The book discusses major topics in complex analysis with applications to number theory. This book is intended as a text

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*English*
*Pages XVI, 287
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*Year 2020*

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- Tarlok Nath Shorey

*Table of contents : Front Matter ....Pages i-xvi Introduction and Simply Connected Regions (Tarlok Nath Shorey)....Pages 1-23 The Cauchy Theorems and Their Applications (Tarlok Nath Shorey)....Pages 25-71 Conformal Mappings and the Riemann Mapping Theorem (Tarlok Nath Shorey)....Pages 73-98 Harmonic Functions (Tarlok Nath Shorey)....Pages 99-116 The Picard Theorems (Tarlok Nath Shorey)....Pages 117-132 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation Theorem and the Gamma Function (Tarlok Nath Shorey)....Pages 133-185 The Riemann Zeta Function and the Prime Number Theorem (Tarlok Nath Shorey)....Pages 187-236 The Prime Number Theorem with an Error Term (Tarlok Nath Shorey)....Pages 237-250 The Dirichlet Series and the Dirichlet Theorem on Primes in Arithmetic Progressions (Tarlok Nath Shorey)....Pages 251-269 The Baker Theorem (Tarlok Nath Shorey)....Pages 271-281Back Matter ....Pages 283-287*

Infosys Science Foundation Series in Mathematical Sciences

Tarlok Nath Shorey

Complex Analysis with Applications to Number Theory

Infosys Science Foundation Series Infosys Science Foundation Series in Mathematical Sciences

Series Editors Irene Fonseca, Carnegie Mellon University, Pittsburgh, PA, USA Gopal Prasad, University of Michigan, Ann Arbor, USA Editorial Board Manindra Agrawal, Indian Institute of Technology Kanpur, Kanpur, India Weinan E, Princeton University, Princeton, USA Chandrashekhar Khare, University of California, Los Angeles, USA Mahan Mj, Tata Institute of Fundamental Research, Mumbai, India Ritabrata Munshi, Tata Institute of Fundamental Research, Mumbai, India S. R. S Varadhan, New York University, New York, USA

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Tarlok Nath Shorey

Complex Analysis with Applications to Number Theory

123

Tarlok Nath Shorey Department of Natural Sciences and Engineering National Institute of Advanced Studies Bengaluru, Karnataka, India

ISSN 2363-6149 ISSN 2363-6157 (electronic) Infosys Science Foundation Series ISSN 2364-4036 ISSN 2364-4044 (electronic) Infosys Science Foundation Series in Mathematical Sciences ISBN 978-981-15-9096-2 ISBN 978-981-15-9097-9 (eBook) https://doi.org/10.1007/978-981-15-9097-9 © Springer Nature Singapore Pte Ltd. 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microﬁlms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional afﬁliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

To Our Grandson Aarav

Preface

The text of the present book is based on my lectures at the School of Mathematics, Tata Institute of Fundamental Research (TIFR), Mumbai, India, during 1976–1977, at the Department of Mathematics, Punjab University, India, during 1977–1979 and at the Department of Mathematics, Indian Institute of Technology Bombay, India, during 2011–2016. I am indebted to my colleague Late Prof. R. R. Simha for discussions during my above lectures on Complex Analysis at TIFR. The book is divided into two parts. The ﬁrst part (Chaps. 1–6) is on Complex Analysis, and the second part (Chaps. 7–10) covers some of the topics in Number Theory, where the theory on Complex Analysis included in the ﬁrst part ﬁnds its relevance and applications. In Chap. 1, basic notions like connectedness, extended complex plane, complex integral, winding number, homotopic paths and simply connected regions are introduced. It is shown in Chap. 1 that the winding number at a point along two closed homotopic paths is equal, and the complement of a simply connected region in the extended complex plane is connected. The former statement is extended in Chap. 2 by proving that the integrals of an analytic function in a simply connected region along two closed homotopic paths are equal, and this implies the Cauchy theorem for closed paths in a simply connected region. This is also extended by a different method to the cycles homologous to zero in an open set in Chap. 2 where well-known theorems like Maximum modulus principle, Open mapping theorem, Inverse function theorem, Rouché theorem and Jensen inequality are derived from the Cauchy theorems. Regarding the latter statement that a complement of a simply connected region is connected, its converse is also valid. In fact, the Riemann mapping theorem is proved in Chap. 3, and it implies that a region is simply connected if and only if its complement in the extended plane is connected. Groups of automorphisms of the open unit disc and the upper half plane are also determined in Chap. 3. Harmonic functions are introduced in Chap. 4 where it is proved that a region X is simply connected if and only if every harmonic function in X has a harmonic conjugate in X: The Weierstrass factorisation theorem and the Hadamard factorisation theorem are proved in Chap. 6 leading to the gamma function and the Stirling formula.

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The second part begins with the Riemann Zeta function fðsÞ. We prove in Chap. 7 that fðsÞ has no zero on the line r ¼ 1. Further, we show that the Prime Number Theorem is equivalent to the non-vanishing of fðsÞ on the line r ¼ 1 by proving the Wiener-Ikehara theorem. Thus, a complete proof of the Prime Number Theorem has been given in this chapter. We give another proof in the next chapter by following the original and classical method of Hadamard and de la Vallée Poussin. In fact, we prove a stronger version with an error term in Chap. 8. Further, we give in Chap. 7 an analytic continuation of fðsÞ in C and prove its functional equation. Further, we P 1 show that q jqj ¼ 1 where q runs through the non-trivial zeros of fðsÞ and give an account of well-known conjectures in the theory of fðsÞ; in particular, we show that the Riemann hypothesis implies Lindelöf hypothesis. The proofs of both the results depend on the Borel-Carathéodry lemma proved in Chap. 5, where it has been applied for a proof of the Little Picard theorem and the Great Picard theorem. In Chap. 9, we consider the Dirichlet series. An important class of the Dirichlet series is the Dirichlet L-function Lðs; vÞ where v is a Dirichlet character. We prove their non-vanishing at s ¼ 1 for all the Dirichlet characters v and derive the Dirichlet theorem that there are inﬁnitely many primes in an arithmetic progression. In the last Chap. 10, we prove the Baker theorem that the linear independence of logarithms of algebraic numbers over rationals implies their linear independence over algebraic numbers. The proof depends on Cauchy residue theorem and an estimate for the number of zeros of an exponential polynomial in a disc proved in Chap. 2 justifying to include Chap. 10 in this book. A list of exercises is given at the end of each chapter; hints are provided for some of them, and some chapters contain examples with complete solutions. This book has 13 ﬁgures. I thank Bidisha (HRI) for drawing these ﬁgures and Bidisha, Divyum (BITS-Pillani) and Saranya (IIITG) for their remarks on a draft of the book. I thank Bidisha, Sandhya (NIAS), Sneh (IISc) and Veekesh (IMSc) for typing the manuscript and Saranya for her interest in this project from the beginning. There was no type-setting in the manuscript, and R. Thangadurai has kindly agreed to undertake this essential, difﬁcult and time-consuming task to bring the manuscript to the present state. Further, he carried out the never ending job of changes and corrections for ﬁnalising the draft of the book. I am indebted to his generous contributions. Further, I thank R. Tijdeman, T. N. Venkataramana and Michel Waldschmidt for their valuable remarks and suggestions. I am indebted to Late Professor Baldev Raj, Director NIAS, for his interest in this project; NIAS for excellent facilities and INSA for ﬁnancial support. Further, I thank the referees for their useful remarks. Finally, I thank my wife Savita for her support when I was working on this book. Bangalore, India

Tarlok Nath Shorey

Contents

1

2

Introduction and Simply Connected Regions . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Connectedness . . . . . . . . . . . . . . . . . . . . . . 1.3 Extended Complex Plane . . . . . . . . . . . . . . 1.4 Distance Between Non-intersecting Compact and Closed sets . . . . . . . . . . . . . . . . . . . . . . 1.5 Complex Integral . . . . . . . . . . . . . . . . . . . . 1.6 Homotopic Paths . . . . . . . . . . . . . . . . . . . . 1.7 Results on Simply Connected Regions . . . . . 1.8 Notation for Denoting Constants . . . . . . . . . 1.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .

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The Cauchy Theorems and Their Applications . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 The Cauchy Theorem and the Cauchy Integral Formula for Convex Open Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 An Account of Some Basic Results on Analytic Functions 2.4 The Cauchy Theorem for Closed Paths in A Simply Connected Region . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 The Cauchy Integral Formula and the Cauchy Theorem for Cycles Homologous To Zero in an Open Set and the Cauchy Residue Theorem . . . . . . . . . . . . . . . . . . 2.6 Argument Principle, Open Mapping Theorem, Maximum Modulus Principle, the Rouché Theorem and The Jensen Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 The Phragmen–Lindelöf Method: Maximum Modulus Principle in an Unbounded Strip and the Hadamard Three-Circle Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Existence of an Analytic Branch of Logarithm of Non-vanishing Analytic Functions . . . . . . . . . . . . . . . .

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2.9 The Jensen Formula . . . . . . . . . . . . . . . . . 2.10 An Estimate for the Number of Zeros of an Polynomial in a Disc . . . . . . . . . . . . . . . . 2.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 3

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Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The Cauchy–Riemann Equations . . . . . . . . . . . . . . . . . . . 4.3 Deﬁnition and Examples of Harmonic Functions . . . . . . . 4.4 Harmonic Conjugate of a Harmonic Function in a Simply Connected Region . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Maximum Principle for Harmonic Functions Satisfying MVP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 The Dirichlet Problem for Open Discs . . . . . . . . . . . . . . . 4.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Conformal Mappings and the Riemann Mapping Theorem 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Examples of Explicit Conformal Mappings . . . . . . . . . 3.3 Automorphisms of the Open Unit Disc . . . . . . . . . . . . 3.4 Automorphisms of the Upper Half Plane . . . . . . . . . . . 3.5 The Riemann Mapping Theorem and More General Theorem 3.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Lemmas for the Proof of Theorem 3.11 . . . . . . . . . . . . 3.7 Proof of Theorem 3.11 . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The Picard Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 The Borel and Carathéodory Lemma and Other Results for the Picard Theorems . . . . . . . . . . . . . . . . . . . . . . . 5.3 Proof of the Schottky Theorem 5.2 . . . . . . . . . . . . . . . 5.4 Proofs of the Little Picard Theorem 5.1 and the Great Picard Theorem 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Weierstrass Factorisation Theorem, Hadamard’s Factorisation Theorem and the Gamma Function . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Inﬁnite Products . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 The Weierstrass Elementary Factors . . . . . . . . . . . . 6.4 The Weierstrass Factorisation Theorem . . . . . . . . . 6.5 Hadamard’s Factorisation Theorem . . . . . . . . . . . . 6.6 Extension of the Weierstrass Factorisation Theorem for an Arbitrary Region . . . . . . . . . . . . . . . . . . . . .

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Representation of Meromorphic Functions by Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Applications of the Weierstrass Factorisation Theorem . The Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . Integral Representation for CðzÞ . . . . . . . . . . . . . . . . . . The Bernoulli Numbers and the Bernoulli Polynomials . The Euler–Maclaurin–Jacobi Sum Formula . . . . . . . . . . The Stirling Formula . . . . . . . . . . . . . . . . . . . . . . . . . . The Beta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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The Riemann Zeta Function and the Prime Number Theorem . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 The Euler Product for fðsÞ . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Applications of the Euler Product for fðsÞ . . . . . . . . . . . . . 7.4 The Abel Summation Formula and Integral Representations

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7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 8

ðsÞ ........................................ for ffðsÞ Analytic Continuation of fðsÞ in r [ 0 and Its Non-vanishing on the Line r ¼ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Estimates for fðsÞ and f0 ðsÞ . . . . . . . . . . . . . . . . . . . . . . . . . Introduction to the Prime Number Theorem . . . . . . . . . . . . . Equivalence of PNT and the Non-vanishing of fðsÞ on the Line r ¼ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lemmas for the Proof of the Wiener-Ikehara Theorem 7.18 . Proof of Theorem 7.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Analytic Continuation of fðsÞ in C and Functional Equation for fðsÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Function lðrÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Main Conjectures in the Theory of the Riemann Zeta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The Prime Number Theorem with an Error Term . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Positive Lower Bound for jfð1 þ itÞj and Zero-Free Region for fðsÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 An Equivalent Version of Prime Number Theorem 8.1 with Error Term in Terms of w1 ðxÞ . . . . . . . . . . . . . . 8.4 Integral Representation for w1 ðxÞ . . . . . . . . . . . . . . . . 8.5 Proof of Theorem 8.1 . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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The Dirichlet Series and the Dirichlet Theorem on Primes in Arithmetic Progressions . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Convergence of the Dirichlet Series . . . . . . . . . . . . . . . . . 9.3 Multiplication of Two Dirichlet Series . . . . . . . . . . . . . . . 9.4 Another Proof of Theorem 7.9 That fð1 þ itÞ 6¼ 0 . . . . . . . 9.5 Characters of Finite Abelian Groups . . . . . . . . . . . . . . . . 9.5.1 Properties of Characters . . . . . . . . . . . . . . . . . . . 9.6 The Dirichlet Characters . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 The Dirichlet L-Functions . . . . . . . . . . . . . . . . . . . . . . . . 9.8 Non-vanishing of Lð1; vÞ . . . . . . . . . . . . . . . . . . . . . . . . . 9.9 The Dirichlet Theorem on Primes in Arithmetic Progression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.10 The Prime Number Theorem in an Arithmetic Progression 9.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10 The Baker Theorem . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . 10.2 The Thue-Siegel Lemma . . . . . . . 10.3 Proof of the Baker Theorem 10.1 10.4 Proof of Corollary 10.2 . . . . . . . . 10.5 An Application of (10.1.1) . . . . . 10.6 Exercises . . . . . . . . . . . . . . . . . .

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References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285

About the Author

Tarlok Nath Shorey is a distinguished professor at the National Institute of Advanced Studies, Indian Institute of Science, Bengaluru, India. Earlier, he taught at the Department of Mathematics, Indian Institute of Technology Bombay, India. He was associated with the Tata Institute of Fundamental Research (TIFR), Mumbai, India, for a period of 42 years. Professor Shorey has done signiﬁcant work on transcendental number theory and Diophantine equation. In 1987, he was awarded the Shanti Swarup Bhatnagar Prize for Science and Technology—India’s highest science award—in the Mathematical Sciences category. He has coauthored a book, Exponential Diophantine Equations, and has more than 142 research publications to his credit. He is fellow of the Indian National Science Academy (INSA), Indian Academy of Sciences (IASc) and The National Academy of Sciences (NASI).

xiii

Symbols

XnY dðA; BÞ Y Y @Y ° lð°Þ X Dða; rÞ Dða; rÞ D H N Z Q R C SL2 ðRÞ Ind° ðaÞ Resðf ; aÞ HðXÞ MðXÞ Entire AutðGÞ E P ðzÞ CðzÞ Bk Bk ðxÞ /a ðzÞ CðsÞ

The complement of set Y in set X Distance between sets A and B, Exercise 1.20 Closure of Y Interior of Y Boundary of Y The range of curve ° Length of a curve °, p. 10 open set Open disc with centre a and radius r Closed disc with centre a and radius r Dð0; 1Þ Upper Half plane Positive rational integers Ring of rational integers Field of rational numbers Field of real numbers Field of complex numbers Section 3.4 Index of a with respect to ° Residue of f at a The set of all holomorphic (analytic) functions in X The set of all meromorphic functions in X Analytic in C Group of one-to-one analytic functions from an open set G onto itself Weierstrass elementary function Gamma function k-th Bernoulli number k-th Bernoulli polynomial za Automorphism /a ðzÞ ¼ 1 az of D Gamma function

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fðsÞ v vðmod mÞ nð s Þ Lðs; vÞ lðrÞ rc ra r ¼ rc r ! 1þ r ! 1 :¼ ½x fxg

Symbols

Riemann zeta function Character Dirichlet Character ðmod mÞ (7.11.11) L-function Section 7.12 Abscissa of convergence Abscissa of absolute convergence Line of convergence r tends to 1 from right r tends to 1 from left means both sides are same by deﬁnition the greatest integer x the fractional part of x

Arithmetic functions: pn lðnÞ Euler constant KðnÞ; dðnÞ; ra ðnÞ pðxÞ, #ðxÞ, wðxÞ xðnÞ; /ðnÞ w1 ðxÞ pðx; m; kÞ f ðxÞ ¼ gðxÞ þ OðhðxÞÞ f ðxÞ ¼ gðxÞ þ oðhðxÞÞ

Section 1.1 Section 7.2 limn!1 1 þ 12 þ þ 1n log n Section 7.3 Section 7.7 Section 7.14 Section 8.1 Section 9.10 Section 1.8 Section 1.8

Chapter 1

Introduction and Simply Connected Regions

1.1 Introduction We work with a metric space which we understand as a complex plane, unless otherwise specified. The letter will denote an open set in the metric space. We introduce the notion of connectedness in Sect. 1.2 and we show in Theorem 1.1 that an open set in a metric space is a disjoint union of open connected sets which we call regions. Further, we prove Theorem 1.3 in Sect. 1.3 that the extended complex plane is a metric space homeomorphic to the Riemann sphere. We introduce curves and paths, complex integral over paths, index of a point with respect to a closed path, homotopic paths, simply connected regions and give their basic properties in Sects. 1.4 and 1.5. Then we prove in Theorem 1.6 that the index of a point with respect to two -homotopic closed paths in are equal whenever the point lies outside and we apply it to prove in Theorem 1.9 that the complement of a simply connected region in the extended complex plane is connected . We shall prove in Sect. 3.5 the Riemann mapping theorem which implies the converse of the above statement i.e a region is simply connected if its complement in the extended complex plane is connected. Hence a region is simply connected if and only if its complement in the extended complex plane is connected. Thus a region is simply connected if and only if it is without holes. This is a very transparent criterion to determine whether a region is simply connected or not. For example, it implies that the unbounded strip {z | a < Re(z) < b} for given real numbers a and b is simply connected. Now we state some notation which we shall follow throughout the tract. We denote by C the complex plane with usual topology. Further, we write R for the real line and H for the upper half plane {z | Im(z) > 0}. We observe that C is a metric space with a metric given by d(z 1 , z 2 ) = |z 1 − z 2 | for z 1 , z 2 ∈ C. For a ∈ C and r > 0, © Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9_1

1

2

1 Introduction and Simply Connected Regions

D(a, r ) =

z |z − a| < r

denotes the open disc with centre at a and radius r. We write D for the open unit disc. Further, D (a, r ) =

z 0 < |z − a| < r

is the punctured disc with centre at a and radius r . We observe that it is an open set. Also we write D(a, r ) = z |z − a |≤ r the closed disc with centre at a and radius r . For A ⊆ X , we write Ac for the complement of A in X and ∂ X for the boundary of X and p1 < p2 < · · · for the sequence of (positive) prime numbers arranged in the increasing order. We use bijection for oneone onto mapping. A continuous bijection from X onto Y with continuous inverse is called homeomorphism from X onto Y . If there is homeomorphism from X onto Y , then we say that X and Y are homeomorphic.

1.2 Connectedness Let X be a metric space and E ⊆ X . We begin with a definition of connectedness. Definition A set E is connected if E cannot be written as a disjoint union of two non-empty relative open subsets of E. Thus E = A ∪ B with A ∩ B = ∅ and A, B open in E implies that either A = ∅ or B = ∅. Otherwise E = A ∪ B is called a separation E into open sets. For example, union E of two disjoint open discs A and B is not connected since E = A ∪ B = (A ∪ B) ∩ E = (A ∩ E) ∪ (B ∩ E), where A ∩ E and B ∩ E are non-empty, disjoint and relatively open in E. As in C, an open connected set in a metric space is called a region. Definition A maximal connected subset of E is called a component of E. For a ∈ E, let C(a) be the union of all connected subsets of E containing a. We observe that a ∈ C(a) since {a} is connected and E=

a∈E

We give some properties of C(a). (i) C(a) is connected

C(a).

1.2 Connectedness

3

The proof is by contradiction. Let C(a) = A ∪ B be a separation of C(a) into open sets. We may assume that a ∈ A and b ∈ B. Then, since b ∈ C(a) and C(a) is the union of all connected subsets of E containing a, there exists E 0 ⊆ E such that E 0 ⊆ C(a) is connected and a ∈ E 0 , b ∈ E 0 . Thus E 0 = E 0 ∩ C(a) = E0 ∩ (A ∪ B) = (E0 ∩ A) ∪ (E0 ∩ B) implies that either E 0 ∩ A = ∅ or E 0 ∩ B = ∅. This is a contradiction since a ∈ E 0 ∩ A and b ∈ E 0 ∩ B. Thus every component of E is of the form C(a) with a ∈ E. (ii) The components of E are either disjoint or identical Let a, b ∈ E. Assume that C(a) ∩ C(b) = ∅. Then we prove that C(a) = C(b). Let x ∈ C(a) ∩ C(b). Then x ∈ C(a). Since C(a) is connected, we derive that C(a) ⊆ C(x). Then a ∈ C(x) which implies C(x) ⊆ C(a) since C(x) is connected. Thus C(a) = C(x). Similarly C(b) = C(x) and hence C(a) = C(b). (iii) The components of an open set are open Let E be an open set. It suffices to show that C(a) with a ∈ E is open. Let x ∈ C(a). Then C(x) = C(a) by (ii). Since x ∈ E and E is open, there exists r > 0 such that D(x, r ) ⊆ E. In fact D(x, r ) ⊆ C(x) since D(x, r ) is connected containing x. Thus x ∈ D(x, r ) ⊆ C(a) and hence C(a) is open. By combining (i), (ii) and (iii), we conclude Theorem 1.1 An open set in a metric space is a disjoint union of regions. For points P0 , P1 , . . . , Ps in the complex plane, we write [P0 , P1 , . . . , Ps ] for the polygonal path obtained by joining P0 to P1 , P1 to P2 , . . ., Ps−1 to Ps by line segments. Now we give a criterion which is easy to apply for showing that the sets in the plane are connected. Theorem 1.2 Let E be a non-empty open subset of C. Then E is connected if and only if any two points in E can be joined by a polygonal path that lies in E. Proof Assume that E is connected. Since E = ∅, let a ∈ E. Let E 1 be the subset of all elements of E that can be joined to a by a polygonal path. Let E 2 be the complement of E 1 in E. Then E = E 1 ∪ E 2 with E 1 ∩ E 2 = ∅, a ∈ E 1 . It suffices to show that both E 1 and E 2 are open subsets of E. Then E 2 = ∅ since E is connected and a ∈ E 1 . Thus every point of E can be joined to a by a polygonal path that lies in E. Hence any two points of E can be joined by a polygonal path that lies in E via a. First, we show that E 1 is open. Let a1 ∈ E 1 . Then a1 ∈ E and since E is open, we find r1 > 0 such that D(a1,r1 ) ⊆ E. Any point of D(a1,r1 ) can be joined to a1 and hence to a by a polygonal path that lies in E since a1 ∈ E 1 . Thus a1 ∈ D(a1 , r1 ) ⊆ E 1 . Next, we show that E 2 is open. Let a2 ∈ E 2 . Again we find r2 > 0 such that

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1 Introduction and Simply Connected Regions

D(a2 , r2 ) ⊆ E since E is open. Now, as above, we see that no point of this disc can be joined to a as a2 ∈ E 2 and hence a2 ∈ D(a2 , r2 ) ⊆ E 2 . Now we assume that if any two points of E can be joined by a polygonal path in E and we show that E is connected. Let E = E1 ∪ E2 be a separation of E into open sets. There is no loss of generality in assuming that there exist points a1 ∈ E 1 and a2 ∈ E 2 such that χ(t) = ta1 + (1 − t)a2 with 0 < t < 1 is an open segment from a2 to a1 lying in E. Let V = {t ∈ (0, 1)|χ(t) ∈ E 1 } and W = {t ∈ (0, 1)|χ(t) ∈ E 2 }. We see that V and W are open in (0, 1). Further, we have separation of the open interval (0, 1) into open sets (0, 1) = V ∪ W, V ∩ W = ∅. Since a1 ∈ E 1 and E 1 is open, there exists r3 > 0 with D(a1 , r3 ) ⊆ E 1 . This implies V = ∅. Similarly W = ∅. Hence the interval (0, 1) is not connected. This is a contradiction.

1.3 Extended Complex Plane By adjoining a point ∞, which we call the point at ∞, to C we get the extended complex plane C∞ = C ∪ {∞}. Next, we introduce the Riemann sphere S = (x1, x2, x3 ) ∈ R3 x12 + x22 + x32 = 1 and f : S → C∞ given by f ((x1 , x2 , x3 )) = and

x1 + i x2 if (x1 , x2 , x3 ) = (0, 0, 1) 1 − x3

1.3 Extended Complex Plane

5

f ((0, 0, 1)) = ∞. We write N = (0, 0, 1) and we call N the north pole of S. Further the function f is called stereographic projection. First, we show that f is one-one. Let (x1 , x2 , x3 ) and (x1 , x2 , x3 ) be distinct elements of S such that f ((x1 , x2 , x3 )) = f ((x1 , x2 , x3 )). Then (x1 , x2 , x3 ) and (x1 , x2 , x3 ) are different from N , and therefore x1 + i x2 x + i x2 = 1 := z ∈ C. 1 − x3 1 − x3 Thus

|z|2 =

(1.3.1)

x12 + x22 x 2 + x22 = 1 . 2 (1 − x3 ) (1 − x3 )2

Since (x1 , x2 , x3 ) and (x1, x2, x3 ) are in S, we have

|z|2 =

1 − x32 1 − x32 = (1 − x3 )2 (1 − x3 )2

implying |z|2 = Therefore

1 + x3 1 + x3 = . 1 − x3 1 − x3

|z|2 − 1 = x3 , |z|2 + 1

|z|2 − 1 = x3 . |z|2 + 1

Thus x3 = x3 and hence x1 = x1 , x2 = x2 by (1.3.1). Next, we show that f is onto. If z = ∞, we take N ∈ S so that f (N ) = ∞. Thus, we may suppose that z ∈ C. We take x1 =

z+z z−z |z|2 − 1 , x , x . = = 2 3 |z|2 + 1 i(|z|2 + 1) |z|2 + 1

(1.3.2)

Then x12 + x22 + x32 =

(z + z)2 − (z − z)2 + (|z|2 − 1)2 4|z|2 + (|z|2 − 1)2 = =1 (|z|2 + 1)2 (|z|2 + 1)2

implying (x1 , x2 , x3 ) ∈ S. Further, we check that f ((x1 , x2 , x3 )) = z. Geometric interpretation of the Stereographic projection. Let z = x + i y be a point in the complex plane and we identify it by (x, y, 0) in R3 . We consider the line joining z to N in R3 . Its parametric representation is t N + (1 − t)z, −∞ < t < ∞.

6

1 Introduction and Simply Connected Regions N Z

S

z

C

Fig. 1.1 Stereographic projection

The points on this line are ((1 − t)x, (1 − t)y, t), −∞ < t < ∞. Thus, the line intersects S if and only if (1 − t)2 x 2 + (1 − t)2 y 2 + t 2 = 1 which we re-write as (1 − t)|z|2 = (1 − t)x 2 + (1 − t)y 2 = 1 + t. Hence t=

|z|2 − 1 . |z|2 + 1

Thus, the line intersects S at Z given by |z|2 − 1 |z|2 − 1 |z|2 − 1 2x 2y |z|2 − 1 1− 2 x, 1 − 2 y, = , , |z| + 1 |z| + 1 |z|2 + 1 |z|2 + 1 |z|2 + 1 |z|2 + 1 z−z |z|2 − 1 z+z . = , , |z|2 + 1 i(|z|2 + 1) |z|2 + 1

We summarise the above procedure as follows: Let z = (x, y) be a point in the complex plane. The line passing through (x, y, 0) and (0, 0, 1) intersects S exactly at one point Z = (x1 , x2 , x3 ) given by (1.3.2) and f ((x1 , x2 , x3 )) = z. We call (x1 , x2 , x3 ) the spherical coordinates of z and (0, 0, 1) the spherical coordinates of ∞ (Fig. 1.1).

1.3 Extended Complex Plane

7

Distance between points in the extended complex plane. Let z ∈ C∞ and z ∈ C∞ . x2, x3 ) be spherical coordinates of z and z , respectively. Then Let (x1, x2, x3 ) and (x1, we define d(z, z ) = (x1 − x1 )2 + (x2 − x2 )2 + (x3 − x3 )2 . Thus d 2 (z, z ) = (x1 − x1 )2 + (x2 − x2 )2 + (x3 − x3 )2 = 2 − 2(x1 x1 + x2 x2 + x3 x3 ). (1.3.3) Now by (1.3.2) x1 x1 + x2 x2 + x3 x3 =

(z + z)(z + z ) (z − z)(z − z ) (|z|2 − 1)(|z |2 − 1) − + (|z|2 + 1)(|z |2 + 1) (|z|2 + 1)(|z |2 + 1) (|z|2 + 1)(|z |2 + 1)

=

(|z|2 − 1)(|z |2 − 1) + 2zz + 2zz (|z|2 + 1)(|z |2 + 1)

=

(|z|2 + 1)(|z |2 + 1) − 2(|z|2 + 1) − 2(|z |2 + 1) + 4 + 2zz + 2zz . (|z|2 + 1)(|z |2 + 1)

Further 2(|z|2 + 1) + 2(|z |2 + 1) − 4 − 2zz − 2zz = 2(|z|2 + |z |2 − zz − zz ) = 2|z − z |2 .

Hence 2(x1 x1 + x2 x2 + x3 x3 ) = 2 −

4|z − z |2 , (|z|2 + 1)(|z |2 + 1)

which, together with (1.3.3), implies d 2 (z, z ) =

4|z − z |2 . (|z|2 + 1)(|z |2 + 1)

Hence d(z, z ) =

2|z − z | (|z|2 + 1)(|z |2 + 1)

.

Let z = ∞. Then x1 = 0, x2 = 0 and x3 = 1 and by (1.3.3) d 2 (z, ∞) = 2 − 2x3 = 2 − 2 Hence d(z, ∞) =

|z|2 − 1 |z|2 + 1 2

|z|2 + 1

.

=

4 . +1

|z|2

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1 Introduction and Simply Connected Regions

Thus C∞ is a metric space. Further, the above metric d induces the following topology / E, then E is open in C∞ if and only if it is open in C. on C∞ . Let E ⊆ C∞ . If ∞ ∈ If ∞ ∈ E, then E is open in C∞ if and only its complement in C∞ is a closed and bounded subset of C. Finally, we check that f and f −1 are continuous to conclude the following result. Theorem 1.3 The extended complex plane is a metric space homeomorphic to the Riemann sphere given by f −1 .

1.4 Distance Between Non-intersecting Compact and Closed sets We prove the following. Theorem 1.4 For a compact set A and a closed set B in the complex plane with A ∩ B = ∅, there exists δ > 0 such that |a − b| ≥ δ for every a ∈ A and b ∈ B. Proof We observe that A ⊆ B c and B c is open. Here B c denotes the complement of B in the complex plane. Therefore, for every a ∈ A, there exists δa > 0 such that D(a, δa ) ⊆ B c . Thus A⊆

D(a, δa ) ⊂ B c .

a∈A

Let δ = inf δa where the infimum is taken over all a ∈ A. Since A is compact, we know that δ > 0, see [[14], p. 78]. Therefore |a − b| ≥ δ for a ∈ A and b ∈ B.

1.5 Complex Integral Let a be a complex number and f be a complex valued function defined in a neighbourhood of a. If the derivative of f at a lim

z→a

f (z) − f (a) z−a

1.5 Complex Integral

9

exists, we denote it by f (a) and we say that f is analytic at a or f is holomorphic at a. Let f be defined in . If f is analytic at every point of , we say that f is analytic in or f is holomorphic in . We denote by H () the set of all functions holomorphic in . A function holomorphic in C is called an entire function. By a curve γ, we mean a continuous function γ from some closed interval [α, β] into C. The interval [α, β] is called its parameter interval. Thus γ is given by γ(t) with α ≤ t ≤ β. We shall also sometimesthat γ is given by z(t) with α ≤ t ≤ β. write We denote by γ ∗ the range γ(t) α ≤ t ≤ β of γ. We observe that γ ∗ is compact and connected since it is a continuous image of a compact and connected interval [α, β]. If a curve γ is piecewise continuously differentiable, then γ is called a path (with parameter interval [α, β]). Thus, there are finitely many α = s0 < s 1 < . . . < s n = β such that the restriction of γ to [s j−1 , s j ] with 1 ≤ j ≤ n are continuously differentiable. We observe that the left-hand derivative of s1 , . . . , sn−1 need not coincide with the right-hand derivative at s1 , . . . , sn−1 , respectively. The path γ is closed if its initial point γ(α) coincides with its end point γ(β). Integral along a path. Let γ be a path with parameter interval [α, β]. Assume that f is continuous on γ ∗ . Then we define

γ

f (z)dz =

β

α

f (z(t))z (t)dt

where γ is given by z(t) with α ≤ t ≤ β. When γ is closed path, then integration over γ is understood to be in the anticlockwise direction, unless otherwise mentioned. Here we notice that the integral on the right-hand side is the Riemann integral since z (t) is a bounded function of t in [α, β] with at most finitely many discontinuities. Properties of the Integral (i) Let φ : [a, b] → [α, β] be continuous, strictly increasing and onto function. Further assume that φ is continuously differentiable. Then φ(a) = α, φ(b) = β and φ([a, b]) = [α, β]. Let γ be a path with parameter interval [α, β]. Then γ1 = γ ◦ φ is a path with parameter interval [a, b]. Let f be continuous on γ ∗ . Then f is also continuous on γ1∗ and we have

γ1

f (z)dz =

γ

f (z)dz.

We call φ a change of parameter function. (ii) Let γ1 and γ2 be paths such that the end point of γ1 coincides with the initial point of γ2 . Then, after suitable re-parametrisation, we get a path γ by following first γ1 and then γ2 . By (i), we have

γ

f (z)dz =

γ1

f (z)dz +

γ2

f (z)dz,

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1 Introduction and Simply Connected Regions

where f is continuous on γ1∗ ∪ γ2∗ . We write γ = γ1 + γ2 . For paths γ1 , γ2 , . . . , γn such that the end points of γ j coincides with the initial point of γ j+1 with 1 ≤ j < n, the path γ = γ1 + γ2 + · · · + γn is defined similarly. (iii) Let γ be a path with parameter interval [0, 1]. We define γ1 (t) = γ(1 − t) for 0 ≤ t ≤ 1. Then γ1 is called a path opposite to γ. We have

γ

f (z)dz = −

γ1

f (z)dz,

where f is continuous on γ ∗ . (iv) Let γ be a path with parameter interval [α, β] and f be continuous on γ ∗ . Then

f (z)dz = γ

α

β

f (z(t))z (t)dt ≤ max∗ | f (z)| z∈γ

β

α

| z (t) | dt

= l(γ) max | f (z) |, ∗ z γ

where l(γ) =

α

β

|z (t)|dt is the length of γ.

Let γ be a closed path. Then the complement of γ ∗ in metric space C∞ is open. Thus it is a disjoint union of regions by Theorem 1.1. We say that these regions are determined by γ in C∞ . There is only one region determined by γ which is unbounded and we call it the unbounded region determined by γ. We observe that it contains ∞. The regions determined by γ in C∞ and the regions determined by γ in C are identical except that the unbounded region determined by γ in C does not contain ∞. For a ∈ C and a ∈ / γ ∗ , we write 1 Indγ (a) = 2πi

γ

dz dz, z−a

and Indγ (a) is called the index of a with respect to γ. This is also called the winding number of a with respect to γ. Further Indγ (a) satisfies the following very useful property given by the next result. Theorem 1.5 For a closed path γ and = C \ γ ∗ , we have Indγ (a) ∈ Z for a ∈ . Further Indγ (a) is constant on each region of determined by γ and it is equal to zero on the unbounded region of determined by γ. The proof of the first assertion is clear if we assume that it is possible to define arg(z − a) uniquely in a region containing γ. Then Indγ (a) is the number of times γ wraps around a. That is why it is also called the winding number of γ around a.

1.5 Complex Integral

11

Proof By definition Indγ (a) =

1 2πi

γ

1 1 dz = z−a 2πi

β

α

z (t) dt z(t) − a

where γ is a closed path given by z = z(t) with α ≤ t ≤ β and z(t) = a for α ≤ t ≤ β. We consider

t z (t) dt. h(t) = α z(t) − a We prove that h(β) is a multiple of 2πi and this implies Indγ (a) is an integer. Since z(t) is piecewise continuously differentiable, we see that the integral on the right exists for α ≤ t ≤ β. Further h(t) is continuous on [α, β] and h (t) =

z (t) z(t) − a

for all but finitely many t ∈ [α, β]. Now we observe that the derivative of e−h(t) (z(t) − a) vanishes for all but finitely many t ∈ [α, β]. This implies that e−h(t) (z(t) − a) = c in [α, β], where c is a constant since the function on the left is continuous in [α, β]. By putting t = α and t = β, we have e−h(α) (z(α) − a) = e−h(β) (z(β) − a) / γ ∗ and h(α) = 0. which implies eh(β) = 1 since z(α) = z(β), α ∈ The function Indγ (z) is integer valued on . Further it is continuous on . Therefore for any component C of , we see that Indγ (C) is a connected set of integers and hence it consists of a single element. ≥ |a| and |z| > l(γ) . Then Next we take z in the unbounded region such that |z| 2 π |z| |z − a| ≥ |z| − |a| ≥ 2 and | Indγ (z) |≤

1 2 l(γ) < 1. 2π | z |

This implies Indγ (z) = 0 on the unbounded region, since Indγ (z) is integer valued and constant on the unbounded region as already proved. Now, we introduce chains and cycles which are more general than paths and closed paths, respectively, and we shall integrate over them. Definitions. (i) Let γ1 , . . . , γn be paths such that γi∗ ⊂ for 1 ≤ i ≤ n. Let be a formal sum given by ˙ · · · +γ ˙ n. (1.5.1) = γ1 +

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1 Introduction and Simply Connected Regions

Then we say that is a chain in . If there exists a representation of a chain such that each γi is a closed path in , then we say that is a cycle in . By a combinatorial argument, it can be shown that a chain is a cycle if and only if in any representation of , the initial and end points of γi are identical in pairs. (ii) Let f be continuous on γ1∗ ∪ γ2∗ ∪ · · · ∪ γn∗ . Then we define

f (z)dz =

n

k=1

f (z)dz.

γk

(iii) Let γ1 , γ2 , . . . , γn and δ1 , δ2 , . . . , δm be chains in . Then we define ˙ 2+ ˙ · · · +γ ˙ n = δ1 +δ ˙ 2+ ˙ · · · +δ ˙ m γ1 +γ if

n

k=1

γk

f (z)dz =

m

j=1

δj

f (z)dz

for every function f continuous on γ1∗ ∪ γ2∗ ∪ · · · ∪ γn∗ ∪ δ1∗ ∪ δ2∗ ∪ · · · ∪ δm∗ . (iv) For a cycle given by (1.5.1) and a ∈ , we define Ind (a) = Indγ1 (a) + · · · + Indγn (a).

1.6 Homotopic Paths Let γ0 and γ1 be closed paths in such that I = [0, 1] is the parameter interval for both. Then γ0 and γ1 are -homotopic if there exists a continuous function H :I×I → such that H (s, 0) = γ0 (s), H (s, 1) = γ1 (s) for 0 ≤ s ≤ 1 and H (0, t) = H (1, t) for 0 ≤ t ≤ 1. We have taken the interval [0, 1] as the parameter interval for both γ0 and γ1 . We can define homotopic paths when γ0 and γ1 have different parameter intervals. Let γ0 and γ1 be paths in with parameter intervals [α0 , β0 ] and [α1 , β1 ], respectively. For i = 0, 1, let φi : [0, 1] → [αi , βi ] be given by

1.6 Homotopic Paths

13

φi (t) = βi t + (1 − t)αi and we observe that γ0 ◦ φ0 and γ1 ◦ φ1 are closed paths such that [0, 1] is the parameter interval for each of them. Then we define that γ0 and γ1 are -homotopic if γ0 ◦ φ0 and γ1 ◦ φ1 are -homotopic. Equivalence relation in homotopic paths. Consider all closed paths in with parameter interval [0, 1]. Denote this set by . For γ0 , γ1 ∈ , we say γ0 ∼ γ1 if γ0 and γ1 are -homotopic. (i) Reflexive: γ0 ∼ γ0 by taking H (s, t) = γ0 (s) for 0 ≤ s ≤ 1, 0 ≤ t ≤ 1. (ii) Symmetric: Let γ0 ∼ γ1 . Then there exists a continuous function H : I×I → satisfying H (s, 0) = γ0 (s), H (s, 1) = γ1 (s) and H (0, t) = H (1, t) for 0 ≤ s ≤ 1, 0 ≤ t ≤ 1. Now we take H1 (s, t) = H (s, 1 − t) to observe that H1 (s, 0) = γ1 (s), H1 (s, 1) = γ0 (s) and H1 (0, t) = H1 (1, t). Thus γ1 ∼ γ0 . (iii) Transitive: Let γ0 , γ1 , γ2 ∈ with γ0 ∼ γ1 and γ1 ∼ γ2 . Then there exist H1 : I × I → , H2 : I × I → satisfying H1 (s, 0) = γ0 (s), H1 (s, 1) = γ1 (s), H1 (0, t) = H1 (1, t) and H2 (s, 0) = γ1 (s), H2 (s, 1) = γ2 (s), H2 (0, t) = H2 (1, t). We define H (s, t) =

if 0 ≤ t ≤ 21 H1 (s, 2t) H2 (s, 2t − 1) if 21 ≤ t ≤ 1.

