3) and Theorem (2.6.4). Ex. 2. J:°O_1 I -n I = IzI E"O 1 n < oo. Then, use Theorem (2.6.6). §7. Ex. 2. Using (2.7.3), we have d(P, P') = 21z - ZII/V/(1 ++ Iz12)(1 + Iz'I2). Letting Iz'I -+ oo, we get d(P, N) = 2/ 1 + IzI2. Ex. 3. Setting z = x + iy, by (2.7.3) we have T1 = 2x/(1 + IzI2), T2 = 2y/(l + IzI2), 1 + T3 = 2IzI2/(1 + IzI2), and 1 - T3 = 2/(1 + IzI2). A circle in the complex plane is written as aIzI2 + bx + cy + d = 0. Multiplying this by 2/(1 + IzI2), one gets bT1 + cT2 + (a - d)T3 + a + d = 0. This is the equation of a hyperplane in R3, and its intersection with a sphere is a circle.

§8. Ex. 1. Let e' be a unit element, too. Then, e' = e e' = e. Let b' be the inverse of a. Then, bab' = eb' = b', and ab' = e, so that b = b'.

Problems. 1. Use Taylor series expansions about z = 0. All are 1.

2. suplfl=n,a.ndinflfI=0. 3. As far as the author knows, the proof requires Lebesgue integration. If the reader has not learned it, he may just read this proof as a story, keeping this fact in mind. Assume that there is a subsequence {f, }OV°_ 1 which converges at every point of 10, 27r]. Set f (x) = lim f" (x). Lebesgue's bounded convergence theorem implies that for f (t)dt = lim fo f" (t)dt. A simple computation yields fo f" (t)dt 0. Thus, fox f (t)dt - 0. By Lebesgue's theorem on differentiation and integration, fo f (t)dt is differentiable almost everywhere, and equals f (x) almost everywhere. Therefore, f (x) = 0 almost everywhere, and so fo " If(t) I dt = 0. It follows again from Lebesgue's bounded convergence theorem that lim fo ff" I f(t)Idt = 0. On the other hand, I

f0 " I sin ntl dt = n fo n" I sin t l dt = fo n I sin tl dt = 2 fo sin tdt = 4. This is a contradiction. 4.

i) nP/(n+ 1)P , 1. ii) 1/ " IfI" = 1/IqI" - oc. iii) log(n!)'/"2 = n =

log j < n f, +' log xdx = n (n + 1) (log(n + 1) - 1) -+ 0. The radius of n convergence is 1. iv) j. 1

5. By the change of variable z -' (z, we may assume that ( = 1. Set sn = En=0 a and s = limn-00 sn = Ln 0 an. For z E i(1), f (z)/(1 - z) _ En0 z" n 0 anz" = E o snz", and then f (z) - s =(I - Z) En 0(3n - s)2".

For z E D (1; cos r) there is a positive constant K (= 2 sec r = 2/ cos r) such that 11-zl < K(1-Izl). For an arbitrary e > 0, there is an n9 E N such that Isn-sl
no. Then, If (z) - sl 5 11 - zI (En° 0 Is,, - sl - IzI" + En=na+1 EIZIn) < (Ell 0 Isn - si + E En 0 I z i") = 11- ZI Ell, Isn - 81 + fl 1- z I/(1- Izl) < -z Z I 1- zI En°_0 Is,, - sl + EK. Taking z so that 11 - zI E"°_o Isn - sI < E, we have

f(z)-sI 0 so that rl M(r1) = r3M(r3). Then, let p/q -' or. rZ M(r2) < ri M(r1). Take the logarithms of both sides, and substitute a = (log M(r1) - log M(r3))/(logr3 - log r1).

= Z i Jo' e"nie''de = 2xs fo ' `ae-1 t & _ d(. If Ial < 1, f (()/(1 - a() is holomorphic in a neighborhood of Ishri fc(o;1) A(l), and so the integral equals f (o). If IaI > 1, use 1/(1-a() = 1/(-1/((-1/a). 11. We have -L fc(o;1) z- o dz

12. Use the mean value theorem.

1

233

HINTS AND ANSWERS

13. y-x+2xy. 14. f(z) = zez.

15. Use the Poisson integral.

16. Use t,(z) = (z - i)/(z + i), and (3.6.7). 17. Note that Re -1F2,-117 =y/fit-z2.

Chapter 4 §2.

Ex. 4. Let C be the union of circles around every point of E. Then, fcw=0.

Ex. 5. Let a (reap., bµ) be the zeros (resp., poles) of f of order m (resp., zm. fl(z-b',)"". Here, set nµ). Then, g(z) = f(z) if a, = oo, and (z - b )"- = z-"- if b, = oo. Then, g(z) is a non-zero constant. Ex. 6. Let g(z) be the right side of the last equation. Then, h(z) = f (z) -g(z)

is holomorphic on C. Thus, h(z) __ c E C. Comparing the Laurent series expansion of f o z about i = 0 with that of g o z, one sees that h(oo) = 0. Ex. 7. Res(±i; f) = 1/2. Ex. 8. Res(tai; f) _ +i/4a3. Ex. 9. Res(0; f) = E 01/n!(n + 1)!, Res(oo; f) Eri o 1/n!(n + 1)!. §3. Ex. 2. Set w = (f'(z)/ f (z))dz. Let M (resp., P) be the number of zeros (reap., poles) of f, counting multiplicities. Then, M - P = Ea c Res(a; w) = 0. Ex. 3. z =2.,/w- +w+En°-_1(-1)"-1(2n-1)!!2-2n+1 21r{(2a2\-

1)/ a - 1 - 2a}.