We check that H is continuous on I × I and satisfies H (s, 0) = H1 (s, 0) = γ0 (s), H (s, 1) = H2 (s, 1) = γ2 (s), H (0, t) = H (1, t). Hence γ0 ∼ γ2 . Definition If γ0 ∈ is -homotopic to a constant path, then we say that γ0 is null homotopic in . If is a region and γ0 ∈ is null homotopic, then γ0 is homotopic to every constant path in . For observing this, let γ1 = a and γ2 = b be constant paths in . Since is connected, there is a path z(t) with 0 ≤ t ≤ 1 with z(0) = a and z(1) = b. Then we take H (s, t) = z(t) for 0 ≤ s ≤ 1 to derive that γ1 ∼ γ2 . Now the assertion follows since ∼ is transitive.

14

1 Introduction and Simply Connected Regions

1.7 Results on Simply Connected Regions Definition If is connected and every closed path in is null homotopic in , then we say that is simply connected. Now we turn to proving the following result. Theorem 1.6 Let 0 and 1 be closed paths in each having parameter interval [0, 1]. Assume that they are -homotopic. Then / . Ind0 (a) = Ind1 (a) for a ∈ As an immediate consequence of Theorem 1.6, we have Corollary 1.7 Let be simply connected region and be a cycle in . Then Ind (a) = 0 for a ∈ / . ˙ 2+ ˙ · · · +γ ˙ n , where each γi is a closed path in and a ∈ / . Proof Let = γ1 +γ Since is simply connected, we see that each γi is null homotopic in . Therefore, we derive from Theorem 1.6 that Indγi (a) = 0 for 1 ≤ i ≤ n and the assertion follows. Now we introduce the following definition. Definition Let be a cycle in satisfying Ind (a) = 0 for a ∈ / . Then we say that is homologous to zero in . Thus every cycle in a simply connected region is homologous to zero in . The proof of Theorem 1.6 depends on the following result. Lemma 1.8 Let γ0 , γ1 be closed paths in with parameter intervals [0, 1]. Let α ∈ C. Assume that | γ1 (s) − γ0 (s)| < |α − γ0 (s)| for 0 ≤ s ≤ 1.

(1.7.1)

Then Indγ0 (α) = Indγ1 (α). / γ1∗ . If α ∈ γ0∗ , then α = γ0 (s) Proof First we derive from (1.7.1) that α ∈ / γ0∗ and α ∈ for some 0 ≤ s ≤ 1 and then by (1.7.1), |γ1 (s) − γ0 (s)| < |γ0 (s) − γ0 (s)|. This is a contradiction. If α ∈ γ1∗ , then α = γ1 (s) for some 0 ≤ s ≤ 1 and by (1.7.1) |γ1 (s) − γ0 (s)| < |γ1 (s) − γ0 (s)|

1.7 Results on Simply Connected Regions

15

which is again a contradiction. Now, since α ∈ / γ0∗ , α ∈ / γ1∗ , Indγ0 (α) and Indγ1 (α) are defined. We consider γ1 (s) − α for 0 ≤ s ≤ 1. (1.7.2) γ(s) = γ0 (s) − α We see that γ is a curve since α ∈ / γ0∗ . Further γ (s) =

(γ0 (s) − α)γ1 (s) − (γ1 (s) − α)γ0 (s) . (γ0 (s) − α)2

(1.7.3)

Now we check that γ is a closed path since γ0 and γ1 are closed paths. We have γ(s) − 1 =

γ1 (s) − γ0 (s) γ1 (s) − α −1= . γ0 (s) − α γ0 (s) − α

Therefore by (1.7.1) |γ(s) − 1 |< 1 for 0 ≤ s ≤ 1. Thus γ ∗ ⊆ D(1, 1). Therefore 0 lies in the unbounded region determined by γ and hence Indγ (0) = 0 by Theorem 1.5. Now 0 = Indγ (0) =

1 2πi

γ

dz 1 = z 2πi

1 0

γ (s) ds. γ(s)

Further, by (1.7.2) and (1.7.3), we have (γ0 (s) − α)γ1 (s) − (γ1 (s) − α)γ0 (s) γ1 (s) γ0 (s) γ (s) = = − . γ(s) (γ0 (s) − α)(γ1 (s) − α) γ1 (s) − α γ0 (s) − α Therefore 1 2πi

1 0

1 γ1 (s) ds = γ1 (s) − α 2πi

0

1

γ0 (s) ds, γ0 (s) − α

and hence Indγ0 (α) = Indγ1 (α). Proof of Theorem 1.6 Proof Since 0 and 1 are -homotopic, there exists a continuous function H :I×I → such that

16

1 Introduction and Simply Connected Regions

H (s, 0) = 0 (s), H (s, 1) = 1 (s), H (0, t) = H (1, t)

(1.7.4)

for 0 ≤ s ≤ 1, 0 ≤ t ≤ 1. Being a continuous image of a compact set I × I , we see that H (I × I ) is compact subset of and further the complement of in C is closed. Therefore, by Theorem 1.4, there exists δ > 0 such that |z − H (s, t)| > 2δ for (s, t) ∈ I × I, z ∈ / .

(1.7.5)

Since H is uniformly continuous on I 2 , these exists a positive integer n such that |H (s, t) − H (s , t )| < δ whenever |s − s |+ | t − t |≤

(1.7.6)

1 . n

For 0 ≤ k ≤ n, we have closed curves H (s, nk ) with 0 ≤ s ≤ 1 by (1.7.4). By (1.7.4)– (1.7.6), we shall approximate these curves by closed paths such that the assumption (1.7.1) of Lemma 1.8 is satisfied for any two paths corresponding to two consecutive values of k. Then the assertion follows from Lemma 1.8. Let 0 ≤ k ≤ n be given. For 1 ≤ i ≤ n, we shall define i i −1 ≤s≤ n n

γk,i (s) with such that γk,i

i −1 n

=H

i −1 k , n n

i i k , γk,i =H , . n n n

(1.7.7)

We observe that there is bijection τ from i−1 , i onto [0, 1] given by τ (s) = 1 − n n (i − ns). Therefore the line segment from γk,i i−1 to γk,i ni is given by n (1 − τ (s))γk,i

for

i−1 n

i −1 n

+ τ (s)γk,i

i −1 k i k i = (1 − τ (s))H , + τ (s)H , n n n n n

≤ s ≤ ni . Now we define for

i−1 n

γk,i (s) = (1 − τ (s))H

≤s≤

i −1 k , n n

Thus

i n

+ τ (s)H

i k , n n

.

i −1 i ≤s≤ n n i k i−1 k is a line segment from H n , n to H n , n . Further we see from (1.7.7) γk,i (s) with

(1.7.8)

1.7 Results on Simply Connected Regions

17

i i i k = γk,i+1 =H , for 1 ≤ i ≤ n. n n n n

γk,i

(1.7.9)

Let = D H i−1 , nk , δ . By (1.7.5), we observe that ⊆ . Also H ni , nk ∈ n by (1.7.6). Therefore γk,i (s) ∈ ⊆ with

i i −1 ≤s≤ n n

since is convex. We define γk on [0,1] such that i −1 i , = γk,i for 1 ≤ i ≤ n. γk n n

(1.7.10)

By (1.7.9), we observe that γk is well defined and further, it is clear that γk is a path in satisfying k γk (0) = γk,1 (0) = H 0, n k γk (1) = γk,n (1) = H 1, n

and

by (1.7.10) and (1.7.7) with i = 1 and i = n. Since H 0, nk = H 1, nk by (1.7.4), we see that γk is a closed path. Now we show that γk (s) − H s, k < δ for 0 ≤ s ≤ 1, 0 ≤ k ≤ n. (1.7.11) n For 1 ≤ i ≤ n and observe that

i−1 n

≤ s ≤ ni , we consider 1 = D H s, nk , δ . By (1.7.6), we

H

i −1 k , n n

∈ 1 , H

i k , n n

∈ 1 .

Therefore γk,i (s) ∈ 1 since 1 is convex. Thus γk,i (s) − H s, k < δ for i − 1 ≤ s ≤ i . n n n Since it holds for every i with 1 ≤ i ≤ n, the assertion follows. Let a ∈ / . Then we derive from (1.7.5) and (1.7.11) that for 0 ≤ s ≤ 1, we have k k ≤ a − γk (s) + γk (s) − H s, ≤| a − γk (s) | +δ. 2δ < a − H s, n n

18

1 Introduction and Simply Connected Regions

Thus | a − γk (s) |> δ for 0 ≤ s ≤ 1.

(1.7.12)

≤ s ≤ ni with By (1.7.10), (1.7.8) and (1.7.6), we obtain for 0 ≤ k < n and i−1 n 1 ≤ i ≤ n that i k i k+1 | γk (s) − γk+1 (s) | ≤ τ (s) H , −H , n n n n i −1 k i − 1 k + 1 , −H , + (1 − τ (s)) H n n n n < (τ (s) + (1 − τ (s))) δ = δ. (1.7.13) Further by (1.7.4) and (1.7.11) with k ∈ {0, n}, we have | γ0 (s) − 0 (s) |=| γ0 (s) − H (s, 0) |< δ

(1.7.14)

| γn (s) − 1 (s) |=| γn (s) − H (s, 1) |< δ.

(1.7.15)

and We observe that the assumptions of Lemma 1.8 with α = a are satisfied by (1.7.12)– (1.7.15) for pairs (γ j−1 , γ j ) with 1 ≤ j ≤ n, (γ0 , 0 ) and (γn , n ). Therefore we conclude from Lemma 1.8 that Indγn (a) = Indγn−1 (a) = · · · = Indγ1 (a) = Indγ0 (a) and Indγ0 (a) = Ind0 (a), Indn (a) = Ind1 (a). Hence Ind0 (a) = Ind1 (a). Finally, we apply Theorem 1.6 and Corollary 1.7 to prove that the complement of a simply connected region in the extended plane is connected. Theorem 1.9 Let be simply connected. Then C∞ \ is connected. We derive from Theorem 1.5 that every cycle in is homologous to 0 in whenever C∞ \ is connected. This is more general than Corollary 1.7 in view of Theorem 1.9. Proof Assume that C∞ \ is not connected. Let C∞ \ = A ∪ B

1.7 Results on Simply Connected Regions

19

Fig. 1.2 Cycle with non-zero index

A

a

Γ

B

be a disconnection of C∞ \ into closed sets, see Exercise 1.4. Since C∞ \ is closed, we see that A and B are closed in C∞ . We may assume that ∞ ∈ B. Then ∞∈ / A. If A is not bounded, then for every n ≥ 1 there exists xn ∈ A such that |xn | > n. Since lim xn = ∞ and A is closed, we see that ∞ ∈ A. This is a contradiction. n→∞

Thus A is bounded. Therefore A is compact and by Theorem 1.4, there exists δ > 0 such that |x − y| > 2δ for x ∈ A, y ∈ B. (1.7.16) We partition the plane with horizontal lines such that the distance between consecutive ones is equal to δ. We also partition the plane with vertical lines such that the distance between the consecutive ones is equal to δ. Thus we have covered the plane with squares Q of sides δ. We denote by δ Q the boundary of Q with anticlockwise direction. Let a ∈ A. Then a ∈ / and there is one and only one square containing a. Further, we can cover the plane in such a way that a is in the centre of this square (Fig. 1.2). Since A is compact, there are only finitely many squares which intersect A. Let =

∂Q j

j

where the formal sum is taken over j such that Q j has non-empty intersection with A. This is a finite sum and each ∂ Q j is a closed path.√Thus is a cycle. Since the distance between any two points in a square is at most 2δ < 2δ and each summand in intersects with A, we see from (1.7.16) that does not meet B. Any side of a square which meets A is a common side of two squares. Further, the direction of this side in one square is opposite to the direction of the side in the other square and it gets cancelled in the sum . Thus there are rectangles R0 , R1 , . . . , Rm such that a lies inside R0 , a lies outside each of R1 , R2 , . . . , Rm and = ∂ R0 + ∂ R1 + · · · + ∂ Rm

(1.7.17)

with ∂ R j ∩ A = ∅ and ∂ R j ∩ B = ∅ for 0 ≤ j ≤ m. Thus ∂ R j with 0 ≤ j ≤ m and hence lie in . Since Ind∂ R0 (a) = 0 and Ind∂ R j (a) = 0 for 1 ≤ j ≤ m, we derive from (1.7.17) that Ind (a) = 1. On the other hand, we derive from Corollary

20

1 Introduction and Simply Connected Regions

1.7 that Ind (a) = 0 since is simply connected, lies in and a ∈ / . This is a contradiction.

1.8 Notation for Denoting Constants It is not always convenient to mention the constants in a proof. Let f, g and h be complex valued functions defined on X ⊆ C. Then f (x) = g(x) + O(h(x)) if there exists constant c such that | f (x) − g(x)| ≤ c|h(x)| for x ∈ X. If there is ambiguity from the context, we write f (x) = g(x) + O(h(x)) on X. If g(x) = 0, then we write

f (x) = O(h(x))

and sometimes f (x) = Oθ (h(x)) if the constant implied by Oθ depends only on θ. We also use the notation f (x) h(x) for f (x) = O(h(x)) and f (x) θ h(x) for f (x) = Oθ (h(x)). Further we write f (x) = g(x) + o(h(x)) if lim

x→∞,x∈X

| f (x) − g(x)| = 0. |h(x)|

and f (x) = o(h(x)) if g(x) = 0.

1.9 Exercises 1.1 Prove that |z| ≤ |Re(z)| + |Im(z)| ≤ for every z ∈ C.

√ 2|z|

1.9 Exercises

21

1.2 Let ω1 , ω2 ∈ C such that such that

ω1 is not real. Then there exists a constant C > 0 ω2

|n 1 ω1 + n 2 ω2 |2 ≥ C(|n 1 |2 + |n 2 |2 ) for all n 1 , n 2 ∈ Z with (n 1 , n 2 ) = (0, 0). ω2 (Hint: For arg = θ with 0 < θ < π, show that ω1 |n 1 ω1 + n 2 ω2 |2 = (n 1 ω1 + n 2 ω2 )(n 1 ω1 + n 2 ω2 ) ≥ (1 − | cos θ|)a(n 21 + n 22 ),

1.3

1.4

1.5 1.6

where a = min (|ω1 |, |ω2 |).) 1 tends to zero uniformly in z = x + i y with 0 ≤ x ≤ 1 and Show that sin πz |y| → ∞. (Hint: Use sin z = (ei z − e−i z )/2i.) Show that a set E is connected if and only if it cannot be written as a disjoint union of two non-empty relative closed subsets of E (Otherwise E = A ∪ B is called a separation of E into closed sets). Show that the union of two regions is a region if and only if they have a common point. A set E is called star shaped if there exists z 0 ∈ E such that every point in E can be joined to z 0 by a line lying in E. A set E is called convex if any two points in E can be joined by a line that lies in E. Show that

(a) A convex set is star shaped. (b) An open star shaped set and an open convex set is a region. 1.7 Assume that A and B are closed sets in a metric space such that A ∪ B and A ∩ B are connected. Then show that A is connected. 1.8 Prove that the closure of a connected set is connected and derive that components of closed sets are closed. 1.9 Compute the components of Q. 1.10 Give an example to show that the assumption E non-empty open set in Theorem 1.2 is necessary. 1.11 Show that the extended plane is compact. 1.12 Let n ∈ Z and γ(t) = a + e2πint with 0 ≤ t ≤ 1. Then show that Indγ (a) = n. 1.13 Let γ1 and γ2 be closed paths given by circles (x − 1)2 + y 2 = 1 and (x − 2)2 + y 2 = 4 in anticlockwise direction. Then show that = γ1 + γ2 is a closed path and Ind (1) = 2, Ind (3) = 1, Ind (5) = 0. 1.14 Show that a star shaped open set is simply connected. Then derive that open convex set is simply connected. 1.15 Let be the set obtained from C by deleting all real numbers ≤ 0. Then show that is simply connected and log z = log |z| + iargz

22

1 Introduction and Simply Connected Regions

where −π < argz ≤ π is analytic in . This is called the principal logarithmic branch (or principal branch of logarithm) of z. Unless otherwise specified, we understand that the branch of logarithm is principal in . 1.16 Derive from Corollary 1.7 as well as from Theorem 1.9 that for 0 < R2 < R1 and a ∈ C, the set z R2 < |z − a| < R1 is not simply connected. 1.17 Show that a real-valued analytic function in a region is constant. (Hint. Allow z = a + h tend to a through real values h and through purely imaginary values in the definition of analytic function at a point in the beginning of §1.5.) 1.18 Let u be a continuous function in [−π, π]. Show that

f (z) =

π

−π

eit + z u(eit )dt eit − z

is analytic in D(0, r ) with 0 ≤ r < 1. 1.19 Let γ be a path. Then prove the following statements. (a) Suppose f ∈ H () and f is continuous in . Then

γ

f (z)dz = f (b) − f (a)

where a and b are initial and end points, respectively, of γ. (b) If γ is closed, then

z n dz = 0 for n ≥ 0. γ

(c) If γ is closed and 0 ∈ / γ ∗ , then

z n dz = 0 for n = −2, −3, . . . . γ

1.20 Let X = (X, d) be a metric space. For subsets A and B of X , we define the distance d(A, B) between sets A and B as d(A, B) = inf inf d(x, y). x∈A y∈B

Let K be compact and C be a closed subset of X such that K ∩ C = ∅. Then show that there exist k ∈ K and c ∈ C such that d(K , C) = d(k, c). (Hint: Show that d(x, C) = d({x}, C) is a continuous function of x ∈ X .) 1.21 Let φ be a real-valued function defined in (a, b). Assume that φ((1 − t)x + t y) ≤ (1 − t)φ(x) + tφ(y)

(1.9.1)

1.9 Exercises

23

whenever x ∈ (a, b), y ∈ (a, b) and 0 < t < 1. Then φ is called convex function in (a, b). Show that (a) A function φ in (a, b) is convex if and only if φ(x) ≤

x − x1 x2 − x φ(x1 ) + φ(x2 ) x2 − x1 x2 − x1

(1.9.2)

for a < x1 < x < x2 < b. (b) If φ is convex in (a, b), then φ is continuous in (a, b). (c) Let φ(x) be given in [0, 1] by φ(x) = 0 if x ∈ [0, 1) and φ(1) = 1. Then φ satisfies (1.9.1) when x ∈ [0, 1] and y ∈ [0, 1] but it is not continuous in [0, 1]. (d) Let f be a real-valued differentiable function in (a, b) such that f is increasing in (a, b). Then f is convex in (a, b). Derive that e x is convex in (−∞, ∞). (Hint: By using Mean value theorem and f increasing, we get f (u) − f (x) f (y) − f (u) ≤ u−x y−u for x < u < y.)

Chapter 2

The Cauchy Theorems and Their Applications

2.1 Introduction We prove in Sect. 2.2 the Cauchy theorem and the Cauchy integral formula for closed paths in an open convex set. We extend in Sect. 2.4 the method of proof of Theorem 1.6 to prove in Theorem 2.5 that

γ0

f (z)dz =

γ1

f (z)dz,

where γ0 and γ1 are closed paths in such that they are -homotopic and f ∈ H (). 1 such that a ∈ / . This also implies the This implies Theorem 1.6 with f (z) = z−a Cauchy theorem for closed path in such that it is null-homotopic in . In particular, we derive in Corollary 2.6 that the Cauchy theorem is valid for closed paths in a simply connected region since every closed path in a simply connected region is null-homotopic. Further, we prove in Theorem 2.11 the Cauchy theorem for all cycles homologous to zero in an open set and this extends Corollary 2.6 by a different method. Further, Theorem 2.11 implies the Cauchy residue theorem 2.13. In fact, we derive Theorem 2.11 from Theorem 2.10 which is the Cauchy integral formula for cycles homologous to zero in an open set. Moreover, we apply in Sect. 2.6 the Cauchy theorems to prove important and useful theorems like Argument principle, Open mapping theorem, Inverse function theorem, Maximum modulus principle, the Rouché theorem, existence of an analytic branch of logarithm of non-vanishing analytic functions and the Jensen formula. The Maximum modulus principle need not be valid in an unbounded region. In Sect. 2.7, we prove the Maximum modulus principle in an unbounded strip by the Phragmen–Lindelöf method and derive the Hadamard three-circle theorem. We give another application of this method which implies μ(σ) ≤ 21 (1 − σ) for 0 ≤ σ ≤ 1 in Sect. 7.12. Further, we prove an estimate for the number of zeros of an exponential polynomial in a disc in Sect. 2.10 and we shall apply it in Chap. 10 for a proof of the Baker theorem. We refer to [1, 4, 5, 12, 19, 23, 24, 29, 31, 33] for the topics in this chapter and for further studies and related topics.

© Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9_2

25

26

2 The Cauchy Theorems and Their Applications

2.2 The Cauchy Theorem and the Cauchy Integral Formula for Convex Open Sets Definition Let f be continuous in . Then f has a primitive function F in if there exists F ∈ H () such that f = F in . Theorem 2.1 (The Cauchy–Goursat theorem) Let be a closed triangle in and p ∈ . Let f be continuous in and analytic in \ { p}. Then where ∂ denotes the boundary of .

∂

f (z)dz = 0

It is clear that Theorem 2.1 implies that γ f (z)dz = 0 for all triangular paths in an open set whenever f ∈ H () and we say that the Cauchy theorem is valid for all triangular paths in . Proof Put J = ∂ f (z)dz. Let L be the length of ∂ and write 0 = . By joining the middle points of the sides of the triangle , we obtain triangles 01 , 02 , 03 , 04 , as shown in Fig. 2.1, such that J=

4 i=1

∂0i

f (z)dz

and the length of ∂i is equal to 2−1 L for 1 ≤ i 0 ≤ 4. Then there exists i 0 with 1 ≤ i 0 ≤ 4 such that f (z)dz ≥ 4−1 |J |. ∂0i0 Put 0i0 = 1 . Proceeding similarly, we obtain a sequence of triangles ⊃ 1 ⊃ · · · ⊃ n ⊃ . . . such that ≥ 4−n |J | for n ≥ 0 f (z)dz (2.2.1) ∂n

and the length of ∂n is equal to 2−n L. Therefore, there exists z 0 such that z 0 ∈ n for every n ≥ 0, [29], p. 7. First, we consider the case p ∈ / . We observe that f (z) is analytic at z 0 since p = z 0 . Then for > 0 there exists δ > 0 depending only on such that A

Fig. 2.1 Bisection of triangle D B

F E

C

2.2 The Cauchy Theorem and the Cauchy Integral Formula for Convex Open Sets

| f (z) − f (z 0 ) − f (z 0 )(z − z 0 )| < |z − z 0 |

27

(2.2.2)

whenever |z − z 0 | < δ. By Exercise 1.19 (b), we have

∂n

f (z)dz =

∂n

( f (z) − f (z 0 ) − f (z 0 )(z − z 0 ))dz for n ≥ 0.

(2.2.3)

Let n 0 be the least positive integer such that 2−n 0 L < δ. Then for z ∈ n 0 , we have |z − z 0 | < 2−n 0 L < δ. Now we derive from (2.2.2) that the absolute value of the integral on the right-hand side of (2.2.3) with n = n 0 is at most 4−n 0 L 2 and hence f (z)dz ≤ 4−n 0 L 2 . (2.2.4) ∂n0 By combining (2.2.1) with n = n 0 and (2.2.4), we get |J | ≤ L 2 . This is true for every > 0 and hence J = 0. Thus we may suppose that p ∈ . Let be a triangle formed by ordered triple {a, b, c} and we write = {a, b, c}. First we prove that J = 0 when p is a vertex of , say p = a. We may assume that a, b and c are not colinear otherwise the assertion follows immediately. Let > 0 and we take x ∈ [a, b], y ∈ [a, c] such that |x − a| < , |y − a| < . We observe that f (z)dz + f (z)dz + f (z)dz. (2.2.5) J= ∂{a,x,y}

∂{x,b,y}

∂{b,c,y}

Further, the last two integrals are equal to zero since a does not lie in triangles {x, b, y} and {b, c, y}. Since f is continuous on compact set {a, x, y}, we observe that | f (z)| is bounded by a constant K on {a, x, y}. Therefore, the absolute value of the first integral on the right-hand side of (2.2.5) is at most 4K which tends to zero as approaches to zero. Now it remains to show that J = 0 when p is not a vertex of . Then f (z)dz + f (z)dz + f (z)dz J= ∂{a,b, p}

∂{b,c, p}

∂{c,a, p}

and, as proved above in this paragraph, all the integrals on the right-hand side are equal to zero. Hence J = 0. We derive an analogue of the above result for closed paths in an open convex set. Theorem 2.2 Let be a convex open set and p ∈ . Let f be continuous in and analytic in \ { p}. Then f has a primitive in and γ

for any closed path γ in .

f (z)dz = 0

28

2 The Cauchy Theorems and Their Applications

Proof Let

F(z) =

[ p,z]

f (ζ)dζ for z ∈

where [ p, z] denotes the line joining from p to z. We observe that [ p, z] lies in since is convex. Let z 0 ∈ and F(z) − F(z 0 ) = f (ζ)dζ − f (ζ)dζ = f (ζ)dζ [ p,z]

[ p,z 0 ]

[z 0 ,z]

by Theorem 2.1. Further F(z) − F(z 0 ) 1 − f (z 0 ) = z − z0 z − z0

[z 0 ,z]

( f (ζ) − f (z 0 ))dζ.

Let ε > 0. Since f is continuous at z 0 , there exists δ > 0 such that | f (ζ) − f (z 0 )| < ε whenever |z − z 0 | < δ. Thus, the absolute value of the right-hand side is less than ε whenever |z − z 0 | < δ. This implies that F is analytic at z 0 and F (z 0 ) = f (z 0 ). Since z 0 is an arbitrary point of , we have F ∈ H () and f = F in and γ

f (z)dz = 0 by Exercise 1.19 (a).

Now we derive the Cauchy integral formula from the above result. Theorem 2.3 (The Cauchy integral formula for convex open sets) Let γ be a closed path in an open convex set and f ∈ H (). Then for a ∈ and a ∈ / γ ∗ , we have 1 2πi

γ

f (z) dz = Indγ (a) f (a). z−a

Proof We define for z ∈ g(z) =

f (z)− f (a) z−a

f (a)

if z = a if z = a.

We observe that g is continuous in and g ∈ H ( \ {a}). Now we apply Theorem 2.2 with f = g and p = a to conclude γ

g(z)dz = 0.

Then, since a ∈ / γ ∗ , we have 1 2πi

γ

f (z) f (a) dz = z−a 2πi

γ

dz = Indγ (a) f (a). z−a

2.2 The Cauchy Theorem and the Cauchy Integral Formula for Convex Open Sets

29

Theorem 2.4 (The Cauchy theorem for open convex sets) Let γ be a closed path in an open convex set and f ∈ H (). Then γ f (z)dz = 0 and f has primitive function in . Proof This follows immediately from Theorem 2.2.

2.3 An Account of Some Basic Results on Analytic Functions We assume that the readers are familiar with expansion of analytic functions in power series and meromorphic functions in Laurent series. We give an account of these results together with some consequences which we shall use in this tract for the convenience of the readers and for later reference. We include proofs or hints for proofs of some of them. We refer to [1, 12, 23] for this section. Section 2.3 (i) Representation of an analytic function by power series, the Liouville theorem and the Fundamental theorem of algebra. The series ∞

cn (z − a)n with a ∈ C, cn ∈ C

(2.3.1)

n=0

is called a power series around a and let R be given by 1 = lim sup | cn |1/n . R n→∞ Then the series converges absolutely in |z − a| < R, uniformly in |z − a| ≤ r < R where 0 < r < R and diverges in |z − a| > R. The number R is called the radius of convergence, the disc |z − a| < R is called the disc of convergence and the circle |z − a| = R is called the circle of convergence of the power series (2.3.1). A power series can be differentiated and integrated term wise for an arbitrary number of times within its disc of convergence. If f ∈ H () and a ∈ , then we derive from Theorem 2.3 that there exists ρ > 0 such that f (z) has power series (2.3.1) in D(a, ρ) and f has derivatives of all orders at z = a given by f (n) (a) =

n! 2πi

|z−a|=r 0 such that f ∈ H (D(a, δ)) where it has no zero other than a. Therefore, we say that the zeros of a non-zero analytic function are isolated. This implies the following useful criterion for determining when two analytic functions are equal in a region . Let f (z) and g(z) be analytic in a region . Suppose that

2.3 An Account of Some Basic Results on Analytic Functions

31

there exists a subset S of with a limit point in and f (z) = g(z) for z ∈ S. Then f (z) = g(z) for z ∈ . This is known as the Identity theorem for holomorphic functions. Section 2.3 (iii) The Morera theorem. As already mentioned after the statement of Theorem 2.1, the Cauchy theorem for triangular paths in an open set is valid. A converse of the above assertion is also valid. Let f be continuous in and γ

f (z)dz = 0

for any triangular path in . Then f ∈ H (). It suffices to prove the theorem when is an open disc. Let a ∈ be fixed and we define F(z) =

[a,z]

f (ζ)dζ, z ∈

where [a, z] is a line in from a to z. As in the proof of Theorem 2.2, we conclude that F ∈ H () and F = f on . Further F ∈ H () by (2.3.3) and hence f ∈ H (). Section 2.3 (iv) Limit function of uniformly convergent sequence of analytic functions. Let f n ∈ H () for n ≥ 1 and assume that lim f n (z) = f (z)

n→∞

uniformly on compact subsets of . Then f (z) ∈ H () by the Morera theorem. In fact, we have lim f n(k) (z) = f (k) (z) for all k ≥ 1

n→∞

uniformly on compact subsets of . An analogue of this result is also valid for series and integrals. Section 2.3 (v) Laurent series. Let 0 ≤ R1 < R2 , a ∈ C and f (z) be analytic in the annulus R1 < |z − a| < R2 . Then f (z) has series expansion called the Laurent series around a given by f (z) =

∞

cn (z − a)n ,

n=−∞

where cn =

1 2πi

|ζ−a|=r, R1 0 such that c−n = 0. Let m be the greatest positive integer such that c−m = 0. Then f has a pole of order m at z = a and we have f (z) =

∞

cn (z − a)n in 0 < |z − a| < R2 ,

n=−m

and

−1

cn (z − a)n is called the principal part of f at the pole a. If m = 1, then

n=−m

f has a simple pole at z = a. If f is analytic in except for poles, then we say that f is meromorphic in . Thus f is meromorphic at a if f is either analytic at a or f has a pole at a. Essential singularity: If there are infinitely many positive integers n such that c−n = 0, then f has essential singularity at z = a. For example, f (z) = e1/z has essential singularity at z = 0. We give a proof of the following result, due to CasoratiWeierstrass, for functions having essential singularity: Suppose that f is analytic in D (z 0 , r ) and it has essential singularity at z = z 0 . Then f (D (z 0 , r )) is dense in the complex plane, i.e. f (D (z 0 , r )) = C. Proof The proof is by contradiction. Assume that there exists > 0, δ > 0 and w ∈ C such that | f (z) − w| > for z ∈ D (z 0 , δ). (2.3.4)

2.3 An Account of Some Basic Results on Analytic Functions

33

We write D for D (z 0 , δ) and D for D(z 0 , δ). Let 1 for z ∈ D . f (z) − w

g(z) =

(2.3.5)

Then g(z) is analytic in D where |g(z)| < −1 by (2.3.4). Now we derive from Exercise 2.6 (a) that g has analytic continuation to D and we denote again the extended function by g. Let g(z 0 ) = 0. By (2.3.5), we have f (z) =

1 + w in D (z 0 , δ), g(z)

and we define F(z) =

f (z) 1 +w g(z)

if z = z 0 , if z = z 0 .

Then F(z) is analytic in D and hence f has removable singularity at z = z 0 . Let g(z 0 ) = 0. Assume that g(z) has zero of order m at z = z 0 . Then g(z) = (z − z 0 )m g1 (z)

(2.3.6)

where g1 ∈ H (D) and g1 (z 0 ) = 0. Also g1 (z) has no zero in D by (2.3.4). Then h := g11 ∈ H (D). Combining (2.3.4) and (2.3.6), we get f (z) − w = (z − z 0 )−m h(z) in D. Further h(z) =

∞

bn (z − z 0 )n with b0 = 0 in D

n=0

since h has no zeros in D. Hence f (z) has a pole of order m at z = z 0 . Thus f (z) has either removable singularity at z = z 0 or has pole at z = z 0 . This is a contradiction.

2.4 The Cauchy Theorem for Closed Paths in A Simply Connected Region Let [P0 , P1 , . . . , Ps ] be a polygon path. If P0 = Ps , then it is called a closed polygon. If P0 , P1 , . . . , Ps ∈ where is an open convex set, then [P0 , P1 , . . . , Ps ] lies in . We prove the following extension of Theorem 1.6.

34

2 The Cauchy Theorems and Their Applications

Theorem 2.5 Let f ∈ H (). Assume that γ0 and γ1 are closed paths in such that γ0 and γ1 are -homotopic. Then we have γ0

f (z)dz =

γ1

f (z)dz.

1 is analytic in whenever a ∈ / , we see that Theorem 1.6 follows from Since z−a 1 the above result with f (z) = z−a , γ0 = 0 and γ1 = 1 . Another immediate consequence of Theorem 2.5 is the following extension of Theorem 2.2 where open convex set has been replaced by a simply connected region.

Corollary 2.6 Let f (z) be analytic in a simply connected region . Then γ

f (z)dz = 0

for any closed path γ in . Let γ be a closed path in a simply connected region and f ∈ H (). Then γ is -homotopic to constant path γ0 in . Now we apply Theorem 2.5 with γ1 = γ to derive that f (z)dz = f (z)dz = 0, γ

γ0

which is the assertion of Corollary 2.6. Finally, we apply Theorem 2.5 to regions which are not simply connected. Corollary 2.7 Let 0 < R1 < R2 and f be analytic in R1 ≤ |z − a| ≤ R2 . Then

f (z)dz =

|z−a|=R1

f (z)dz. |z−a|=R2

There exists such that f ∈ H () and contains R1 ≤ |z − a| ≤ R2 . Further, we observe that the circles |z − a| = R1 and |z − a| = R2 are -homotopic. Hence the assertion follows from Theorem 2.5 with γ0 : |z − a| = R1 and γ1 : |z = a| = R2 . Proof of Theorem 2.5 Since γ0 and γ1 are -homotopic, there exists H : I2 → such that H is continuous and H (s, 0) = γ0 (s), H (s, 1) = γ1 (s) for 0 ≤ s ≤ 1

(2.4.1)

2.4 The Cauchy Theorem for Closed Paths in A Simply Connected Region

35

and H (0, t) = H (1, t) for 0 ≤ t ≤ 1.

(2.4.2)

Since H (I 2 ) is a compact subset of and C \ is closed, we derive from Theorem 1.4 that there exists δ > 0 such that |H (s, t) − z| > 2δ for 0 ≤ s ≤ 1, 0 ≤ t ≤ 1, z ∈ / .

(2.4.3)

Since H is uniformly continuous on I 2 , there exists a positive integer n such that for 0 ≤ s, s , t, t ≤ 1, we have

|H (s, t) − H (s , t )| < δ whenever

(2.4.4)

4 . n2

(2.4.5)

⊂ ,

(2.4.6)

(s − s )2 + (t − t )2 ≤ For 0 ≤ j < n and 0 ≤ k < n, we write Z jk = H

j k , n n

P jk = [Z jk , Z j+1,k , Z j+1,k+1 Z j,k+1 , Z jk ] and Q k = [Z 0k , Z 1k , . . . , Z nk ]. We observe that P jk and Q k are closed polygons in by (2.4.2). We give a sketch of the proof of Theorem 2.5. We shall use the above inequalities as in the proof of Theorem 1.6. We derive from them that P jk with 0 ≤ j < n and 0 ≤ k < n are closed polygons lying in open discs contained in . Therefore, we derive from Theorem 2.4 that f (z)dz = 0. (2.4.7) P jk

Further, we shall calculate n−1

P jk = Q k − Q k+1 for 0 ≤ k < n.

(2.4.8)

j=0

By combining (2.4.7) and (2.4.8), we get

f (z)dz = Qk

f (z)dz for 0 ≤ k < n Q k+1

(2.4.9)

36

2 The Cauchy Theorems and Their Applications

and we derive again by Theorem 2.4 as above that

γ0

f (z)dz =

f (z)dz and Q0

f (z)dz =

γ1

f (z)dz.

(2.4.10)

Qn

Now the assertion follows immediately by combining (2.4.9) and (2.4.10). Let 0 ≤ j < n and 0 ≤ k < n. It is clear that P jk is a closed polygon. Further we observe from (2.4.6), (2.4.4) and (2.4.5) that P jk is a closed polygon whose end points are contained in D(Z jk , δ). Since D(Z jk , δ) is convex, we derive from (2.4.3) that P jk ⊂ D(Z jk , δ) ⊆ . Therefore, by Theorem 2.4, we get (2.4.7). Next we assume (2.4.8) and then complete the proof of Theorem 2.5. By (2.4.7) and (2.4.8), we get 0=

n−1

f (z)dz =

P0k +···+Pn−1,k

P jk

j=0

f (z)dz =

Q k −Q k+1

f (z)dz

implying (2.4.9) and hence

f (z)dz = Q0

f (z)dz = · · · = Q1

f (z)dz. Qn

Therefore it suffices to prove (2.4.10). We prove the first and the proof for the second is similar. Let 0 ≤ j < n. For nj ≤ t ≤ j+1 , we consider n γ0 j + [Z j+1,0 , Z j0 ], where

(2.4.11)

γ0 j = γ0 | j , j+1 . n

n

since We observe that (2.4.11) is a closed path with parameter interval nj , j+1 n Z j0 = H nj , 0 = γ0 nj and Z j+1,0 = H j+1 by (2.4.6) and , 0 = γ0 j+1 n n (2.4.1). Further it lies in D(Z j0 , δ) by (2.4.1), (2.4.4) and (2.4.5). Therefore, we derive from Theorem 2.2 that f (z)dz = 0 γ0 j +[Z j+1,0 ,Z j0 ]

and thus

2.4 The Cauchy Theorem for Closed Paths in A Simply Connected Region

37

Zj+1,k+1 Zj+2,k+1

⏐ ⏐ ⏐ ⏐

Zj,k+1

Zj+1,k Zj+2,k

Zj,k

Fig. 2.2 Contour for proof of (2.4.8)

f (z)dz =

γ0 j

Therefore

n−1 j=0

f (z)dz. [Z j0 ,Z j+1,0 ]

γ0 j

f (z)dz =

n−1

f (z)dz

j=0 [Z ,Z j0 j+1,0 ]

and hence

γ0

f (z)dz =

f (z)dz. Q0

Finally we prove (2.4.8). We consider P jk + P j+1,k with 0 ≤ j ≤ n − 2. We find that (see Fig. 2.2) P jk + P j+1,k = [Z jk , Z j+1,k , Z j+2,k ] + [Z j+2,k+1 , Z j+1,k+1 , Z j,k+1 ] +[Z j+2,k , Z j+2,k+1 ] + [Z j,k+1 , Z jk ]. Proceeding similarly, we have n−1

P jk = [Z 0k , Z 1k , . . . , Z nk ] + [Z n,k+1 , . . . , Z 0,k+1 ]

j=0

+[Z nk , Z n,k+1 ] + [Z 0,k+1 , Z 0k ]. By (2.4.6) and (2.4.2)

38

2 The Cauchy Theorems and Their Applications

k k = H 0, = Z 0k Z nk = H 1, n n and

k+1 k+1 = H 0, = Z 0,k+1 . Z n,k+1 = H 1, n n

Hence

n−1

P jk = Q k − Q k+1 .

j=0

This proves Theorem 2.5.

2.5 The Cauchy Integral Formula and the Cauchy Theorem for Cycles Homologous To Zero in an Open Set and the Cauchy Residue Theorem Before we state the Cauchy theorems that we shall prove in this section, we give the following result required for their proofs. Lemma 2.8 Let f be analytic in and g be defined on × by ⎧ ⎨ f (z) − f (w) , if z = w g(z, w) = z−w ⎩ f (z), if z = w.

(2.5.1)

Then g(z, w) is continuous on × . Proof The continuity of g(z, w) at non-diagonal points of × follows immediately. Therefore, we need to only prove the continuity of g(z, w) at diagonal points of × . Let a ∈ . Since is open and f ∈ H () by (2.3.3), we observe that for > 0 there exists r > 0 such that D(a, r ) ⊂ and

| f (z) − f (a)| < for z ∈ D(a, r ).

(2.5.2)

For z, w ∈ D(a, r ), we have by Exercise 1.9,

f (ζ)dζ,

f (w) − f (z) = L

where L is the line in D(a, r ) from z to w given by ζ(t) = (1 − t)z + tw with 0 ≤ t ≤ 1.

(2.5.3)

2.5 The Cauchy Integral Formula and the Cauchy Theorem …

39

Then

f (ζ)dζ =

1

0

L

1

f (ζ(t))ζ (t)dt = (w − z)

f (ζ(t))dt.

(2.5.4)

0

Now we observe from (2.5.1), (2.5.3) and (2.5.4) that

1

g(z, w) =

f (ζ(t))dt if z = w,

0

which implies that g(z, w) − g(a, a) =

1

( f (ζ(t)) − f (a))dt if z = w.

0

Further, the absolute value of the integral on the right-hand side is < by (2.5.2) and hence |g(z, w) − g(a, a)| < whenever z, w ∈ D(a, r ) with z = w. If z = w, the preceding inequality is also valid by (2.5.1) and (2.5.2). Now, we state the following result. Lemma 2.9 (The Fubini theorem) Let I and J be intervals not necessarily bounded and f be a measurable function from I × J into C such that

| f (x, y)|dy < ∞.

dx I

Then

J

I

f (x, y)dy =

dx J

f (x, y)d x.

dy J

I

Now we are ready to prove the Cauchy integral formula for cycles homologous to zero in an open set. Theorem 2.10 Let f ∈ H () and be a cycle in homologous to zero in . Then 1 2πi

f (w) dw = f (z) Ind (z) for z ∈ \ ∗ . w−z

(2.5.5)

Proof Let g be defined on × by (2.5.1). Then g is continuous on × by Lemma 2.8. Let 1 g(z, w)dw for z ∈ . h(z) = 2πi It suffices to show that

h(z) = 0 for z ∈ \ ∗

since then by (2.4.11), we have

(2.5.6)

40

2 The Cauchy Theorems and Their Applications

1 2πi

f (z) f (w) dw = w−z 2πi

dw = f (z) Ind (z). w−z

For this, we first show that h(z) ∈ H (). Next, we construct an entire function φ(z) such that φ(z) = h(z) for z ∈ and lim φ(z) = 0 for z ∈ C. Now we conclude |z|→∞

from the Liouville theorem, see Sect. 2.3 (i), that φ(z) = 0 for z ∈ C which, in particular, implies (2.5.6). Let z ∈ and z n ∈ such that lim z n = z. Denote by A the set obtained by n→∞

∗ ∗ adjoining z to the sequence {z n }∞ n=1 . Then A × is compact since A and are ∗ compact. Therefore g is uniformly continuous on A × . Then

lim g(z n , w) = g(z, w)

n→∞

uniformly on ∗ . Now lim h(z n ) =

n→∞

1 2πi

lim

n→∞

g(z n , w)dw =

1 2πi

lim g(z n , w)dw =

n→∞

1 2πi

g(z, w)dw = h(z).

Hence h is continuous on . Next, we show that h ∈ H () for which there is no loss of generality in assuming that is an open disc. Let be any triangular path in . Then 1 h(z)dz = g(z, w)dw dz. 2πi ∂ ∂ Since g(z, w) is continuous on ∂ × ∗ which is compact, we see that g(z, w) is bounded on ∂ × ∗ and hence

∂

γ

|g(z, w)|dw dz < ∞.

Therefore, we derive from the Fubini theorem (Lemma 2.9) that ∂

h(z)dz =

1 2πi

∂

g(z, w)dz dw.

Since g(z, w) ∈ H () for each w ∈ by (2.5.1) and is an open disc, we derive that g(z, w)dz = 0 for w ∈ ∂

by Theorem 2.2 and Lemma 2.8. Hence ∂

h(z)dz = 0.

2.5 The Cauchy Integral Formula and the Cauchy Theorem …

41

Now we conclude from the Morera theorem, see Sect. 2.3 (iii), that h ∈ H (). Finally we prove (2.5.5). Let 1 = z ∈ C \ ∗ Ind (z) = 0 . We observe from Sect. 1.2 that 1 is union of components of C \ ∗ in which Ind vanishes and the components of C \ ∗ are open. Therefore 1 is open. Further 1 ⊇ C \ since is homologous to zero in . Therefore ∪ 1 = C. Let

1 h 1 (z) = 2πi

f (w) dw for z ∈ 1 . w−z

Since f is continuous on and z ∈ / ∗ , we check that h 1 ∈ H (1 ). Further for z ∈ ∩ 1 , we have f (w) 1 dw − f (z) Ind (z) = h 1 (z). h(z) = 2πi w − z Therefore, the function φ defined by φ(z) =

h(z) for z ∈ h 1 (z) for z ∈ 1

is entire. Let 2 be the unbounded component of determined by . Then 2 ⊂ 1 since Ind vanishes on 2 . Therefore lim φ(z) = lim h 1 (z) = 0

|z|→∞

|z|→∞

where the right most equality follows since f (w) is bounded on ∗ and |w − z| ≥ |z| 2 whenever w ∈ and |z| → ∞. Hence φ(z) is identicaly zero in C by the Liouville theorem, see Sect. 2.3 (i). In particular h(z) = 0 for z ∈ giving (2.5.6). Theorem 2.10 implies the Cauchy theorem, stated below, for cycles homologous to zero in an open set. Theorem 2.11 Let be a cycle homologous to zero in and f ∈ H (). Then

f (z)dz = 0.

42

2 The Cauchy Theorems and Their Applications

It is clear that Theorem 2.11 implies Corollary 2.6 since a closed path in a simply connected region is homologous to zero. Proof Let a ∈ \ ∗ and we write F(z) = (z − a) f (z). By Theorem 2.10, we have 1 F(z) dz = F(a) Indγ (a) = 0. 2πi z − a Since the left-hand side is equal to

1 2πi

f (z)dz, the assertion follows.

Finally, we obtain the following extension of Theorem 2.5. Theorem 2.12 Let 0 and 1 be cycles in such that / Ind0 (a) = Ind1 (a) for a ∈ and f ∈ H (). Then

(2.5.7)

0

f (z)dz =

1

f (z)dz.

For -homotopic closed paths 0 and 1 , we see that (2.5.7) holds by Theorem 1.6 and hence Theorem 2.12 implies Theorem 2.5. Proof Let = 0 \1 . By (2.5.7), we observe that / Ind (a) = Ind0 (a) − Ind1 (a) = 0 for a ∈ implying is homologous to zero in . Therefore, we derive from Theorem 2.11 with = 0 \ 1 that f (z)dz = 0

implying

0

f (z)dz =

1

f (z)dz.

Another consequence of Theroem 2.11 is the following result known as the Cauchy residue theorem. Theorem 2.13 Let f be meromorphic in an open set where it has poles at finitely many distinct points a1 , a2 , . . . , am . Let γ be a closed path in not passing through any ak with 1 ≤ k ≤ m such that it is homologous to zero in . Then 1 2πi

γ

f (z)dz =

m k=1

Indγ (ak )Res( f ; ak ).

2.5 The Cauchy Integral Formula and the Cauchy Theorem …

43

By taking = D(a, R) and γ : |z − a| = r where 0 < r < R such that |ak − a| < r for 1 ≤ k ≤ m, we derive from Theorem 2.13 that 1 2πi

|z−a|=r

f (z)dz =

m

Res( f ; ak )

k=1

since Indγ (ak ) = 1 for 1 ≤ k ≤ m. Further, the assumptions of Theorem 2.13 are satisfied if is simply connected. Therefore, the assertion of Theorem 2.13 is valid if is simply connected. Proof of Theorem 2.13 Let νk = Indγ (ak ) for 1 ≤ k ≤ m.

(2.5.8)

Let r1 , r2 , . . . , rm be positive real numbers such that ¯ l , rl ) = ∅ for 1 ≤ k, l ≤ m, k = l. ¯ k , rk ) ∩ D(a D(a For 1 ≤ k ≤ m, let γk be given by γk (t) = ak + rk e−2πiνk t , 0 ≤ t ≤ 1

(2.5.9)

and we put G = \ {a1 , a2 , . . . , am }. We observe that G is an open set, f ∈ H (G) and γ1 , γ2 , . . . , γm are closed paths lying in G. Let ˙ 1+ ˙ · · · +γ ˙ m. (2.5.10) = γ +γ We show that is homologous to zero in G. Let a ∈ / G. Then either a ∈ / or a ∈ , a = ak for some 1 ≤ k ≤ m. In the first possibility, we see that Indγ (a) = 0 ¯ k , νk ). Further, in / D(a by assumption and Indγk (a) = 0 for 1 ≤ k ≤ m since a ∈ the second possibility, Indγ (ak ) = νk by (2.5.8), Indγk (ak ) = −νk by (2.5.9) and / G in either Indγl (ak ) = 0 for l = k. Therefore, we conclude that Ind (a) = 0 for a ∈ of the cases and thus is homologous to zero in G. Now we conclude from Theorem 2.11 with = G that f (z)dz = 0

which, together with (2.5.10), implies that γ

f (z)dz = −

m k=1

γk

f (z)dz.

44

2 The Cauchy Theorems and Their Applications

Let 1 ≤ k ≤ m. Since f ∈ H (D (ak , rk )), the Laurent series expansion of f (z) at z = ak is given by ∞ bν (z − ak )ν , ν=−∞

where b−1 = Res( f ; ak ). Further the series converges uniformly on |z − ak | = rk . Therefore, we can integrate it term by term, see Sect. 2.3 (iv). Hence γ

where

γk (z

1 2πi

f (z)dz = −

∞ m k=1 ν=−∞

bν

γk

(z − ak )ν dz,

− ak )ν dz = 0 if ν = −1. Thus we have

γ

m

f (z)dz = −

Res( f ; ak )Indγk (ak ) =

k=1

m

Indγ (ak )Res( f ; ak )

k=1

since Indγk (ak ) = −νk = −Indγ (ak ) by (2.5.9) and (2.5.8).

The Cauchy residue theorem is valid even when there are infinitely many isolated points ak s. Then, by following exactly the same proof, we have 1 2πi

γ

f (z)dz =

∞

Indγ (ak ) Res( f ; ak ).

k=1

But we should show that the sum on the right-hand side is finite. For this, we prove that Indγ (ak ) = 0 for all but finitely many ak . Let A be the set of all a ∈ C \ γ ∗ such that Indγ (a) = 0. Let a ∈ A. By Theorem 1.4, there exists r > 0 such that D(a, r ) ∩ γ ∗ = ∅. We observe that Indγ vanishes on D(a, r ) by Theorem 1.5 since D(a, r ) is connected and Indγ (a) = 0. Thus D(a, r ) ⊆ A and hence A is open. Then C \ A is closed. Further it is bounded since A contains the unbounded region determined by γ. Thus C \ A is compact, and therefore it contains only finitely many ak ’s as they are isolated points. Hence the above sum is finite. We have shown in Theorem 2.4 that every function analytic in an open convex set has a primitive in . In the next result, we prove that this is also the case even when is simply connected and more generally, for regions where every closed path is homologous to zero in the region. Theorem 2.14 Let f ∈ H () where is region and a ∈ . Assume that every closed path in is homologous to zero in . Then f has primitive F in and it is given by F(z) =

[a,z]

f (ζ)dζ, z ∈

where [a, z] denotes a path from a to z in .

2.5 The Cauchy Integral Formula and the Cauchy Theorem …

45

Proof By Theorem 2.11, we observe that F is independent of the choice of the path. Further, as in the proof of Theorem 2.2, we show that F ∈ H () and F = f in . We close this section with the following example which we shall use in the proof of the Jensen formula in Sect. 2.9. Example 2.1 We show that 1 2π

2π

log |1 − eiθ |dθ = 0.

0

Let log z be the principal branch of logarithm of z defined on C \ (−∞, 0] where it is analytic, see Exercise 1.15. Let = {z | Re(z) < 1} and observe that is convex ∈ H () since Re(1 − z) > 0 open set. Let h(z) = log(1 − z) for z ∈ . Then h(z) z and h(0) = 0. Let 0 < δ < 1/2 be sufficiently small. We consider where 1 (θ) = eiθ with δ ≤ θ ≤ 2π − δ and γ is the arc inside the unit open disc with 1 as centre and passing through the points eiδ and e−iδ . Further we take = 1 + γ. Thus is a closed path in . Let ρ be the radius of the circle with centre 1 and containing γ (Fig. 2.3). Then ρ2 = |1 − eiδ |2 = (1 − cos δ)2 + sin2 δ δ = 2 − 2 cos δ = 4 sin2 2 implying ρ = 2 sin

δ 2

> 0 since ρ > 0. Since

δ 2

< sin δ < δ, we have

δ δ δ = 2 < ρ < 2 = δ. 2 4 2

(2.5.11)

We derive from Theorem 2.4 that 1 2πi

Fig. 2.3 Keyhole contour

h(z) = 0. z

Γ1 O

γ

eiδ 1 e−iδ

46

2 The Cauchy Theorems and Their Applications

Further Re

1 h(z) 1 2π−δ h( (θ)) 1 = Re 1 (θ)dθ 2πi 1 z 2πi δ 1 (θ) 1 2π−δ log(1 − eiθ ) iθ ie dθ = Re 2πi δ eiθ 2π−δ 1 log |1 − eiθ |dθ. = 2π δ

Therefore 0=

1 2π

2π−δ

δ

log |1 − eiθ |dθ + Re

1 h(z) . 2πi γ z

Thus it suffices to show that 1 δ→0 2πi

lim

γ

h(z) = 0. z

For z ∈ γ ∗ , we derive from (2.5.11) that |z| = |z − 1 + 1| ≥ 1 − |z − 1| = 1 − δ >

1 , 2

π 2 π 1 + < 2 log | log(1 − z)| ≤ max log |1 − z)|, log |1 − z|−1 + = log 2 δ 2 δ

by taking δ sufficiently small and l(γ) ≤ πρ < πδ. Hence 1 2 log(1/δ) h(z) 1 2 log(1/δ) ≤ πδ < →0 2πi 2π 1/2 1/δ γ z as δ tend to zero.

2.6 Argument Principle, Open Mapping Theorem, Maximum Modulus Principle, the Rouché Theorem and The Jensen Formula We begin with an immediate consequence of the Cauchy residue Theorem 2.13. Theorem 2.15 (Argument principle) Let f be meromorphic in . Let a j and bk be the set of all points in where f has zeros and poles, respectively. Let γ be a closed path homologous to zero in not passing through any a j and bk . Then

2.6 Argument Principle, Open Mapping Theorem, Maximum Modulus …

1 2πi

47

f (z) Indγ (a j ) − Indγ (bk ), dz = f (z) j k

γ

where the zeros and poles are counted with their multiplicities in the sums on the right-hand side. Here the terms in each of the sums are zero except for finitely many. The name refers to the interpretation of left-hand side as Ind (0) if = f ◦ γ and f ∈ H (). Proof Let n j be the order of zero of f at a j and pk be the order of pole at bk . We observe that

f f

has simple poles at a j and bk such that Res

f ; aj f

= n j , Res

f ; bk f

= − pk ,

see Exercise 2.7 (a), (b). Therefore, we conclude from Theorem 2.13 with f replaced by ff and its version for infinitely many isolated points that 1 2πi

γ

f (z) dz = n j Indγ (a j ) − pk Indγ (bk ), f (z) j k

where the terms in each of the sums on the right-hand side are zero except for finitely many. The assertion follows immediately. Corollary 2.16 Let f (z) be analytic in and α ∈ C. Let z j = z j (α) be all the zeros of f (z) − α in . Assume that γ is a closed path in not passing through any z j (α). Put = f ◦ γ. Then

Indγ (z j (α)) = Ind (α)

j

where the sum is taken over all zeros of f (z) − α counted with multiplicity. Proof We observe that Indγ (z j (α)) is defined since γ does not pass through any / ∗ otherwise f (γ(t0 )) − α = 0 for some t0 . Therefore γ(t0 ) = z j (α). Further α ∈ z j (α) for some j. This is a contradiction. Thus Ind (α) is also defined. We apply Theorem 2.15 with f (z) replaced by f (z) − α. We get 1 2πi

γ

f (γ) dz = Indγ (z j (α)). f (z) − α j

The left-hand side is equal to 1 2πi

1 0

1 f (γ(t))γ (t) dt = f (γ(t)) − α 2πi

1 0

1 (t) dt = (t) − α 2πi

dz = Ind (α). z−α

48

2 The Cauchy Theorems and Their Applications

We derive the following very useful result from Corollary 2.16. Theorem 2.17 Let f be non-constant function which is analytic at z 0 and f (z 0 ) = w0 . Assume that f (z) − w0 has a zero of order n. Then there exists ε0 > 0 such that for 0 < ε < ε0 , we have δ > 0 depending only on ε so that every element in D(w0 , ε) is assumed by f exactly n times in each of D(z 0 , δ ) with δ ≤ δ. Proof We take ε1 > 0 such that f is analytic in D(z 0 , ε1 ) and f (z) − w0 has no zeros in 0 < |z − z 0 | < ε1 . Let ε0 = ε21 and 0 < ε < ε0 . Then 2ε < ε1 and thus f is analytic in D(z 0 , 2ε) and f (z) − w0 has no zeros in 0 < |z − z 0 | ≤ ε. We apply Corollary 2.16 with α = w0 , = D(z 0 , 2ε) and γ : |z − z 0 | = ε and = f ◦ γ to derive Indγ (z j (w0 )) = n Indγ (z 0 ) = n, Ind (w0 ) = j

where the sum is taken over all the zeros of f (z) − w0 counted with multiplicity. / ∗ and ∗ is compact, we see from Theorem 1.4 that there exists δ > 0 Since w0 ∈ such that D(w0 , δ) ∩ ∗ = ∅.

/ ∗ , and therefore γ does not pass through Let 0 < δ ≤ δ and a ∈ D(w0 , δ). Then a ∈ any z j (a). Consequently, we conclude again from Corollary 2.16 that Ind (a) =

Indγ (z j (a)),

j

where the sum is taken over all the zeros of f (z) − a counted with multiplicity. By Theorem 1.5, we have Ind (a) = Ind (w0 ) = n since a ∈ D(w0 , δ). Hence

Indγ (z j (a)) = n

j

by Corollary 2.16. The assertion follows since Indγ (z j (a)) = 1 if |z j (a) − z 0 | < ε and 0 otherwise. For the next result, we introduce the following definition. Definition Let f be defined on . Then f is called an open mapping if f (U ) is open for every open set U of . Now we derive from Theorem 2.17 the following result. Theorem 2.18 (Open mapping theorem) Let f be non-constant and analytic in a region . Then f is an open mapping.

2.6 Argument Principle, Open Mapping Theorem, Maximum Modulus …

49

Proof Let U be an open set in . We show that f (U ) is open. Let w0 ∈ f (U ). Then there exists z 0 ∈ U such that f (z 0 ) = w0 . Since U is open, there exists δ > 0 such that D(z 0 , δ) ⊆ U and we shall take δ sufficiently small. By Theorem 2.17, there exists ε > 0 such that for every element in D(w0 , ε) is assumed from D(z 0 , δ). Then w0 ∈ D(w0 , ε) ⊆ f (D(z 0 , δ)) ⊆ f (U ) and hence f (U ) is open.

Next we derive from Theorem 2.18 very important Maximum modulus principle. For this, we need the following definition. Definition Let f be defined on and a ∈ . Then | f | has a local maximum at a if there exists δ > 0 such that D(a, δ) ⊆ and | f (a)| ≥ | f (z)| for every z ∈ D(a, δ). Further, we say that | f | has no local maximum in if | f | does not have local maximum at every point of . Theorem 2.19 (Maximum modulus principle) Let f be non-constant and analytic in a region . Then | f | has no local maximum in . Proof Let f be non-constant and analytic in . It suffices to show that f has no local maximum at every point of . Let z 0 ∈ and let U be an open disc in containing z 0 . We show that there exists z 1 ∈ U such that | f (z 1 )| > | f (z 0 )| and then the assertion follows. By Theorem 2.18, we see that f (U ) is an open set containing f (z 0 ). Then there exists δ1 > 0 such that D( f (z 0 ), δ1 ) ⊆ f (U ). Further we can find w ∈ D( f (z 0 ), δ1 ) such that |w| > | f (z 0 )| and w = f (z 1 ) for some z 1 ∈ U . Thus | f (z 1 )| = |w| > | f (z 0 )|. Now we give another version of the Maximum modulus principle. Theorem 2.20 Let be a bounded open set. Let f ∈ H () and f be continuous ¯ Assume that f is non-constant. Then for every a ∈ , we have on . | f (a)| < max | f (z)|. z∈δ

¯ We observe that K is compact. Since f is continuous on K , Proof Let K = . there exists z 0 ∈ K such that | f (z 0 )| = max | f (z)|. z∈K

If z 0 ∈ , then | f | has a local maximum at z 0 . Therefore by Theorem 2.19, we may assume that z 0 ∈ ∂. Let a ∈ and | f (a)| = | f (z 0 )|. Then | f | has a local maximum at a. This is not possible again by Theorem 2.19. Hence | f (a)| < | f (z 0 )| = max | f (z)|. z∈∂

50

2 The Cauchy Theorems and Their Applications

The boundedness of in Theorem 2.20 is necessary. Thus the Maximum modulus principle is not valid in unbounded open sets. We elaborate it by the following example. Let π π = z = x + iy − < y < 2 2 and f (z) = exp(exp(z)). Then is unbounded and f x ± π i = exp exp x ± π i = | exp (±i exp(x)) | = 1. 2 2 x

Thus max | f (z)| = 1. On the other hand, f (x) = ee → ∞ as x tends to infinity z∈∂

through positive reals. Now we derive from Theorems 2.17 and 2.18 the following two results. Theorem 2.21 Let f ∈ H () and f be one-one. Then f (z) = 0 for z ∈ and f −1 ∈ H ( f ()). Proof Let z 0 ∈ and f (z 0 ) = w0 . Assume that f (z 0 ) = 0. Then f (z) − w0 has a zero at z = z 0 of order n > 1. Therefore, every value in a neighbourhood of w0 is assumed n times by f . This contradicts that f is one-one. Let w, w1 ∈ f (). Then there exists z, z 1 ∈ such that f (z) = w, f (z 1 ) = w1 . By writing ψ = f −1 , we have ψ(w) = z, ψ(w1 ) = z 1 . By Open mapping theorem, we see that ψ is continuous on H ( f ()). Therefore z → z 1 as w → w1 . Thus ψ (w1 ) = lim

w→w1

1 ψ(w) − ψ(w1 ) z − z1 = . = lim z→z 1 f (z) − f (z 1 ) w − w1 f (z 1 )

Let f ∈ H () and z 0 ∈ . Then we say that f is locally invertible at z 0 if there exists r > 0 such that D(z 0 , r ) ⊆ and f −1 ∈ H ( f (D(z 0 , r ))). Theorem 2.22 (Inverse function theorem) Let f ∈ H () and z 0 ∈ with f (z 0 ) = 0. Then f is locally invertible at z 0 . Proof Let f (z 0 ) = w0 . Then f (z) − w0 has a simple root at z = z 0 since f (z 0 ) = 0. By Theorem 2.17, there exist ε > 0 and δ > 0 such that for a ∈ D(w0 , ε) there exists unique z ∈ D(z 0 , δ) with f (z) = a. We take U = f −1 (D(w0 , ε)). Then U is an open set, f ∈ H (U ) and f is one-one on U. Now we apply Theorem 2.21 with = U to derive f −1 ∈ H ( f (U )). Another application of Corollary 2.16 is the following result giving the number of zeros of a polynomial in a disk.

2.6 Argument Principle, Open Mapping Theorem, Maximum Modulus …

51

Theorem 2.23 (The Rouché theorem) Let γ be a closed path homologous to zero on . Assume that Indγ (α) ∈ {0, 1} for α ∈ \ γ ∗ and let 1 ⊆ \ γ ∗ be such that Indγ (α) = 1 for α ∈ 1 . Let f ∈ H () and g ∈ H () satisfy | f (z) − g(z)| < | f (z)| for z ∈ γ ∗ .

(2.6.1)

Then N f = Ng where N f and Ng denote the number of zeros of f and g, respectively, in 1 . Proof By (2.6.1), we see that γ does not pass through any zero of f. Now we apply Corollary 2.16 with α = 0 and = f = f ◦ γ. Then Ind f (0) = N f .

(2.6.2)

Further, we also observe from (2.6.1) that γ does not pass through any zero of g. Therefore we derive again from Corollary 2.16, as above, that Indg (0) = Ng

(2.6.3)

with g = g ◦ γ. By re-writing (2.6.1) as | f (t) − g (t)| < | f (t)| for 0 ≤ t ≤ 1, we conclude from Lemma 1.8 with α = 0 that Ind f (0) = Indg (0).

(2.6.4)

Hence we derive from (2.6.2), (2.6.3) and (2.6.4) that N f = Ng .

Example 2.2 Determine the numbers of zeros of z 87 + 36 z 57 + 71z 4 + z 3 − z + 1 inside |z| = 1. Solution. We take g(z) = z 87 + 36 z 57 + 71 z 4 + z 3 − z + 1,

f (z) = 71 z 4 .

Then for |z| = 1, we have | f (z) − g(z)| = z 87 + 36 z 57 + z 3 − z + 1| ≤ 1 + 36 + 1 + 1 + 1 < 71 = | f (z)|. Hence, we conclude from Theorem 2.23 with = C, γ : |z| = 1 and 1 = D(0, 1) that g(z) has four zeros inside |z| = 1 counted according to multiplicity.

52

2 The Cauchy Theorems and Their Applications

2.7 The Phragmen–Lindelöf Method: Maximum Modulus Principle in an Unbounded Strip and the Hadamard Three-Circle Theorem It has already been pointed out after the proof of Theorem 2.20 that the Maximum modulus principle need not be valid in unbounded regions. We prove a version of this principle for an unbounded strip, see Theorem 2.24 and Corollary 2.25. Further, we derive from it the Hadamard three-circle theorem. The proof depends on the Phragmen-Lindelöf method. This method also enables us to prove that the function is constant in a region whenever we restrict its growth in the region. We need not necessarily assume that it is bounded in the region for concluding that it is constant as is the case with the Liouville theorem. For example, an entire function f (z) is 1 constant if | f (z)| ≤ 1 + |z| 2 for z ∈ C, see Ex. 2.2 (ii). Theorem 2.24 For given a, b ∈ R, let = (x + i y)a < x < b . Let f be continuous on , f ∈ H () and | f (z)| < B for z ∈ and fixed B > 0. For a ≤ x ≤ b, let M(x) = sup{| f (x + i y)|| − ∞ < y < ∞}. Then we have (M(x))b−a ≤ (M(a))b−x (M(b))x−a for a ≤ x ≤ b.

(2.7.1)

As an immediate consequence of Theorem 2.24, we derive the following Maximum modulus principle for unbounded strips. Corollary 2.25 Suppose that the assumptions of Theorem 2.24 are satisfied. Further suppose that f is not constant. Then | f (z)| < max (M(a), M(b)) for z ∈ . Proof Assume that M(a) = M(b). Then by Theorem 2.24, we have M(x)b−a < (max (M(a), M(b)))b−x+x−a = (max (M(a), M(b)))b−a for a ≤ x ≤ b. Therefore M(x) < max (M(a), M(b)) for a ≤ x ≤ b and the assertion follows. Let M(a) = M(b). Then, by Theorem 2.24 as above, we have M(x) ≤ M(a) for a ≤ x ≤ b. Let z 0 ∈ . Then there is r > 0 such that

2.7 The Phragmen–Lindelöf Method: Maximum Modulus …

53

D(z 0 , r ) ⊂ . By Theorem 2.20, we have | f (z 0 )| < max | f (z)| ≤ M(x) ≤ M(a) |z−z 0 |=r

for some a < x < b. This implies again the assertion of Corollary 2.25.

Proof of Theorem 2.24 For > 0, we consider f + in place of f if M(a) = 0 and f − in place of f if M(b) = 0 and let tend to zero to observe that there is no loss of generality in assuming that M(a) > 0 and M(b) > 0. For z ∈ , we write z = x + i y with a ≤ x ≤ b. Suppose that the assertion is valid for all functions f satisfying the assumptions of Theorem 2.24 together with M(a) = M(b) = 1 and we prove it completely. Let b−z

z−a

g(z) = (M(a)) b−a (M(b)) b−a . We observe that g(z) is analytic in C and it has no zero in C. Further b−x

x−a

|g(z)| = (M(a)) b−a (M(b)) b−a

(2.7.2)

and the exponents on the right-hand side of (2.7.2) lie in [0, 1]. Therefore |g(z)| ≥ τ for z ∈ where τ = min(1, M(a)) min(1, M(b)) > 0 and |g(a + i y)| = M(a), |g(b + i y)| = M(b). Now we consider h(z) = M(b) = 1 that sup

−∞ 0, we have f (x + i y) = O e|y| for a ≤ x ≤ b.

(2.7.7)

Assume that f (a + i y) = O((|y| + 1)k1 ), f (b + i y) = O((|y| + 1)k2 ).

(2.7.8)

Then f (x + i y) = O((|y| + 1)k(x) ) for a ≤ x ≤ b uniformly in a ≤ x ≤ b where k(x) is a linear function in x assuming the values k1 and k2 at x = a and x = b, respectively. We observe that k(x) is convex in (a, b). Proof The assertion is immediate if y = 0 and we assume that y > 0 as the proof for y < 0 is similar. Let = {z = x + i y | a ≤ x ≤ b, y > 0} . First, we prove (2.7.8) with k1 = k2 = 0. Let M (a) = sup | f (a + i y)|, M (b) = sup | f (b + i y)| 0 h by Theorem 2.4. If k < h, we derive from Corollary 2.7 that 1 )···(z−αk−1 ) dz, the integral in the sum on the right-hand side is equal to |z|=R (z−α (z−α1 )···(z−αh ) where the circle |z| = R includes γ and the above integral tends to zero as R tends to infinity. Thus P(z)dz 1 . πh = 2πi γ (z − α1 ) · · · (z − αh ) Further

1 2πi

γ

( f (z) − P(z))dz =0 (z − α1 ) · · · (z − αh )

by Theorem 2.4 since f (z) ≡ P(z) (mod α1 , . . . , αn ). Then 1 πh = 2πi

γ

f (z)dz (z − α1 ) · · · (z − αh )

for

1 ≤ h ≤ n.

(2.10.1)

2.10 An Estimate for the Number of Zeros of an Exponential Polynomial in a Disc ∞

We say that a power series

r =0

αr z r is majorised by power series

∞ r =0

65

βr z r if |αr | ≤

βr for r ≥ 0. Let z 0 ∈ C with z 0 = 0, f (z) = e zz0 , f (z) ≡ P(z) (mod α1 , . . . , n αn ), P(z) = ph z h−1 and λ1 , . . . , λn be a sequence of positive real numbers. Then h=1

F(z 0 ) =

n

a k e αk z 0

k=1

= =

n k=1 n

ak P(αk ) ak

k=1

=

n

ph

h=1

=

n

ph αkh−1

h=1 n

ak αkh−1

k=1

n

h−1 ph λh λ−1 (0). h F

h=1

Therefore |F(z 0 )| ≤

n

| ph |λh

h=1

max

1≤h≤n

|F (h−1) (0)| λh

z0 ∈ C

for

(2.10.2)

since the inequality follows immediately if z 0 = 0. We write f (z) = e

z0 z

=

∞

ar z r

with

ar =

r =0

and g(z) = e|z0 |z =

∞

br z r

with

br =

r =0

z r0 r!

|z 0 |r . r!

Thus |ar | = |br | for r ≥ 0. Let β1 = β2 = · · · = βn = and g(z) ≡ Q(z)(mod β1 , β2 , . . . , βn ). Then n Q(z) = τh (z − )h−1 , (2.10.3) h=1

where τh =

|z 0 |h−1 |z0 | g (h−1) () = e . (h − 1)! (h − 1)!

(2.10.4)

66

2 The Cauchy Theorems and Their Applications

For u ∈ γ, we have αh −1 1 α1 ··· 1 − = As u −s−h , = u −h 1 − (u − α1 ) · · · (u − αh ) u u s=0 ∞

where As is the sum of all products α1s1 · · · αshh with s1 + · · · + sh = s and si ≥ 0. Then, by (2.10.1), we have

πh =

1 2πi

∞ γ

ar u r

∞

r =0

∞ As u −s−h du = ar +h−1 Ar .

Similarly τh =

(2.10.5)

r =0

s=0

∞

br +h−1 Br

(2.10.6)

r =0

where Bs are all products of β1s1 · · · βnsn with s1 + · · · + sn = s and si ≥ 0. Now we derive from (2.10.5) and (2.10.6) that |πh | ≤ τh for 1 ≤ h ≤ n since |αr | ≤ βr and |ar | = br for r ≥ 0. This implies from (2.10.1) and (2.10.3) that P(z) is majorised n−1 by Q 1 (z) = τk (z + )k since |αr | ≤ for 1 ≤ r ≤ n. By (2.10.4), we have k=0

Q 1 (z) = e|z0 |

n−1 |z 0 |k k=0

k!

(z + )k :=

n

qh z h−1 .

h=0

We take λh = (h − 1)! for 1 ≤ h ≤ n. Then n h=1

| ph |λh ≤

n

τh (h − 1)! =

h=1

n

Q (h−1) (0) 1

h=1

= e|z0 | ≤ e|z0 |

n−1

|z 0 |k

k−h+1 (k − h + 1)!

k=0

h=1

n−1

n−1 h

|z 0 |k

k=0

≤ e(|z0 |+1)

h=0 n−1 k=0

Therefore we obtain from (2.10.2) that

n

|z 0 |k .

h!

2.10 An Estimate for the Number of Zeros of an Exponential Polynomial in a Disc

|F(z 0 )| ≤ e(|z0 |+1)

n−1

|z 0 |k max | 1≤h≤n

k=0

F (h−1) (0) |. (h − 1)!

67

(2.10.7)

¯ R). We Let w1 , . . . , wm be all the zeros of F(z) taken with multiplicity in D(0, write M(R) = max |F(z)| and |z|=R

G(z) =

F(z) . (z − w1 ) · · · (z − wm )

We observe that G(z) is entire and |(z − w1 ) · · · (z − wm )| ≤ (2R)m

|z| ≤ R

for

and |(z − w1 ) · · · (z − wm )| ≥ (3R)m

for

|z| = 4R.

By Maximum modulus Theorem 2.20, we derive max |G(z)| ≥

M(R) (2R)m

max |G(z)| ≤

M(4R) . (3R)m

|z|=R

and |z|=4R

Therefore

M(R) M(4R) ≤ max |G(z)| ≤ max |G(z)| ≤ . m |z|=R |z|=4R (2R) (3R)m

This implies that

m log

M(4R) . M(R)

M(4R) Thus it suffices to show log (n + R). M(R) By Theorem 2.20, there exists z 0 with |z 0 | = 4 such that M(z 0 R) = max |F(z)|. |z|≤4R

We have F(z R) =

n k=1

By (2.10.7), we derive

ak ewk Rz .

68

2 The Cauchy Theorems and Their Applications M(4R) = F(z 0 R) ≤ e5R

4n − 1 4−1

n R j−1 F ( j−1) (0) 5R 4 − 1 ≤ e max M(R) 1≤ j≤n ( j − 1)! 4−1

by the Cauchy inequalities, see Sect. 2.3 (i). This implies log (n + R).

M(4R) M(R)

2.11 Exercises 2.1 Let f and g be entire functions such that | f (z)| = |g(z)| for z ∈ C. Then show that gf is a constant function. 2.2 Let f be entire function such that | f (z)| ≤ A + B|z|k for z ∈ C, where A, B ∈ C and k are positive integers. Then show that f is a polynomial of degree at most k. Further prove that f (z) is constant if the above inequality holds with k = 21 for z ∈ C. (Hint: Estimate | f (n) (0)|.) 2.3 Let f be analytic in the unit disc such that | f (z)| ≤ (1 − |z|)−1 . Show that the coefficients in the expansion f (z) =

∞

an z n

n=0

satisfy |an | ≤ e for n ≥ 1. 2.4 Let f (z) be an entire function. Assume that there exists an open disc where f does not assume some value. Then show that f is constant. (Hint: Apply the Liouville theorem.) 2.5 Let δ > 0 and {an }∞ n=0 be a sequence of positive real numbers approaching ∞ to zero. Prove that the series an z n is uniformly convergent in |z| ≤ 1 and n=0

|z − 1| ≥ δ. (Hint: Let Tn (z) =

n

ak z and Sn (z) = k

k=0

Tn (z) = an Sn (z) −

z k . Then use

k=0 n−1 k=0

by summation by part.)

n

Sk (z)(ak+1 − ak )

2.11 Exercises

69

2.6 (a) Let f be bounded and analytic in D (a, r ). Then show that f has a removable singularity at z = a. (b) Let f ∈ H ( \ {a}). Suppose that there exists a sequence {rn }∞ n=1 of positive real numbers such that lim rn = 0 and f is bounded on the circles n→∞ |z − a| = rn with n ≥ 1. Then show that f has a removable singularity at z = a. (Hint: Show that the coefficients cn with n < 0 in the Laurent series of f (z) around z = a vanish.) 2.7 (a) Let f be analytic at z = a and it has a zero of order m at z = a. Then show f has a simple pole at z = a with residue equal to m. that f (b) Let f be analytic in D (a, r ) for some r > 0 and assume that it has a pole of order m at z = a. Then show that ff has a simple pole at z = a with residue equal to −m. 1 (c) Calculate the residue of 2 at z = 0 by using series expansion for z sin z sin z. (d) Suppose f has pole of order m ≥ 1 at z = a. Then Res( f ; a) =

1 g (m−1) (a), (m − 1)!

where g(z) = (z − a)m f (z). 2.8 Let n be an even integer. Apply the Cauchy residue theorem to evaluate

2π

(cos θ)n dθ =

0

2.9 (a) Show that

2π

0

2πn! . 2n ((n/2)!)2

2π dθ =√ . 2 + sin θ 3

(Hint: The integral is equal to |z|=1

(z − i(−2 +

√

dz

√ √ 3))(z − i(− 2 − −3))

by substituting eiθ = z and apply the Cauchy residue theorem.) (b) Show that 2π (cos3 θ + sin2 θ)dθ = π. 0

2.10 Let f be acomplex-valued function. We say that f has an isolated singularity at ∞ if f 1z has an isolated singularity at z = 0. Similarly f is holomorphic (meromorphic) at ∞ if f 1z is holomorphic (meromorphic) at 0. If f 1z has

70

2 The Cauchy Theorems and Their Applications

a removable singularity at 0, then it can be defined as a holomorphic function in a neighbourhood of zero and we say that f is holomorphic at ∞. Further, we call f holomorphic (meromorphic) on C∞ if f is holomorphic (meromorphic) on C and f 1z is holomorphic (meromorphic) at zero. If f is holomorphic (meromorphic) on C∞ , then show that f is constant (rational function). 2.11 For 0 < a < 1, show that

∞

π t a−1 dt = . 1+t sin πa

0

∞

eax d x by putting t = e x and apply the e x +1 0 eaz dz where is the rectangle Cauchy residue theorem for computing z e +1 with vertices R, R + 2πi, −S + 2πi, −S such that R and S are positive.) 2.12 Suppose f ∈ H () where ⊇ D(0, 1). Assume that | f (z)| = 2 for |z| = 1 and | f (0)| = 1. Then show that f has a zero in the open unit disc. (Hint: Apply Maximum modulus principle.) 2.13 Give an example of a region such that its complement in C is infinite and every bounded holomorphic function on is constant. (Hint: Consider C\N.) 2.14 Let be bounded region. Assume that f ∈ H () and f is continuous on ∪ ∂. Then show that (Hint: The integral is equal to

∂( f ()) ⊆ f (∂). (Hint: Apply Open mapping theorem.) 2.15 Let D = D(0, 1) and let f ∈ H (D). Assume that f is not constant and f (z) = 1 for |z| = 1. Then show that D ⊆ f (D). (Hint: Let |w0 | < 1 and we consider g(z) = f (z) − w0 . Apply the Rouché theorem to show that f and g have the same number of zeros in D. Therefore, it suffices to show that f has at least one zero in D. Let z 0 ∈ D be fixed and we consider g1 (z) = f (z) − f (z 0 ). Derive again by the Rouché theorem that f and g1 have the same number of zeros in D and hence f has at least one zero in D since g1 has a zero at z = z 0 in D.) 2.16 Find a necessary and sufficient condition that az 2 + bz + c with a = 0 is one to one in |z| < 1. 2.17 Let D = D(0, 1) and suppose that f ∈ H () where ⊇ D and | f (z)| < 1 for |z| = 1. Then find the number of fixed points of f in D. (Hint: Apply the Rouché theorem to show that g(z) = f (z) − z and h(z) = −z have same number of zeros in D.) 2.18 Let m < n be positive integers. Show that the polynomial 1+z+ has exactly n zeros in D.

zm z2 + ··· + + 3z n 2! m!