§4. Ex. 1. i) 21r/(1 - a2). ii) 1r/2 a +a. iii)

Ex. 2. i) 0. ii) a/2'. iii)

iv) 7r(2n - 2)!/22n-'((n - 1)!)2. Ex. 3. i) 2irie_ / '2+'/2/3 + 1rie-' /3. ii) (ir cos a)/2. iii) a/2e. iv) or/4. Ex. 4. i) a/2sin(a7r/2). ii) a(a + 1) ... (a + n - 2)a/(n - 1)!sinair. iii) it/2a1+0 cos(air/2). Ex. 5. i) (1 + log a)/2a. ii) (log2 a - log2 b)/2(a - b). iii) -wr/4.

Problems

00 1. i) En=-00

(-1)' sin l

-

n

1

VL.k=maxio.n}

-1 °oos1

t

T

.

ii) sin &

cos 1 00 n

(

1

sin 1

z-1 - 2

1

z-1

-

+

nz-2n-1

111) En--O( + z-1 + z-1 +1 When n > 0, the pole is at co, and the residue is 0; when n = -1, the pole is at 0, and the residues are 1 at 0 and -1 at oo; when n < -1, the pole is at 0, and the residues are 0 at 0 and at oo. ii) The poles are 7r/2 + nzr, n E Z, and the residues are all I. iii) No pole. The residues are 1 at 0, and -1 at oo. iv) The poles are -2 with residue -2, and -3 with residue 3. The residue at oo is 1. v) The poles are -1 with residue 6, and oo with residue -6. 3. i) 7r. ii) 21ri/(n + 1)!. iii) When n >_ 0 and m > 0, 0; when n < 0 and

n. 2. i)

m

0, 0 for m < -n - 2, and 27ri(_ 1)(-1)m+n+l for in >_ -n - 1; when

nm 2, 1z41 - 61zI - 3 > 1, and so there is no solution there. b)m+n-1

Thus, there are four solutions in !zI < 2. On jzI = 1, 16zl - Iz° + 312! 2, so that

HINTS AND ANSWERS

234

there is one solution there. Thus, there are three solutions in R(1, 2).

5. On IzI = R', IzI - If (z) I > IzI - If(z)I >_ R' - OR. Hence, for R' < R sufficiently close to R, the number of zeros of z and z - f (z) in A(M) are the same. Thus, there is one z E A (R) such that z - f (z) = 0. 6. Use Rouche's theorem on I z I = 1. 7.

Let f (zo) = 0. Apply Rouche's

theorem for f (z) and for f (z) + (f. (z) - f(z)) = f. (z) on a small circle Iz - zol = b. 8. i) ( (a + 1)/a - 1)7r/2.

ii)

iii) 21rie ' /3 - 7ri/3. iv) 27r(1 - a/ Q - 1). v) 7r/2ea. Vi) 7r/aea.

7r (cos '1 7r + b sin 2 7r) / sin air.

vii)

Use the curve in Figure 101. -7r tan 2 ir. viii) Use the equality log3 z - (log z + 27ri )3

= -127rlogz+

6ai loge z - 8iri. 3 7r + Z47r3. ix) Use the curve in Figure 101. -7r2/4.

FIGURE 101

Chapter 5 §1. Ex. 1. Use Theorem (2.8.9). Ex. 2. Apply the reflection principle to extend f meromorphically to all C. Hence, f is rational and holomorphic in C, so that f is polynomial.

Ex. 3. Let f (z) = E a,, z". The first condition implies a" E R. The second condition implies that a" = 0 for even n. Thus, f (z) is odd. §3. Ex. 3. Use Theorem (3.5.14).

Problems 1. Glue two C along [0,1]. 5. Let If (P) l = max{ I f 11 = M. Show that {Q E X ; if (Q) j = M} is open and closed.

7. The length is log R_r. The area is 47rr2/(R2 - r2).

Chapter 6 §2. Ex. 1. Use 7j'(z) = (z - i)/(z + i), which maps H biholomorphically onto A(1), and maps geodesics to geodesics. Ex. 3. dH(z7, z2) = 1og{(1 + Izi - z2I/Izi - z2I)/(1 - Iz1 - z2I/I1I - Z21))§3- Ex. 2. For arbitrary zz E A(1), j = 1,2, such that ir(z,) = wj, by Theorem (6.3.10). For an arbitrary e > 0, there is doitl(zl,z2) >_ dD(wt, a piecewise C' curve C C D from wi to w2 such that LD(C) < dD(wl, w2) + C. Fix zi as above. Let e be the lifting of C with the initial point z1. Let z2

be the terminal point of C. Then, 7r(z2) = w2, and LD(C) = L,1(,)(C) (cf. (6.3.2)). Therefore, DA(l)(zl, z2) < L,&(1 (C) = LD(C) < dD(wi, w2)+e. Hence,

HINTS AND ANSWERS

235

do(l)(zl, z2) < dD(wi, w2) §4. Ex. 1. Using Lemma (2.2.7) or (7.1.12), find countable open coverings, U,, C V,, of D such that {fn I V,, }' is normal. Then, apply the diagonal argument used in the proof of Theorem (2.2.6). Ex. 2. Apply Schwarz' lemma to g o f -I : 0(1) A(1). Ex. 3. Use the uniqueness of Theorem (6.4.4). Ex. 4. Parametrize the parabola y2 = 4c2(x + c2) by x = t2 - c2, y = 2d. Set

z = x + iy = (t + ic)2. Replace t by a complex variable w, and take fz such that = i. Set w = f - ic. As w = t E R runs from -oo to oo, y(t) with z = z(t) = x(t) + iy(t) runs from -oc to oo. Thus, w = f - is is the required mapping. §5. Ex. 1. f(z) _ (z - 1)/(z + 1).