2.11 Exercises

71

2.19 Determine the number of zeros of 2z 5 − 6z 2 + z + 1 in 1 ≤ |z| ≤ 2. (Hint: Denote by g(z) the polynomial in the exercise. Apply the Rouché theorem to find the number of zeros of g(z) in |z| ≤ 2 as well as in |z| < 1.) 2.20 Let f (z) be analytic in a neighbourhood of z 0 where it has a zero of order N . Then show that f (z) = (g(z)) N for some function g(z) which is analytic in a neighourhood of z 0 and g (z 0 ) = 0. 2.21 Let f be meromorphic but not analytic in C. Show that e f (z) is not meromorphic in C. 2.22 Prove that there is no analytic branch of logarithm defined on C\{0}. (Hint: Let T = C\{0} and G be the subset of T obtained by deleting all the negative real numbers from T . Let Logz be the principal branch of logarithm of z. Denoting by f (z) analytic branch of logarithm on G, we see from Theorem 2.28 that f (z) and Log z differ only by an integral multiple of 2πi. This contradicts that f is continuous at z = −1.) ¯ 2.23 Let f ∈ H ( D(0, R)) and n(r ) denote the number of zeros of f (counted with multiplicity) of f (z) in D(0, r ) with 0 < r < R. Assume that f (0) = 0. Then show that 2π R 1 dr = n(r ) log | f (r eiθ )|dθ − log | f (0)|. r 2π 0 0 (Hint: Let z 1 , . . . , z n with |z 1 | ≤ |z 2 | ≤ · · · ≤ |z n | be all the zeros of f (z) in D(0, R) and for 1 ≤ k ≤ N , let n k (z) = 1 if |z| < rk and 0 otherwise. In view of the Jensen formula, it suffices to show that R R dr log = n(r ) z r k 0 k=1

N

and the left-hand side is equal to N k=1

R |z k |

dr = r

R 0

N

dr n k (r ) = r k=1

R 0

dr n(r ) r

.

Chapter 3

Conformal Mappings and the Riemann Mapping Theorem

3.1 Introduction We begin this chapter with a definition of conformal mapping. Definition Let 1 , 2 and be open sets. A one-one analytic mapping f from 1 onto 2 such that f −1 ∈ H (2 ) is called an analytic homeomorphism of 1 onto 2 . A one-one analytic mapping from 1 onto 2 is called a conformal mapping of 1 onto 2 . Thus analytic automorphism of is one-one analytic function of onto itself with analytic inverse, and conformal mapping of is one-one analytic function of onto itself. By Theorem 2.21, we see that a conformal mapping of 1 onto 2 is an analytic homeomorphism of 1 onto 2 , and conformal mapping of is an automorphism of . We say that 1 and 2 are conformally equivalent whenever there is a conformal mapping f from 1 onto 2 (or between 1 and 2 ) and we write 1 ∼ 2 , where ∼ is an equivalence relation. If 1 ∼ 2 , then f (z) = 0 for z ∈ 1 and some authors take this as the definition of conformal mapping which is less restrictive (see [23, 31]). For example, f (z) = z 2 is not one-one but f vanishes nowhere in C\{0}. If f (z) = 0 at z = z 0 , then f preserves angles at z 0 . This implies that for lines L 1 and L 2 starting at z 0 , the angle between f (L 1 ) and f (L 2 ) at f (z 0 ) is the same, in size and orientation, made by L 1 and L 2 at z 0 . We shall not consider this more general definition of conformal mapping. We always mean, as is the general practice among complex analysts (see [29]) that conformal mapping is one-one onto analytic function. The set of all automorphisms of is a group under composition of mappings and we denote it by Aut(). For λ ∈ C with |λ| = 1, we observe that z → λz is an automorphism of D = D(0, 1) and this automorphism is called a rotation. In Sect. 3.2, we give some explicit examples of conformal mappings. We begin section Sect. 3.3 with the Schwarz lemma which is a very useful application of the Maximum modulus principle. Further we apply the Schwarz lemma to give a characterisation of all the automorphisms of the open unit disc. In Sect. 3.4, we use © Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9_3

73

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3 Conformal Mappings and the Riemann Mapping Theorem

this characterisation to determine all the automorphisms of the upper half plane. In Sects. 3.5–3.7, we prove the Riemann mapping theorem that every simply connected region different from C is conformally equivalent to the open unit disc. Moreover we prove the uniqueness of this conformal mapping. We refer to [1, 12, 23] for the topics in this chapter and for further studies and related topics.

3.2 Examples of Explicit Conformal Mappings (i) Let 1 = H and 2 = D. We show that 1 and 2 are conformally equivalent. For this, we take i −z for z ∈ 1 . F(z) = i +z Let z ∈ 1 and we write z = x + i y with y > 0. Then F(z) =

−x − i(y − 1) . x + i(y + 1)

Thus |F(z)|2 = F(z)F(z) =

x 2 + (y − 1)2 x 2 + (y + 1)2

and hence F(z) ∈ 2 since y > 0. Further we observe that F is analytic in 1 since −i ∈ / 1 . Also F is one-one. Therefore it remains to show that F is onto. Let w ∈ 2 and we write w = u + iv with u 2 + v 2 < 1. We find z ∈ 1 such that F(z) = w. = w and then Thus i−z i+z z=i

1−w 1+w

=i

1 − u − iv 1 + u + iv

Therefore Im(z) =

=i

(1 − u − iv)(1 + u − iv) . (1 + u + iv)(1 + u − iv)

1 − u 2 − v2 >0 (1 + u)2 + v 2

and hence z ∈ 1 . (ii) For a positive integer n, let π 1 = z ∈ C 0 < arg z < n and 2 = H. We show that 1 and 2 are conformally equivalent. For this, we take f (z) = z n for z ∈ 1 .

3.2 Examples of Explicit Conformal Mappings

75

We write z = r eiθ with 0 < θ < πn . Then f (z) = r n einθ . Thus Im( f (z)) = r n sin nθ > 0 since 0 < θ < πn and hence f is a function from 1 into 2 . We show that f is one-one. Let z 1 , z 2 ∈ 1 with f (z 1 ) = f (z 2 ) and z 1 = r1 eiθ1 , z 2 = r2 eiθ2 with 0 < θ1 ≤ θ2

0, x 2 + y 2 < 1 be the upper half disc and 2 = u + iv u > 0, v > 0 be the first quadrant. Then we show that f (z) =

1+z 1−z

is a conformal mapping of 1 onto 2 . Let z = x + i y ∈ 1 . Then |x| < 1, y > 0, x 2 + y 2 < 1 and f (z) =

1 + x + iy 2y (1 + x + i y)(1 − x + i y) 1 − (x 2 + y 2 ) +i . = = 1 − x − iy (1 − x − i y)(1 − x + i y) (1 − x)2 + y 2 (1 − x)2 + y 2

Since Re( f (z)) > 0 and Im( f (z)) > 0, we see that f (z) ∈ 2 . It is clear that f is / 1 and further f is one-one. Thus it remains to show that analytic in 1 since 1 ∈ f is onto. Let w = u + iv ∈ 2 . Then u > 0, v > 0 and we shall find z ∈ 1 such = w and then that f (z) = w. Thus 1+z 1−z z=

(u − 1) + iv ((u − 1) + iv)((u + 1) − iv) u 2 + v2 − 1 w−1 2v = = = + i . w+1 (u + 1) + iv ((u + 1) + iv)((u + 1) − iv) (u + 1)2 + v 2 (u + 1)2 + v 2

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3 Conformal Mappings and the Riemann Mapping Theorem

Thus Im(z) > 0 since v > 0. Further |z| < 1 since |w − 1| < |w + 1| whenever w lies in the first quadrant and hence z ∈ 1 . (iv) Let 1 = H and 2 = u + iv 0 < v < π . For showing 1 and 2 are conformally equivalent, we consider f (z) = log z, z ∈ 1 , where the branch of logarithm is principal. Then f ∈ H (1 ), f is one-one and we show that f is onto. Let w = u + iv ∈ 2 . Then 0 < v < π and we take z = ew = eu eiv = eu (cosv + i sinv). Thus Im(z) = eu sinv > 0 since 0 < v < π. Therefore z ∈ 1 and f (z) = ω. Hence f is a conformal mapping between 1 and 2 .

3.3 Automorphisms of the Open Unit Disc The proofs of this section and also of subsequent sections in this chapter depend on the well-known Schwarz lemma. Lemma 3.1 Let f be analytic in D where f satisfies | f (z)| ≤ 1, f (0) = 0. Then | f (z)| ≤ |z| for |z| < 1 and | f (0)| ≤ 1. If | f (z)| = |z| for some |z| < 1 or | f (0)| = 1, then f (z) = cz in |z| < 1 for some constant c whose absolute value is 1. Proof Let ⎧ ⎨ f (z) , if z = 0 g(z) = z ⎩ f (0), if z = 0. Then g is analytic in |z| < 1 since f (0) = 0. For 0 ≤ r < 1, we derive from Theorem 2.20 that 1 1 max | f (z)| ≤ . max |g(z)| = max |g(z)| = |z|≤r |z|=r r |z|=r r Letting r tend to 1, we get |g(z)| ≤ 1 in |z| < 1. Then

3.3 Automorphisms of the Open Unit Disc

77

| f (z)| ≤ |z| in |z| < 1. Assume that either | f (z 0 )| = |z 0 | for some |z 0 | < 1 or | f (0)| = 1. Then g(z) = 1 for some |z| < 1. Therefore g is constant in D of absolute value 1 by Theorem 2.19. Then f (z) = cz in |z| < 1 where c is a constant with |c| = 1. We write T = z |z| = 1 . For a ∈ D, we consider function φa from C∞ into C∞ given by z−a . (3.3.1) φa (z) = 1 − az We observe that 1 1 1 1 = ∞, φ−a − = ∞, φa (∞) = − , φ−a (∞) = . φa a a a a The function φa satisfies the following properties. Lemma 3.2 Let a ∈ D. Then (i) (ii) (iii) (iv) (v)

φa is one-one, onto and the inverse of φa is φ−a and φa (a) = 0. φa is analytic in D(0, a1 ) containing D(0, 1). φa (T ) = T. φa (D) = D. 1 φa (0) = 1 − |a|2 , φa (a) = 1−|a| 2.

Proof (i) We show φ−a ◦ φa (z) = z, φa ◦ φ−a (z) = z for z ∈ C. We prove the first and the proof for the latter is similar. If z = ∞, then φ−a

1 =∞ ◦ φa (∞) = φ−a − a

and if z = a1 , then φ−a ◦ φa

1 1 = φ−a (∞) = a a

by (3.3.2). Therefore we may assume that z = φ−a ◦ φa (z) = φ−a

z−a 1 − az

since aa = |a|2 < 1 for a ∈ D.

=

1 a

z−a 1−az

1+

and z = ∞. Now +a

z−a a 1−az

=

z(1 − aa) = z, 1 − aa

(3.3.2)

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3 Conformal Mappings and the Riemann Mapping Theorem

(ii) We observe that φa (z) is analytic in C except at z = a1 and a1 > 1. Let 1 < r < 1 . Then φa (z) is analytic in D(0, r ). Since D ⊆ D(0, r ), the assertion follows. |a| (iii) For t ∈ R, eit − a eit − a . = φa (eit ) = 1 − aeit eit (e−it − a) Therefore |φa (eit )| = 1. Thus φa (T ) ⊆ T. Similarly φ−a (T ) ⊆ T implying T ⊆ φa (T ). Hence φa (T ) = T. (iv) This follows immediately from (iii). (v) We have φa (z) =

(1 − az) − (z − a)(−a) 1 − aa 1 − |a|2 = = . (1 − az)2 (1 − az)2 (1 − az)2

Thus

φa (0) = 1 − |a|2

and φa (a) =

1 − |a|2 1 = . (1 − |a|2 )2 1 − |a|2

Now we derive from Lemmas 3.1 and 3.2, the following result. Lemma 3.3 Let f (z) be non-constant and analytic in D where | f (z)| < 1. Let α ∈ D with f (α) = a. Then | f (α)| ≤

1 − |a|2 . 1 − |α|2

Moreover equality occurs only when f = φ−a ◦ (cφα ) in D, for some constant c whose absolute value is 1.

3.3 Automorphisms of the Open Unit Disc

79

Proof We consider g(z) = φa ◦ f ◦ φ−α (z) in |z| < 1. Then by Lemma 3.2 (i), (ii) and (iv), we see that g(z) is analytic in |z| < 1 satisfying |g(z)| < 1 for |z| < 1 and g(0) = φa ◦ f ◦ φ−α (0) = φa ◦ f (α) = φa (a) = 0. Thus the assumptions of Lemma 3.1 are satisfied and hence we conclude that |g (0)| ≤ 1.

(3.3.3)

Now we compute g (0) by using Lemma 3.2 (v) g (0) = (φa ◦ f ) (φ−α (0)) φ−α (0) = (1 − |α|2 )(φa ◦ f ) (α) = (1 − |α|2 )φa ( f (α)) f (α) = (1 − |α|2 )φa (a) f (α) =

1 − |α|2 f (α). 1 − |a|2

By (3.3.3), we get | f (α)| ≤

1 − |a|2 . 1 − |α|2

Suppose that the above relation holds with equality sign. Then |g (0)| = 1. Now we derive from Lemma 3.1 that there exists a constant c with |c| = 1 such that g(z) = cz for |z| < 1. Therefore φa ◦ f ◦ φ−α (z) = χc (z) for |z| < 1 where χc (z) = cz. Thus f = φ−a ◦ χc ◦ φα = φ−a ◦ cφα in D. Now we use Lemma 3.3 to characterise all automorphisms of D carrying given α ∈ D to origin. Theorem 3.4 Let f be an automorphism of D and α ∈ D such that f (α) = 0. Then f = cφα in D where c is a constant of absolute value 1.

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3 Conformal Mappings and the Riemann Mapping Theorem

We observe that rotation is an automorphism of D. On the other hand, Theorem 3.4 with α = 0 states as follows. Corollary 3.5 All the automorphisms of D carrying the centre to centre are given by rotations. Proof of Theorem 3.4. Let h be the inverse of f . Then h( f (z)) = z for z ∈ D, h(0) = α.

(3.3.4)

By Theorem 2.21, we see that h ∈ H (D), h (z) = 0 for z ∈ D. By differentiating both sides in (3.3.4), we get h ( f (z)) f (z) = 1 for z ∈ D. By putting z = α, we have h ( f (α)) f (α) = h (0) f (α) = 1. Now we derive from Lemma 3.3 with a = 0 that | f (α)| ≤

1 , |h (0)| ≤ 1 − |α|2 1 − |α|2

and hence | f (α)| =

1 . 1 − |α|2

Now we apply again Lemma 3.3 to conclude that f = φ0 ◦ cφα = cφα in D where c is a constant of absolute value 1.

We give the following example before closing this section. Example 3.1 Let f ∈ H (D) and | f (z)| < 1 for z ∈ D. Assume that f has two distinct fixed points in D. Then f (z) = z for z ∈ D. Solution. Let p and q be distinct fixed points of D. We may assume that p = q and q = 0. First, we consider the case p = 0. We have f (q) = q implying that

3.3 Automorphisms of the Open Unit Disc

81

| f (q)| = |q| and derive from the Schwarz lemma 3.1 that f (z) = λz, where λ is a constant of absolute value 1. Putting z = q, we get q = λq implying that λ = 1 and hence f (z) = z for z ∈ D. Next we consider the case p = 0. Let F(z) =

p−z for z ∈ D. 1 − pz ¯

We observe that F(z) = φ− p (−z), and therefore F(z) ∈ H (D) such that F is oneone and |F(z)| < 1 for D. Now we consider g(z) = F ◦ f ◦ F(z) for z ∈ D. Then g(z) ∈ H (D) and |g(z)| < 1 for z ∈ D. Further F ◦ F(z) = F

p−z 1 − pz ¯

=

p− 1−

p−z 1− pz ¯ p−z p¯ 1− pz ¯

=

(1 − p p)z ¯ (1 − pz) ¯ p− p+z = =z 1 − pz ¯ − p( ¯ p − z) 1 − p p¯

since p ∈ D. Therefore, F(z) = F −1 (z) for z ∈ D. We have g(0) = F ◦ f (F(0)) = F( f ( p)) = F( p) = 0 and g(F(q)) = F ◦ f ◦ F(F(q)) = F ◦ f (q) = F(q). Therefore, g has two distinct fixed points in D, namely, 0 and F(q) = 0 since p = q. Therefore, we derive from the earlier case p = 0 that g(z) = z for z ∈ D. Then f (z) = F ◦ F(z) = z for z ∈ D.

3.4 Automorphisms of the Upper Half Plane Denote by G = SL2 (R) the set of all 2 × 2 matrices M=

ab c d

(3.4.1)

with a, b, c, d ∈ R such that its determinant ad − bc is equal to 1. It is clear that G is a group under matrix multiplication. For M ∈ G, we denote f M (z) =

az + b . cz + d

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3 Conformal Mappings and the Riemann Mapping Theorem

For M ∈ G and N ∈ G, we check by direct computation that fMN = fM ◦ fN .

(3.4.2)

We prove that the group Aut(H) of automorphisms of H is given by { f M | M ∈ G} . First we show that { f M | M ∈ G} ⊆ Aut(H). Lemma 3.6 Let M ∈ G. Then f M ∈ Aut(H). Proof Let M ∈ G be given by (3.4.1). It is clear that f M is analytic in H since / H. Let z ∈ H with z = x + i y. Then y > 0 and − dc ∈ ax + b + iay az + b = Im cx + d + icy cz + d (cx + d)ya − (ax + b)yc (ax + b + i ya)(cx + d − i yc) = = Im (cx + d)2 + c2 y 2 (cx + d)2 + c2 y 2 y = >0 (cx + d)2 + c2 y 2

Im( f M (z)) = Im

since ad − bc = 1. Thus f M is analytic function of H into H. Let f M (z 1 ) = f M (z 2 ). Then (ad − bc)z 1 = (ad − bc)z 2 implying z 1 = z 2 , since ad − bc = 0. Therefore f M is one-one and we show that f M is onto to complete the proof of Lemma 3.6. Let N ∈ G be the inverse of M. Then d −b N= −c a and we check that for z ∈ H fM

dz − b −cz + a

= f M ( f N (z)) =

dz−b a −cz+a +b dz−b c −cz+a +d

= z,

since ad − bc = 1.

Finally, we prove the following theorem. Theorem 3.7 We have Aut(H) = { f M | M ∈ G}. The proof depends on the following two results and Corollary 3.5. Lemma 3.8 Let z, w ∈ H. Then there exists M ∈ G such that f M (z) = w. Proof Let z = x + i y ∈ H. Then y > 0. Let b and c = 0 be real numbers which we shall choose later. We put M1 =

0 −c−1 c 0

, M2 =

1b 01

3.4 Automorphisms of the Upper Half Plane

83

and M = M2 M1 . It is clear that M1 , M2 ∈ G and thus M ∈ G. By (3.4.2), we have f M (z) = f M2 ◦ f M1 (z) = f M2

1 − 2 c z

.

(3.4.3)

y 1 = 2 2 =1 Im − 2 c z c |z|

We observe that

by choosing c suitably. Further we write −

1 =χ+i c2 z

(3.4.4)

where χ ∈ R. Then, we see from (3.4.2) that f M (z) = f M2 (χ + i) = χ + b + i. We choose b = −χ and hence

f M (z) = i.

Let w ∈ H. Similarly, as above, there exists M3 ∈ G such that f M3 (w) = i. Let M4 = M3−1 M. Then M4 ∈ G and f M4 (z) = f M3−1 ( f M (z)) = f M3−1 (i) = w

by (3.4.2). Lemma 3.9 For real number θ, let Mθ = and F(z) = Then

cos θ − sin θ sin θ cos θ

i −z for z ∈ H. i +z

(3.4.5)

F ◦ f Mθ ◦ F −1 (z) = e−2iθ z for z ∈ D.

Proof By example (i) in Sect. 3.2, we see that F is a conformal mapping of H onto D. For θ ∈ R, let λθ (z) = e−2iθ z for z ∈ D. It suffices to show that F ◦ f Mθ = λθ ◦ F in H

84

3 Conformal Mappings and the Riemann Mapping Theorem

since then for z ∈ D, we have F ◦ f Mθ ◦ F −1 (z) = F ◦ f Mθ (F −1 (z)) = λθ ◦ F(F −1 (z)) = λθ (z) = e−2iθ z. Now for z ∈ H F ◦ f Mθ (z) = F( f Mθ (z)) = F

z cos θ − sin θ z sin θ + cos θ

=

i− i+

z cos θ−sin θ z sin θ+cos θ z cos θ−sin θ z sin θ+cos θ

z(i sin θ − cos θ) + (i cos θ + sin θ) z(i sin θ + cos θ) + (i cos θ − sin θ) −ze−iθ + ie−iθ i −z = e−2iθ F(z). = = e−2iθ zeiθ + ieiθ i +z

=

Proof of Theorem 3.7. By Lemma 3.6, it suffices to show that Aut(H) ⊆ G. Let f ∈ Aut(H). Then there exists β ∈ H such that f (β) = i. By Lemma 3.8, there exists N ∈ G such that f N (i) = β. We put g = f ◦ fN .

(3.4.6)

Then g(i) = f ( f N (i)) = f (β) = i. Now we consider F ◦ g ◦ F −1 where F is given by (3.4.5). We observe that F ◦ g ◦ F −1 ∈ Aut(D) and F ◦ g ◦ F −1 (0) = F(g(i)) = F(i) = 0. Hence we conclude from Corollary 3.5 that there exists θ ∈ R such that F ◦ g ◦ F −1 = tθ in D where

(3.4.7)

tθ (z) = e−2iθ z for z ∈ D.

Further, we see from Lemma 3.9 that F ◦ f Mθ ◦ F −1 = tθ in D.

(3.4.8)

By combining (3.4.6), (3.4.7) and (3.4.8), we have f ◦ f N = g = f Mθ . Thus by (3.4.2) f = f Mθ N −1 and Mθ N −1 ∈ G.

3.5 The Riemann Mapping Theorem and More General Theorem 3.11

85

3.5 The Riemann Mapping Theorem and More General Theorem 3.11 In Sect. 3.2, we gave examples of regions that are conformally equivalent. In this section, we prove a general theorem. Theorem 3.10 (The Riemann mapping theorem) Let = C be a simply connected. Then is conformally equivalent to D. In view of the Liouville theorem, see Sect. 2.3(i), the assumption = C is necessary. Theorem 3.10 is a consequence of the following more general result. Theorem 3.11 Let = C be a region. Assume that for every f ∈ H () with 1 ∈ H (), there exists g ∈ H () such that f (z) = g 2 (z) for z ∈ . Then is f conformally equivalent to D. By Corollary 2.30, the assumption of Theorem 3.11 is satisfied whenever = C is simply connected. Therefore, Theorem 3.11 implies Theorem 3.10 and the following criterion for determining whether a region is simply connected or not simply connected. Corollary 3.12 A region is simply connected if and only if its complement in the extended plane is connected. Proof Let be a region. By Theorem 1.9, it remains to show that is simply connected whenever C∞ \ is connected. Let = C. Then the assumptions of Theorem 3.11 are satisfied by Corollary 2.29 and we derive that is homeomorphic z is a homeomorphism from onto D. Hence there to D. If = C, then z → 1+|z| exists a homeomorphism ψ from onto D in either of the cases. Let γ be a closed path in . We define H from I × I into given by H ((s, t)) = ψ −1 (tψ(γ(s))) for 0 ≤ s ≤ 1, 0 ≤ t ≤ 1. It is clear that H is continuous since γ, ψ and ψ −1 are continuous. Further H (s, 0) = ψ −1 (0), H (s, 1) = γ(s) for 0 ≤ s ≤ 1 and H (0, t) = H (1, t) since γ is closed. Thus γ is null-homotopic in and hence is simply connected.

3.6 Lemmas for the Proof of Theorem 3.11 Lemma 3.13 Let be an open set. Then there exists sequence {K n }n≥1 of compact ˚ ˚ sets such that = ∞ n=1 K n and K n ⊆ K n+1 for n ≥ 1 where K n+1 denotes the

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3 Conformal Mappings and the Riemann Mapping Theorem

interior of K n+1 . Further for a compact set K in , we have K ⊆ K n for some n ≥ 1. Proof For n ≥ 1, let K n be given as z ∈ K n if and only if z ∈ , |z| ≤ n and |z − a| ≥ 1 for every a ∈ / . It is clear that K n ⊆ and it is bounded. Further, it is closed n since it is an intersection of two closed sets {z | |z| ≤ n} and {z | d(z, C − ) ≥ n1 }. Therefore K n is compact. It is clear that K n ⊆ K n+1 . In fact, we prove K n ⊆ K˚ n+1 . For this, we show that (3.6.1) D(z, r ) ⊆ K n+1 for z ∈ K n where r =

1 1 − . Let z ∈ K n and ζ ∈ D(z, r ). Then n n+1 |ζ − z|

0 such that D(z, ρ) ⊆ . Let N be 1 the least integer greater than or equal to max M, ρ . Then |z| ≤ M ≤ N and for a∈ / 1 |z − a| ≥ ρ ≥ . N Therefore z ∈ K N and thus ⊆

∞

K n . Hence =

n=1

⊆

∞

n=2

∞

n=1

K˚ n ⊆

∞

n=1

K˚ n

K n . Now

3.6 Lemmas for the Proof of Theorem 3.11

since K n ⊆ K˚ n+1 . Also

∞

87

K˚ n ⊆ and hence =

n=1

∞

K˚ n .

n=1

Let K be a compact subset of . Then K ⊂

∞

K˚ n

n=1

is an open cover of K . Therefore K ⊂

P

K˚ n

n=1

for some P. Thus K ⊂ K1 ∪ K2 · · · ∪ K P = K P . Lemma 3.14 (Hurwitz) Let be a region and { f n }n≥1 be a sequence such that f n ∈ H () converges uniformly on compact subsets of . Assume that f n with n ≥ 1 are one-one. Then the limit function f of the sequence { f n }n≥1 is either constant or one-one on . Proof We see from Sect. 2.3(iv) that f ∈ H (). We may assume that f is not oneone and then there exist z 1 , z 2 ∈ such that z 1 = z 2 with f (z 1 ) = f (z 2 ). For n ≥ 1, we define gn (z) = f n (z) − f n (z 1 ) and g(z) = f (z) − f (z 1 ) so that gn ∈ H (), g ∈ H (), gn (z 2 ) = 0 since f n is one-one and g(z 2 ) = 0. Further, we may suppose that g is not constant in otherwise f is constant in and the assertion follows. Since the zeros of g are isolated, see Sect. 2.3(ii), there exists r > 0 such that |z 1 − z 2 | > r and g(z) = 0 in 0 < |z − z 2 | ≤ r. Let γ be a circle with z 2 as centre and r as radius. Then there exists δ > 0 such that |g(z)| > δ for z ∈ γ ∗ . Also |gn (z) − g(z)|

0 and compact subset K ⊆ there exists δ > 0 depending only on and K such that | f (z 1 ) − f (z 2 )| < for f ∈ F and z 1 , z 2 ∈ K with |z 1 − z 2 | < δ. The limit of the subsequence in the above definition (i) belongs to H () but it need not belong to F. We prove the following result on uniformly bounded family F. Lemma 3.15 (Montel) Let F be a family of H () with region. Assume that F is uniformly bounded on compact subsets of . Then F is a normal family. Proof First, we show that the family F is equicontinuous on compact subsets of . ◦ By Lemma 3.13, we find compact subsets K n with n ≥ 1 such that K n ⊆ K n+1 for ∞

K n and every compact subset of is contained in K n for some n. Let n ≥ 1, = n=1 ◦ c ◦ )c is closed and K n (K n+1 ) = ∅, n ≥ 1 be an integer. Since K n is compact, (K n+1 we derive from Theorem 1.4 that there exists δn > 0 such that

3.6 Lemmas for the Proof of Theorem 3.11

89

◦ |z 1 − z 2 | > 2δn for z 1 ∈ K n , z 2 ∈ / K n+1 .

Thus

◦ D(z, 2δn ) ⊆ K n+1 ⊆ K n+1 for z ∈ K n .

Let z ∈ K n , z ∈ K n with |z − z | < δn . Let γ be a circle with centre at z and radius 2δn . Thus z lies inside the circle γ and γ ∗ ⊆ K n+1 . Let f ∈ F. Then we derive from Theorem 2.3 that f (ζ) f (ζ) 1 1 1 1 1 dζ dζ − dζ = f (ζ) − 2πi γ ζ − z 2πi γ ζ − z 2πi γ ζ − z ζ − z f (ζ) z − z = dζ. 2πi γ (ζ − z )(ζ − z )

f (z ) − f (z ) =

We observe that |ζ − z | = 2δn and |ζ − z | = |ζ − z + z − z | > 2δn − δn = δn . Since F is uniformly bounded on compact subsets of , there exists a constant M(K n+1 ) depending only on K n+1 such that | f (z ) − f (z )|

0 and δn < δn . δ = δ(n) = + M(K n+1 ) Thus for |z − z | < δ, we have M(K n+1 ) M(K n+1 ) M(K n+1 ) < . |z − z | < δ= δn δn + M(K n+1 ) Therefore we have

| f (z ) − f (z )| <

(3.6.3)

for f ∈ F and z , z ∈ K n and |z − z | < δ. Thus the family F is equicontinuous on compact subsets of since every compact subset K of is contained in K n for some n, and therefore (3.6.3) is valid for all f ∈ F and z , z ∈ K and |z − z | < δ. Hence F is equicontinuous on compact subsets of . Let { f m } be a sequence in F. We show that it has a subsequence which converges uniformly on compact subsets of . Let E be a countable dense set of . For example, we take E to be the set of all points of with rational coordinates. We arrange the elements of E as w1 , w2 , . . . Since { f m (w1 )} is a bounded sequence

90

3 Conformal Mappings and the Riemann Mapping Theorem

by assumption, the sequence { f m } has a subsequence { f m1 } such that { f m1 (w1 )} converges. Since it is bounded, the sequence { f m1 } has a subsequence { f m2 } such that { f m2 (w2 )} converges. Proceeding recursively, we see that for i ≥ 1 there is a subsequence { f mi } of { f m,i−1 } such that { f mi (wi )} converges. Here we write f m0 for f m . Now we consider the diagonal sequence { f mm }. This converges at all w1 , w2 , . . . We show that the diagonal sequence converges uniformly on K n with n ≥ 1. Then it converges uniformly on all compact subsets K of since K ⊆ K n for some n by Lemma 3.13. For n ≥ 1 and δ = δ(n) as above, we have Kn ⊆

D(z, δ).

(3.6.4)

D(z, δ)

(3.6.5)

z∈K n

Then Kn ⊆

z∈E∩K n

since E ∩ K n is dense in K n . Since K n is compact, we observe that open cover (3.6.5) admits a finite subcover. Thus there exist z 1 , z 2 , . . . , z p ∈ E ∩ K n such that K n ⊆ D(z 1 , δ) ∪ D(z 2 , δ) · · · ∪ D(z p , δ). For > 0, there exists M depending only on and K n such that | frr (z i ) − f ss (z i )| ≤

(3.6.6)

for r ≥ M, s ≥ M and 1 ≤ i ≤ p. Let z ∈ K n . Then z ∈ D(z i , δ) for some i with 1 ≤ i ≤ p by (3.6.5). Further, by (3.6.3) and (3.6.6), we have | frr (z) − f ss (z)| = | frr (z) − frr (z i ) + frr (z i ) − f ss (z i ) + f ss (z i ) − f ss (z)| ≤ | frr (z) − frr (z i )| + | frr (z i ) − f ss (z i )| + | f ss (z) − f ss (z i )| < + + = 3 wherever r ≥ M, s ≥ M. Thus { f mm } converges uniformly on K n . Since every compact subset of is contained in K n for some n by Lemma 3.13, we conclude that { f mm } converges uniformly on compact subsets of . Let z 0 ∈ and be the set of all ψ ∈ H () such that ψ is one-one on and ψ() ⊆ D. Our aim is to prove that contains an element which is onto. We show such that that for ψ ∈ there exists ψ1 ∈ |ψ1 (z 0 )| > |ψ (z 0 )|

(3.6.7)

whenever the assumptions of Theorem 3.11 are satisfied and ψ is not onto. Next we consider

3.6 Lemmas for the Proof of Theorem 3.11

91

η = sup |ψ (z 0 )| : ψ ∈ and prove that the supremum is assumed for some ψ0 ∈ has to be an onto function.

(3.6.8) . Then it is clear that ψ0

In the next two lemmas, we suppose that the assumptions of Theorem 3.11 are satisfied. In the first one, we show that is non-empty and in the second, we prove (3.6.7) for every ψ ∈ which is not onto. Lemma 3.16 Let satisfy the assumptions of Theorem 3.11. Then is non-empty. Proof Since = C, let w0 ∈ C such that w0 ∈ / . We consider the function f (z) = z − w0 .

(3.6.9)

/ . Thus 1/ f ∈ H (). We observe that f ∈ H () and f has no zero in since w0 ∈ Therefore, by our assumption, there exists g ∈ H () such that f (z) = g 2 (z) for z ∈ .

(3.6.10)

Let a ∈ . By Open mapping theorem 2.18, we observe that g() is open containing g(a). Therefore there exists r > 0 such that D(g(a), r ) ⊆ g(). Now we show that D(−g(a), r ) ∩ g() = ∅. Suppose there exists z 1 ∈ such that g(z 1 ) ∈ D(−g(a), r ). Thus |g(z 1 ) + g(a)| < r i.e. | − g(z1 ) − g(a)| < r. Then −g(z 1 ) ∈ D(g(a), r ) ⊆ g(). Therefore there exists z 2 ∈ such that − g(z 1 ) = g(z 2 ).

(3.6.11)

By squaring both sides of (3.6.11), we derive from (3.6.9) and (3.6.10) that z 1 − w0 = (−g(z 1 ))2 = (g(z 2 ))2 = z 2 − w0 implying z 1 = z 2 . Thus g(z 1 ) = 0 by (3.6.11). Then z 1 − w0 = 0. This is a contradic/ . Hence D(−g(a), r ) ∩ g() = φ. Now we consider tion since z 1 ∈ and w0 ∈ ψ(z) =

r for z ∈ . g(z) + g(a)

92

3 Conformal Mappings and the Riemann Mapping Theorem

We observe that ψ(z) ∈ H () and |ψ(z)| ≤ 1 for z ∈ since |g(z) + g(a)| ≥ r for z ∈ . In fact |ψ(z)| < 1 for z ∈ by Theorem 2.19. Further ψ ∈ H () and implies g 2 (z ) = g 2 (z ), and therefore z = z one-one on , since ψ(z ) = ψ(z ) by (3.6.10) and (3.6.9). Hence ψ ∈ . Lemma 3.17 Let z 0 ∈ and satisfies the assumptions of Theorem 3.11. Let ψ ∈ satisfying such that ψ() is a proper subset of D. Then there exists ψ1 ∈ (3.6.7). Proof Let ψ ∈ . Since ψ() is a proper subset of D, there exists α ∈ D such that α∈ / ψ(). We consider φα ◦ ψ, where φα is given by (3.3.1). We recall that φα is an automorphism of D. For z ∈ , we observe that φα ◦ ψ(z) = φα (ψ(z)) =

ψ(z) − α =0 1 − αψ(z)

only when ψ(z) = α which is not the case since α ∈ / ψ(). Therefore φα ◦ ψ ∈ H () has no zero in . Then, by our assumption, there exists g ∈ H () such that g 2 (z) = φα ◦ ψ(z) for z ∈ .

(3.6.12)

By writing s(w) = w2 for w ∈ D, we write (3.6.12) as s ◦ g = φα ◦ ψ in .

(3.6.13)

Then we see from (3.6.13) that ψ(z 1 ) = ψ(z 2 ) For z 1 = z 2 in , let g(z 1 ) = g(z 2 ). since φα is one-one. Therefore g ∈ . Let ψ1 = φβ ◦ g with g(z 0 ) = β.

(3.6.14)

We observe from (3.6.12) that ψ1 (z 0 ) = φβ (g(z 0 )) = φβ (β) = 0. By (3.6.13) and (3.6.14) ψ = φ−α ◦ s ◦ g = φ−α ◦ s ◦ φ−β ◦ ψ1 = F ◦ ψ1 , where F = φ−α ◦ s ◦ φ−β . By chain rule

by (3.6.15). Thus

ψ (z 0 ) = F (ψ1 (z 0 ))ψ1 (z 0 ) = F (0)ψ1 (z 0 ) |ψ (z 0 )| = |F (0)||ψ1 (z 0 )|

and hence it suffices to show that |F (0)| < 1.

(3.6.15)

3.6 Lemmas for the Proof of Theorem 3.11

93

We observe that F(D) ⊆ D. Let F(0) = a. Then by Lemma 3.3 with α = 0, we have |F (0)| ≤ 1 − |a|2 . Suppose |F (0)| = 1. Then a = 0 and the above relation with equality sign holds. Now we conclude again from Lemma 3.3 with α = 0 that F(z) = λz for z ∈ D where λ is a constant of absolute value 1. This is not possible since F is not one-one as s(w) = w2 . Hence |F (0)| < 1 and the proof is complete.

3.7 Proof of Theorem 3.11 Let z 0 ∈ and be the set of all ψ ∈ H () such that ψ is one-one on and ψ() ⊆ D. Assume that satisfies the assumptions of Theorem 3.11. Then the assertions of Lemmas 3.16 and 3.17 are valid. by Theorem 2.21. Then We observe that |ψ (z 0 )| > 0 for ψ ∈ we see from such that (3.6.8) and Lemma 3.16 that η > 0. There exists a sequence {ψn } in lim |ψn (z 0 )| = η > 0.

n→∞

(3.7.1)

We observe that |ψ(z)| < 1 for ψ ∈ . In particular, is uniformly bounded on compact subsets of . Therefore is a normal family by Lemma 3.15. Hence the above sequence {ψn } has a subsequence which we denote again by {ψn }, and which converges uniformly on compact subsets of satisfying (3.7.1). Let lim ψn (z) := h(z) for z ∈

n→∞

(3.7.2)

converge uniformly on compact subsets of . Then h ∈ H () by Sect. 2.3(iv) and |h(z)| < 1 for z ∈ by Theorem 2.19. Further lim ψn (z) = h (z) for z ∈

n→∞

converges uniformly on compact subsets by Sect. 2.3(iv). Therefore lim |ψn (z 0 )| = |h (z 0 )|

n→∞

which, together with (3.7.1), implies |h (z 0 )| = η > 0.

(3.7.3)

94

3 Conformal Mappings and the Riemann Mapping Theorem

Thus h is not constant on . Now we derive from Lemma 3.14 that h is oneone on . Thus h ∈ . Hence we conclude from (3.6.8) and Lemma 3.17 that h() = D. Finally, we give a more precise version of Theorem 3.11. Theorem 3.18 Assume that satisfies the assumptions of Theorem 3.11 and z 0 ∈ . Then there exists unique function f from onto D such that f ∈ H (), f is one-one and f (z 0 ) = 0 with f (z 0 ) > 0. Proof Let z 0 ∈ . First, we prove the uniqueness of f. Suppose that there is another g. Then we consider χ1 = g ◦ f −1 . We observe that χ1 is an automorphism of D such that χ1 (0) = g( f −1 (0)) = g(z 0 ) = 0. Therefore, we derive from Corollary 3.5 that g ◦ f −1 (z) = λz for z ∈ D where λ is a constant of absolute value 1. Denoting tλ (z) = λz for z ∈ D, we see that g ◦ f −1 = tλ in D and thus g = tλ ◦ f in . Therefore g(z) = tλ ( f (z)) = λ f (z) for z ∈ . Differentiating both sides with respect to z at z = z 0 , we have g (z 0 ) = λ f (z 0 ). Since g (z 0 ) > 0 and f (z 0 ) > 0, we see that λ > 0. Hence λ = 1 since |λ| = 1. Thus g = f . Next, we turn to show the existence of f. Let h be given by (3.7.2). We recall that h is a conformal mapping from onto D. Let h(z 0 ) = δ = 0. Then we consider χ = φδ ◦ h ∈

.

We observe that χ (z 0 ) = (φδ ◦ h) (z 0 ) = φδ (h(z 0 ))h (z 0 ) = φδ (δ)h (z 0 ) = by Lemma 3.2 (v). Therefore we see from (3.7.3)

h (z 0 ) 1 − |δ|2

3.7 Proof of Theorem 3.11

95

|χ (z 0 )| > |h (z 0 )| = η. This contradicts (3.7.3) and (3.6.8). Hence h(z 0 ) = 0. By Theorem 2.21, we see that h (z 0 ) = 0. Let |h (z 0 )| = h(z 0 )eiα and θα (z) be given by

θα (z) = eiα z for z ∈ D.

Then

θα (z) = eiα .

Now we consider χ1 = θα ◦ h ∈

. We have

χ1 (z 0 ) = θα (h(z 0 )) = eiα h(z 0 ) and χ1 (z 0 ) = (θα ◦ h) (z 0 ) = θα (h(z 0 ))h (z 0 ) = eiα h (z 0 ) = |h (z 0 )| > 0.

Hence f = χ1 satisfies the assertion of Theorem 3.18. Example 3.4. (a) Let F ∈ H (), F() ⊆ D and z 0 ∈ . Then |F (z 0 )| ≤ | f (z 0 )|,

(3.7.4)

where f = f z0 is the analytic homeomorphism given by Theorem 3.18. (b) Let F ∈ H () and F() ⊆ D. Assume that there exists z 0 ∈ such that F(z 0 ) = 0. Then (3.7.4) with equality sign implies that F(z) = λ f (z) for z ∈ ,

(3.7.5)

where λ is a constant of absolute value 1. (c) Let F ∈ H () and F() = D. Then the assertion of (b) holds. Proof (a) The proof depends on Lemma 3.1 as in the proof of Lemma 3.3. Let g(z) = φ F(z0 ) ◦ F ◦ f −1 (z) for z ∈ D.