Ex. 2. f (z) = (iz + 1)/(-iz + 1). Ex. 3. It follows from Theorem (6.5.7) and Theorem (3.6.11). §6. Ex. 1. If D 0 C, C, C', then C \ D contains at least three points, which may be assumed to be 0, 1, oc. Set Do = C \ (0, 1, co}. The inclusion mapping

t: D 3 z - z E Do is holomorphic, and has a lifting c o a: D -' 0(1), where 0(l) is the universal covering of Do (Theorem (6.6.7)). Thus, there is a bounded non-constant holomorphic function on D. It follows from Theorem (6.4.6) that D is biholomorphic to A(1). §7. Ex. 1. Considering f - ia, we may assume a = 0. Using (2.8.12), we have lip o f (z) I < 1 for f E F, z E D. Thus, {,I o f; f E F} is normal, and so .F is normal. Ex. 2. Let f (z) be a non-constant entire function. Assume that f (z) does not take three values 0,1, oo. Set f,, (z) = f (vz), v = 1, 2,.... Then, f does not take 0, 1, oo, and hence If, } is a normal family. Since f (0), (f,, } contains a subsequence, converging uniformly on compact subsets. On the other hand, E IanI2lvl2n - oo (v -' oc), where f (z) a,, z". This zx ff * I implies that any subsequence of does not converges uniformly on C(0; 1), and so a contradiction. Ex. 3. Set f,,(z) = e1z, v = 1, 2, ... , which do not take the values 0, oc. Since ff,(0) = 1 and f,,(0) = v, any subsequence of does not converge uniformly on compact subsets.

Problems 1. 2 log 3. log 25 9

.

: A(1) -+ Dj be the universal coverings, and let F : i (1) -+ A (l) be a lifting of f oirl. Take zo E 0(1) such that lr,(zo) = a. The assumption implies f `gt (zo) = gI (zo ), and by Theorem (6.2.3) F'g1 = gI at all points. Hence, 2. Let i r k

f`hD, = hD, 3. Use Exercise 2 of §3, and take a geodesic in the universal covering 0(1).

Take W E C so that the cardinality of f -'(w) is the maximum. Set fI(w) = {zJ}1_I. Then, f'(z,) 34 0. There are neighborhoods A(w;co) of w 4.

and U; of zJ such that f I Ui : U; - A (w; eo) are biholomorphic. We may assume

HINTS AND ANSWERS

236

that U, are bounded. Take R > 0 such that A(R) D U,UJ. By the choice of w, If (z) - wI > co on C \ A(R). Hence, in a neighborhood of oo, 1/(f (z) - w) is bounded and holomorphic. By Riemann's extension Theorem (5.1.1), f (z) has at most a pole at oo. Thus, f is rational. 5. Define a biholomorphic mapping f : A(1) - (z E C; Re z > 0) by f (z) _ (1 +z)/(1- z). The real axis in 0(1) is mapped by f onto the real positive axis. Thus, f (D) = {z; Re z > 0, Im z > 0}, and w = f(z)2 has the required property.

6. Composeg: DE) z-+z" =09=)'Q E {wEA(1);1mw>0} (log1=0) with the one obtained in 5, and then with ty(w) = (w - i)/(w + i). 7. Set z = e'B. Then w = 2 cos 6. We see that C(0;1) is mapped onto 1-2,2].

Using the branch f = 1, we have z = (w - w - 4)/2, which expands to Thus, z = 0 is corresponding to oo. Thus, A(1) is mapped z = w+Y + .

univalently onto D. One sees also that i5\ d(1) is mapped to D. 10. Suppose that (f,, } does not converge uniformly on a compact subset K C D. Then, there is an Eo > 0 such that for every N E N there are vN, µN > N and zN E K with I f,,,, (zN) - fMN (zN) I > co. Run N = 1,2,..., and choose h, and zN - zo E K. Thus, we have subsequences to get f,,N - 9, fµN jg(zo) - h(zo)j > co. On the other hand, g = h on E, and hence g = h. This is a contradiction. 11. Apply (3.6.7). 12. Let a : A(1) D be the universal covering, and let f : 0(1) A(1) be a lifting of f . We may assume that a(0) = zo and j(0) = 0. By the assumption and

Lemma (6.3.11), f E Aut(O(1)), and f(z) = e2ztOz,0 < 6 < 1. If 0 E Q, then for some m E N f"' = f o . o f (m times) = ido(l). Thus, f"' =idD, and f -' = E N, v = 1, 2, ... , so f"'' 1. Let 6 rxf Q. Take an increasing sequence m,, < that elm x'e -+ 1. Applying Theorem (6.7.6) to choose a subsequence, we may

assume that f'- - idD and f" -' - g. Then, f o g = idD. 13. The first half follows immediately from Theorem (6.5.7). Then, using Schwarz' reflection principle, extend f to f E Aut(C). One sees that f (z) _ az + b (a 96 0). Thus, Dl is similar to D2. 17. Use the little and big Picard theorems.