(3.7.6)

Then g(z) ∈ H (D) and g(D) ⊆ D. Further g(0) = φ F(z0 ) ◦ F( f −1 (0)) = φ F(z0 ) (F(z 0 )) = 0. Now we derive from Lemma 3.1 that |g (0)| ≤ 1.

(3.7.7)

96

3 Conformal Mappings and the Riemann Mapping Theorem

Also, as in the proof of Lemma 3.3, we calculate g (0) =

F (z 0 ) 1 . 1 − |F(z 0 )|2 f (z 0 )

(3.7.8)

By combining (3.7.7) and (3.7.8), we have F (z 0 ) 2 f (z ) ≤ 1 − |F(z 0 )| ≤ 1. 0 (b) Let z 0 ∈ such that F(z 0 ) = 0 and (3.7.4) holds with equality sign. Then |g (0)| = 1 by (3.7.8) and we derive from Lemma 3.1 that (3.7.5) holds where λ is a constant of absolute value 1. (c) Since F() = D, there exists z 0 ∈ such that F(z 0 ) = 0. Then (c) follows immediately from (b).

3.8 Exercises 3.1 The image of a simply connected region under conformal mapping is simply connected. Give an example to show that the continuous image of a simply connected region need not be simply connected. 3.2 (a) Show that Aut(D) and Aut(H) are non-commutative isomorphic groups. (b) Show that an entire function taking values in H is constant. (c) Prove that the regions 1 = {z ∈ C | Re(z) > 0, Im(z) > 0} and 2 = {z ∈ C | Re(z) < 0, Im(z) > 0, |z| < 1} are conformally equivalent. (Hint: Use function F given in Sect.3.2(i).) 3.3 Show that the function f (z) = − 21 z + 1z is a conformal mapping from the upper half disc z = x + i y |z| < 1, y > 0 onto the upper half plane. (Hint: If f (z) = w, then z 2 + 2wz + 1 = 0 which has two distinct roots whenever w = ±1 and their product is equal to 1.) 3.4 Find a conformal mapping between D and the right half plane {z | Re(z) > 0}. (Hint: Use the function f given in Sect. 3.2(iii).) 3.5 Show that the function f (z) = sin z takes the upper half plane conformally onto the half strip w = u + iv | − π2 < u < π2 , v > 0 . (Hint: Observe that sin z =

1 1 ei z − e−i z =− iw + , w = ei z 2i 2 iw

and apply Ex 3.3.) 3.6 Prove that there is no conformal mapping from the punctured disc {z ∈ C | 0 < |z| < 1} onto the annulus {z | 1 < |z| < 2}.

3.8 Exercises

97

(Hint: Let D = D(0, 1) and D = D (0, 1) and A be the annulus in the exercise. Assume that there exists a conformal mapping from D onto A. Then there exists g ∈ H (D), g D = f and g(D) = A which implies that g(z) = g(0) for some z ∈ D . Further D has open discs V containing z and W containing 0 such that V ∩ W = ∅. Now apply Open mapping theorem.) 3.7 (a) Does there exists an analytic function from D into D such that f 21 = 43 and f 21 = 23 ? (b) Find a function f analytic in D such that | f (z)| < 1 for z ∈ D and f (0) = 21 and f (0) = 43 . 3.8 Let f (z) be analytic in |z| < 1 where | f (z)| ≤ 1. Assume that f (z) has a zero at z 0 of order m with |z 0 | < 1. Then show that | f (0)| ≤ |z 0 |m . (Hint: Apply the Schwarz lemma to g(z) =

f (z) (φz0 (z))m (m) f (z 0 ) (z−z 0 )m m! (φz0 (z 0 ))m

if z = z 0 otherwise

in |z| < 1.) 3.9 Let f (z) be analytic in |z| < R where | f (z)| ≤ M and f (0) = 0. Then show that M |z| for |z| < R. | f (z)| ≤ R If the equality holds for some z with |z| < R, then f (z) = λz where λ is a constant of absolute value 1. 3.10 (a) Let f (z) be analytic in |z| < R where | f (z)| < M. Let f (z 0 ) = w0 with |z 0 | < R and |w0 | < M. Then M( f (z) − w0 ) ≤ R(z − z 0 ) . R 2 − zz M 2 − w f (z) 0 0 f (Rz) in |z| < 1. Apply the Schwarz lemma to h(z) = M φ wM0 ◦ g ◦ φ− zR0 (z) in |z| < 1.) (b) (The Schwarz -Pick lemma) Derive from Ex 3.12 (a) the following: Assume that f ∈ H (D) with | f (z)| < 1 for z ∈ D. Then (Hint: Let g(z) =

1 | f (z)| ≤ . 1 − | f (z)|2 1 − |z|2 3.11 Let f be an entire function such that | f (z)| = 1 whenever |z| = 1. Then prove that f (z) = cz n for some n ≥ 0 and constant c with |c| = 1.

98

3 Conformal Mappings and the Riemann Mapping Theorem

3.12 Denote by GL+ 2 (R) the set of all matrices M given by (3.4.1) such that a, b, c, d ∈ R with ad − bc >0. This is a group under matrix multiplication az + b ab is called ∈ GL+ containing SL2 (R). For M = 2 (R), f M (z) = c d cz + d linear fractional transformation from H onto H. Let f = id be a linear fractional transformation from H onto H such that f ◦ f = id. Prove that f has a unique fixed point in H. 3.13 Show that the family of all analytic functions z + a1 z 2 + · · · + an−1 z n + · · · with |an | ≤ n on the unit disc is a normal family. 3.14 Let { f n }∞ n=1 be a uniformly bounded sequence of analytic functions on . Assume that for each z ∈ , the sequence { f n (z)}∞ n=1 converges. Then show converges uniformly on compact subsets of . that { f n }∞ n=1 be a sequence of analytic functions in a region such that it 3.15 Let { f n (z)}∞ n=1 converges uniformly on compact subsets of to f (z) which is not identically zero. Assume that f has a zero of order N at z 0 ∈ . Then show that there exists ρ > 0 such that for large n, f n (z) has exactly N zeros in D(z 0 , ρ) counted with multiplicity and these zeros converge to z 0 as n tends to infinity. (Hint: Choose ρ > 0 and δ > 0 such that D(z 0 , ρ) ⊂ , | f (z)| > δ in D (z, ρ) f (z) (z) and fnn (z) converges uniformly to ff (z) on |z − z 0 | = ρ. Now apply Argument principle.) 3.16 Let { f n (z)}∞ n=1 be a sequence of analytic functions in a region such that it converges uniformly on compact subsets of . Assume that f n (z) attains each value at most m times in counted with multiplicity. Show that either f (z) is constant or f (z) attains each value at most m times in counted with multiplicity. 3.17 Show that no two of the regions C\{0}, C and D are conformally equivalent.

Chapter 4

Harmonic Functions

4.1 Introduction We introduce the Cauchy–Riemann equations in Sect. 4.2 and the harmonic functions in Sect. 4.3 and show that real and imaginary parts of an analytic function are harmonic. We prove the existence of a harmonic conjugate of a harmonic function in a simply connected region in Sect. 4.4 where we also prove its converse. We introduce continuous functions with Mean Value Property in a region in Sect. 4.5 and prove in Sect. 4.5 Maximum principle for harmonic functions in a region satisfying MVP in and in Sect. 4.5 and Sect. 4.6 that such functions characterise harmonic functions in . The proof of this characterisation of harmonic functions in terms of continuous functions with Mean Value Property depends on the maximum principle for these functions which is contained in Sect. 4.5 and solution of the Dirichlet Problem for open disc which we prove in Sect. 4.6. We refer to [1, 12, 19, 23] for the topics in this chapter and for further studies and related topics.

4.2 The Cauchy–Riemann Equations Theorem 4.1 Let f (z) be analytic at z 0 = x0 + i y0 and f (z) = u(x, y) + iv(x, y) where z = (x, y). Then ∂u ∂v ∂u ∂v (x0 , y0 ) = (x0 , y0 ), (x0 , y0 ) = − (x0 , y0 ). ∂x ∂y ∂y ∂x

(4.2.1)

The Eq. (4.2.1) are called the Cauchy–Riemann equations. Proof We have f (z 0 ) = lim

h→0

f (z 0 + h) − f (z 0 ) . h

© Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9_4

99

100

4 Harmonic Functions

For h ∈ R and h = 0, we re-write the right-hand side as f (z 0 + h) − f (z 0 ) f (x0 + h + i y0 ) − f (x0 , y0 ) = h h u(x0 + h, y0 ) − u(x0 , y0 ) v(x0 + h, y0 ) − v(x0 , y0 ) = +i . h h Letting h tend to zero such that h ∈ R and h = 0, we get f (z 0 ) = u x (x0 , y0 ) + ivx (x0 , y0 ), ∂u ∂v (x0 , y0 ) and vx (x0 , y0 ) = ∂x (x0 , y0 ). Similarly, letwhere we write u x (x0 , y0 ) = ∂x ting h tend to zero such that h is purely imaginary, we get

f (z 0 ) = −iu y (x0 , y0 ) + v y (x0 , y0 ). Now the assertion follows by comparing the real and imaginary parts in the above two expressions for f (z 0 ). Let u = u(x, y) be a real-valued function. For z = x + i y, we write u(z) = u(x, y).

(4.2.2)

Theorem 4.2 Let u = u(x, y) and v = v(x, y) be real-valued functions defined on with continuous partial derivatives satisfying Cauchy–Riemann equation (4.2.1). Then f (z) = u(z) + iv(z) is analytic in . Proof Let z = x + i y ∈ . By the mean value theorem, there exist numbers s1 and t1 with |s1 | < |s| and |t1 | < |t| such that u(x + s, y + t) − u(x, y) = (u(x + s, y + t) − u(x, y + t)) + (u(x, y + t) − u(x, y)) = u x (x + s1 , y)s + u y (x, y + t1 )t.

(4.2.3)

Let φ(s, t) = u(x + s, y + t) − u(x, y) − u x (x, y)s + u y (x, y)t .

(4.2.4)

Then we derive from (4.2.3) and (4.2.4) that t φ(s, t) s u x (x + s1 , y + t) − u x (x, y) + u y (x, y + t1 ) − u y (x, y) . = s + it s + it s + it

Then

φ(s, t) =0 s→0 s + it lim

t→0

(4.2.5)

4.2 The Cauchy–Riemann Equations

101

since u x and u y are continuous at (x, y). We re-write (4.2.4) as u(x + s, y + t) − u(x, y) = u x (x, y)s + u y (x, y)t + φ(s, t).

(4.2.6)

Similarly v(x + s, y + t) − v(x, y) = vx (x, y)s + v y (x, y)t + ψ(s, t), where lim

s→0 t→0

ψ(s, t) = 0. s + it

(4.2.7)

(4.2.8)

Now for z = x + i y, we see from (4.2.6) and (4.2.7) that u(x + s, y + t) − u(x, y) + i v(x + s, y + t) − v(x, y) f (z + s + it) − f (z) = s + it s + it φ(s, t) + iψ(s, t) =u x (z) + ivx (z) + s + it

by using the Cauchy–Riemann equations (4.2.1). Therefore lim

s+it→0

f (z + s + it) − f (z) = u x (z) + ivx (z) s + it

by (4.2.5) and (4.2.8). Hence f is analytic at z.

4.3 Definition and Examples of Harmonic Functions Definition Let (x0 , y0 ) ∈ R2 and u be a real-valued function defined in a neighbourhood of (x0 , y0 ). Then u is harmonic at (x0 , y0 ) if (i) u is continuous at (x0 , y0 ). (ii) u has continuous partial derivatives of the first and the second order at (x0 , y0 ) satisfying (4.3.1) u x x (x0 , y0 ) + u yy (x0 , y0 ) = 0, ∂u (x0 , y0 ) . The Eq. (4.3.1) is called the Laplace equation. where u x y (x0 , y0 ) = ∂∂y ∂x Further u is called harmonic in if it is harmonic at every point of . Remarks (i) We identify the elements (x, y) of R2 with x + i y of C and it will be clear from the context whether we are taking (x, y) or x + i y. (ii) We understand (4.2.2) for any real-valued function as u = u(x, y). (iii) If u is harmonic in , then u + c for any constant c is harmonic in . Let f ∈ H () be given by

102

4 Harmonic Functions

f (z) = u(x, y) + iv(x, y).

(4.3.2)

Theorem 4.3 Let f ∈ H (). Then Re( f ) and Im( f ) are harmonic in . Proof The proof depends on f ∈ H () and f ∈ H (), see (2.3.3). Let f be given by (4.3.2). First, we prove that u and v have continuous partial derivatives of orders 0, 1 and 2 at every point of . We prove the assertion for u and the proof for v is similar. Let (x0 , y0 ) ∈ and z 0 = x0 + i y0 . Then we see from (4.3.2) that u is continuous at (x0 , y0 ) since f is continuous at z 0 . Further, as in the proof of Theorem 4.1, we have f (z 0 ) = u x (x0 , y0 ) + ivx (x0 , y0 ) = v y (x0 , y0 ) − iu y (x0 , y0 ).

(4.3.3)

By differentiating both the equations in (4.3.3), we have u x and u y that are continuous at (x0 , y0 ) since f (z) is continuous at z 0 . Next, we have f (z 0 ) = u x x (x0 , y0 ) + ivx x (x0 , y0 ) = v yx (x0 , y0 ) − iu yx (x0 , y0 ) = vx y (x0 , y0 ) − iu x y (x0 , y0 ) = −u yy (x0 , y0 ) − iv yy (x0 , y0 ). This implies u has continuous partial derivative of order 2 at (x0 , y0 ) since f (z) is continuous at z 0 . Since (x0 , y0 ) is an arbitrary point of , we conclude that u has continuous partial derivatives of order 0, 1 and 2 at every point of . Differentiating the first equation in (4.2.1) with respect to x and the second in (4.2.1) with respect to y, we get u x x (x0 , y0 ) = v yx (x0 , y0 ), vx y (x0 , y0 ) = −u yy (x0 , y0 ), which implies (4.3.1) since v yx (x0 , y0 ) = vx y (x0 , y0 ), see [10], p. 154. Hence u is harmonic in . Example 4.1 (Identity theorem for harmonic functions) Let u be harmonic in a region and let V be a non-empty open subset of such that u = 0 in V . Then u = 0 in . Proof Let u be harmonic in . For z ∈ with z = x + i y, we consider g(z) = u x (x, y) − iu y (x, y). We observe that u x and −u y are defined in and they satisfy Cauchy–Riemann equations in since u is harmonic in . Therefore we derive from Theorem 4.2 that g is analytic in . Further, by assumption, u x and −u y vanish on V and therefore g = 0 on V . Then g = 0 on by Identity theorem for holomorphic functions, see Sect. 2.3 (ii). Then u x = u y = 0 in which implies that u is constant in . Then the assertion follows immediately.

4.4 Harmonic Conjugate of a Harmonic Function in a Simply Connected Region

103

4.4 Harmonic Conjugate of a Harmonic Function in a Simply Connected Region Definition Let u be harmonic in a region . Then v is called a harmonic conjugate of u in if (i) v is harmonic in . (ii) There exists f ∈ H () such that f = u + iv in . Let u be harmonic in a region . Assume that v and v1 are harmonic conjugates of u in . Then there exist f ∈ H () and f 1 ∈ H () such that f 1 = u + iv1 in .

f = u + iv,

Then v − v1 = −i( f − f 1 ) ∈ H () is real valued. Therefore v and v1 differ by a constant, see Ex 1.17. Next we determine harmonic conjugate of a harmonic function in an open disc or in the whole complex plane. The method depends on the following well-known results from Integral calculus, see [10], p. 128–129 and we shall use them without reference. I Let f be integrable on [a, b] and

x

F(x) =

f (t)dt for a ≤ x ≤ b.

a

Then F(x) is continuous in [a, b]. If f is continuous at x0 ∈ [a, b], then F (x0 ) = f (x0 ). II Let f be integrable on [a, b]. If there exists a differentiable function F on [a, b] such that F = f . Then

b

f (t)dt = F(b) − F(a).

a

Theorem 4.4 Let = D(0, R) where 0 < R ≤ ∞. Let u be harmonic in . Then there exists a harmonic conjugate of u in . Proof It suffices to find a real-valued function v = v(x, y) satisfying (i) v has continuous partial derivatives at every point of . (ii) u and v satisfy the Cauchy–Riemann equations u x = v y and u y = −vx at every point of .

104

4 Harmonic Functions

Then u + iv = f ∈ H () by Theorem 4.2. Now we see from Theorem 4.3 that v will be a harmonic conjugate of u in . For (x, t) ∈ , we have by the first equation in (ii) u x (x, t) = v y (x, t). We integrate both sides with respect to t along a vertical line from 0 to y. We have

y

y

v y (x, t)dt =

0

u x (x, t)dt.

0

Thus

y

v(x, y) − v(x, 0) =

u x (x, t)dt.

0

By putting v(x, 0) = h(x), we have v(x, y) =

y

u x (x, t)dt + h(x).

0

We determine h(x) such that the second equation in (ii) is satisfied. By substituting v(x, y) in the second equation in (ii), we have u y (x, y) = −

∂ ∂x y

y

u x (x, t)dt − h (x)

0

=− u x x (x, t)dt − h (x) 0 y = u yy (x, t)dt − h (x) 0

= u y (x, y) − u y (x, 0) − h (x) by (4.3.1). Therefore

h (x) = −u y (x, 0)

which is satisfied if

x

h(x) = −

u y (s, 0)ds + C,

0

where C is any constant. Then

y

v(x, y) = 0

u x (x, t)dt −

x

u y (s, 0)ds + C.

0

We check that v satisfies (i) and (ii) and hence v is a harmonic conjugate of u in .

4.4 Harmonic Conjugate of a Harmonic Function in a Simply Connected Region

105

Example 4.2 Let u(x, y) = y 3 − 3x 2 y. We observe that u is continuous at (x, y) and it has partial derivatives of orders 1 and 2 at (x, y). Further u x x = −6y, u yy = 6y implying u x x + u yy = 0. Therefore u is harmonic in C. Further u x (x, t) = −6xt, u y (s, 0) = −3s 2 and we derive from Theorem 4.4 that y v(x, y) = −6xtdt − 0 3

x

−3s 2 ds + C

0

= x − 3x y 2 + C is harmonic conjugate of u in C for any constant C. The corresponding analytic function is f (z) = y 3 − 3x 2 y + i(x 3 − 3x y 2 + C) for z = x + i y. Thus f (z) = i(x 3 + C) for z = x + i y with y = 0 and hence f (z) = i(z 3 + C) for z ∈ C, by the Identity theorem for holomorphic functions, see Sect. 2.3 (ii).

Now we show that the assertion of Theorem 4.4 is valid for all simply connected regions. In fact, we prove Theorem 4.5 A region is simply connected if and only if every harmonic function in has a harmonic conjugate in . We shall use the following results in the proof of Theorem 4.5. Lemma 4.6 Let u = u(x, y) and v = v(x, y) be harmonic function in a region . For (x, y) ∈ , let R = R(x, y) = Then R is harmonic in .

1 log((u(x, y))2 + (v(x, y))2 ). 2

106

4 Harmonic Functions

Proof It is clear that R is continuous and it has continuous partial derivatives of orders 1 and 2 at every point of . We show that R satisfies the Laplace equation at every point of . At (x, y) ∈ , we have Rx =

uu x + vvx , u 2 + v2

Ry =

uu y + vv y u 2 + v2

and (u 2 + v 2 )2 (Rx x + R yy ) = (u 2 + v 2 )(u 2x + vx2 + u 2y + v 2y ) − 2(uu x + vvx )2 − 2(uu y + vv y )2

by using u x x + u yy = 0 and vx x + v yy = 0. Simplifying, we get (u 2 + v 2 )2 (Rx x + R yy ) = u 2 vx2 + u 2 v 2y + v 2 u 2x + v 2 u 2y − (u 2 u 2y + u 2 u 2x + v 2 v 2y + v 2 u 2x ) −2uvu x u y − 2uvvx v y =0

by using the Cauchy–Riemann equations. Lemma 4.7 Let = C\{0}. For z ∈ with z = x + i y, let u(x, y) = log |z| =

1 log(x 2 + y 2 ). 2

Then u is harmonic in . Proof We observe that u is continuous in where it has continuous partial derivatives of orders 1 and 2 since ux = and uxx =

x y , uy = 2 x 2 + y2 x + y2

y2 − x 2 x 2 − y2 , u = . yy (x 2 + y 2 )2 (x 2 + y 2 )2

The latter equation implies that (4.3.1) is satisfied at every point of . Hence u is harmonic in . Lemma 4.8 Let D1 and 1 be an open discs. Let F be holomorphic function from D1 into 1 and u be harmonic in 1 . Then u ◦ F is harmonic in D1 . Proof Let F(z) = A(x, y) + i B(x, y) for z = x + i y ∈ D1 . Since u is harmonic in 1 and D1 is a disc, we derive from Theorem 4.4 that there exists G ∈ H (1 ) such that

4.4 Harmonic Conjugate of a Harmonic Function in a Simply Connected Region

107

G(z) = φ(x, y) + iψ(x, y) for z = x + i y ∈ 1 where φ(x, y) = u(x, y). Then, for z = x + i y ∈ D1 , we have G ◦ F(z) = G(A(x, y) + i B(x, y)) = φ(A(x, y), B(x, y)) + iψ(A(x, y), B(x, y)) = u(A(x, y), B(x, y)) + iψ(A(x, y), B(x, y)) and Re(G ◦ F(z)) = u(A(x, y), B(x, y)) = u ◦ F(z). Now we conclude that u ◦ F is harmonic in D1 by Theorem 4.3.

Proof of Theorem 4.5 Assume that is simply connected and let u be harmonic in . We show that u has a harmonic conjugate in . We may assume that = C otherwise the assertion follows from Theorem 4.4. Then, by the Riemann mapping theorem 3.10, there exists an analytic homeomorphism F from D onto . Let z 0 ∈ D. Then F(z 0 ) ∈ and there exist 0 < s < r < 1 such that F(D(z 0 , s)) ⊆ D(F(z 0 ), r ) ⊆ . In Lemma 4.8, we take D1 = D(z 0 , s), 1 = D(F(z 0 ), r ) and F is holomorphic function from D1 into 1 . Since 1 ⊆ , we see that u is harmonic in 1 . Let uoF = u 1 . Then u 1 is harmonic in D1 by Lemma 4.8. In particular, u 1 is harmonic at z 0 . Since z 0 is an arbitrary point of D, we see that u 1 is harmonic in D. Now we derive from Theorem 4.4 that there exist v1 harmonic in D and f 1 ∈ H (D) such that f 1 = u 1 + iv1 in D. Then

f 1 ◦ F −1 = u + iv1 ◦ F −1 in

and f 1 ◦ F −1 ∈ H (). Hence we conclude from Theorem 4.3 that v1 ◦ F −1 is harmonic conjugate of u in . Let be a region and we assume that every harmonic function in has a harmonic conjugate in . We show that is simply connected. We may assume that = C otherwise the assertion follows since C is simply connected. It suffices to show that for every f ∈ H () with 1f ∈ H (), there exists g ∈ H () such that f (z) = g 2 (z) for z ∈ . Then is conformally equivalent to D by Theorem 3.11. This implies, as in the proof of Corollary 3.12, that every closed path in is null-homotopic. Hence is simply connected. Let f ∈ H () with 1f ∈ H (). we put Re( f ) = u, Im( f ) = v.

108

4 Harmonic Functions

Then u and v are harmonic in by Theorem 4.3. For x + i y ∈ , we put R(x, y) = log | f (x + i y)| =

1 log (u(x, y))2 + (v(x, y))2 2

(4.4.1)

which is defined since f (z) = 0 for z ∈ and R(x, y) is harmonic in by Lemma 4.6. Then, by our assumption, there exists a harmonic function S in and g1 ∈ H () such that (4.4.2) g1 = R + i S in . Let

Then

h(z) = eg1 (z) for z ∈ . f (z) h(z)

∈ H () and

(4.4.3)

f (z) h(z) = 1 for z ∈

f (z) by (4.4.1), (4.4.2) and (4.4.3). Therefore h(z) is constant in by Open mapping Theorem 2.18. Then f (z) = ceg1 (z) = eg1 (z)+c1 ,

where c and c1 are constants. By putting g(z) = e

g1 (z)+c1 2

,

we see that g(z) ∈ H () and f (z) = (g(z))2 for z ∈ .

4.5 Maximum Principle for Harmonic Functions Satisfying MVP We begin with the following definition. Definition Let u be real-valued continuous function in a region . Then u has Mean Value Property (MVP) in if for every a ∈ , we have 1 u(a) = 2π

2π

u(a + r eiθ )dθ

0

whenever D(a, r ) ⊆ . We derive from Theorem 4.4 the following result. Theorem 4.9 Let u be harmonic in a region . Then u satisfies MVP in .

4.5 Maximum Principle for Harmonic Functions Satisfying MVP

109

Proof Let a ∈ with D(a, r ) ⊆ . There exists an open disc E such that D(a, r ) ⊆ E ⊆ and we derive from Theorem 4.4 that u has a harmonic conjugate in E. Therefore there exists f ∈ H (E) such that u = Re( f ). Now f (a) =

1 2πi

|z−a|=r

f (z) dz z−a

by Theorem 2.3. By putting z − a = r eiθ with 0 ≤ θ ≤ 2π, we have f (a) =

1 2πi

2π 0

f (a + r eiθ )ir eiθ 1 dθ = iθ re 2π

2π

f (a + r eiθ )dθ.

0

By comparing the real parts on both the sides, we get 1 u(a) = 2π

2π

u(a + r eiθ )dθ.

0

This holds for every a ∈ whenever D(a, r ) ⊆ .

Next we prove the Maximum principle for the continuous functions with MVP and in particular for harmonic functions. Theorem 4.10 Let u be real-valued continuous function in a region and assume that u has MVP in . Suppose that there exists a ∈ such that u(z) ≤ u(a) for all z ∈ . Then u is constant in . Proof We assume that u is not constant in . Let u be continuous in a region satisfying MVP in and there exists a ∈ such that u(z) ≤ u(a) for z ∈ . We consider A = {z ∈ |u(z) = u(a)}. We may assume that A = ∅ since a ∈ A. It suffices to show that A is both open and closed. Then A = since is connected and hence u is constant in . Let z ∈ A. Then there exists a sequence {z n }∞ n=1 with z n ∈ A such that lim z n = z. n→∞

Since u is continuous, we have lim u(z n ) = u(z). But u(z n ) = a for n ≥ 1 since n→∞

z n ∈ A. Therefore u(z) = u(a) which implies that z ∈ A. Thus A ⊆ A and hence A is closed. Now we show that A is open. Let z 0 ∈ A and there exists r with D(z 0 , r ) ⊆ / A. such that D(z 0 , r ) is not contained in A. Then there exists b ∈ D(z 0 , r ) and b ∈ Thus

110

4 Harmonic Functions

u(b) < u(a) = u(z 0 ). Since u is continuous, there exists s > 0 such that u(z) < u(a) for z ∈ D(b, s). Let |b − z 0 | = ρ < r. Then there exists an arc on the circle |z − z 0 | = ρ containing b of positive length where u(z) < u(z 0 ) and u(z) ≤ u(a) = u(z 0 ) elsewhere on the circle. Therefore 2π 1 u(z 0 + ρeiθ )dθ < u(z 0 ). 2π 0 On the other hand 1 2π

2π

u(z 0 + ρeiθ )dθ = u(z 0 )

0

since u satisfies MVP by assumption. This is a contradiction.

We give another version of the Maximum principle which is an immediate consequence of Theorem 4.10. Corollary 4.11 Let be a bounded region. Assume that u is a non-constant realvalued continuous function defined on and u has MVP in . Then there exists a ∈ ∂ such that u(z) < u(a) for z ∈ . Proof Since u is continuous on and is compact, there exists a ∈ such that u(z) ≤ u(a) for z ∈ . If a ∈ , we derive from Theorem 4.10 that u is constant in and hence in . This is a contradiction. Therefore a ∈ ∂. If u(z 0 ) = u(a) for some z 0 ∈ , then u(z) ≤ u(z 0 ) for all z ∈ , which is again not possible by Theorem 4.10. The Minimum principle is an immediate consequence of the Maximum principle. Theorem 4.12 (a) Let u be a real-valued continuous function with MVP in a region . Suppose that there exists a ∈ such that u(z) ≥ u(a) for all z ∈ . Then u is constant in . (b) Assume that is a bounded region. Let u be non-constant real-valued continuous function on and u has MVP in . Then there exists a ∈ ∂ such that u(z) > u(a) for z ∈ . Proof (a) We have −u(a) ≥ −u(z) for z ∈ and the assumptions of Theorem 4.10 are satisfied with u replaced by −u. Now we derive from Theorem 4.10 that −u and hence u is constant in . (b) The assertion follows similarly from Theorem 4.11 with u replaced by −u.

4.5 Maximum Principle for Harmonic Functions Satisfying MVP

111

The converse of Theorem 4.9 is also valid. Theorem 4.13 Let u be a real-valued continuous function with MVP in a region. Then u is harmonic in . Thus a continuous function u in a region has continuous partial derivatives of orders 1 and 2 satisfying the Laplace equation at all points of whenever u satisfies MVP in . The proof of Theorem 4.13 depends on Theorems 4.11, 4.12(b) and the following solution of the Dirichlet Problem for open discs. Theorem 4.14 Let a ∈ C, ρ > 0 and f be real-valued continuous function defined on the circle |z − a| = ρ. Then there exists unique real-valued continuous function u in D(a, ρ) such that u is harmonic in D(a, ρ) and u(z) = f (z) for |z − a| = ρ. Theorem 4.14 has been extended to simply connected regions. Now, assuming Theorem 4.14, we give a proof of Theorem 4.13 and we postpone the proof of Theorem 4.14 to the next section. Lemma 4.15 Theorem 4.14 implies Theorem 4.13. Proof Let u be a real-valued continuous function with MVP in . Let a ∈ . Since is open, there exists ρ > 0 such that D(a, ρ) ⊆ . It suffices to show that u is harmonic in D(a, ρ). Then u is harmonic at a and the assertion follows since a is an arbitrary point in . Since D(a, ρ) ⊆ , we see that u is continuous in D(a, ρ) and it has MVP in D(a, ρ). By Theorem 4.14, there exists a real-valued continuous function v in D(a, ρ) such that v is harmonic in D(a, ρ) and such that u(z) = v(z) if |z − a| = ρ. Further v has MVP in by Theorem 4.9. Next we consider g = u − v in D(a, ρ).

(4.5.1)

(4.5.2)

We observe that g is real-valued continuous function in D(a, ρ) and it has MVP in D(a, ρ). Further g(z) = 0 if |z − a| = ρ (4.5.3) by (4.5.2) and (4.5.1). Assume that g is not a constant function. Then g(z) < 0 in D(a, ρ) by Corollary 4.11 and g(z) > 0 in D(a, ρ) by Theorem 4.12 (b). This is a contradiction. Therefore g is a constant function c in D(a, ρ). In fact c = 0 since ¯ g is continuous in D(a, ρ) and zero on |z − a| = ρ. Hence u = v is harmonic in D(a, ρ).

112

4 Harmonic Functions

4.6 The Dirichlet Problem for Open Discs We begin with the Poisson kernel which we shall use in the proof of Theorem 4.14. Definition For 0 ≤ r < 1 and 0 ≤ θ ≤ 2π, the function Pr (θ) =

∞

r |n| einθ

(4.6.1)

n=−∞

is called the Poisson kernel. We understand that 00 = 1 in the sum on the right-hand side of (4.6.1) so that Pr (θ) = 1 if r = 0. We calculate Poisson kernel in the next result. Lemma 4.16 For 0 ≤ r < 1 and 0 ≤ θ ≤ 2π, we have

1 + r eiθ Pr (θ) = Re 1 − r eiθ

=

1 − r2 . 1 − 2r cos θ + r 2

(4.6.2)

Proof For 0 ≤ |z| < 1, we have ∞ 1+z = (1 + z)(1 − z)−1 = (1 + z)(1 + z + z 2 + · · · ) = 1 + 2 zn . 1−z n=1

Here the rearrangement of terms of the series is permissible since the series is absolutely convergent. This is also the case with the subsequent series appearing in the proof of this lemma. By putting z = r eiθ with 0 ≤ r < 1 in (4.6.1), we have ∞ 1 + r eiθ = 1 + 2 r n einθ . 1 − r eiθ n=1

Now

1 + r eiθ Re 1 − r eiθ

= 1+2

∞ n=1

= 1+

∞ n=1

Further

and

r n cos nθ = 1 +

r n einθ +

∞

r n einθ + e−inθ

n=1 −1 n=−∞

r |n| einθ = 1 +

∞

r |n| einθ = Pr (θ).

n=−∞ n =0

1 + r eiθ (1 + r eiθ )(1 − r e−iθ ) 1 − r 2 + 2ir sin θ = = 1 − r eiθ |1 − r eiθ |2 |1 − r eiθ |2

4.6 The Dirichlet Problem for Open Discs

113

|1 − r eiθ |2 = 1 − 2r cos θ + r 2 .

Therefore Pr (θ) = Re

1 + r eiθ 1 − r eiθ

=

1 − r2 . 1 − 2r cos θ + r 2

(4.6.3)

The Poisson kernel satisfies the following properties. Lemma 4.17 (a) For 0 ≤ r < 1, we have Pr (θ) > 0 for 0 ≤ θ ≤ 2π and Pr (θ) is periodic with period 2π. Further

1 2π

π −π

(b) Let δ > 0. Then

Pr (θ)dθ = 1.

lim Pr (θ) = 0

(4.6.4)

r →1−

uniformly in θ with δ ≤ |θ| ≤ π. Proof (a) It is clear that Pr (θ) > 0 for 0 ≤ θ ≤ 2π and periodic with period 2π by (4.6.3). By integrating both sides in (4.6.1), we get

π −π

Pr (θ) = =

π

∞

r |n| einθ dθ

−π n=−∞ π ∞ |n|

r

n=−∞

−π

einθ dθ = 2π

π since the series converges uniformly in θ and −π einθ = 2π if n = 0 and 0 otherwise. (b) Let δ > 0 and 0 < r < 1. We may assume that |θ| ≤ π2 otherwise the assertion follows immediately from (4.6.3). By differentiating both sides with respect to θ in (4.6.3) and putting θ = t, we have Pr (t) = Then

−(1 − r 2 )2r sin t . (1 − 2r cos t + r 2 )2

Pr (t) < 0 for δ ≤ t ≤

π . 2

Thus Pr (θ) ≤ Pr (δ) for δ ≤ θ ≤ Since Pr (θ) = Pr (−θ) by (4.6.3), we get

π . 2

114

4 Harmonic Functions

Pr (θ) ≤ Pr (δ) for δ ≤ |θ| ≤

π . 2

Since lim− Pr (δ) = 0, we derive that lim− Pr (θ) = 0 uniformly in δ ≤ |θ| ≤ π2 . r →1

r →1

Proof of Theorem 4.14 We claim that there is no loss of generality in assuming that a = 0 and ρ = 1. Suppose that the assertion of Theorem 4.14 is valid with a = 0 and ρ = 1. Let f be real-valued continuous function on |z − a| = ρ. Then we consider g(z) = f (a + ρz) for |z| = 1.

(4.6.5)

We observe that g is continuous on |z| = 1. Then there exists real-valued continuous function v(z) in D and harmonic in D such that v(z) = g(z) for |z| = 1.

Let u(z) = v

z−a ρ

(4.6.6)

for z ∈ D(a, ρ).

(4.6.7)

Then u is real-valued continuous function in D(a, ρ) and harmonic in D(a, ρ) such that u(z) = f (z) for (4.6.7), (4.6.6) and (4.6.5).

|z − a| = ρ by combining Let M = max | f (eiφ )| | |φ| ≤ 2π . We prove Theorem 4.14 with u(r eiθ ) =

π

Pr (θ − φ) f (eiφ )dφ if 0 ≤ r < 1, 0 ≤ θ ≤ 2π if r = 1, 0 ≤ θ ≤ 2π. f (e )

1 2π

−π iθ

(4.6.8)

Let 0 ≤ r < 1. We show that u is real part of an analytic function and then it is harmonic in D by Theorem 4.3. By (4.6.8) and (4.6.3), we have u(r eiθ ) =

1 2π

π

f (eiφ )Re

−π

1 + r ei(θ−φ) 1 − r ei(θ−φ)

dφ.

We observe that u(r eiθ ) = Re(g(z)) with z = r eiθ where g(z) =

1 2π

π

−π

f (eiφ )

eiφ + z eiφ − z

dz,

(4.6.9)

which is analytic in D, see Ex 1.18. Therefore u is harmonic in D by Theorem 4.3 and in particular, it is continuous in D. Further u(eiθ ) = f (eiθ ) for 0 ≤ θ ≤ 2π by (4.6.8). Now we show that u is continuous on |z| = 1. There exists M such that |u(eiθ )| = | f (eiθ )| ≤ M for 0 ≤ θ ≤ 2π

4.6 The Dirichlet Problem for Open Discs

115

since f is continuous on |z| = 1. Further f (eiθ ) with 0 ≤ θ ≤ 2π is uniformly continuous. Therefore for > 0, there exists δ > 0 such that |u(eiθ ) − u(eiφ )| = | f (eiθ ) − f (eiφ )| <

(4.6.10)

whenever |θ − φ| ≤ δ. Let A be an arc of the circle |z| = 1 with eiθ as the centre of the arc and subtending an angle δ at the origin. Then |θ − φ| ≤ δ whenever eiφ ∈ A. Thus it suffices to show that for any eiθ with 0 ≤ θ ≤ 2π, we have |u(r eiθ ) − u(eiθ )| < 2 whenever r → 1− .

(4.6.11)

By (4.6.8), we have 1 2π

u(r eiθ ) =

π

−π

Pr (θ − φ1 )u(eiφ1 )dφ1 for 0 ≤ r < 1.

By putting θ − φ1 = φ, we get for 0 ≤ r < 1 u(r eiθ ) =

1 2π

π+θ

−π+θ

Pr (φ)u(ei(θ−φ) )dφ =

1 2π

π −π

Pr (φ)u(ei(θ−φ) )dφ

since the integrand is periodic with period 2π. By Lemma 4.17(a), we have π π 1 1 i(θ−φ) Pr (φ)u(e )dφ − Pr (φ)u(eiθ )dφ u(r e ) − u(e ) = 2π −π 2π −π π 1 Pr (φ) u(ei(θ−φ) ) − u(eiθ ) dφ = 2π −π 1 Pr (φ) u(ei(θ−φ) ) − u(eiθ ) dφ = 2π |φ| 0, the absolute value of the first integral is at most 2π

π

−π

Pr (φ)dφ =

by Lemma 4.17 (a) and the absolute value of the second integral is at most 2M max Pr (φ) < 2M δ≤|φ|≤π

2M

when r → 1− by Lemma 4.17 (b) and hence (4.6.11) follows.

116

4 Harmonic Functions

It remains to show that u is unique satisfying the assertion of Theorem 4.14. Let v be a continuous function in D such that v is harmonic in D and v(z) = f (z) for |z| = 1. Now we consider the function w = u − v. Then w is harmonic in D, and therefore it has MVP in D by Theorem 4.9. Since w = 0 on |z| = 1, we conclude from Corollary 4.11 and Theorem 4.12 (b), as in the proof Lemma 4.15, that w = 0 in D. Hence v = u.

4.7 Exercises 4.1 Let f (z) = z 2 . Derive from Theorem 4.2 that f (z) is differentiable everywhere and f (z) = 2z. 4.2 Let z = x + i y. Show that the Cauchy–Riemann equations are satisfied at the origin for f (z) but f (0) does not exist if √ (i) f (z) = |x y| if z = 0 and f (0) = 0 x y 2 (x + i y) if z = 0 and f (0) = 0. (ii) f (z) = x 2 + y4 4.3 Show that the following functions are harmonic in the plane and find their harmonic conjuagates. (i) x y (ii) x y + 3x 2 y − y 3 x (iii) 2 x + y2 (iv) e−y sin x 4.4 (The Liouville theorem for harmonic functions) A bounded harmonic function in the plane is constant. (Hint: Apply Theorem 4.9.) 4.5 Prove that

i +z and u(0, 1) = 0 u(x, y) = Re i −z is harmonic in the unit disc, vanishes on the boundary and is not bounded in D. 4.6 Let {u n }∞ n=1 be a sequence of harmonic functions in D and assume that it converges uniformly on compact subset of D. Then show that the limit function is harmonic. (Hint: Use (4.6.8) and (4.6.9).) 4.7 If u is harmonic in a region , then prove that partial derivatives of u of all orders in are harmonic.