19. We may assume that U = A(1) and P = 0. We also may assume that f omits 0, 1, oo. Let z,, E A`(1) be such that Iz.-I < 1/2, limz = 0, and a. If a = oo, we consider 1/ If (z), so we may assume that a E C. lim f

Set !v(() = f (ZvC), v = 1, 2, .... Then, fy : W(2) - C \ {0, 1}, V = 1, 2, ... , form a normal family by Theorem (6.7.6). Hence, we may assume that U-11 converges uniformly on C(0;1). Set f (z) = F a,z" (Laurent series). Then

E"

a,, z" z", v = 1, 2, ... , converge uniformly on C(O;1). There is an M > 0 I2" < M. Thus, for n < 0, an = 0. such that 2* fo" I f,.(e'e)I2d9 = E l o

HINTS AND ANSWERS

237

Chapter 7 §2. Ex. 1. Let g(z) denote the right side. It follows that g(z + 1) = g(z). Set

f (z) = it cot az - g(z). Looking at the principal part at z = n E Z, one sees that f (z) is entire. We want to show the boundedness of f (z). By making use of the periodicity, f (-z) = - f (z), and of f (z) = f (z), it suffices to show that f (z) is bounded on the set of z = x + iy, 0 :5 x _ 1. Since I cot lrzj
_ 1. Note that lg(z)I < 1+f(1+ +ioo1 z2n < y2) + / (1 + 2y) fi C1 + f(1 + 2y) fo y3dt = 2y)/ (; + 1-1 2 +y dt 1+2fF°O_1(2+y)/{(n-1)2+y2}

e-

C1 + v/ 2-(l +2y)2ir/y < C2 < oo. Therefore, f (z) is constant. The constant term of the expansion of f (z) about z = 0 is 0, and hence f (z) = 0. Ex. 2. Differentiate both sides of the equation obtained in Ex. 1. Check that termwise differentiation is possible.

Ex. 3. Let z = e2"'8 with 0 E Q. For 0 < r < 1, f (re2A`s) is, up to finite terms, the sum of r"' + r("+1)} + - . Thus, lim,._1 f (re2"'B) = +oo. Since such points are dense in the boundary of d(l), f (z) cannot be analytically continued over any boundary point of 0(1).

Ex. 4. There is a sequence of points zn E H such that it accumulates to an arbitrary point of R, and )(z,) = A(z1). §4. Ex. 1. We have log+ E,"_1 xj < log+ n max{x,} < ma(log+ x,} + log n log+ xi + log n.

Ex. 3 i) N(r, f) = 0. m(r, f) = A fp/2rcos0d0 = . ii) n(r,1/(f -a)) _ + 0(1), and hence N(r, l/(f - a)) = x + O(logr). It follows from i) that m(r,1I (f - a)) = O(log r).

Ex. 4. I/a. Ex. 5. Use Ex. 3.

§5. Ex. 1. cost = lln _oa (1 - na+x 2) ez/(na+n/2) Ex. 2. i) E 1/Jan I" is convergent if and only if f W 1/xµ (logx)2µdx converges. Thus, it converges for 0 < 14 < 1, and diverges for p > 1. Weierstrass' product 11 rim 1(1 - z/n!). is = 0. 2 1 - n 1ogn ) e`/"(lOg") . ii) Ex. 3. Take Weierstrass' product associated with an = nl/P, n = 1, 2, ... .

Ex. 4. It follows from i) and ii) that O(n) = (n - 1)! for n E N. Condition iii) implies that for 0 < x1 /< x2 < X3 < X4, (1og0(x2)- log¢(x1))/(x2 -TI) -:5 (log 0(X3) - log 0(x2))/(x3 - X2) < (logO(x4) - log¢(x3))I(x4 - x3). For

0 < x < 1, set x1 = n - 1 < X2 = n < x3 = n + x < x4 = n + 1, and substitute them in the above expression: log(n - 1) S (log ¢(n + x) - log(n -1)!)/x log n.

Combine this with O(n+x) = (n+x-1) .. xO(x) to get (n-1)t(n-1)!/x(x+ 1) < O(x) < n=(n - 1)!/x(x + n - 1). Therefore, (n/(n + x))4(x) < n=n!/x(x + 1)...(x + n) Ianl-" < oo, (7.4.21) implies

'dt that f °° n(t)t-"-'dt < oo. For all larger > 0, we have 1 > fZr (1 - 2-°)n(r)/ar". Setting r = Ianl, one obtains n/lan1° < C (a' constant). Hence, pl < po. If p, = oo, then po = oo. Suppose that p, < oo. For an n(t)t_'

arbitrary f3 > pl, take p, 0 small enough so that (1 + e)(logn)/Iog Ianl < 3. Thus, jani-a < n-1-`, and E Ian10 < 00. We have p, >_ po. 4. i) Use (7.4.2). ii) Use (7.4.2), >loglz3l = Elog{1 - (1 - Iz,l)}, and

Theorem (2.6.4). ii) Usenj(z-a,)/aj(ajz-1) = la3I{l+z(1-Ia,I2)/aj(ajz-1)}. 5. Let g(z) be a branch of log f (z) on C. Then, f (z) = e9(z>. By the assumption, IReg(z)I = loglf(z)l < O(IzIP+'). Using (3.6.7), one sees that Ig(z)I = O(IzIP+'). Thus, T(r,g) = O(logr). Lemma (7.4.13) implies that g(z) is a polynomial. Let p be the degree of g(z). Then, log M(r, f) = corP(1 + 0(1)) with co54 0, and so p = p. 6. The order of eaz+P(z) is 1. Suppose it has only finitely many zeros. Then, eaz + P(z) = Q(z)e"`z, where Q(z) is a polynomial and µ E C'. Differentiate both sides deg P + 1 times. Then, ea= = R(z)e"', where R(z) is a polynomial. Hence, A = µ, and P(z) = (Q(z) - 1)e'Z, and so Q = 1. Thus, P =- 0, which is a contradiction. >p_' converges absolutely, and so is ]l(1 +p-Z). The 7. Since Rez > 0, uniform convergence on compact subsets is proved in the same way. Let 2 =

P1 < P2 < ... be primes. Then, (1 - pi s)-'(1 -

p2:)-'

... (1 - Pnz)-' _

HINTS AND ANSWERS z+P12z+...)(1+p2 2

(1+P1

'

Z

+P2 +P1 n=1

22

+P2 -Z

+ (pp) +

2s+...) ... (1+p;z +p,-2z .

239

1+plZ+

Letting n -, oo, we get n(1 -pnz) _

n- = ((z)-

8. This follows from Theorem (7.4.9), (7.4.8), and (2.8.5).

9. Let Q[w1,w2] be the period parallelogram of Q. Take a E C so that the boundary of E = a+Q[w1,w2] contains no point of aj +fl nor bj +52. Thus, we

may assume that aj and bj are contained in the interior of E. Eaj - E bj = = dz. Using the periodicity, we have sai fee z 'L z dz = mw1 + nw2 E sai fOE z 75T

n.

10. Set v = 1/(w - e4) and e = ej - e4, j = 1,2,3. Then, f dw/ P-(w) _ f-1

Cl

v?...(1 _esv)

dv.

11. The meromorphic function f (z) = p(z) - ap(z) -,Q has the period group 1, and has degree 3. Let aj, j = 1, 2, 3, be zeros of f . Since the poles of f are on Il, a1 + a2 + a3 E fl by problem 9. For arbitrary u, v, there are suitable a,,0 such that f (u) = f (v) = 0. Then, necessarily f (-u - v) = 0. Therefore, if u+v+w E !Q, there are a, Q such that f (u) = f (v) = f (w) = 0. The determinant of a system of linear equations with non-trivial solutions must be 0. 12. The dynamical equation is d29/dt2 = -(g/R) sin 0. Multiplying both sides

by dO/dt and integrating, we have (dO/dt)2 = (49/R)((Lb//)2 - sine 0/2), where vo = R0'(0), 0(0) = 0 (the initial condition). Set 0 < k = vo/ f4R < 1 (to avoid the rotation). Letting x = k sin 0/2, we get (dx/dt)2 = (g/R)(1 x2)(1- k2x2). Thus, t = V19-/R fo dx/ (1 -x2)(1 - k2x2), which is an elliptic integral.

References

In writing this book the author has referred to a number of books already published. Some of them are listed below:

[l] A. Hurwitz and R. Courant, Funktionentheorie, Springer Verlag, Berlin, 1929. [21

[31

L. Ahlfors, Complex Analysis, McGraw-Hill, Auckland et al., 1979; Japanese Translation by K. Kasahara, Gendaisugakusha, Tokyo, 1982. H. Cartan, Theorie Elementaire de Functions Analytiques d'Une ou Plusieures Variables Complexes, Hermann, Paris, 1961; Japanese Translation by R. Takahashi, Iwanami Shoten, Tokyo, 1965. K. Kasahara, Complex Analysis-Analytic Functions of One Variable (in Japanese), Jikkyoshuppansha, Tokyo, 1978. Y. Komatsu, Introduction to Analysis [I] (in Japanese), Hirokawa Shoten, Tokyo, 1962.

Y. Komatsu, Function Theory (in Japanese), Asakura Shoten, Tokyo, 1960.

Y. Komatsu, Exercises in Function Theory (in Japanese), Asakura Shoten, Tokyo, 1960. R. Takahashi, Complex Analysis (in Japanese), University of Tokyo Press, Tokyo, 1990. T. Ochiai and J. Noguchi, Geometric Function Theory in Several Complex Variables (in Japanese), Iwanami Shoten, Tokyo, 1984; English

Translation by Noguchi and Ochiai, Amer. Math. Soc., Providence, Rhode Island, 1990.

For the readers who want to advance to a further study on complex analysis, the author would Wce to give a list of books, which is not intended to be complete. For Riemann surfaces, I recommend the following.

[10] H. Weyl, Die Idee der Riemannsche Fliiche, B.G. Teubner, Stuttgart, 1913; Japanese Translation by J. Tamura, Iwanami Shoten, 1974. (This is a famous classical book, settling the concept of Riemann surfaces.)