Chapter 5

The Picard Theorems

5.1 Introduction We know that e z is analytic in C and it never vanishes. Thus e z omits the value 0 and none else. In fact, it is the case that any non-constant entire function omits at most one value. Theorem 5.1 (The Little Picard theorem) Let f (z) be an entire function. Suppose that there exist two distinct values that f does not assume. Then f is constant in C. This is an immediate consequence of the following result of Schottky. Theorem 5.2 Let 0 < R < R1 and f (z) be an analytic function in |z| ≤ R1 , where it is never equal to 0 or 1. (a) Then there exists a constant K depending only on f (0) such that | f (z)| ≤ exp K

R14 (R1 − R)4

(b) Let δ > 0 be such that δ < | f (0)|

δ. Then K depends only on δ. Theorem 5.2 is of independent interest and we show that it implies Theorem 5.1. Let f be an entire function in |z| ≤ R, where it omits the values a and b with a = b and f (z) − a g(z) = . b−a © Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9_5

117

118

5 The Picard Theorems

Then g(z) is entire and it does not assume the values 0 and 1. Let 0 < R < R1 . Then we derive from Theorem 5.2 that R14 for |z| ≤ R, |g(z)| ≤ exp K 1 (R1 − R)4 where K 1 is a constant depending only on g(0). By taking R1 = 2R, we see that |g(z)| ≤ 16K 1 for |z| ≤ R. Letting R tend to ∞, we conclude that g is a bounded entire function and then it is constant by the Liouville theorem, see Sect. 2.3 (i). Therefore f is constant. Another application of Theorem 5.2 states as follows. Theorem 5.3 (The Great Picard theorem) Let ρ > 0, z 0 ∈ C and f (z) be analytic in 0 < |z − z 0 | < ρ. Suppose that there are two distinct values which are assumed by f only finitely many times in 0 < |z − z 0 | < ρ. Then z 0 is not an essential singularity of f . In particular, the assertion is valid if f omits two distinct values in 0 < |z − z 0 | < ρ. Further the theorem implies that in any deleted neighbourhood of essential singularity of f , it assumes every value, except possibly one, infinitely many times. This is a considerable sharpening of a theorem of Casorati–Weierstrass, proved in Sect. 2.3 (v), which states that the image of every deleted neighbourhood of z 0 under f is dense in C if f has an essential singularity at z 0 . In fact, we shall use this result in our proof of Theorem 5.3. Further Theorem 5.3 implies the following improvement of Theorem 5.1. Corollary 5.4 Let f be an entire function which is not a polynomial. Then f assumes every complex value, with possibly one exception, infinitely many times. It is clear that Corollary 5.4 implies Theorem 5.1 since a non-constant polynomial assumes every value by the Fundamental theorem of algebra, see Sect. 2.3 (i). For a 1 proof of Corollary 5.4, we observe that g(z) = f has an essential singularity z at z = 0 and then the assertion follows from Theorem 5.3. We shall give lemmas for the proof of Theorem 5.2 in Sect. 5.2. Further we prove Theorem 5.2 in Sect. 5.3 and Theorem 5.3 in Sect. 5.4. We refer to [12] and [31] for the topics in this chapter and for further studies and related topics.

5.2 The Borel and Carathéodory Lemma and Other Results for the Picard Theorems The proof of Theorem 5.2 depends on two lemmas which we prove in this section. The first one is due to Borel and Carthéodory. It is of independent interest and it has several applications. We shall apply it in Sect. 6.5 for a proof of the Hadamard

5.2 The Borel and Carathéodory Lemma and Other Results for the Picard Theorems

119

factorisation theorem, in Sect. 7.11 for showing that the Riemann Zeta function has 1 infinitely many zeros ρ such that = ∞ where the sum is taken over all non|ρ| ρ trivial zeros and in Sect. 7.13 for deriving Lindelöf hypothesis from the Riemann hypothesis. Lemma 5.5 (Borel and Carthéodory) Let z 0 ∈ C, 0 < r < R and f (z) be analytic in D(z 0 , R) given by f (z) =

∞

cn (z − z 0 )n for z ∈ D(z 0 , R).

n=0

Let U be a real number such that Re( f (z)) ≤ U for z ∈ D(z 0 , R). Then |cn | ≤

(5.2.1)

2(U − Re( f (z 0 )) for n ≥ 1. Rn

(5.2.2)

¯ 0 , r ), we have Further for z ∈ D(z | f (z) − f (z 0 )| ≤

2r (U − Re( f (z 0 ))). R −r

(5.2.3)

Proof By considering the function f (z + z 0 ) in place of f (z), we may assume that z 0 = 0. We write in |z| < R ∞ f (z) = cn z n n=0

and φ(z) = U − f (z) = U −

∞

cn z n =

n=0

∞

bn z n

(5.2.4)

n=0

where b0 = U − c0 , bn = −cn for n ≥ 1 and β0 := Re(b0 ). Then for n ≥ 0, we see from (2.3.2) that bn =

1 2πi

|z|=r

φ(z) dz. z n+1

By putting z = r eiθ with −π < θ ≤ π, we get for n ≥ 0

(5.2.5)

120

5 The Picard Theorems

1 bn = 2πi

π −π

φ(r eiθ )ir eiθ r −n = r n+1 ei(n+1)θ 2π

π −π

φ(r eiθ )e−inθ dθ.

From now onwards in the proof of this lemma, we write without reference φ(r eiθ ) = P(r, θ) + i Q(r, θ) := P + i Q where P(r, θ) and Q(r, θ) are real-valued functions. We have 1 2π

bn r n =

π

(P + i Q)e−inθ dθ for n ≥ 0.

−π

(5.2.6)

Next, we derive from Theorem 2.4 and r < R that for n ≥ 1 π 1 rn rn π 0= φ(z)z n−1 dz = φ(r eiθ )ieinθ dθ = (P + i Q)einθ dθ. 2πi |z|=r 2πi −π 2π −π By taking conjugates on both sides, we have 0=

1 2π

π

−π

(P − i Q)e−inθ dθ

which, together with (5.2.6), implies that 1 bn r = π

π

n

−π

P(r, θ)e−inθ dθ for n ≥ 1.

Now we take absolute values on both sides to get |bn |r n ≤

1 π

π −π

|P(r eiθ )|dθ for n ≥ 1.

But by (5.2.4) and (5.2.2), we have P(r eiθ ) = Re(φ(r eiθ )) = U − Re( f (r eiθ )) ≥ 0. Therefore |bn |r n ≤

1 π

π −π

P(r eiθ )dθ for n ≥ 1.

We recall from (5.2.4) that φ(r eiθ ) =

∞ n=0

bn r n (cos nθ + i sin nθ).

(5.2.7)

5.2 The Borel and Carathéodory Lemma and Other Results for the Picard Theorems

121

Therefore ∞ P(r, θ) = Re φ(r eiθ ) = r n (Re(bn ) cos nθ − Im(bn ) sin nθ) n=0

and hence by (5.2.5), we have 1 2π

π

since −π

π

−π

P(r, θ)dθ = β0 .

cos nθdθ = 0 for n ≥ 1 and

(5.2.7) that

π −π

sin nθdθ = 0 for n ≥ 0. Then we see from

|bn |r n ≤ 2β0 for n ≥ 1. Letting r tend to R, we get from (5.2.5) that |cn | = |bn | ≤

2β0 for n ≥ 1. Rn

Now for |z| ≤ r < R, we have ∞ ∞ ∞

r n r n . | f (z) − f (0)| = cn z ≤ |bn |r n ≤ 2β0 = 2β0 R R −r n=1 n=1 n=1 Putting β0 = Re(b0 ) = U − Re( f (0)) by (5.2.5) in the above inequalities, we get (5.2.2) and (5.2.3). It is convenient to derive the following immediate consequence of Lemma 5.5. Corollary 5.6 Let 0 < r < R and f (z) be analytic in |z| ≤ R. Let B(R) = max Re( f (z)), M(r ) = max | f (z)|. |z|≤R

Then M(r ) ≤

|z|=r

2r R +r B(R) + | f (0)|. R −r R −r

Proof We apply Lemma 5.5 with z 0 = 0, U = B(R) and −Re( f (0)) ≤ | f (0)|. We derive from (5.2.3) that | f (z)| ≤ Hence

2r (B(R) + | f (0)|) + | f (0)| for |z| ≤ r. R −r

122

5 The Picard Theorems

M(r ) ≤

R +r 2r B(R) + | f (0)|. R −r R −r

Lemma 5.7 Let M > 0 and C > 0. Let φ(r ) be defined in 0 ≤ r ≤ R1 satisfying

and

0 ≤ φ(r ) ≤ M for 0 ≤ r ≤ R1

(5.2.8)

√ C φ(R) for 0 < r < R ≤ R1 . φ(r ) ≤ (R − r )2

(5.2.9)

Then there exists an absolute constant A such that φ(r ) ≤

AC 2 for 0 < r < R1 . (R1 − r )4

Proof By (5.2.9) and (5.2.8), we have √ C M for 0 < r < R ≤ R1 . φ(r ) ≤ (R − r )2 Let 0 < r < r1 < r2 ≤ R1 . Then, by (5.2.9) with R = r1 and R1 = r2 , we have φ(r ) ≤

C(φ(r1 ))1/2 (r1 − r )2

φ(r1 ) ≤

C(φ(r2 ))1/2 . (r2 − r1 )2

and

Thus C φ(r ) ≤ (r1 − r )2

C (r2 − r1 )2

21

1

(φ(r2 )) 22 .

Proceeding inductively for 0 < r < r1 < · · · < rn−1 < rn ≤ R1 , we get from (5.2.8) φ(r ) ≤

C (r1 − r )2

C (r2 − r1 )2

21

...

C (rn − rn−1 )2

1 2n−1

1

M 2n .

We take r0 = r1 , r1 = Then

1 1 1 (R1 + r ), r2 = (R1 + r1 ), . . . , rn = (R1 + rn−1 ). 2 2 2

5.2 The Borel and Carathéodory Lemma and Other Results for the Picard Theorems

r j − r j−1 =

1 (r j−1 − r j−2 ) for j ≥ 2 2

and r1 − r0 = Thus for j ≥ 1 r j − r j−1 =

1 2 j−1

1 (R1 − r ). 2

(r1 − r ) =

and therefore (r j − r j−1 )2 = Hence

123

1 (R1 − r ), 2j

1 (R1 − r )2 . 4j

n 1 j 1+ 1 +···+ n−1 1 1 1 2 φ(r ) ≤ C 1+ 2 +···+ 2n−1 (R1 − r )−2 2 4 j=1 2 j−1 M 2n .

Letting n tend to infinity, we have φ(r ) ≤

AC 2 for 0 < r < R1 (R1 − r )4

where κ

A = 4 with κ =

∞ j=1

j < ∞. 2 j−1

Let 0 < R < ρ < R1 . From now onwards in this section, we assume that f (z) is analytic in |z| ≤ R1 , where it is not equal to 0 or 1 so that the assumptions of Theorem 5.2 are satisfied. In fact, we can find R1 > R1 such that f (z) is analytic in |z| < R1 where it omits the values 0 and 1. Now we derive from Theorem 2.28 that there exist g1 (z) and g2 (z) analytic in |z| < R1 such that g1 (z) = log f (z), g2 (z) = log(1 − f (z)),

(5.2.10)

where each logarithm has its principal value at z = 0. It suffices to prove Theorem 5.2 (a) with K depending only on | log g1 (0)| and | log g2 (0)|. Therefore, we may assume that R1 exceeds a sufficiently large number depending only on | log g1 (0)| and | log g2 (0)| otherwise the assertion follows immediately. For 0 < r ≤ R1 , we put B1 (r ) = max Re(−g1 (z)). |z|≤r

Further we put M1 (r ) = max |g1 (z)|, M2 (r ) = max |g2 (z)| |z|=r

|z|=r

124

5 The Picard Theorems

and M(r ) = max(M1 (r ), M2 (r )).

(5.2.11)

We have by Theorem 2.20, B1 (r ) = − min Re(g1 (z)) = − min log | f (z)| = max log |z|≤r

|z|≤r

|z|=r

1 . | f (z)|

(5.2.12)

First, we give a sketch of the proof of Theorem 5.2. The proof depends on applying Corollary 5.6 twice. In the first application, we give an upper bound for M1 (R) in terms of B1 (ρ). This estimate implies an estimate for M1 (R) if B1 (ρ) ≤ 1, see Lemma 5.8 with ρ close to R1 . Thus we may suppose that B1 (ρ) > 1. Then we see from (5.2.12) that there exists z with |z | = ρ such that log | f (z1 )| = B1 (ρ) > 1. Thus | f (z )| < e−1 < 21 which implies that | log(1 − f (z ))| ≤ 2| f (z )| < 1 where logarithm has principal value. We observe that log(1 − f (z )) = g2 (z ) − 2nπi for some integer n. Since f (z) = 0 in |z| ≤ R1 and g2 (z) has principal value at z = 0, we see from (5.2.10) that g2 (z) is not an integral multiple of 2πi in |z| ≤ R1 . Therefore g2 (z) − 2nπ is analytic with no zeros in |z| ≤ R1 . Now we derive from Theorem 2.28 that there exists h(z) analytic in |z| ≤ R1 such that h(z) = log(g2 (z) − 2nπi) where the logarithm has principal value at z = 0. Then max |h(z)| ≥ |h(z )| ≥ B1 (ρ) − log 2. |z|=ρ

On the other hand, we apply Corollary 5.6 to get an upper bound for the left-hand side in terms of log M2 (R1 ). Combining the above two applications of Corollary 5.6, we find that M1 (R)/ log M2 (R1 ) is bounded by a number depending only on R and R1 . Further, by interchanging g1 (z) and g2 (z), we derive that M(R)/ log M(R1 ) is bounded in terms of R and R1 which, together with Lemma 5.7, implies the assertion of Theorem 5.2. Now we give a complete proof of Theorem 5.2. We bound M1 (R) in terms of B1 (ρ) as follows. Lemma 5.8 We have M1 (R) ≤

2ρ (B1 (ρ) + |g1 (0)|). ρ− R

Proof By Corollary 5.6 with f (z) = g1 (z), r = R and R = ρ, we have M1 (R) ≤ since R < ρ.

2R ρ+ R 2ρ B1 (ρ) + |g1 (0)| ≤ (B1 (ρ) + |g1 (0)|) ρ− R ρ− R ρ− R

5.2 The Borel and Carathéodory Lemma and Other Results for the Picard Theorems

125

Next we bound B1 (ρ) in terms of log M2 (R1 ) as given in the next result. Lemma 5.9 We have 2R1 1 log max |g2 (0)|, + 3M2 (R1 ) + 2 + π + 1 . B1 (ρ) ≤ R1 − ρ |g2 (0)| (5.2.13) Proof The right-hand side of the above inequality is at least 2R1 >1 R1 − ρ since R1 > ρ. Therefore, we may assume that B1 (ρ) > 1. Then we see from (5.2.12) that there exists z with |z | = ρ such that log

1 = B1 (ρ) > 1. | f (z )|

Then | f (z )| < e−1

1 since g2 (z) has principal value at z = 0. Therefore we see from (5.2.16) that |h(0)| ≤ log |g2 (0) − 2nπi| + π ≤ log(|g2 (0)| + M2 (R1 ) + 1) + π. Hence we always have |h(0)| ≤ log max |g2 (0)|,

1 |g2 (0)|

+ M2 (R1 ) + 1 + π.

(5.2.19)

By combining (5.2.17), (5.2.18) and (5.2.19), we have 2R1 1 log max |g2 (0)|, + 3M2 (R1 ) + 2 + π . |z|=ρ R1 − ρ |g2 (0)| (5.2.20) On the other hand, we see from (5.2.15) and (5.2.14) that max |h(z)| ≤

max |h(z)| ≥ |h(z )| = | log(g2 (z ) − 2nπi)| ≥ | log |g2 (z ) − 2nπi|| |z|=ρ

=

1 log |g2

(z )

− 2nπi|

≥ log

1 − log 2 | f (z )|

and hence max |h(z)| ≥ B1 (ρ) − log 2. |z|=ρ

(5.2.21)

By combining (5.2.20) and (5.2.21), we get (5.2.13). Lemma 5.10 Let 0 < R < R1 . Then √ K 1 R12 M(R1 ) M(R) ≤ (R1 − R)2 where K 1 is a number depending only on | log g1 (0)| and | log g2 (0)|. Proof Let R < ρ < R1 . By combining Lemmas 5.8 and 5.9, we have M1 (R) ≤

4ρR1 1 log max |g2 (0)|, + 3M2 (R1 ) + 2 + |g1 (0)| + π + 1 . (R1 − ρ)(ρ − R) |g2 (0)|

5.2 The Borel and Carathéodory Lemma and Other Results for the Picard Theorems

127

By considering g2 (z) in place of g1 (z) and g1 (z) in place of g2 (z), the above inequality is valid if the suffixes 1 and 2 of g1 and g2 are interchanged. Thus M(R) ≤

4ρR1 (log M(R1 ) + K 2 ) , (R1 − ρ)(ρ − R)

where K 2 > 2. Further K 2 and the subsequent letter K 3 , K 4 , K 5 are numbers dependR + R1 . Then ing only on | log g1 (0)| and | log g2 (0)|. We put ρ = 2 M(R) ≤

16R12 (log M(R1 ) + K 2 ) (R1 − R)2

since ρ < R1 . We may assume that K 2 < log M(R1 ) otherwise the assertion follows. Then 32R12 log M(R1 ). M(R) < (R1 − R)2 Since M(R1 ) > e K 2 > e2 , we see from R < R1 that √ 32R12 M(R1 ) M(R) < . (R1 − R)2

5.3 Proof of the Schottky Theorem 5.2 (a) Let 0 < r < s ≤ R1 . Assume that f omits the values 0 and 1 in |z| ≤ R1 . By Lemma 5.10 with R = r and R1 = s, we get √ √ K 3 s 2 M(s) K 3 R12 M(s) M(r ) < ≤ . (s − r )2 (s − r )2 By taking R = s, φ(r ) = M(r ), M = M(R1 ) and C = K 3 R12 , we have √ C φ(s) for 0 < r < s ≤ R1 φ(r ) < (s − r )2 and we conclude from Lemma 5.10 that φ(r ) ≤

AK 32 R14 for 0 < r < R1 (R1 − r )4

(5.3.1)

128

5 The Picard Theorems

where A is an absolute constant. Since φ(r ) = M(r ) ≥ log max | f (z)| , |z|=r

we derive from (5.3.1) with r = R < R1

K 4 R14 | f (z)| ≤ exp (R1 − R)4

for |z| ≤ R

where K 4 = AK 32 . (b) The number K in Theorem 5.2 (a) depends only on | log g1 (0)| and | log g2 (0)|. Let δ > 0 be such that δ < | f (0)|

δ. δ

We observe that 0 < δ < 1. It suffices to show that | log g1 (0)| and | log g2 (0)| are bounded above by a number depending only on δ. We had | log g1 (0)| ≤ log max | f (0)|,

1 | f (0)|

+ π ≤ log

1 +π δ

and | log g2 (0)| ≤ log max |1 − f (0)|,

since

1 |1 − f (0)|

1 1 1 + π ≤ log max 1 + , +π ≤ +π δ δ δ

1 1 1 ≤ log 1 + . log max 1 + , δ δ δ

Hence the proof of Theorem 5.2 is complete.

5.4 Proofs of the Little Picard Theorem 5.1 and the Great Picard Theorem 5.3 It is already shown in Sect. 5.1 that Theorem 5.2 implies Theorem 5.1. Now we derive Theorem 5.3 from Theorem 5.2. The proof is by contradiction. We may assume that f has an essential singularity at z 0 and f omits two distinct values a and b in (z)−a in place of f (z), we may suppose that a = 0 |z − z 0 | < ρ. By considering f b−a and b = 1. Further, by taking f (ρz + z 0 ) in place of f (z), we may assume that z 0 = 0 and ρ = 1. Let = {u + iv|u < 0} .

5.4 Proofs of the Little Picard Theorem 5.1 and the Great Picard Theorem 5.3

129

Let e be a function from to D = D (0, 1) given by e(ω) = eω for ω = u + iv ∈

since eω = eu eiv ∈ D by u < 0. We observe that e is onto. Further, we write g = f ◦ e on and

g() = f (e()) = f (D ).

(5.4.1)

Since f (z) is analytic in 0 < |z| < 1 and f has an essential singularity at z = 0, we derive from the theorem of Casorati–Weierstrass (see Sect. 2.3 (v)) that there exists a sequence {z n }∞ n=1 with z n ∈ D and 1 > |z 1 | > |z 2 | > |z 3 | > · · · such that lim |z n | = 0

n→∞

and | f (z n ) − 2|

− 1 = . 2 2 2 2

Thus δ < |h(0)|

δ for δ = . δ 5

Therefore we can take K an absolute constant by Theorem 5.2 (b). Let |ω − ωn | ≤ 2π. We write

ω = ωn + ω with |ω | ≤ 2π. Since

g(ω) = g(ωn + ω ) = h(ω ), we have |g(ω)| ≤ K for |ω − ωn | ≤ 2π. We take ω = ωn + iv with −π < v ≤ π. Then g(ω) = f (e(ω)) = f (e(ωn + iv)) = f (z n eiv ) by (5.4.2) and hence | f (z)| ≤ K for |z| = |z n | whenever n is sufficiently large. Since |z n | tend to zero with n, we conclude that f has a removable singularity at z = 0, see Ex 2.6(b). This is a contradiction. Example 5.1 Let f be an entire function which is not translation. Then f ◦ f has a fixed point. Proof The proof is by contradiction. We may assume that f is not translation and f ◦ f has no fixed point. Now we consider the function g(z) =

f ( f (z)) − z . f (z) − z

If f has a fixed point z = z 0 , then f ◦ f (z 0 ) = f ( f (z 0 )) = f (z 0 ) = z 0 and this is a contradiction. Therefore f has no fixed point. Then g(z) is entire, never zero and does not take the value 1. Now we derive from Theorem 5.1 that there exists a constant c such that f ( f (z) − z) = c for z ∈ C f (z) − z and c = 1 is a constant. Rewriting, we have

5.4 Proofs of the Little Picard Theorem 5.1 and the Great Picard Theorem 5.3

131

f ( f (z) − z) = c( f (z) − z) for all z ∈ C. By differentiating both sides, we get f (z)( f ( f (z)) − c) = 1 − c.

(5.4.3)

Since c = 1, we see from (5.4.3) that f ◦ f does not take the value c. If f ◦ f (z 0 ) = 0 for some z 0 , then (5.4.3) with z = f (z 0 ) implies that c = 1. Therefore, f ◦ f is an entire function which misses the values 0 and c. Now we apply again Theorem 5.1 to conclude that f ◦ f is constant in C. Then f is constant by (5.4.3) and hence f (z) = az + b. We may assume that a = ±1 otherwise the assertion follows. Further a = 1 since f is not translation. Finally a = −1 since otherwise f ( f (z)) = f (−z + b) = −(−z + b) + b = z and the assertion is valid. Example 5.2 Let = C\{0} and f be analytic automorphism of . Then f is of the form c1 z or c1 /z where c1 is a constant. Proof By Theorem 5.3 and f is injective, we derive that f (z) and f (1/z) do not have essential singularity at z = 0. Therefore f is a rational function by Ex 2.10. Since f is holomorphic in , we have f (z) =

p(z) for some j ≥ 1 zj

where p(z) is a polynomial in z with p(0) = 0. We derive from Theorem 2.21 and f is injective that f (z) = 0 in . Therefore the polynomial z j p (z) − j z j−1 p(z) has no zero in and hence, by the Fundamental theorem of algebra, see Sect. 2.3 (i), the polynomial is equal to c2 z n for some integer n ≥ 1 and constant c2 . Thus f (z) = c2 z k where k = n − 2 j. Then the case k = −1 is excluded by Ex 2.22. Further f (z) = c2

z k+1 + c3 k+1

(k = −1)

where c3 is a constant. Then k ∈ {0, −2} since f is injective and hence the assertion follows.

5.5 Exercises 5.1 Let f be one-one entire function. Then show that f (z) = az + b for some a, b ∈ C and a = 0.

132

5 The Picard Theorems

(Hint: It suffices to show that there exists an integer n ≥ 0 such that f (z) zn ≤ M

(5.5.1)

for some constant M as this will imply f is a polynomial by Ex 2.2 and then the assertion follows from the Fundamental theorem of algebra. Apply Theorem 5.3 and Ex 2.10 to g(z) = f ( 1z ) to conclude (5.5.1).) 5.2 (Landau theorem) Let a, b, α ∈ C satisfy a = b and α = 0. Let f (z) be analytic in |z| ≤ R such that f (0) = α and f (0) = β. Then there exists a number K = . K (α, a, b) such that f takes at least one of the values a or b whenever R ≥ 2K |β| (Hint: Apply Theorem 5.2 and the integral representation for f (0).) 5.3 Let = C\{0, 1} and f ∈ H (). Suppose that f is not constant. Then show that f = qp where p and q are polynomials with roots in {0, 1} and they do not have a common root. 5.4 Let 0 < r < R and f (z) be analytic in |z| < R. Let U ≥ 0 be a real number such that Re( f (z)) ≤ U for |z| < R. Then show that for ν > 0 max | f (ν) (z)| ≤ |z|=r

2ν!R (U − Re( f (0))). (R − r )ν+1

5.5 (a) Is it possible to prove (5.2.3) without the term Re( f (0)) on its right-hand side? (b) Give an example to show that in the assertion (5.2.3) of Lemma 5.5, it is not 2r possible to replace the factor R−r on the right-hand side of (5.2.3) by one which does not tend to infinity as r → R. (Hint: Consider f (z) = −i log(1 − z) and 0 < r < R < 1.) (c) Give an example to show that (5.2.2) is not valid with 2 is replaced by a smaller constant on its right-hand side. z .) (Hint: Consider f (z) = 1+z

Chapter 6

The Weierstrass Factorisation Theorem, Hadamard’s Factorisation Theorem and the Gamma Function

6.1 Introduction Analogous to infinite series, we consider infinite products in Sect. 6.2 and we prove their properties which are needed for a proof of the Weierstrass factorisation theorem. If a1 , . . . , am ∈ C, we have f (z) = (z − a1 ) . . . (z − am ), which is entire and vanishes at each a j with 1 ≤ j ≤ m. The Weierstrass factorisation theorem is an extension of the above statement when {an }∞ n=1 is an infinite sequence. Such an extension is not possible for all infinite sequences {an }∞ n=1 . For example, should have no limit point if we wish to construct a non-zero entire the set {an }∞ n=1 function. On the other hand, it is possible to construct an entire function which has zeros precisely at all positive integers. For a proof of the Weierstrass factorisation theorem, we shall need infinite products and the Weierstrass elementary factors which we introduce and prove their properties in Sects. 6.2 and 6.3. Further we give a proof of the Weierstrass factorisation theorem and the Hadamard factorisation theorem in Sects. 6.4 and 6.5, respectively. In fact, the assertion of the Weierstrass factorisation theorem continues to be valid if C is replaced by an open set. We prove this extension in Sect. 6.6 where we derive that M() is the quotient field of H () where M() is the set of all meromorphic functions in . On the other hand, the Weierstrass factorisation theorem gives precise information on the factors and this is valuable for the study of entire functions of finite order. This is also required for the factorisations of sin πz in Sect. 6.8 and the gamma function (z) in Sects. 6.9, 6.10 and 6.13 where we prove well-known properties of (z). Next we prove the Stirling formula for (z) in Sect. 6.13. This is very useful formula as it implies that (z) tends to zero very rapidly in a strip. The proof depends on the Euler–Maclaurin–Jacobi formula which we prove in Sect. 6.12 and the Bernoulli polynomials considered in Sect. 6.11. We also give in Sect. 6.13 good upper and lower bounds for (z) when z > 0, see Sect. 6.14. © Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9_6

133

134

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

We prove the Mittag-Leffler theorem on representation of meromorphic function by partial fractions in Sect. 6.7 and represent cot πz by partial fractions. Finally, the beta function is introduced in Sect. 6.14. We write n 0 , n 1 , n 2 , . . . for positive integers and 1 1 we recall that the Euler constant is given by lim 1 + + · · · + − log n . We n→∞ 2 n refer to [1, 12, 13, 19, 21, 23, 28, 31, 33] for the topics in this chapter and for further studies and related topics.

6.2 Infinite Products For n ≥ 1, let z n ∈ C and pn =

n

z k . If lim pn exists and is finite, then we say n→∞

k=1

that the infinite product

∞

zn

(6.2.1)

n=1

converges. If lim pn = p

n→∞

is finite, then we say that the infinite product converges to the value p and we write ∞

z n = p.

n=1

Remark If z n 0 = 0, then pn = 0 for n ≥ n 0 and p = 0. Thus, the infinite product converges to zero. The product may also be equal to zero even when all its terms are non-zero. For example, for 0 < |a| < 1, let z n = a and then pn = a n and hence p = 0. The nth term of a convergent infinite series converges to zero as n tends to infinity. Analogously, we show that the nth term of a non-zero convergent infinite product tends to 1 with n. Lemma 6.1 Assume that the infinite product (6.2.1) converges to a non-zero value. Then lim z n = 1. n→∞

Proof We write zn = and then

pn pn−1

6.2 Infinite Products

135

lim z n =

n→∞

lim pn

n→∞

lim pn−1

=1

n→∞

since (6.2.1) converges to a non-zero limit.

We see from Lemma 6.1 that Re(z n ) > 0 for sufficiently large n in a non-zero convergent infinite product (6.2.1). We shall always take the principal branch of logarithm and we understand that log 0 = −∞. We prove the following result. Lemma 6.2 The infinite product (6.2.1) converges to a non-zero limit if and only if ∞ log z n is finite. n=1

The above result enables us to transform the questions on infinite products to those of infinite series which we are already familiar with. Proof Assume that

∞

log z n is convergent. We put

n=1 n

sn =

log z k for n ≥ 1.

(6.2.2)

k=1

Then lim sn = s,

n→∞

where s is finite. This implies lim esn = es = 0

(6.2.3)

n→∞

since the exponential function is continuous. Also, we see from (6.2.2) that esn =

n

zk .

(6.2.4)

k=1

By (6.2.4) and (6.2.3), we get Next, we assume that

∞

∞

z n = es = 0.

n=1

z n = p = 0. Then lim

n→∞

n=1

lim log

n→∞

pn p

= 0, lim arg n→∞

pn = 1. Therefore p pn p

= 0.

(6.2.5)

136

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

Now, by (6.2.2), we get log

and log

pn p

pn+1 p

= sn − log p + 2πi h n

(6.2.6)

= sn+1 − log p + 2πi h n+1 ,

(6.2.7)

where h n and h n+1 are integers. Thus 2πi (h n+1 − h n ) = arg

pn+1 p

− arg

pn p

− arg(z n+1 )

by (6.2.7), (6.2.6) and (6.2.2). Therefore, for n ≥ n 0 = n 0 (), we see from (6.2.5) that 2π|h n+1 − h n | ≤ + + π. 2 2 By taking 0 < < π, we derive that h n+1 = h n for n ≥ n 0 and hence h n = h n 0 := h for n ≥ n 0 . Therefore, we derive from (6.2.6) and (6.2.5) that lim sn = log p − 2πi h,

n→∞

and hence

∞

log z n = log p − 2πi h.

n=1

Thus, the series converges.

In fact, we have proved in Lemma 6.2 that if the series converges to s then the infinite product converges to es and if the infinite product converges to p then the series converges to log p − 2πi h for some integer h. This additional information will be useful in our applications of Lemma 6.2. This lemma suggests to define absolute convergence of infinite products as follows. Definition The infinite product (6.2.1) converges absolutely if there exists N > 0 ∞ such that | log z n | < ∞. n=N

We derive from Lemma 6.2 that an absolutely convergent infinite product converges to a non-zero limit, and further we see from Lemma 6.2 that rearrangement of terms in (6.2.1) is permissible without affecting its value, see Exercise 6.1.

6.2 Infinite Products

Next, we replace

137 ∞

| log z n | < ∞ by

∞

|z n − 1| < ∞ in the definition of abso-

n=1

n=N

lute convergence of an infinite product. We prove the following. Lemma 6.3 The infinite product (6.2.1) converges absolutely if and only if

∞

|z n −

n=1

1| < ∞. Proof Assume that

∞

|z n − 1| < ∞. Then there exists n 1 such that |z n − 1|

1 such that f (z) = f 1 (z) . . . f n 7 −1 (z)g(z), where g(z) is never zero in D(a, r ). Now the assertion follows immediately. (b) We derive from (6.2.9), as in the proof of Lemma 6.5, that there exists N > 1 such that for all z ∈ K , we have

140

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

f (z) = f 1 (z) · · · f N −1 (z)G(z), where G=

∞

(6.2.14)

f n (z)

n=N

satisfies

| f n (z) − 1| < 1

(6.2.15)

n≥N

by (6.2.12). Then log G(z) =

∞

log f n (z) + 2πi h,

n=N

where h ∈ Z and the logarithm is principal. We choose a branch Log G(z) of logarithm for G(z) such that Log G(z) = log G(z) − 2πi h =

∞

log f n (z),

n=N

where the series converges uniformly on K by (6.2.15). Therefore, term-wise differentiation of the series is permissible. Now, by taking logarithmic derivative (first taking logarithm and then differentiating) on both sides, we have ∞ G (z) f n (z) = , G(z) f (z) n=N n

see Sect. 2.3 (iv). Hence, we conclude (6.2.13) by taking logarithmic derivatives on both the sides in (6.2.14).

6.3 The Weierstrass Elementary Factors We begin with definition of the Weierstrass elementary factors. Definition Let E 0 (z) = 1 − z and zP z2 + ··· + for an integer P > 0. E P (z) = (1 − z) exp z + 2 P Then we get the following.

(6.3.1)

6.3 The Weierstrass Elementary Factors

141

Lemma 6.8 For an integer P ≥ 0 and |z| ≤ 1, we have |E P (z) − 1| ≤ |z| P+1 . Proof The assertion is valid for P = 0 by definition of E 0 (z). Therefore, we suppose that P ≥ 1. Let ∞ E P (z) = 1 + ak z k (6.3.2) k=1

be the power series of E P (z) around z = 0. We differentiate the power series termwise which is justified since E P (z) is entire. We have E P (z) =

∞

kak z k−1 .

k=1

Also, by differentiating both sides of (6.3.1), we get E P (z)

zP z2 + ··· + (1 − z)(1 + z + · · · + z P−1 ) − 1 = exp z + 2 P 2 zP z + ··· + . = −z P exp z + 2 P

Therefore

∞ k=1

kak z

k−1

zP z2 + ··· + = −z exp z + z P P

.

(6.3.3)

By comparing the coefficients of z k on both sides in (6.3.3), we get ak = 0 for 1 ≤ k ≤ P and ak ≤ 0 for k ≥ P + 1.

(6.3.4)

Now in |z| ≤ 1, we derive from (6.3.2), (6.3.3) and (6.3.4) that ∞ ∞ ∞ ∞ P+1 k k−(P+1) |z| |E P (z) − 1| = ak z = ak z ≤ |z| P+1 |ak | = −|z| P+1 ak k=P+1 k=1 k=P+1 k=P+1

and 0 = E P (1) = 1 +

∞

ak = 1 +

k=1

∞

ak ,

k=P+1

which imply that |E P (z) − 1| ≤ |z| P+1 .

142

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

6.4 The Weierstrass Factorisation Theorem Let {an }∞ n=1 be a sequence of non-zero complex number such that lim |an | = ∞. n→∞

Then elements of the sequence {an }∞ n=1 need not be distinct but there is no an which repeats in the sequence infinitely many times. Let {Pn }∞ n=1 be a sequence of nonnegative integers such that ∞ r Pn +1 < ∞ for every r > 0. |an | n=1

(6.4.1)

Now we show that such a sequence {Pn }∞ n=1 always exists. Let r > 0 and Pn = n − 1 for n ≥ 1. Since |an | tends to infinity with n, there exists positive integer n 8 such that |an | > 2r for n > n 8 . Therefore, the above series is bounded by n8 ∞ ∞ r n r Pn +1 r n ≤ + |an | |an | 2r n=1 n=1 n=n +1 8

=

n8 n=1

r |an |

n

∞

+

2−n < ∞.

n=n 8 +1

Now we derive from Theorem 6.7 (a) and Lemma 6.8 the following result. ∞ Lemma 6.9 Let {an }∞ n=1 be a sequence as above and {Pn }n=1 be a sequence of non-negative integers satisfying (6.4.1). Then

F(z) =

∞

E Pn

n=1

z an

(6.4.2)

is entire and the infinite product (6.4.2) converges absolutely and uniformly on compact subsets of C. Further, the zeros of F are given by an with n ≥ 1 and the multiplicity of zero of F at an is equal to the number of times an occurs in the sequence {am }∞ m=1 . Proof We show that

∞ z E P − 1 n a n

n=1

converges uniformly on compact subsets of C. Let K be a compact subset of C. Then there exists r > 0 such that |z| ≤ r whenever z ∈ K . Since |an | tends to infinity with n, there exist positive integers n 9 and n 10 such that |an | ≥ 2r for n > n 9 and Pn +1 ∞ ∞ ∞ z z ≤ n 10 + E P − 1 ≤ n + 2−n < ∞ 10 n a a n n n=1 n=n +1 n=n +1 9

9

6.4 The Weierstrass Factorisation Theorem

143

by Lemma 6.8. Hence, the assertion follows. Now we apply Lemma 6.5 with f n (z) = E pn ( azn ) and F(z) given by (6.4.2). We conclude that the infinite product in (6.4.2) converges absolutely and uniformly on compact subsets of C. Therefore, F(z) is entire by Sect. 2.3 (iv). Further, we conclude from Theorem 6.7 (a) that the zeros of F are given by an with n ≥ 1 and the multiplicity of the zero of F at an is equal to the number of times an appears in the sequence {am }∞ m=1 . We derive from Lemma 6.9 the following result. Theorem 6.10 (The Weierstrass factorisation theorem) Let f be a non-zero entire function with infinitely many zeros. Let {an }∞ n=1 be the sequence of all non-zero zeros of f repeated according to multiplicity. Assume that the multiplicity of the zero of f at z = 0 is equal to m. Let {Pn }∞ n=1 be a sequence of non-negative integers satisfying (6.4.1). Then there exists an entire function g(z) such that f (z) = z e

m g(z)

∞

E Pn

n=1

z an

.

(6.4.3)

Further, the infinite product on the right-hand side converges absolutely and uniformly on compact subsets of C. The existence of a sequence {Pn }∞ n=1 of non-negative integers satisfying (6.4.1) is already shown. Theorem 6.10 is stated for functions f having infinitely many zeros. In fact, (6.4.3) continues to be valid if f (z) has only finitely many zeros provided that the infinite product in (6.4.3) is replaced by a finite product taken over all the zeros an of f and Pn by 0. This follows immediately from Corollary 2.30. f (z) in place of f (z), we may suppose that m = 0. By zm Lemma 6.9, we derive that F(z) given by (6.4.3) is entire. Further, we observe that f (z) is entire and it has no zero the zeros of f (z) and F(z) are identical. Therefore, F(z) in C. Now we derive from Corollary 2.30 with = C that Proof By considering

f (z) = e g(z) , F(z) where g(z) is entire. Now (6.4.3) follows from (6.4.2). Finally, the infinite product on the right-hand side converges absolutely and uniformly on compact subsets of C by Lemma 6.9.

144

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

6.5 Hadamard’s Factorisation Theorem It is desirable to take Pn in (6.4.1) as small as possible such that Pn is same for all n ≥ 1. This is possible, due to Hadamard, for functions of finite order which we introduce now. Let f (z) be non-zero entire function. For r > 0, let M(r ) = max | f (z)|. |z|=r

We say that f is of finite order if there exists a constant β ≥ 0 such that log M(r ) = O(r β ) as r → ∞.

(6.5.1)

Then there exists unique w ≥ 0 such that (6.5.1) holds when β > w and does not hold when β < w. We observe that (6.5.1) may or may not hold when β = w. The number w is called the order of f . If f is not of finite order, then we say that f is of infinite order w = ∞. For example, w = 0 if f (z) is a polynomial, w = 1 if z f (z) = e z and w = ∞ if f (z) = ee . Assume that f has infinitely many zeros. Then it has infinitely many zeros an with an = 0 for n ≥ 1 such that lim |an | = ∞. We consider the series n→∞

∞

|an |−α with α ≥ 0.

(6.5.2)

n=1

The series is divergent if α = 0. Further there exists unique τ ≥ 0 such that either τ = ∞ or (6.5.2) converges if α > τ and does not converge if α < τ . We observe that (6.5.2) may or may not converge if α = τ . We say that τ is exponent of convergence for the sequence {|an |}∞ n=1 . For example, τ = 1 if f (z) = sin πz. Let τ < ∞ and k + 1 be the least integral value of α for which (6.5.2) converges. Then k ≥ 0 since (6.5.2) with α = 0 is divergent. Further k < τ < k + 1 if τ is not an integer. If τ is an integer, then τ = k or k + 1 according to (6.5.2) with α = k diverges or converges, respectively. We define k = ∞ if τ = ∞. The number k is called the rank of f . We define k = τ = 0 if f has only finitely many zeros. We prove the following. Lemma 6.11 If w < ∞, then τ ≤ w. Proof Let f (0) = 0 and m ≥ 1 be the order of f at z = 0. Then we observe that the exponents of convergence for the functions f (z) and g(z) = f (z)/z m are identical with respect to the sequence {|an |}∞ n=1 and g(0) = 0. Therefore, there is no loss of generality in assuming that f (0) = 0.

6.5 Hadamard’s Factorisation Theorem

145

We put |an | = rn for n ≥ 1 and we apply Example 2.3 with R = 2rn , r = rn and z 0 = 0. We conclude that M(2rn ) . (6.5.3) 2n ≤ | f (0)| Let > 0 and denote by n 11 , n 12 , n 13 positive constants depending only on . Since w < ∞, we have (6.5.4) log M(2rn ) < (2rn )w+ for n ≥ n 11 . By combining (6.5.3) and (6.5.4), we get

and therefore

∞ n=1

n < rnw+2 for

n ≥ n 12 ,

∞

∞

|an |−w−3 =

rn−w−3

0. This implies the assertion of lemma. Let τ < ∞ and k be the rank of f . Put Pn = k for n ≥ 1. Then (6.4.1) is satisfied and we conclude from Theorem 6.10 and Lemma 6.11 that f (z) = z e

m g(z)

∞

Ek

n=1

z an

,

where g(z) is entire, m is the order of f at z = 0 and k ≤ τ ≤ w. Further the infinite product converges absolutely and uniformly on compact subsets of C. We prove the following. Theorem 6.12 (Hadamard’s factorisation theorem) Let f be non-zero entire function of order w < ∞. Assume that {an } be the sequence of all non-zero zeros of f . Let k be the rank of f . Let m be the order of f at z = 0. Then there exists a polynomial h(z) of degree h ≤ w such that max(h, k) ≤ w and f (z) = z m eh(z)

an

Ek

z an

,

(6.5.5)

where the product is taken over all an of the sequence. Furthermore, the product converges absolutely and uniformly on compact subsets of C. Proof By considering f (z)/z m in place of f (z), we may assume that m = 0. By Lemma 6.11, it remains to show in (6.5.5) that entire function h(z) is a polynomial of degree less than or equal to w and we give its proof now.

146

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

It suffices to prove for F(z) = f (z)e−h(0) in place of f (z) in (6.5.5) and we observe that F(0)=1. Therefore, there is no loss of generality in assuming that f (0) = 1, h(0) = 0

(6.5.6)

by (6.5.5) with m = 0. Then we write h(z) =

∞

bν z ν .

(6.5.7)

ν=1

Let R > 0. We rewrite (6.5.5) as

f (z) =

|an | k. Hence bν = 0 for ν > w. This implies that h(z) is a polynomial of degree less than or equal to w.

148

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

Corollary 6.13 Let f be an entire function of finite non-integral order. Then f has infinitely many zeros. Proof Assume that f has only finitely many zeros a1 , . . . , an . Then we derive from Theorem 6.12 that f (z) = e g(z) (z − a1 ) . . . (z − an ) for z ∈ C, where g(z) is a polynomial. Then f (z) and e g(z) have the same orders. But the order of e g(z) is equal to the degree g. Therefore, order of f is equal to the degree g which is integral. This is a contradiction.

6.6 Extension of the Weierstrass Factorisation Theorem for an Arbitrary Region The assertion of Theorem 6.10 is valid for C. The following analogue is valid for any region, but provides less information on the factors. Theorem 6.14 Let be a region. Let {a j }∞ j=1 be a sequence of distinct points in be a sequence of positive integers. Then there with no limit point in and {m j }∞ j=1 exists f ∈ H () such that the zeros of f are given by a j with j ≥ 1. Further, for every j ≥ 1, the multiplicity of the zero of f at a j is equal to m j . There is no loss of generality in assuming that the set a j with j ≥ 1 is infinite, otherwise the assertion follows immediately. First, we consider Theorem 6.14 when all the a j lie in a closed disc containing C\. Lemma 6.15 Let {α j }∞ j=1 be a sequence of distinct points in a region with no limit point in . Suppose that there exists R > 0 such that C\ ⊆ D(0, R)

(6.6.1)

α j ∈ D(0, R) for j ≥ 1.

(6.6.2)

and Assume that {m j }∞ j=1 be a sequence of positive integers. Then there exists g ∈ H () such that the zeros of g are given by α j with multiplicity m j for j ≥ 1 and lim g(z) = 1.

z→∞

(6.6.3)

Proof We form a sequence {z n }∞ n=1 which starts with α1 occurring m 1 times, then α2 occurring m 2 times and so on. Since is open, we observe that C\ is closed. Further, it is bounded by (6.6.1) and hence C\ is compact. Therefore, for every n ≥ 1, there exists wn ∈ C\ such that |z n − ωn | is the distance between z n and

6.6 Extension of the Weierstrass Factorisation Theorem for an Arbitrary Region

149

C\, see Exercise 1.20. Since C\ is compact, we have lim wn = w ∈ C\. n→∞

Further lim z n = z ∈ C\ since |z n | ≤ R for all n and {z n }∞ n=1 has no limit point n→∞ in . Also |z n − wn | ≤ |z n − w| for all w ∈ C\. (6.6.4) In particular, |z n − wn | ≤ |z n − z|, which implies that lim |z n − wn | = 0.

(6.6.5)

n→∞

n −wn For n ≥ 1, we observe that E n zz−w with z ∈ is analytic in since wn ∈ n C\ and it has only one zero at z = z n with multiplicity 1. Let K be a compact subset of . Then we derive from (6.6.5) and Theorem 1.4 there exists n 14 such that for n ≥ n 14 and z ∈ K , we have |z − wn | ≥ 2|z n − wn |

(6.6.6)

by (6.6.5). Therefore, we obtain from Lemma 6.8 that |E n Thus

z n − wn z − wn

− 1| ≤ 2−n−1 for z ∈ K , n ≥ n 14 .

∞

En

n=1

z n − wn z − wn

−1

converges absolutely and uniformly on compact subsets of . Now we derive from Theorem 6.7 (a) that g(z) =

∞ n=1

En

z n − wn z − wn

∈ H ()

has zeros only at z n with multiplicity m n for n ≥ 1. Thus, it remains to show (6.6.3). For this, we observe that |z n | ≤ R by (6.6.1) and |wn | ≤ R by (6.6.2). Therefore, for |z| > R1 where R1 > R z n − wn 2R z − w ≤ R − R. n 1 Let 0 < δ

2 R. Then δ

150

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

z n − wn z − w < δ for |z| > R1 . n Therefore, by Lemma 6.8, we have E n z n − wn − 1 ≤ δ n+1 for |z| > R1 , n ≥ 1. z−w

(6.6.7)

n

For |z| > R1 , we have |g(z) − 1| = | exp

∞

log E n

n=1

where τ=

∞

z n − wn z − wn

log E n

n=1

z n − wn z − wn

− 1| = |eτ − 1|,

and logarithm has principal value. Now we estimate |τ |. For |z| > R1 , we derive from (6.6.6) ∞ ∞ z n − wn z n − wn = log E log 1 + E − 1 |τ | ≤ n n z−w z−w n=1 ∞

En

≤2

n=1

n

z n − wn z − wn

n

n=1

∞ 1 2δ 2 − 1 ≤ 2 < δ n+1 = 1 − δ 2 n=1

since 0 < δ < 13 . Therefore, for |z| > R1 , we have |g(z) − 1| = |eτ − 1| ≤ 2|τ | ≤

4δ 2 . 1−δ

Now we let δ tend to zero for concluding (6.6.3).

Proof of Theorem 6.14 Suppose that the assumptions of Theorem 6.14 are satisfied. Let a ∈ such that a = a j for j ≥ 1. Then, by our assumption, a is not a limit point of the set a j with j ≥ 1. Therefore, there exists r > 0 such that D(a, r ) ⊆ contains no a j for j ≥ 1. Let 1 = − {a}. Then 1 is a region by Theorem 1.2 and we define T on 1 into C given by T (z) = (z − a)−1 for z ∈ 1 .

(6.6.8)

We observe that T ∈ H (1 ) and we write T (1 ) = 2 .

(6.6.9)

6.6 Extension of the Weierstrass Factorisation Theorem for an Arbitrary Region

151

Since a continuous image of a connected set is connected, we see that 2 is connected. In fact, 2 is a region by open mapping Theorem 2.18. Further, we write T (a j ) = α j with j ≥ 1.

(6.6.10)

Thus, α j = (a j − a)−1 ∈ 2 such that {α j }∞ j=1 has no limit point in 2 and we put 1 R = . For z ∈ C with |z| > R, we observe that r 1 1 z=T + a with + a ∈ D (a, r ) ⊆ 1 . z z Thus, by (6.6.9), we have {z||z| > R} ⊂ 2 .

(6.6.11)

Also by (6.6.10), we get |α j | =

1 1 < = R for j ≥ 1 |a j − a| r

(6.6.12)

since D(a, r ) contains no a j . Now we apply Lemma 6.15 with = 2 and α j = T (a j ) for j ≥ 1. In view of (6.6.11) and (6.6.12), we observe that the assumptions of Lemma 6.15 are satisfied. Hence, we derive that there exists g ∈ H (2 ) satisfying (6.6.3) such that the zeros of g are given by α j with multiplicity m j for j ≥ 1. Now we consider (6.6.13) f (z) = g(T (z)) for z ∈ 1 . We observe that f ∈ H (1 ). Further, we see from (6.6.3) that f is bounded in D (a, r ). Therefore, f has a removable singularity at z = a, see Exercise 2.9 (a). Further by (6.6.9), we have f (a j ) = g(T (a j )) = g(α j ) = 0. In fact, the zeros of f are given by a j with multiplicity m j for j ≥ 1 by (6.6.13) since the zeros of g are given by α j with multiplicity m j . We derive from Theorem 6.14 that every meromorphic function in a region can be written as a quotient of analytic functions in . Corollary 6.16 Let f be meromorphic in a region . Then there exist g ∈ H () and h ∈ H () such that g f = . h Proof Let P1 , P2 , . . . be the poles of f such that the order of the pole at P j is r j . By Theorem 6.14, there exists h ∈ H () such that the zeros of h are given

152

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

by P1 , P2 , . . . with multiplicity r1 , r2 , . . . . Therefore, g = f h ∈ H () since it has removable singularities at P1 , P2 , . . . . Hence f = hg . Since the zeros of an analytic function in a field are isolated, we see that H () is an integral domain. Denote by M() the set of all meromorphic functions in . Then M() is a field. In fact, Corollary 6.16 tells that it is the quotient field of H (), see Exercise 6.11.

6.7 Representation of Meromorphic Functions by Partial Fractions In the following result due to Mittag-Leffler, we give representation of an arbitrary meromorphic function by partial fractions. Theorem 6.17 Let {bν }∞ lim bν = ν=1 be a sequence of complex numbers satisfying ν→∞ ∞ and Pν with ν ≥ 1 be polynomials without constant term. Then there exists meromorphic function in C with poles precisely at bν where the principal part is

1 Pν z−b . The most general meromorphic function of this type can be written in ν the form 1 f (z) = Pν − pν (z) + g(z), z − bν ν where pν (z) are suitably chosen polynomials and g(z) is entire. Further, the series converges absolutely and uniformly on any compact set not containing poles.

1 is analytic in Proof We may assume that bν = 0 for ν ≥ 1. Then Pν z−b ν D(0, |bν |) and therefore it has power series expansion Aν0 + Aν1 z + · · · + Aνn ν z n ν + · · · in D(0, |bν |). We shall choose suitable integers n ν for ν ≥ 1, and we take pν (z) = Aν0 + Aν1 z + · · · + Aνn ν z n ν . and max |Pν (ζ)| = Mν . Then in |z| ≤ |b4ν | , we ζ∈C 1 − derive from finite development of power series, see Sect. 2.3 (i), for Pν z − bν pν (z) and |ζ − z| ≥ |ζ| − |z| ≥ |b4ν | for ζ ∈ C that

Let C be the circle with |ζ| =

|bν | 2

6.7 Representation of Meromorphic Functions by Partial Fractions

153

1 1 Pν (ζ)dζ n ν +1 Pν |z| = − p (z) ν 2πi n ν +1 (ζ − z) z − bν C ζ 2|z| n ν +1 ≤ 2Mν . |bν | In fact for |z| ≤

|bν | , we have 4 1 Pν ≤ 2−n ν Mν < 2−ν − p (z) ν z − bν

(6.7.1)

by taking n ν so large that 2n ν > 2ν Mν . Let K be a compact set not containing any bν . Then there exists r > 0 such that |z| ≤ r whenever z ∈ K . Since lim bν = ∞, there exists ν0 such that r ≤ |b4ν | for ν→∞ ν ≥ ν0 . Therefore, (6.7.1) is valid for z ∈ K and ν ≥ ν0 . Hence, the series ∞ ν=0

Pν

1 z − bν

− pν (z)

converges absolutely and uniformly on compact subsets of C not containing any bν . Consequently, the above series is meromorphic in C. Now the assertion of Theorem 6.17 follows immediately. Next, we represent cot πz by partial fractions and we shall need this representations in the next section. Lemma 6.18 We have 1 1 1 π cot πz = + + , z n=0 z − n n

(6.7.2)

where the sum is taken over all non-zero integers. Proof Let K be a compact set not containing any integer. There exists r > 0 such that |z| ≤ r for z ∈ K . We have 1 1 z + = . z−n n (z − n)n For z ∈ K and n > 2r, we observe that |z − n| ≥ n − |z| > n − by (6.7.3) 1 1 2r ≤ . + z − n n n2

(6.7.3) n 2

=

n 2

and hence

154

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

Consequently, the series in (6.7.2) converges absolutely and uniformly on compact subsets not containing any integer. Thus, the right-hand side of (6.7.2) is a meromorphic function which has a simple pole at every integer with residue 1 and it has no pole at any other point. This is also the case with π cot πz. Therefore 1 1 1 + h(z) = π cot πz − − z n=0 z − n n

(6.7.4)

is entire. Since the series converges uniformly on compact subsets not containing any integer, term-wise differentiation of the series is permissible and we get h (z) =

n∈Z

1 π2 . − (z − n)2 sin2 πz

(6.7.5)

It suffices to show that h (z) = 0 for z ∈ C. Then h(z) is constant by Taylor series expansion of h(z) around z = 0. By letting z tend to zero in (6.7.4), we see that h(0) = 0. Hence, h(z) = 0 for z ∈ C implying (6.7.2). Let z = x + i y. By writing n∈Z

1 1 1 = = , 2 2 |z − n| |x + i y − n| (x − n)2 + y 2 n∈Z n∈Z

(6.7.6)

we see that the series in (6.7.5) tends to zero uniformly in 0 ≤ x ≤ 1 if |y| → ∞. π2 by Exercise 1.3. Hence, we derive from (6.7.5) that This is also the case with 2 sin πz h (z) is bounded uniformly in 0 ≤ x ≤ 1 since h (z) is continuous and h (z) tends to zero uniformly in 0 ≤ x ≤ 1 as |y| → ∞. Thus, h is bounded in C since h is periodic with period 1. Hence, h (z) is constant for all z ∈ C. In fact, h (z) = 0 for z ∈ C by letting |y| → ∞. Since we have proved that h is zero in the proof of Lemma 6.18, we also conclude from (6.7.5) that Lemma 6.19

π2 1 = . 2 (z − n)2 sin πz n∈Z

6.8 Applications of the Weierstrass Factorisation Theorem In this section, we give two examples on the factorisation of entire functions. The first example is on entire functions which have zeros precisely at all integers and the second one on entire functions which have zeros precisely at all negative integers.

6.8 Applications of the Weierstrass Factorisation Theorem

155

Example 6.1 Let f (z) be entire which has simple zeros precisely at all integers. f (z) is entire and it has Then the zeros of f (z) and sin πz are identical. Therefore, sin πz no zero. Therefore f (z) = ea(z) sin πz, where a(z) is entire by Corollary 2.30. Now we factorise sin πz = πz

∞ z z en . 1− n n=−∞

(6.8.1)

n=0

For proving the above relation, we take Pn = 1 for n ≥ 1 in Theorem 6.10. Further, for n ≥ 1, let n if n = 0 is even − . an = n+12 if n is odd 2 We observe that the set {an | n ≥ 1} coincides with the set of non-zero integers. Then ∞ ∞

r Pn +1 r 2 =2 0 so that (6.4.1) is satisfied. Then, we conclude from Theorem 6.10 with m = 1 the Product formula for sin πz that sin πz = ze g(z)

∞ z z en , 1− n n=−∞

(6.8.2)

n=0

where g(z) is entire, the product converges absolutely and uniformly on compact subsets of C and rearrangement of terms in the product is permissible. Further, we observe that none of the factors on the right-hand side has a zero in C\Z. Therefore, we derive from Theorem 6.7 (b) that ∞

1 π cot πz = g (z) + + z n=0

1 1 + z−n n

(6.8.3)

for z ∈ C\Z. By (6.8.3) and (6.7.2), we see that g (z) = 0 for z ∈ C\Z. In fact, g(z) is constant in C since g(z) is entire. Hence, g(z) = g for some constant g ∈ C. Then we see from (6.8.2) that eg sin πz = 1 = lim z→0 πz π

156

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

since the infinite product tends to one as z approaches to zero. Thus e g = π and hence (6.8.1) follows. Example 6.2 For n ≥ 1, take an = −n, Pn = 1 in Lemma 6.9. Then ∞ ∞ r Pn +1 r 2 = 0 so that (6.4.1) is satisfied. Then we derive from Lemma 6.9 that G(z) :=

∞

E1

n=1

z −n

=

∞ z −z e n 1+ n n=1

(6.8.4)

is entire and the product converges absolutely and uniformly on compact subsets of C. Thus, G(z) is an entire function with simple zeros at every negative integer and with no zero at any other point. The function G(z) satisfies the following relation. Lemma 6.20

G(z − 1) = zeγ G(z) for z ∈ C,

(6.8.5)

where γ is the Euler constant. Proof We observe that G(z − 1) has simple zero at every non-positive integer and it has no other zeros. This is also the case with the right-hand side of (6.8.5). Since the zeros of G(z−1) and G(z) are identical, we see from Corollary 2.30 that z G(z − 1) = zeγ1 (z) G(z),

(6.8.6)

where γ1 (z) is entire. We show that γ1 (z) is constant. We observe that each of the factors of G(z) and of G(z − 1)/z has no zero in C\Z≤0 where Z≤0 denotes the set of all non-positive integers. Now we apply Theorem 6.7 (b) to derive from (6.8.6) and (6.8.4) that for z ∈ C\Z≤0 ∞ n=1

Further

1 1 − z−1+n n

=

∞ 1 1 1 + γ1 (z) + − . z z+n n n=1

(6.8.7)

6.8 Applications of the Weierstrass Factorisation Theorem ∞ n=1

1 1 − z−1+n n

157

∞ 1 1 1 = −1+ − z z+n n+1 n=1 ∞ ∞ 1 1 1 1 1 + = −1+ − − z z+n n n n+1 n=1 n=1 ∞ 1 1 1 − , = + z n=1 z + n n

which, together with (6.8.7), implies γ1 (z) = 0 for z ∈ C\Z≤0 . In fact, γ1 (z) is constant in C since γ1 (z) is entire. Now we write γ1 for γ1 (z). By putting z = 1 in (6.8.6), we see from (6.8.4) that G(0) = eγ1 G(1) = eγ1

∞ 1 −1 1+ e n. n n=1

Since G(0) = 1 by (6.8.4), we obtain ∞ 1 −1 1 1+ e n = eγ1 . n n=1 By the definition of convergence of infinite product, we have lim

n→∞

Therefore

e 1 + 2 +···+ n n+1 1

1

1

= eγ1 .

1 1 γ1 = lim 1 + + · · · + − log n . n→∞ 2 n

Hence, γ1 is the Euler constant.

6.9 The Gamma Function The gamma function (z) is given by 1 , z H (z)

(6.9.1)

H (z) = eγz G(z)

(6.9.2)

(z) = where

158

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

and γ is the Euler constant. Then we see from (6.9.2) and (6.8.4) the product formula for (z) given by ∞ e−γz z −1 z (z) = 1+ en . (6.9.3) z n=1 n Now we prove the following relations satisfied by (z). Theorem 6.21 We have (a) (z + 1) = z(z). π (b) (z)(1 − z) = . sin πz √ 1 (c) 22z−1 (z) z + = π (2z). 2 The relation (a) is called the functional equation for (z) and (c) is called the duplication formula for (z). The following assertions follow immediately from Theorem 6.21. Corollary 6.22 We have (i) (z) = 0 for z ∈ C. (ii) (n) (n − 1)! for n ≥ 1. = √ 1 = π. (iii) 2 (iv) (z) is a meromorphic function in C with simple pole at z = −n for every n ≥ 0 and it has no other pole. Further the residue of the pole at z = −n is (−1)n for n ≥ 0. n! Theorem 6.21 implies Corollary 6.22 (i) Since G(z) has a zero at every negative integer, we see from (6.9.2) and (6.9.1) that (z) has a pole at every non-positive integer. Thus, we may suppose that z is different from a non-positive integer. Now we conclude from Theorem 6.7 (a) that the infinite product in (6.9.3) is non-zero since none of its term is zero and hence (z) never vanishes. (ii) The proof is by induction on n. The assertion follows for n = 1 since 0! = 1 and (1) = 1 by (6.9.3) with z = 1. We assume that (n) = (n − 1)! for n ≥ 1. We have (n + 1) = n(n) by Theorem 6.21 (a) and hence (n + 1) = n(n) = n!. (iii) By putting z = 21 in Theorem 6.21 (b), we get (( 12 ))2 = π which implies ( 21 ) = √ π since ( 21 ) > 0 by (6.9.3). (iv) Since G(z) has simple zero at non-positive integers and no zero at any other point, we see that (z) is a meromorphic function with simple pole at non-positive integers and it is analytic elsewhere. Let n ≥ 0. Then (6.9.4) lim (z + n + 1) = (1) = 1 z→−n

6.9 The Gamma Function

159

by (ii). On the other hand, we see from Theorem 6.21 (a) that (z + n + 1) = (z + n)(z + n) = (z + n)(z + n − 1)(z + n − 1). Continuing similarly, we get (z + n + 1) = (z + n)(z + n − 1) . . . z(z). Letting z tend to −n on both sides, we see from (6.9.4) that (−1)n n! lim (z + n)(z) = 1 z→−n

and the assertion follows immediately. Proof of Theorem 6.21 (a) By (6.9.2) and (6.8.5) H (z − 1) = eγ(z−1) G(z − 1) = eγ(z−1) zeγ G(z) = zeγz G(z) = z H (z). Therefore, by (6.9.1) (z + 1) =

1 1 = = z(z). (z + 1)H (z + 1) H (z)

(b) We rewrite (6.8.1) as ∞ ∞ ∞ sin πz z2 z z z −z n n 1− 2 =z e e =z 1− 1+ π n n n n=1 n=1 n=1

(6.9.5)

since the product converges absolutely. Thus, by (6.8.2) and (6.8.4), we have sin πz = zG(z)G(−z), π which we rewrite as

π 1 = . sin πz zG(z)G(−z)

(6.9.6)

By (6.9.1) and (6.9.2), we have (z)(−z) =

1 z

1 G(z)

1 (−z)

1 , G(−z)

which, together with (6.9.6) and Theorem 6.21 (a), implies that π = −z(−z)(z) = (z)(1 − z). sin πz

160

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

√ 1 = π by Theorem 6.21 (b). By 2 taking logarithmic derivatives on both sides in (6.9.3), we derive from Theorem 6.7 (b) that ∞ 1 1 1 (z) = −γ − + − , (z) z n=1 n z+n

(c) We recall that (1) = 1 by (6.9.3) and

where the sum converges absolutely and uniformly on compact subsets of C. Therefore, by term-wise differentiation, we obtain d dz

(z) (z)

=

∞ ∞ 1 1 1 + = 4 . z 2 n=1 (z + n)2 (2z + 2n)2 n=0

(6.9.7)

Therefore d dz

(z + 21 )

=

(z + 21 )

∞ n=0

1 (z +

2n+1 2 ) 2

=4

∞ n=0

1 , (2z + 2n + 1)2

which, together with (6.9.7), implies that d dz

(z) (z)

d + dz

(z + 21 )

(z + 21 )

=4

∞ n=0

1 d =2 (2z + n)2 dz

(2z) . (2z)

By integrating twice both sides, we get 1 (z) z + = eaz+b (2z), 2

(6.9.8)

where a and b are constants. By putting z = 21 on both sides in (6.9.8) and using (1) = 1 that √ a π = e 2 +b , which implies that 1 a + b = log π. 2 2 By putting z = 1 both sides in (6.9.8), we get

3 3 = (1) = ea+b (2) = ea+b . 2 2

By Theorem 6.21 (a) and (b), we have

(6.9.9)

6.9 The Gamma Function

Therefore

1√ π 2

161

1 1 1 1√ 3 = +1 = = π. 2 2 2 2 2

= ea+b . Thus a+b =

1 log π − log 2 2

(6.9.10)

and we get from (6.9.9) and (6.9.10) that a = −2 log 2 and b = Hence, we conclude from (6.9.8) that

1 2

log π + log 2.

√ 1 (z) z + = 2−2z 2 π (2z), 2 which implies the assertion. Next we derive from (6.9.3) the following result known as the Euler formula. Theorem 6.23 (The Euler formula for the gamma function) (z) =

∞ 1 z −1 1 z 1+ 1+ z n=1 n n

and (z) = lim

m→∞

(6.9.11)

1.2. . . . (m − 1)m z . z(z + 1) . . . (z + m − 1)

Proof Since the infinite product in (6.9.3) is absolutely convergent, we observe that rearrangement of terms in the infinite product in (6.9.3) is permissible. By (6.9.3), we have ∞ m z −z z −z 1 γz (1+ 21 +···+ m1 −log m)z n = ze e = z lim e e n 1+ 1+ m→∞ (z) n n n=1 n=1

= z lim m −z m→∞

We write m −z =

m z

. 1+ n n=1

m−1

1+

n=1

1 n

−z

=

m n=1

1+

1 n

−z

1+

1 m

z .

Thus m ∞ 1 −z 1 1 −z z

z

1+ 1+ = z lim 1+ 1+ =z . m→∞ (z) n n n n n=1 n=1

162

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

Hence

∞ z −1 1 z 1 1+ 1+ , (z) = z n=1 n n

which proves (6.9.11). Then we have (z) =

m−1 1 z −1 1 z 1.2 . . . (m − 1)m z 1+ 1+ = lim lim . m→∞ z(z + 1) . . . (z + m − 1) z m→∞ n=1 n n (6.9.12)

A direct application of the Euler formula gives another proof of Theorem 6.21. The Euler formula implies Theorem 6.21. (a) By (6.9.11), we have (z + 1) = lim m→∞ (z)

1·2···(m−1)m z+1 (z+1)···(z+m) 1·2···(m−1)m z z(z+1)...(z+m−1)

= lim

m→∞

mz z = lim m→∞ 1 + z+m

z m

= z.

(b) By (z + 1) = z(z), we have (z)(1 − z) = −z(−z)(z).

(6.9.13)

By (6.9.12), we get (z)(−z) = −

−1 −1 ∞ m−1 1 1 z2 z2 lim = − , 1 − 1 − z 2 m→∞ ν=1 ν2 z 2 ν=1 ν2

which, together with (6.9.5), implies that (z)(−z) = −

1 πz . z 2 sin πz

(6.9.14)

Now the assertion follows by combining (6.9.13) and (6.9.14). (c) We consider 22z−1 (z)(z + 21 ) (z) = . (2z) By (6.9.12), we have 1 (m − 1)!)2 m 2z+ 2 22z−1 1 = lim 22z−1 (z) z + m→∞ z(z + 1) . . . (z + m − 1)(z + 1 ) . . . (z + 2 2

1 2

+ m − 1)

6.9 The Gamma Function

163

and 2 m−1 2z −1 1 2z 1 (2m − 1)!(2m)2z 1+ 1+ lim . (2z) = = lim m→∞ 2z(2z + 1) . . . (2z + 2m − 1) 2z m→∞ n n n=1

But 1 1 2z(2z + 1) . . . (2z + 2 m − 1) = 22m z(z + 1) . . . (z + m − 1) z + ... z + + m − 1 . 2 2

Therefore (z) =

22m−1 m 1/2 ((m − 1)!)2 , (2m − 1)!

which is independent of z. Hence ( 21 )(1) √ 1 (z) = = = π 2 (1)

and the assertion follows.

6.10 Integral Representation for (z) We prove that

∞

(z) =

e−t t z−1 dt for Re(z) > 0.

0

Proof Let σ = Re(z) > 0. For n ≥ 1, we define

(z, n) =

n 0

t n z−1 1− t dt. n

By writing tn for t in the integrand, we have

1 z (1 − t)n t z−1 dt. (z, n) = n 0

Integrating by parts, we get

(z, n) = n z

n z

1

(1 − t)n−1 t z dt

0

since σ > 0. Proceeding as above, inductively, we obtain

(6.10.1)

164

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

(z, n) =

1 · 2 · · · (n − 1)n z n(n − 1) · · · 1 nz = . z(z + 1) · · · (z + n) z(z + 1) . . . (z + n)

Now we conclude from the Euler formula (6.9.12) that lim

n→∞

Let

∞

1 (z) =

(z, n) = (z).

(6.10.2)

e−t t z−1 dt = lim

n

n→∞ 0

0

e−t t z−1 dt.

(6.10.3)

We show that 1 (z) = (z).

(6.10.4)

First we prove that 1 (σ) < ∞ for every σ > 0. We write

∞

1 (σ) =

e

−t

t

σ−1

1

dt =

0

e

−t

t

σ−1

∞

dt +

0

e−t t σ−1 dt.

1

There exists t0 = t0 (σ) such that e−t t σ−1 ≤ e−t e 2 = e− 2 for t ≥ t0 = t0 (σ). t

t

Therefore

∞

e

−t

t

σ−1

t0

dt =

1

e

−t

t

σ−1

1 t0

≤

∞

dt +

e−t t σ−1 dt +

1

e−t t σ−1 dt

t0 ∞

e− 2 dt < ∞. t

t0

Further, by writing t = 1/u, we have

1 0

e−t t σ−1 dt =

∞

e− u u −σ+1 u −2 du < 1

1

∞

u −σ−1 du < ∞

1

since σ > 0. Hence, 1 (σ) < ∞ for σ > 0. For n ≥ 1, we write t n for 0 ≤ t ≤ n f n (t) = 1 − n and we see from (6.10.2) and (6.10.1) that (z) = lim

n→∞

(z, n) = lim

n→∞ 0

n

f n (t)t z−1 dt.

(6.10.5)

6.10 Integral Representation for (z)

We observe that

et

1.

(6.11.9)

168

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

Further we see from (6.11.3) and (6.11.6) that ∞

Bk z z + =1+ zk . z e −1 2 k! k=2 Since the left-hand side is an even function of z, we derive that Bk = 0 for k odd ≥ 3.

(6.11.10)

Further we compute, by (6.11.9) and (6.11.10), Bk for 2 ≤ k ≤ 14 as follows: B2 =

1 1 1 1 5 691 7 , B4 = − , B6 = , B8 = − , B10 = , B12 = − , B14 = . 6 30 42 30 66 2730 6

In the next result, we give some properties satisfied by the Bernoulli polynomials. Lemma 6.24 The Bernoulli polynomials Bk (x) with k ≥ 0 satisfy the following: (a) Bk (x) is a monic polynomial of degree k given by Bk (x) =

k k m=0

m

(b)

Bk (1) − Bk (0) =

Bm x k−m .

(6.11.11)

1 if k = 1 0 if k = 1.

(c) Bk (1 − x) = (−1)k Bk (x). (d)

Bk (x) = k Bk−1 (x) for k > 0.

Proof (a) By (6.11.2), (6.11.1) and (6.11.3), we have ∞ ∞ ∞ Bk xk Bk (x) k z xz k k z = z e = z z . k! e −1 k! k! k=0 k=0 k=0 Now the assertion (6.11.11) follows immediately by comparing the coefficients of k z k on both sides. Further we see from (6.11.11) that the coefficient of x in Bk (x) is k B0 = 1 by (6.11.6). Hence, Bk (x) is a monic polynomial of degree k given by 0 (6.11.11). (b) By (6.11.1), we have

6.11 The Bernoulli Numbers and the Bernoulli Polynomials

u(1, z) − u(0, z) =

169

z z ez − z = z. ez − 1 e −1

By differentiating both sides k times, we see from (6.11.4) that Bk (1) − Bk (0) = 1 if k = 1 and 0 otherwise. (c) By (6.11.1), we have u(1 − x, z) =

−z z e(1−x)z = −z e−x z = u(x, −z). ez − 1 e −1

By differentiating both sides k times, we derive from (6.11.4) that Bk (1 − x) = (−1)k Bk (x). (d) By (6.11.1), we get ∂ u(x, z) = zu(x, z), ∂x which we rewrite as

∞ ∞ Bk (x) k Bk−1 (x) k z = z k! (k − 1)! k=1 k=1

by (6.11.2). By comparing the coefficients of z k on both sides, we get Bk (x) Bk−1 (x) = for k > 0, k! (k − 1)! which implies the assertion.

6.12 The Euler–Maclaurin–Jacobi Sum Formula Let b > a and q ≥ 1 be integers. Let f be q times continuously differentiable in [a, b]. Then b n=a+1

where

f (n) =

q Br (r −1) f (−1)r (b) − f (r −1) (a) + Rq , r! a r =1 (6.12.1)

b (−1)q+1 Rq = Bq (x − [x]) f (q) (x)d x. q! a b

f (x)d x +

Proof Let F be q times continuously differentiable in [0, 1]. By (6.11.11) with k = 1 and (6.11.6), we have 1 B1 (x) = B0 x + B1 = x − , B1 (x) = 1. 2

(6.12.2)

170

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

Then

1

1

F(x)d x =

0

0

F(x)B1 (x)d x.

Integrating the right-hand side by parts, we derive from (6.12.2), (6.11.6) and Lemma 6.24 (d) with k = 2 that

0

1

1 F(1) + F(0) − F (x)B1 (x)d x 2 0 F(1) + F(0) 1 0 − = F (x)B2 (x)d x. 2 2 1

F(x)B1 (x)d x =

Further by using Lemma 6.24 (b) with k > 1 and Lemma 6.24 (d) with k = 3, we get

0

1

F (x)B2 (x)d x = B2 (F (1) − F (0)) − = B2 (F (1) − F (0)) −

1

0

1 3

F (x)B2 (x)d x 1 0

F (x)B3 (x)d x.

Now we proceed inductively, as above, for obtaining

1 0

Br (r −1) F(1) + F(0) + F F(x)d x = (−1)r −1 (1) − F (r −1) (0) 2 r ! r =2

(−1)q 1 Bq (x)F (q) (x)d x. (6.12.3) + q! 0 q

Since B1 = − 21 , we get F(1) − F(0) F(1) + F(0) = F(1) − = F(1) + B1 (F(1) − F(0)). 2 2 Then we rewrite (6.12.3) as

F(1) =

1

F(x)d x +

0

(−1)q+1 + q!

q

(−1)r

r =1

1

Br (r −1) F (1) − F (r −1) (0) r!

Bq (x)F (q) (x)d x.

0

For positive integer n with a ≤ n ≤ b, let F(x) = f (n − 1 + x). Then F(x) is q times continuously differentiable in [0, 1] since f is q times continuously differentiable in [a, b]. Then we derive from the above formula that

6.12 The Euler–Maclaurin–Jacobi Sum Formula

n

f (n) =

f (x)d x +

n−1

171

q Br (r −1) f (−1)r (n) − f (r −1) (n − 1) r! r =1 q+1 1 (−1) + Bq (x) f (q) (n − 1 + x)d x. q! 0

Letting n run from a + 1 to b, we get

b

b

f (n) =

f (x)d x +

a

n=a+1

where Rq =

q Br (r −1) f (−1)r (b) − f (r −1) (a) + Rq , r! r =1

1 b (−1)q+1 Bq (x) f (q) (n − 1 + x)d x. q! 0 n=a+1

For n = a + r with 1 ≤ r ≤ b − a, we have

1

Bq (x) f

(q)

(n − 1 + x)d x =

0

1

Bq (x) f (q) (a + r − 1 + x)d x.

0

Putting a + r − 1 + x = y, the above integral is equal to

a+r

a+r −1

Hence

Bq (y − [y]) f (q) (y)dy.

(−1)q+1 Rq = q!

b

Bq (x − [x]) f (q) (x)d x.

a

6.13 The Stirling Formula We derive from (6.12.1) the following result for (z) given by the Stirling formula. Theorem 6.25 Let m be a positive integer. For all complex numbers z different from zero and negative integers, we have

1 1 log (z) = z − log z − z + log 2π + K m (z), 2 2 where logarithm has principal value and

(6.13.1)

172

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

K m (z) =

m j=1

B2 j 1 1 − 2 j−1 (2 j − 1)2 j z 2m

1

∞

B2m (x − [x]) d x. (x + z)2m

(6.13.2)

Proof We check that both the sides in (6.13.1) are analytic functions of z in the region obtained from the complex plane by deleting 0 and the negative real axis. Therefore, by Sect. 2.3 (ii), it suffices to prove (6.13.1) for all real numbers z ≥ z 0 where z 0 > 0 is sufficiently large. By Theorem 6.21 (a) and the Euler formula (6.9.11), we have ∞

z − 1 −1 1 z−1 1+ 1+ (z) = (z − 1)(z − 1) = ν ν ν=1 N z − 1 −1 1+ = lim (N + 1)z−1 . N →∞ ν ν=1 Since (z) has no zero and it has pole at zero and at negative integer and none of the term in the above product vanishes, we derive that log (z) = lim

N →∞

N

ν+z−1 log (z − 1) log N − ν ν=1

.

(6.13.3)

Now we apply the Euler–Maclaurin–Jacobi formula (6.12.1) to the sum on the right-hand side of (6.13.3) with a = 1, b = N , f (x) = log(x + z − 1) − log x and q = 2m. By observing f (r ) (x) = (−1)r −1 (r − 1)!

1 1 − r (x + z − 1)r x

for x ∈ [a, b] and 1 ≤ r ≤ 2m, we see that the assumptions required for the validity of (6.12.1) are satisfied. Hence we conclude N ν=1

log

ν+z−1 ν

= log z + = log z +

N

log

ν=2

N 1

− log z) +

ν+z−1 ν

(log(x + z − 1) − log x)d x +

2m j=1

B2 j (2 j − 1)2 j

1 (log(N + z − 1) − log N 2

1 1 1 − 2 j−1 − 2 j−1 + 1 (N + z − 1)2 j−1 N z

N 1 1 1 dx B2m (x − [x]) − + 2m 1 (x + z − 1)2m x 2m

(6.13.4)

since B1 = − 21 and B3 = B5 = · · · = B2m−1 = 0. Integrating by parts, we have for the second term on the rightmost side of (6.13.4)

6.13 The Stirling Formula

N

173

log(x + z − 1)d x = (N + z − 1) log(N + z − 1) − z log z − N + 1

1

and

N

log x d x = N log N − N + 1.

1

Further the third term on the rightmost side of (6.13.4) tends to − 21 log z as N → ∞. Next the fourth and the fifth terms on the rightmost side of (6.13.4) contribute K m (z) + L m where L m is independent of z as N approaches infinity since

∞ 1

B2m (x − [x]) dx = (x + z − 1)2m

∞ 0

B2m (x − [x]) d x. (x + z)2m

Therefore, we derive from (6.13.3) and (6.13.4) that log (z) = A + lim B(N ), n→∞

1 log z + K m (z) + L m A= z− 2

where

and B(N ) = −(N + z − 1) log(N + z − 1) + (N + z − 1) log N z−1 = −(N + z − 1) log 1 + N satisfying lim B(N ) = −z + 1.

N →∞

Hence

1 log z − z + K m (z) + L m , log (z) = z − 2

(6.13.5)

where L m = L m + 1. Since lim K m (z) = 0, we have z→∞

1 log z + z = L m . log (z) − z − z→∞ 2 lim

It suffices to prove that L m = (6.13.5).

1 2

(6.13.6)

log 2π and then the assertion (6.13.1) follows from

174

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

By putting z = N + 1 and using (N + 1) = N ! by Corollary 6.22 (ii), we have 1 1 log(N + 1) + N + 1 = lim log N ! − N + log N + N log N ! − N + 2 2 N →∞ N →∞

L m = lim

since log(N + 1) = log N + O(1/N ). Therefore lim

N →∞

By putting z =

N! = e Lm . N N N 1/2 e−N

(6.13.7)

1 in both sides of (6.9.5), we get 2 ∞ 1 −1 π 1− 2 = . 4n 2 n=1

Thus N π 4n 2 42N (N !)4 = lim = lim N →∞ N →∞ ((2N )!)2 (2N + 1) 2 (2n − 1)(2n + 1) n=1

(6.13.8)

by rewriting N

(2n − 1)(2n + 1) =

n=1

(2N )!(2N + 1) (2N )!(2N + 1) = . (2 · 4 · · · 2N )2 4 N (N !)2

Now, we derive from (6.13.8) and (6.13.7) that π 42N N 4N e−4N N 2 e4L m 1 = lim = e2L m , N →∞ (2N )4N e−4N 2N (2N + 1)e2L m 2 4 which implies that L m =

1 2

log(2π).