241

REFERENCES

242

(11] K. Iwasawa, Algebraic Functions (in Japanese), Iwanami Shoten, Tokyo, 1952; 2nd ed., 1972; English Translation by G. Kato, Amer. Math. Soc.,

Providence, Rhode Island, 1993. (This is a famous introductory book from algebra.)

[12] R.C. Gunning, Lectures on Riemann Surfaces, Princeton University Press, Princeton, 1966. (Comprehensive lecture notes, using sheaf theory.)

[13] Y. Kusunoki, Function Theory (in Japanese), Asakura Shoten, Tokyo, 1973. (This is a nice introductory book of closed and open Riemann surfaces to the research level.) [14] Y. Imayoshi and M. Taniguchi, An Introduction to Teichmiiller Spaces (in Japanese), Nipponhyoronsha, Tokyo, 1989; English Translation by Y. Imayoshi and M. Taniguchi, Springer-Verlag, Tokyo et al., 1992. (This deals with the deformation theory of the complex structure of Riemann surfaces, which is called Teichmiiller theory.) For value distribution theory, I recommend [15] R. Nevanlinna, Le Theoreme de Picard-Borel et Ia Thdorie de Fonctions Meromorphes, Gauthier-Villars, Paris, 1939. (This is a monograph by the creator of Nevanlinna theory, and it clearly unrolls the development of the mathematical theory in front of your eyes.) (16] W.K. Hayman, Meromorphic Functions, Oxford University Press, Oxford, 1964. (This is an introduction to Nevanlinna theory in one variable, up to the research level.) (17] M. Ozawa, Modern Function Theory I - Theory of Value Distribution (in Japanese), Morikitashuppansha, Tokyo, 1976. (This is a monograph at the research level.) [18] S. Kobayashi, Hyperbolic Manifolds and Holomorphic Mappings, Marcel Dekker, New York, 1970. (This is an introduction to the theory of Kobayashi hyperbolic manifolds by Kobayashi himself, and has had a deep influence on complex analysis since then.) (19] W. Stoll, Value Distribution Theory for Meromorphic Maps, Aspects of Math. E7, Vieweg, Braunschweig, 1985. (20] P.A. Griffiths, Entire Holomorphic Mappings in One and Several Com-

plex Variables, Ann. Math. Studies 85, Princeton University Press, Princeton, 1976. For complex analysis in several variables and the theory of complex manifolds, I recommend

(21] L. Hormander, Introduction to Complex Analysis in Several Variables, van Nostrand, New York, 1966. (This is an introduction to the theory from the viewpoint of elliptic differential equations and functional analysis. This book established one research direction of the theory.)

REFERENCES

243

[22] A. Well, Introduction h. 1'Etude des Varietks kahleriennes, Hermann, Paris, 1958. (This is a comprehensive course in the theory of Ki filer manifolds by one of the greatest mathematicians of this century. Mathematically it is very nourishing. (23] S. Nakano, Complex Function Theory in Several Variables - Differential Geometric Approach (in Japanese), Asakura Shoten, Tokyo, 1982. (This is a good book on function theory on Kiihler manifolds by the author, who is famous for the Kodaira-Nakano vanishing theorem.) (24] T. Nishino, Function Theory of Several Complex Variables (in Japanese), Tokyo University Press, Tokyo, 1997. (The author, who was a student of K. Oka, the founder of the theory, gives a treatise based on Oka's idea and work. An English translation is in preparation.)

Index

Abel's continuity theorem, 46 absolute convergence, 13, 213.5 absolute value, 2 accumulation point, 4 act (action), 132 addition theorem for the pe-function, 226 adjoint harmonic function, 85 analytic, 28 analytic continuation, 117, 124 analytic curve, 121 annulus, 91 arc, 6 arcwise connected, 8 area, 138 argument, 3 argument principle, 100 Ascoli-Arzel theorem, 20 at most a pole, 93 automorphic form, 222 big Picard theorem, 171 biholomorphic mapping, 137 Borel exceptional value, 202 boundary, a boundary correspondence, 152 boundary distance, 22 boundary point, S bounded, 4 11 branch, 76, 126 branch point, 128 canonical product, 245 Casorati-Weierstrass' theorem, 118 Cauchy condition, l3 Cauchy product, 13 Cauchy sequence, 10 Cauchy's integral formula, 73, 77 Cauchy's integral theorem, 70 Cauchy-Riemann equations, 54 characteristic function, 192 circle, 5, 39 circle of convergence, 25 closed, 4 closed curve, 6 closed set, 5 closure, 4 compact, 14 complete, 147, 148 complex coordinate, 2 complex derivative, 49 complex differentiable, 49, 87 complex function, 17

complex number, 1 complex plane, 2 complex Poisson integral, 82 complex torus, 212 conformal mapping, 54 conformal metric, 137 conformal pseudo-metric, 131 conformality, 54 conjugate, 2 connected, 5 connected component, 9 constant curve, 7 continuous, 18 contraction principle, 149, 151 converge, %) 19

converge absolutely, 13, 23 converge uniformly, 19 convergent power series, 25 correspondence of circle to circle, 42 counting function, 197 covering, 5 130 covering mapping, 130 covering transformation group, 133 cross ratio, 47 curve, 6 curvilinear integral, 58 70. 88 deck transformation group, 133 defect, 202 degree, 214 derivative, 49 derived function, 49 differential, 87 88 Dirichlet problem, 8.5 discrete, 5 disk, 3 disk neighborhood, 4,.3-9 135 disk of definition, 124 distance, 3 domain, 5, 87 domain of existence, 191 doubly periodic function, 212 elliptic, 48 elliptic curve, 212

elliptic function, 211212, 224 elliptic integral, 224 end, 7 entire function, 118 equicontinuous, 20 exceptional value, 167 exponent of convergence, 200 245

246

exponential function, 3Q exterior point, 5 finite length, 55 fixed point, 4Z Fubini-Study metric. 138 function element, 124 fundamental domain, 164 fundamental group, 130 fundamental periods, 212 fundamental theorem of algebra, 80 Gaussian curvature, 139 Gaussian plane, 2 general curve, 6 generators. 212 group, 44 Hadamard's three circles theorem, 20harmonic function, 82 Harnack inequality, M hermitian metric, 187 hermitian pseudo-metric, 131 holomorphic, 49 87 13(i holomorphic differential, 88 holomorphic function, 136 holomorphic local coordinate, 135 bolomorphic transformation, 137 holomorphic transformation group, 131

homeomorphism, n 131 homotopic, 63 homotopic to a point, 63

bomotopy, fa homotopy class. 6,3.

Hurwitz's theorem, 141 hyperbolic, 48, 147, 142 hyperbolic distance, 145 hyperbolic geodesic, 1.4fi hyperbolic length, 145, 142

hyperbolic metric, 148 identity theorem, 29 imaginary unit. 1 indefinite integral, Z2 infinite product, 13, 34 infinite product of functions. 36 infinity, 38 initial point, Z injective, U interior point, 3

inverse, 1

INDEX

Laplace equation, 82 Laplacian, 82 lattice, 212 lattice point, 212 Laurent series, 83 length, 54, 1.38 length parameter. 56 length parametrization, 56 lifting, 133 limit. 9 1.1 limit function, 19 line segment, 65 linear transformation, 4_0 141 Liouville'a theorem, 72 Lipschitz' condition, 35

little Picard theorem, 167 logarithmic branch point, 128 logarithmic differential, 20 kowdromic, 48 majorant, 24 majorant test, 24 maximum function, 198 maximum principle, 79 mean value theorem, 81. 84 meromorphic differential, 2 meromorphic function. 94 Mittag-LefHer's theorem, 185 Mobius transformation. 44 modular group, 46 monodromy theorem, 1M Montel, 168 Morera's theorem, ZZ multi-valued, 75

mutual reflections, 42 43 natural boundary, 181 neighborhood, 4 negative variation. 193 Nevanlinna's defect relation. 242 Nevanlinna's first main theorem, 191 Nevanlinna's inequality, 198 non-Euclidean, 141 non-singular, Z. 121 normal. 168 normal family, 153 north pole, 31 number of zeros off surrounded by C, 101 fl-invariant, 21A

inverse function theorem, 1

1-valued, 73.

isolated essential singularity, 93. 171 isolated point, 3

open, 4 open covering, 5 open mapping theorem, 103 open set in, 5

Jensen's formula, 194 Jordan curve, Z Jordan's theorem, 66 Koebe's function, 175 lambda function, 167

orbit. 132 order, 93. 200 order function, 191

INDEX

247

order of urpoint. 1112 order of the branch, 128

Riemann surface of the inverse function,

orientation, 8 parabolic, 48 parameter. 6 parameter change, fi partial sum, 12. 23 partition, 54 partition point, .54 period, 210 period group, 214

Riemann's extension theorem, LIZ Riemann's zeta function, 209

period parallelogram, 212 Picard exceptional value, 241

piecewise continuously differentiable, Z piecewise linear curve (Streckenzug), 9, 85 PoincarE metric. 139. 144, 148

point at infinity, 38 Poiason integral, 82 Poisson kernel, 82 polar coordinate, 3 pole, 93 positive orientation, 127 1112 positive variation, 133

121

Riemann-Stielt jea' integral, 1113 ring domain, 91

rotation number. 7G 3S Rouch6's theorem, 141 Runge increasing covering. 184 Runge's theorem, 1113 Schwarz' reflection principle, 113 Schwarz-Christoffel's formula, 176 Schwarz-Pick's lemma, 151 Scbwarzian derivative, 175 sequence, 2 sequence of complex functions, 13 series, 12

series of functions, 23

series of order change, 13 simple curve, Z simply connected, 63 south pole, 3Z special linear group, 44

power series, 25 power series expansion, 28 preserve the orientation, 157

star-shaped, 114

primitive function, a 88

subgroup, 43

principal congruence subgroup, 46 principal part, 185. principal value, 10-8

subsequence, 111

principle of the permanence of the func-

Taylor series, 28

tional relation, 130 Prinzip von der Gebietstreue (open map-

terminal point, 1

ping theorem), 1113

projective special linear group, 41 proximity function, 132 Puiseux series, 1.115

Puiseux series expansion, lS pull-back, 132

purely imaginary, 2 radius of convergence, 25 ratio of the circumference of a circle to its

diameter. 35 rational function, 94 refinement, 55 reflection, 42 reflection points, 122 reflection principle, 43 relative topology, 5 relatively compact, 12 removable singularity, 118 residue, 95 Riemann mapping theorem, L54 Riemann sphere, 337 52 Riemann surface, 127. 135

stereographic projection, 38 Stolz' domain. 411

sum, a surjective, 13 topology, 4

total variation, 132 transcendental, L12. 174 transitively, 44 trigonometric functions, 31 uniformization of Riemum surfaces, 1311 uniformization theorem, 158 uniformly bounded, 211 uniformly continuous, 18 uniqueness of analytic continuation, LLZ

unit circle, 5 unit disk, 3 univalent, 1112

universal covering, 132 universal covering mapping, 132 upper half plane, 45 variation, 192. 193 w-point, 1.42 Weierstrass' canonical form, 224 Weierstrass' irreducible factor, 213 Weierstrass' M-test, 24 Weierstrass' pe-function, 211 Weierstrass' product, 202. 2115

248

Weierstrass' theorem, 189 zero, 93

INDEX

Symbols

E, a element U union

d(z; OD) boundary distance, 22 SL 2, C), 40 PSL(2, C), 41 Aut(A(1)), 43 Aut(C), 43

intersection

N natural numbers (positive integers)

H = {z E C; Im z > 0} upper half

Z integers

Z+ non-negative integers Q rational numbers R real numbers C complex numbers, 1

plane, 45

Aut(H), 45

SL2;R,45 SL 2, Z), 46

C'=C\{0}

r(n), 46

Aut(- ) automorphism group max maximum min minimum mod congruence

8z = 8/8z, 50

[x) = max{n E Z. n < x} Gauss'

I(d)!, 54 L(C) length of curve, 555

8==8/01,50 (d), 54 L(C; (d)), 54

symbol (n) = V

.01J, 55

nl

v.n -vY

Ck-class k-times continuously dif-

ClJ, 55

ferentiable i imaginary unit, 1 z the (complex) conjugate, 2

R(a;ri,r2) annulus, 91

a,82 R(rl, r2) = R(0; rl, r2), 91 Res(a; f) residue, 95 nb(a; C) rotation number with respect to b 96 rrl (D)p fundamental group, 130 Lh(C), 138 Ah(E), 138

Re real part, 2 Im imaginary part, 2 A(a; r) disk, 3 A(r) disk, 3 0 empty set, 4 intersection, 4 A° complement, 4

geodesic, 146 L(zl, z2) geodesic, 146 hD hyperbolic metric on D, 149 LD(C) = LhD(C), 149 dD hyperbolic distance, L49 C(

A\B={aEA;a 0B},4 A the closure, 4 A interior, 5

OA =A\Aboundary, 5

A'(r) = 0(r) \ {0}, 110

C(a; r) circle, 5 I, [a, b] interval of R, 6

0'=0'(1),170

C(4). C(O: 1-' C) curves, 7

V(O; (d)), 192

lim limit, 10 tg relatively compact, 12

at = max{±x, 0}, 192

AA. (0' (r)), 171 V(O) total variation, 192

249

SYMBOLS

250

V±(O), 1.93

log+ x = log max{x, 11, 196 m(r, f ), 192

n(r,f), 192 N(r, f ), 197 T (r, f ), 197

M(r, f) maximum function, 193 p1 order, 200 E(z; p), 243

I', 207 r(z) Gamma function, 207 ((z) Riemann's zeta function, 209 Q[w1,w21 period parallelogram, 211, 212

11[w1,w2] lattice, 212

GL 2, Z 213 deg f , 214

E', 216 p(z) Weierstrass' p (pe) function, 212

Correction List Introduction to Complex Analysis (Version 1998)

p. 6, 1 8: continuous function = increasing continuous function

p. 17, 15: p. 32, 1 10: F2.-'--2) -z>

a

1

p. 32, 1 6: mean : intermediate p. 36, 1 14: series

sequence

p. 49, 1 12: f(a+z)-f(z)

f(a+h)- f(a)

p. 55, 1 20: (¢(tl) - O(ti_1))2 (01(t3) - 01(171))2 p. 57, 1 5 ti 9: ti), - I . till _1 (3 places) : a change of parameter of p. 66, 1 1: homotopic to la - zol p. 69, T 8: lal p. 72, 13, 15: zl z (2places) la - ail. Let f be a holomorphic functionon on D. The p. 73, 1 3: la -- WI. The p. 76, 1 2: (t2), : (t2) < 0,(

I[ti-1 +E,t, -E)

1). 77, 1 1: V)Ilt.,- 1, t,j 2az p. 83,15: a: p. 85, 1 7: Delete " lt(c'°)

1). 95, 1 7: f ==> f 1 p. 105, 1 13: a,,, a,,, p. 110, In FIGURE 44: y

2ai p. 114, 1 3: 2a p. 124, 1 4: polynomial

1

rational function

p. 129, 1 10: function ; functions

p. 1'30.13: f:D-'D= f:D--.D' p. 139, 1 6: ,_) an(x) p. 159. 10 ti 9: an injective = a bijective

0

p. 173, 1 9: o 1). 175, 1 18: (`tZ) p. 176, 1 1: 1 2zr p. 180, 1 9, 13:1 p. 183, 119:

DEC

/

z

( ,z,,

2 2a 1( (n - 1)! p. 217, 1 13: V(z)2 p. 218, 1 1: z, a,

(z)2

K)

6: polynomials : rational functions p. 229, 1 12: \/-2y/ y/\/2p. 219,

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