Next we turn to asymptotic behaviour of (z). For this, we shall use inverse trigonometric function arctanx for x > 0. We recall that this function satisfies (2.7.9) and 1 π arctan(x) = − arctan 2 x for 0 < x ≤ 1. We begin with an estimate for |K 1 (z)|. Lemma 6.26 Let z = σ + it with z = 0 such that either σ > 0, t = 0 or t = 0. Then 1 if σ > 0, t = 0 , |K 1 (z)| ≤ 8σ1 |t| arctan if t = 0 8|t| σ

6.13 The Stirling Formula

175

where 0 ≤ arctan

|t| = |argz| < π. σ

Proof By (6.13.2) with m = 1, we have K 1 (z) = −

1 2

∞ 0

B2 (x − [x]) B2 1 dx + =− 2 (x + z) 2z 2

0

∞

B2 (x − [x]) − B2 d x. (x + z)2

By (6.11.11) and (6.11.6), we get B2 (x) − B2 = x 2 − x. Therefore |B2 (x − [x]) − B2 | ≤ and hence |K 1 (z)| ≤

1 8

∞ 0

1 , 4

dx . (σ + x)2 + t 2

We may assume that t > 0. Let σ > 0. By putting σ + x = t tan θ, the integral is equal to

π t 1 2 1 dθ = arctan t π2 −arctan σt t σ and the assertion follows. The proof for the case σ ≤ 0 is similar.

Sharper bounds can be obtained by estimating |K m | for larger values of m. We derive from Theorem 6.25 and Lemma 6.26 the estimates for |(z)| in a vertical strip and these are very useful in applications. Corollary 6.27 Let a ≤ b be fixed real numbers and z = σ + it with a ≤ σ ≤ b and |t| ≥ 1. Then (i) √ 1 π 1 πi 1 , (z) = 2π e− 2 |t| |t|σ− 2 ei|t|(log |t|−1) e 2 (σ− 2 ) 1 + O |t| √ 1 − π2 |t| σ− 21 (ii) |(z)| = 2πe 1+O , |t| |t| π

1 (iii) |(z)| = O e− 2 |t| |t|σ− 2 and π

1 1 = O e 2 |t| |t| 2 −σ . (iv) |(z)| Proof Since (z) = (z), we may suppose that t ≥ 1. Further we may assume that t exceeds sufficiently large number depending only on a and b, otherwise Corollary 6.27 follows. We prove (i) which implies immediately (ii), (iii) and (iv).

176

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

By Theorem 6.25 and Lemma 6.26, we have 1 1 1 t log(2π) + (σ + it − ) log(σ + it) − (σ + it) + θ arctan , 2 2 8t σ (6.13.9) where |θ| ≤ 1. The second term on the right-hand side of (6.13.9) is equal to log (z) =

1 t σ − + it , σ 2 + t 2 + i arctan log 2 σ where 1 1 σ2 1 2 2 log( σ + t ) = log t + log 1 + 2 = log t + O 2 2 2 t t and

π σ π σ t arctan = − arctan = − + O σ 2 t 2 t

1 t3

.

Further the last term on the right-hand side of (6.13.9) is equal to θ

1 π σ

1 − arctan =O . 8t 2 t t

Therefore log (z) =

πi 1 log(2π) + 2 2

and hence (z) =

1 π 1 1 σ− − t+ σ− log t + it (log t − 1) + O , 2 2 2 t

√ 1 πi 1 π 1 . 2πe 2 (σ− 2 ) e− 2 t t σ− 2 eit (log t−1) 1 + O t

Now we give lower and upper bounds for (z) which are valid for all z > 0 and they are useful in number theory. We prove the following. Theorem 6.28 For z > 0, we have

2π z z 2π z z 1 ≤ (z) ≤ e 12z . z e z e

(6.13.10)

For z > 0, we observe that (6.13.10) is equivalent to 1 1 −1 1 log z + z − log(2π) ≤ z 0 ≤ log (z) − z − 2 2 12

(6.13.11)

by taking logarithms in (6.13.10). The proof depends on the following identities which we shall prove first and then give a proof of Theorem 6.28.

6.13 The Stirling Formula

177

Lemma 6.29 For n ≥ 0, we have (i) (z + n)−2 = (z + n)−1 − (z + n + 1)−1 + (z + n)−2 (z + n + 1)−1 . (ii) (z + n)−2 (z + n + 1)−1 = 21 (z + n)−2 − 21 (z + n + 1)−2 + 21 (z + n)−2 (z + n + 1)−2 . (iii) (z + n)−2 (z + n + 1)−2 = 13 (z + n)−3 − 13 (z + n + 1)−3 − 13 (z + n)−3 (z + n + 1)−3 . Proof (i) We have (z + n)−2 − (z + n)−2 (z + n + 1)−1 = (z + n)−2 (1 − (z + n + 1)−1 ) = (z + n)−1 (z + n + 1)−1 = (z + n)−1 − (z + n + 1)−1 .

(ii) We have 1 1 (z + n)−2 (z + n + 1)−2 = (z + n)−2 (z + n + 1)−2 (2z + 2n + 1) 2 2 1 = (z + n)−2 (z + n + 1)−2 ((z + n + 1)2 − (z + n)2 ) 2 1 1 = (z + n)−2 − (z + n + 1)−2 . 2 2

(z + n)−2 (z + n + 1)−1 −

(iii) By multiplying both sides by 3(z + n)3 (z + n + 1)3 , we prove 3(z + n + 1)(z + n) = (z + n + 1)3 − (z + n)3 − 1. The right-hand side is equal to (z + n + 1)2 + (z + n)2 + (z + n)(z + n + 1) − 1 = (z + n + 2)(z + n) + (z + n)2 +(z + n + 1)(z + n) = 3(z + n + 1)(z + n).

Proof of Theorem 6.28 The proof depends on (6.9.7) which we rewrite as ∞

d2 log (z) = (z + n)−2 > 0. dz 2 n=0 By Lemma 6.29 (i), the sum in (6.13.12) is equal to z −1 +

∞ (z + n)−2 (z + n + 1)−1 . n=0

(6.13.12)

178

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

Now we apply Lemma 6.29 (ii) to each term of the above sum. We get ∞ ∞ 1 1 (z + n)−2 (z + n + 1)−1 = z −2 + (z + n)−2 (z + n + 1)−1 . 2 2 n=0 n=0

Thus

∞

d2 1 −2 1 −1 z + log (z) = z + (z + n)−2 (z + n + 1)−2 , dz 2 2 2 n=0

and hence

d2 1 log (z) ≥ z −1 + z −2 2 dz 2

since z > 0. Further, by Lemma 6.29 (iii), we have ∞

d2 1 1 1 log (z) = z −1 + z −2 + z −3 − (z + n)−3 (z + n + 1)−3 2 dz 2 6 6 n=0 1 1 ≤ z −1 + z −2 + z −3 . 2 6 Combining the above two inequalities, we get 0≤

1 1 d2 log (z) − z −1 − z −2 ≤ z −3 . 2 dz 2 6

Let F(z) =

1 d log (z) − log z + z −1 dz 2

so that 0 ≤ F (z) ≤

(6.13.13)

1 −3 z . 6

This implies that F is non-decreasing. Further, by integrating from z 0 to z with z > z 0 > 1, we have

z

1 z −3 F (ζ)dζ ≤ ζ dζ. 0≤ 6 z0 z0 Thus 0 ≤ F(z) − F(z 0 ) ≤

1 −2 1 −2 (z − z −2 ) ≤ z . 12 0 12 0

6.13 The Stirling Formula

179

Therefore, F(z) for z > 1 is bounded above and hence lim F(z) = c

z→∞

exists. Further, by letting z tend to infinity and taking z 0 = z, we have −

1 −2 z ≤ F(z) − c ≤ 0. 12

1 log z + z − cz g(z) = log (z) − z − 2

(6.13.14)

Now we define

(6.13.15)

so that we see from (6.13.13) that g (z) = F(z) − c. Further we derive from (6.13.14) that −

1 −2 z ≤ g (z) ≤ 0, 12

which implies as above, lim g(z) = c1

z→∞

exists and 0 ≤ g(z) − c1 ≤

1 −1 z . 12

(6.13.16)

(6.13.17)

Next we consider z+1 1 log + 1 − c. g(z + 1) − g(z) = − z + 2 z

By letting z tend to infinity on both sides, we derive that c = 0. Now

1 g(2z) − g(z) − g z + 2

1 1 = log log 2 − 2z − log z − 2z − 2 2 (z)(z + 21 ) 1 1 1 log z + z log z + − + z− 2 2 2 1 z + 21 1 2 2 −2z (2z) + z log − . = log 1 z 2 (z)(z + ) (2z)

2

180

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

By letting z tend to infinity on both sides, we conclude from (6.13.16) and Theorem 6.21 (c) that −c1 = lim

z→∞

1 1 g(2z) − g(z) − g z + = − log(2π), 2 2

and hence c1 =

1 log(2π). 2

(6.13.18)

Finally, we combine (6.13.15) with c = 0, (6.13.17) and (6.13.18) to conclude (6.13.11).

6.14 The Beta Function Let a and b be complex numbers such that Re(a) > 0 and Re(b) > 0. Then the beta function

1

β(a, b) =

t a−1 (1 − t)b−1 dt

a

converges. By substituting t =

u 1+u

in the above integrand, we get

∞

β(a, b) =

u a−1 du. (1 + u)a+b

0

(6.14.1)

A close connection between the beta function and the gamma function is given by β(a, b) =

(a)(b) . (a + b)

Proof For a complex number z with Re(z) > 0 and positive real number r , we have

∞

e

−r t

0

t

z−1

1 dt = z r

∞

e−s t z−1 ds =

0

(z) . rz

(6.14.2)

By putting r = 1 + u with u > 0 and z = a + b on both sides in (6.14.2), we get

∞ 0

e−(1+u)t t a+b−1 dt =

(a + b) . (1 + u)a+b

(6.14.3)

Now, by (6.14.1), (6.14.3), (6.14.2) and the Fubini theorem, see Lemma 2.9, we conclude

6.14 The Beta Function

181

∞

∞ 1 u a−1 e−(1+u)t t a+b−1 dt du (a + b) 0 0

∞

∞ 1 −t a+b−1 = e t dt e−ut u a−1 du (a + b) 0 0

∞ 1 (a) = e−t t a+b−1 a dt (a + b) 0 t

∞ (a) (a)(b) = e−t t b−1 dt = . (a + b) 0 (a + b)

β(a, b) =

6.15 Exercises 6.1 Show that an absolutely convergent infinite product is convergent. Further prove that rearrangement of terms in an absolutely convergent infinite product is permissible and its value remains unchanged. 6.2 Show that ∞ 1 1 1− 2 = . (a) n 2 n=2 ∞

1 (b) = 2. 1+ n(n + 2) n=1 ∞

1 n (c) for |z| < 1. 1 + z2 = 1 − z n=0 ∞ |z n |2 < ∞. Then show that cos z n converges absolutely. n=1 n=1

∞ n−1 1 + (−1) converges in Re(z) > 1/2 and 6.4 Prove that the infinite product nz

6.3 Assume that

∞

n=1

converges absolutely in Re(z) > 1. 6.5 (a) Let an = 1 +

i n

for n ≥ 1. Show that

n (−1) √ n

i

|an | converges but

n=1

converge. (b) Let an = e

∞

. Show that

∞

(c) Give an example of a sequence {an }∞ n=1 such that (Hint: (c) Take an = 1 +

an does not

n=1

an converges but not absolutely.

n=1

does not converge.

∞

(−1)n i for n ≥ 1.) √ n

∞ n=1

(an − 1) < ∞ but

∞ n=1

an

182

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

z − n − n12 6.6 Show that is analytic in C\Z>0 . z−n n=1 ∞ 6.7 Suppose that {an }∞ lim |an | = ∞, n=1 and {bn }n=1 are sequences such that n→∞ ∞ ∞ z − an is analim |bn | = ∞ and |an − bn | < ∞. Then show that n→∞ z − bn n=1 n=1 lytic in C\{bn }∞ n=1 . ∞ ∞ (1 − 6.8 Let {αn }n=1 be a sequence in D(0, 1) such that αn = 0 for n ≥ 1 and ∞

n=1

αn ) < ∞. For a non-negative integer k, let ∞ αn − z |αn | B(z) = z . 1 − α¯ n z αn n=1 k

Then prove that B(z) is analytic in D(0, 1) where |B(z)| < 1. Further B(z) has no zero except at the point αn with n ≥ 1 and at the origin when k > 0. The function B(z) is called the Blaschke product. (Hint: We may assume that k = 0 and write B(z) =

∞

f n (z). For 0 < r < 1,

n=1

show that | f n (z) − 1| ≤ C(1 − |αn |) where C is a constant and |B(z)| = 1 for |z| = 1.) 6.9 Prove the factorisation z

e z − 1 = ze 2

∞

1+

n=1

z2

. 4π 2 n 2

6.10 Show that 2

∞ 1 z 1 1− . = (s − z)(s + z) (s)2 n=0 n+s Derive cos πz =

∞ n=0

⎛ ⎝1 −

z n+

2 ⎞ 1 2

⎠.

(Hint: Apply Hadamard’s factorisation theorem.) 6.11 For a region , show that H () is an integral domain and M() is the quotient field of H (). (Hint: Apply that the zeros of non-zero analytic functions are isolated and identity theorem for holomorphic functions, see Sect. 2.3 (ii).)

6.15 Exercises

183

6.12 Let f be an entire function of finite order. Then derive from the Hadamard factorisation theorem that f assumes each complex number with one possible exception. Thus, Hadamard’s factorisation theorem implies the Little Picard theorem for functions of finite order. 6.13 For an integer k ≥ 1, prove that ∞ 1 (2π)2k B2k . = (−1)k+1 2k n 2(2k)! n=1

Further derive ∞ ∞ ∞ 1 π2 1 π4 1 π6 = = = , , . n2 6 n=1 n 4 90 n=1 n 6 945 n=1 ∞ 1 is irrational. It has been Remark Since π is irrational, we see that 2k n n=1 ∞ 1 conjectured that with k ≥ 1 is irrational. It remains unproved for 2k+1 n n=1 k ≥ 2 but it has been proved in the case k = 1 by Apéry [2].

(Hint: By taking logarithmic derivative to factorisation of have ∞ ∞ z 2k z cot z = 1 − 2 . (πn)2k n=1 k=1 Also

sin z z

(see (6.8.1)), we

∞ B2k 2k z z cot z = 1 − (−1)k+1 4k (2k)! k=1 ∞

by putting x = 2i z in

xr x = and using B2r +1 = 0 for r ≥ 1 and Br x e −1 r! r =0

B1 = −1/2.) 6.14 Let sk (n) = 1k + 2k + · · · + (n − 1)k . Then prove that (k + 1)sk (n) =

k k+1 i=0

i

Bi n k+1−i for k ≥ 1,

where Bi is ith Bernoulli number. Remark It has been extensively studied when sk (n) is a power, see [[26], p. 182].

184

6 The Weierstrass Factorisation Theorem, Hadamard’s Factorisation …

6.15 Prove that

∞

e−t log t dt = −γ,

0

where γ is the Euler constant.

(1) = −γ derived from (6.9.3) by taking logarithmic derivatives (Hint: Use (1) on both sides.) 6.16 Show that

∞ ∞ (−1)n + e−t t z−1 dt for all (z) = n!(z + n) 0 n=1

z ∈ C.

6.17 Show that

∞ 1 1 (a) for α > 0 and exp(−x α )d x = α α

0 ∞ 2 (b) e−πx d x = 1. −∞

6.18 (Gauss multiplication theorem) For a positive integer n, show that (n−1) n−1 1 1 ... z + = (2π) 2 n 2 −nz (nz). (z) z + n n

(6.15.1)

(Hint: Equation (6.15.1) with n = 2 is given in Theorem 6.21 (c) and the proof is similar. Consider φ(z) =

n nz (z)(z + n1 ) . . . (z + n(nz)

n−1 ) n

.

Show by the Euler formula that φ(z) is constant. Now evaluate (φ(z))2 at z = 1/n by Theorem 6.21 (b).) 6.19 (a) Suppose f and g are entire functions with no common zeros. Then there exist entire functions A and B such that A f + Bg = 1. (b) Suppose f and g are entire functions. Then there exists entire function h such that f = h f 1 and g = hg1 and f 1 , g1 have no common zeros. (Hint: By Theorem 6.17, construct a meromorphic function M satisfying the following: (i) The principle part of M exists only at the zeros of g. 1 is identical at every zero of g.) (ii) The principle part of M and of fg 6.20 (a) Show that ∞ 1 (−1)n π = + 2z . sin(πz) z z2 − n2 n=1

6.15 Exercises

185

(b) Show that

∞

(−1)n (2n − 1) π = . cos(πz) z 2 − (n − 21 )2 n=1

/ R and ω = ωm,n = mω1 + nω2 . Find a mero6.21 Let ω1 , ω2 ∈ C such that ωω21 ∈ morphic function which has a double pole and principal part (z−ω1m,n )2 at every ωm,n with m, n ∈ Z. (Hint: We consider ℘ (z) =

1 1 1 + − 2 ) 2 2 z (z − ω ) ω m,n m,n (m,n)=(0,0)

This is known as the Weierstrass elliptic function. For ω = ωm,n with |ω| > 2|z|, 6|z| we estimate that the absolute value of the term in the series is at most |ω| 3 . Therefore, the series converges uniformly on compact subsets of C not containing 1 < ∞ and this follows from Exercise 1.2.) any ωm,n if |ωm,n |3 (m,n)=(0,0) Remark This is very important function in number theory. Like an exponential function, it also satisfies addition theorem and differential equation. This led to transcendental results for values of this function analogous to those of exponential equation. 6.22 Let f be a non-zero entire function of order w < ∞, τ be the exponent of convergence of the sequence of absolute values of zeros of f and h be the degree of the polynomial in the factorisation (6.5.5) of f . Then show that ω = max(τ , h), the existence of either side implies that of the other. (Hint: It suffices to show that ω ≤ max(τ , h). For this, estimate | f (z)| in (6.5.5) to show that there exists a constant C such that max | f (z)| < C(r h + r θ ) |z|=r

as r → ∞ for every θ > τ .)

Chapter 7

The Riemann Zeta Function and the Prime Number Theorem

7.1 Introduction For a complex number s, we always denote its real part by σ and imaginary part by t. Thus s = σ + it. The Riemann Zeta function is defined as ζ(s) =

∞ 1 in σ > 1. s n n=1

In Sect. 7.2, we show that it has the Euler product implying that it has no zero in σ > 1. In Sect. 7.3, we prove that log ζ(s) is analytic in σ > 1 where logarithm has (s) its principal branch, and further an integral representation for ζζ(s) in σ > 1 is given in Sect. 7.4. We give the analytic continuation of ζ(s) in σ > 0 in Sect. 7.5 where it is shown that ζ(s) has no zero on the line σ = 1. The Prime Number Theorem is stated in Sect. 7.7 where its equivalent versions in terms of ϑ(x) and ψ(x) have also been given. Assuming the Wiener–Ikehara theorem, the equivalence of the Prime Number Theorem and the non-vanishing of ζ(s) on the line σ = 1 is established in Sect. 7.8. Further the Wiener–Ikehara theorem is proved in Sects. 7.9 and 7.10. Thus, we have now a complete proof of the Prime Number Theorem which was proved for the first time by J. Hadamard and de la Vallée Poussin, independently, in 1896. We shall prove the Prime Number Theorem with error term by following their classical proofs in the next chapter. In Sect. 7.11, it is shown that ζ(s) has analytic continuation in the whole complex plane satisfying a functional equation. Further we derive from the functional equation some results on the location of zeros of ζ(s) and finally it is proved in this section that it has infinitely many zeros. In Sect. 7.13, we state well-known conjectures the Riemann hypothesis, the Lindelöf hypothesis and the Density hypothesis on ζ(s) and we prove that the Riemann hypothesis implies the Lindelöf hypothesis. In Sect. 7.12, we introduce well-known function μ(σ) the 1 1 ≤ which is equivalent to related to the Lindelöf hypothesis and show that μ 2 4 © Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9_7

187

188

7 The Riemann Zeta Function and the Prime Number Theorem

1 1 + it = O(|t| 4 + ) for > 0. We refer to [3, 6, 11, 17, 18, 29, 32, 33] for the 2 topics in this chapter and for further studies and related topics.

ζ

7.2 The Euler Product for ζ(s) We recall the Riemann Zeta function ζ(s) = Here

∞ 1 where s = σ + it with σ > 1. s n n=1

n −s = n −σ n −it = n −σ e−it log n ,

where logarithm has principal value. We observe that |n −s | = n −σ . Therefore ∞ ∞ 1 du |ζ(s)| ≤ ≤1+ σ n uσ 1 n=1 and for δ > 0 with σ ≥ 1 + δ, we have

∞

|ζ(s)| ≤ 1 + 1

du . u 1+δ

Hence, the series for ζ(s) converges uniformly in σ ≥ 1 + δ. Further it follows from Theorem 1.4 that every compact set in σ > 1 is contained in σ ≥ 1 + δ for some δ > 0. Therefore, the series for ζ(s) converges uniformly on compact subsets of σ > 1. Since the terms of the series are analytic in C, we derive from Sect. 2.3 (iv) that ζ(s) is analytic in σ > 1. Euler considered ζ(s) for real s and used it for showing that 1 = ∞, p p where the sum is taken over all primes. In particular, there are infinitely many primes as proved by Euclid. It was Riemann who considered ζ(s) as a function of complex variable and it turned out to be central in the studies of the Prime Number Theory and in fact in several important directions of mathematics. The connection of ζ(s) with primes is given by the following theorem.

7.2 The Euler Product for ζ(s)

189

Theorem 7.1 (The Euler identity) ζ(s) =

1 −1 in σ > 1, 1− s p p

(7.2.1)

where the infinite product on the right-hand side is absolutely convergent. By Corollary 6.6, we have Corollary 7.2 ζ(s) = 0 in σ > 1. We shall derive Theorem 7.1 from a more general result which will also be applied for deriving analogous result for L-functions. A complex-valued function defined on the set N of all positive integers is called an arithmetic function. An arithmetic function f is called multiplicative if f (mn) = f (m) f (n), whenever m and n are positive integers such that (m, n) = 1. If the above relation holds for all positive integers m and n, then f is called completely multiplicative. Theorem 7.3 Let a(n) be a multiplicative arithmetic function and assume that ∞

|a(n)| < ∞.

(7.2.2)

n=1

Then

∞

a(n) =

(1 + a( p) + a( p 2 ) + · · · ), p

n=1

where the product is taken over all primes p. Further the infinite product on the right-hand side is absolutely convergent. Proof For N ≥ 1, let PN =

N

(1 + a( pi ) + a( pi2 ) + · · · ),

i=1

where pi denotes the ith prime. We multiply all the terms in the product on the right-hand side of PN and collect the terms obtained. Since a(n) is multiplicative, these are of the form a( p1ν1 ) · · · a( pnνn ) = a( p1ν1 p2ν2 · · · pnνn ), where νi ≥ 0 areintegers. By the fundamental theorem of arithmetic, PN = a(n), where the sum is taken over all positive integers n whose greatest prime factor

190

7 The Riemann Zeta Function and the Prime Number Theorem

≤ PN . This is permissible since PN is a finite product of series which are absolutely ∞ convergent by (6.2.2). Thus a(n) − PN = a(n), where the sum is taken n=1

over all positive integers n such that n is divisible by at least one prime exceeding N . In fact such an n itself exceeds N and hence ∞ a(n) − PN ≤ |a(n)|. n≥N +1

n=1

Let > 0. By (7.2.2), there exists N0 = N0 () > 0 such that

|a(n)| < for N ≥ N0 ,

n≥N +1

∞ a(n) − PN < for N ≥ N0 .

and thus

n=1

Hence

∞

a(n) =

(1 + a( p) + a( p 2 ) + · · · ),

(7.2.3)

p

n=1

where the product is taken over all the primes p. For showing that the infinite product on the right-hand side of (7.2.3) is absolutely convergent, we consider s( p) = a( p) + a( p 2 ) + · · · . We observe that

|s( p)| ≤

p≤x

by (7.2.2). Since

∞ (|a( p)| + |a( p 2 )| + · · · ) ≤ |a(n)| < ∞ p≤x

n=1

|s( p)| is a series of positive terms with all its partial sums

p

uniformly bounded, we derive that

|s( p)| < ∞,

p

and hence the product on the right-hand side of (7.2.3) is absolutely convergent by Lemma 6.3 with 1 + s( p) if n = p prime, zn = 1 otherwise.

7.2 The Euler Product for ζ(s)

191

We derive from Theorem 7.3 its analogue for completely multiplicative functions. Corollary 7.4 Let a(n) be a completely multiplicative function satisfying (7.2.2). Then ∞ a(n) = (1 − a( p))−1 , p

n=1

where the product is absolutely convergent. Proof By Theorem 7.3, we have ∞

a(n) =

(1 + a( p) + a( p 2 ) + · · · ) =

p

n=1

Since

∞

(1 + a( p) + (a( p))2 + · · · ).

p

|a( pr )| < ∞ by (7.2.2), we observe that |a( p)| < 1 since a(n) is com-

r =1

pletely multiplicative and hence ∞

a(n) =

(1 − a( p))−1

p

n=1

such that the product is absolutely convergent by Lemma 6.3.

Proof of Theorem 7.1 Let σ > 1 and a(n) = n −s . We observe that a(n) is completely multiplicative and ∞

|a(n)| =

n=1

∞

n −σ < ∞.

n=1

Now (7.2.1) follows from Corollary 7.4.

As already mentioned, the proof of Theorem 7.3 and hence of Theorem 7.1 depends on the fundamental theorem of arithmetic. In fact, Theorem 7.1 is equivalent to the fundamental theorem of arithmetic and it may be viewed as an analytic analogue of the fundamental theorem of arithmetic. The left-hand side of the Euler identity is the value of an analytic function ζ(s), whereas the right-hand side is an infinite product taken over all primes. This leads us to study Prime Number Theory using the Riemann Zeta function, thus laying the foundation of the analytic number theory, i.e. applications of analysis to certain problems in number theory. Example 7.1 We have ∞

μ(n) 1 in σ > 1, = ζ(s) ns n=1

192

7 The Riemann Zeta Function and the Prime Number Theorem

where μ(n) is the Möbius function given by ⎧ ⎨ 1 if n = 1 0 if there exists a prime p with p 2 |n μ(n) = ⎩ (−1)k if n = P1 · · · Pk where P1 , . . . , Pk are distinct primes. Solution. By Theorems 7.1 and 7.3 with a(n) =

μ(n) , we have ns

∞ 1 1 μ(n) 1− s = = . ζ(s) p ns p n=1

7.3 Applications of the Euler Product for ζ(s) As stated earlier in the beginning of Sect. 7.2, we give a proof of Euler on the infinitude of primes. More precisely, we prove the following. Theorem 7.5

1 = ∞, p p

where the sum is taken over all primes. Proof Let σ > 1. By taking logarithms on both the sides in (7.2.1), we have log ζ(σ) =

p

We observe that 0

1 and therefore log ζ(σ) tends to infinity as σ tends to 1+ . Hence, the assertion follows from (7.3.1). Remark Let a > 0 and b > 0 be integers such that (a, b) = 1. For showing p≡b (mod a)

1 =∞ p

on the lines of above proof, we need to consider more general series than ζ(s), namely, L-functions and their values at s = 1. We shall turn to them in Chap. 9. For a complex number z, let σz (n) = d z where the sum is taken over positive d|n

divisors of n. We prove the following. Theorem 7.6 (The Ramanujan identity) For a, b ∈ C and σ > max(1, 1 + Re(a), 1 + Re(b), 1 + Re(a + b)), we have

(7.3.2)

∞

ζ(s)ζ(s − a)ζ(s − b)ζ(s − a − b) σa (n)σb (n) = . ζ(2s − a − b) ns n=1

(7.3.3)

Proof First, we consider the right-hand side of the above identity. Since σ > 1 + Re(a + b) by (7.3.2), there exists > 0 such that σ > 1 + Re(a + b) + 3. Further we observe that there exists n 0 = n 0 () > 0 such that for n ≥ n 0 |σa (n)| ≤ n Re(a) d(n) < n Re(a)+ , where d(n) denotes the number of positive divisors of n. Similarly |σb (n)| < n Re(b)+ . Here we have used d(n) < n for n ≥ n 0 , see Exercise 7.2. Therefore, for n ≥ n 0

194

7 The Riemann Zeta Function and the Prime Number Theorem

σa (n)σb (n) n Re(a+b)+2 1 ≤ < 1+ . ns nσ n Thus, the series on the right-hand side in (7.3.3) is absolutely convergent. We check b (n) is multiplicative, see Exercise 7.3. Then we derive from Theorem 7.3 that σa (n)σ ns σa (n)σb (n) with a(n) = that the right-hand side of (7.3.3) is equal to ns

∞ σa ( p m ) σb ( p m ) . (7.3.4) p ms p m=0 It remains to show that the above product (7.3.4) is equal to the left-hand side of (7.3.3). We observe from (7.3.2) that 2σ > 1 + Re(a + b). This is clear if 1 + Re(a + b) ≤ 0, otherwise 2σ > 2(1 + Re(a + b)) > 1 + Re(a + b) by (7.3.2). Then we derive from Theorem 7.1 that the left-hand side of (7.3.3) is equal to p

(1 − p −2s+a+b) . (1 − p −s )(1 − p −s+a )(1 − p −s+b )(1 − p −s+a+b )

By writing p −s = z, the above product equals p

(1 − pa+b z 2 ) A B C D = + + + , (1 − z)(1 − pa z)(1 − p b z)(1 − pa+b z) 1−z 1 − pa z 1 − pb z 1 − pa+b z

where A= C=

,

B=

− pa , (1 − pa )(1 − p b )

− pb , (1 − pa )(1 − p b )

D=

pa+b . (1 − pa )(1 − p b )

1 (1 −

pa )(1

−

pb )

Thus, the left-hand side of (7.3.3) is equal to p

1 (1 − pa )(1 − p b )

1 pa pb pa+b − − + a b 1−z 1− p z 1− p z 1 − pa+b z

.

Since z = p −s , we have max(|z|, | pa z|, | p b z|, | pa+b z|) = max( p −σ , pa−σ , p b−σ , pa+b−σ ) < 1 by (7.3.2). Therefore, the above product is equal to p

∞ 1 1 − p (m+1)a − p (m+1)b + p (m+1)(a+b) z m , a b (1 − p )(1 − p ) m=0

which is equal to (7.3.4) since σa ( p m ) =

p(m+1)a −1 , pa −1

σb ( p m ) =

p(m+1)b −1 pb −1

and

7.3 Applications of the Euler Product for ζ(s)

195

1 − p (m+1)a − p (m+1)b + p (m+1)(a+b) = 1 − p (m+1)a 1 − p (m+1)b . For n ≥ 1, let (n) =

0 if n is not a prime power or n = 1 log p if n = p m .

(7.3.5)

Then Theorem 7.7 We have log ζ(s) =

∞ p

m=1

1 mp ms

is analytic in σ > 1 and ∞

−

ζ (s) (n) = in σ > 1, ζ(s) ns n=1

where the logarithm has principal branch. Proof Let L(s) =

p

Now

∞

1 ms mp m=1

in σ > 1.

∞ ∞ 1 1 1 1 ≤ + + · · · ≤ < ∞. mp ms σ 2σ σ p p n p p m=1 n=1

Thus, the above series converges absolutely in σ > 1 and therefore rearrangement of terms is permissible. Further it converges uniformly in σ ≥ 1 + δ for every δ > 0. Then L(s) converges uniformly on compact subsets of σ > 1. Therefore, L(s) is analytic function in σ > 1, see Sect. 2.3 (iv). Moreover, ζ(s) = e L(s) in σ > 1 and L(2) = log ζ(2). Therefore, L(s) = log ζ(s) by Theorem 2.28 where logarithm has principal value. Hence ∞ 1 log ζ(s) = ms mp p m=1 is analytic in σ > 1. Further we write A(n) = Thus

if n = p m 0 if n is not a prime power. 1 m

(7.3.6)

196

7 The Riemann Zeta Function and the Prime Number Theorem

log ζ(s) =

∞ A(n) in σ > 1. ns n=1

Since the series converges uniformly in σ ≥ 1 + δ for every δ, term-wise differentiation is permissible. Therefore ∞ A(n) log n ζ (s) in σ > 1 =− ζ(s) ns n=1

by Sect. 2.3 (iv). Hence, by (7.3.6) and (7.3.5), we have ∞

−

ζ (s) (n) = in σ > 1. ζ(s) ns n=1

(7.3.7)

7.4 The Abel Summation Formula and Integral (s) Representations for ζζ(s) First we state The Abel summation formula. Let 0 ≤ λ1 ≤ λ2 ≤ · · · be a sequence of real numbers such that λn → ∞ as n → ∞ and let a1 , a2 , . . . be a sequence of complex numbers. For x ≥ 0, let an A(x) = λn ≤x

and f (x) be a complex-valued function. Then k

an f (λn ) = A(λk ) f (λk ) −

n=1

k−1

A(λn )( f (λn+1 ) − f (λn )).

(7.4.1)

n=1

If f has continuous derivative in [λ1 , ∞) and x ≥ λ1 , then

an f (λn ) = A(x) f (x) −

λn ≤x

x

λ1

A(t) f (t)dt.

(7.4.2)

Proof We write A(λ0 ) = 0 and k n=1

an f (λn ) =

k n=1

(A(λn ) − A(λn−1 )) f (λn ) = A(λk ) f (λk ) +

k−1 n=1

A(λn ) ( f (λn ) − f (λn+1 )) .

ζ (s) ζ(s)

7.4 The Abel Summation Formula and Integral Representations for

197

This proves the first assertion. Let k be the largest integer such that λk ≤ x. Then the sum in the right-hand side of (7.4.1) is equal to k−1

A(λn )

n=1

λn+1 λn

f (t)dt =

k−1

λn+1

λn

n=1

A(t) f (t)dt =

λk λ1

A(t) f (t)dt

since f has continuous derivative in [λ1 , ∞). Also A(λk ) f (λk ) = A(x) f (x) − (A(x) f (x) − A(λk ) f (λk )) = A(x) f (x) − Hence

an f (λn ) =

λn ≤x

k

x

λk

A(t) f (t)dt.

an f (λn ) = A(x) f (x) −

n=1

x λ1

A(t) f (t)dt.

(s) Integral representation for ζζ(s) We show ∞ ζ (s) ψ(t) − =s dt in σ > 1 ζ(s) t s+1 1

where for x ≥ 0 ψ(x) =

p

(7.4.3)

log p for x ≥ 0.

pm ≤x

Proof It suffices to use here the weak estimate for ψ(x) < 2x(log x)2 . This follows since the number of possibilities for p is at most x and for each p, there are at x < 2 log x possibilities for m. For all n ≥ 1, we apply the Abel summation most log log 2 an = ψ(x) and f (t) = t −s with σ > formula with λn = n, an = (n), A(x) = n≤x

1. Then we derive from (7.4.2) that x (n) ψ(t) ψ(x) = s +s dt. s n x t s+1 1 n≤x Letting x → ∞, we conclude from (7.3.7) that − since

ζ (s) =s ζ(s)

∞ 1

ψ(t) dt in σ > 1 t s+1

198

7 The Riemann Zeta Function and the Prime Number Theorem

ψ(x) 2x(log x)2 → 0 as x → ∞. xs ≤ xσ

7.5 Analytic Continuation of ζ(s) in σ > 0 and Its Non-vanishing on the Line σ = 1 Theorem 7.8 There exists analytic function H (s) in σ > 0 except at s = 1 where it has a simple pole with residue 1 and H (s) = ζ(s) in σ > 1. In fact H (s) =

s −s s−1

∞ 1

{u}du in σ > 0. u s+1

(7.5.1)

The function H (s) is called the analytic continuation of ζ(s) in σ > 0 and we shall denote it again by ζ(s). Proof Let σ > 1. By the Abel summation formula with an = 1, λn = n, A(x) = [x] and φ(u) = u −s , we see from (7.4.2) that the formula x x x 1 [u] du {u}du [x] 1 {x} = + s du = − + s − s s s s+1 s−1 s s n x x x u s+1 1 u 1 u 1 n≤x x 1 {u}du s {x} − = − − s . (7.5.2) s−1 s s − 1 (s − 1)x x u s+1 1 Letting x → ∞, we have ζ(s) = since

s −s s−1

1 1 x s−1 = x σ−1 ,

∞ 1

{u}du u s+1

(7.5.3)

{x} 1 xs < xσ

tend to zero as x tends to infinity. We show that the integral on the right-hand side of (7.5.1) is analytic in σ > 0. For x ≥ 1 and σ > 0, let Fx (s) = 1

We observe that

x

{u}du du. u s+1

7.5 Analytic Continuation of ζ(s) in σ > 0 and Its Non-vanishing on the Line σ = 1

Fx (s + h) − Fx (s) − h

−{u} log u du u s+1 1

x 1 1 − u s+1 − log u u s+h+1 − s+1 = {u} du. h u 1

199

x

(7.5.4)

Let > 0. There exists δ > 0 such that for |h| < δ, we have 1 − 1 − log u u s+h+1 u s+1 − s+1 < . h u x Therefore, the absolute value of (7.5.4) is less than whenever |h| < δ and hence

Fx (s) =

x

1

−{u} log u du in σ > 0. u s+1

Let δ > 0. For σ ≥ δ and y > x ≥ 1 with x, y ∈ N , we have |Fy (s)

−

Fx (s)|

=

y x

y ∞ {u} log u log udu du du ≤ ≤ →0 δ s+1 1+δ u u x x u 1+ 2

as x → ∞ uniformly in σ ≥ δ > 0. Thus, Fx (s) is a sequence of analytic functions such that ∞ −{u} log u du lim Fx (s) = x→∞ u s+1 1 converges uniformly in σ ≥ δ > 0 for every δ > 0. Therefore, by Sect. 2.3 (iv), we derive that ∞ {u} log u du u s+1 1 is analytic in σ > 0. Hence, we conclude from (7.5.1) that H (s) is analytic in σ > 0 except at s = 1 where it has simple pole with residue 1. Further H (s) = ζ(s) in σ > 1 by (7.5.3). We recall from Corollary 7.2 that ζ(s) = 0 in σ > 1. Now we are ready to prove Theorem 7.9 ζ(1 + it) = 0. Proof By Theorem 7.8, we may suppose that t = 0. Let σ > 1. By Theorem 7.7 log ζ(s) =

∞ p

Now

∞

cn 1 = , ms mp ns m=1 n=2

cn ≥ 0.

200

7 The Riemann Zeta Function and the Prime Number Theorem

cn cn cn = σ e−it log n = σ (cos(t log n) − i sin(t log n)) . ns n n Therefore Re Now

c n

ns

=

cn cos(t log n). nσ

∞ ∞ cn cn log |ζ(s)| = Re(log ζ(s)) = Re cos(t log n). = s σ n n n=2 n=2

Therefore log |ζ 3 (σ)ζ 4 (σ + it)ζ(σ + 2it)| = 3 log |ζ(σ)| + 4 log |ζ(σ + 2it)| + log |ζ(σ + 2it)| ∞ cn = (3 + 4 cos(t log n) + cos(2t log n)). nσ n=2

Further for real θ, we observe that 3 + 4 cos θ + cos 2θ = 3 + 4 cos θ + 2 cos2 θ − 1 = 2(cos θ + 1)2 ≥ 0. Therefore, since cn ≥ 0, we derive that |ζ 3 (σ)ζ 4 (σ + it)ζ(σ + 2it)| ≥ 1. We rewrite the above inequality as ζ(σ + it) 4 3 ζ(σ + 2it) (σ − 1) ≥ 1. ((σ − 1)ζ(σ)) σ−1 If ζ(1 + it) = 0, then the left-hand side of the above inequality tends to zero as σ → 1+ since ζ(s) has a simple pole at s = 1. This is a contradiction. Hence ζ(1 + it) = 0. The proof of Theorem 7.9 depends on the inequality 3 + 4 cos θ + cos 2θ ≥ 0 for real θ. In Chap. 9, we shall give a different proof which does not depend on this relation but depends on the Ramanujan identity given in Theorem 7.6.

7.6 Estimates for ζ(s) and ζ (s) We begin with an estimate for ζ(s) in a region asymptotically close to the line σ = 1. Lemma 7.10 Let α ≥ 0. There exist positive numbers u 1 = u 1 (α) and t1 ≥ 2 depending only on α such that for

7.6 Estimates for ζ(s) and ζ (s)

201

σ > σ0 = 1 −

α with t ≥ t1 , log t

(7.6.1)

we have |ζ(s)| < u 1 log t. Proof Assume (7.6.1). We may suppose that t1 ≥ 2 is sufficiently large depending only on α such that σ > σ0 > 21 . Further we may assume that σ ≤ 2, otherwise |ζ(s)| ≤

∞ 1 < ζ(2) nσ n=1

and the assertion follows. By subtracting (7.5.2) from (7.5.3), we have ∞ 1 1 {u}du {x} ζ(s) − = . + − s n s (s − 1)x s−1 xs u s+1 x n≤x Thus |ζ(s)| ≤

∞ 1 du 1 1 + + + (2 + t) σ0 σ0 −1 σ0 σ0 +1 n t x x u x n≤x

(7.6.2)

since 2 ≥ σ > σ0 . Further we estimate

∞

(2 + t) x

since σ0 >

1 2

du u σ0 +1

=

−σ0 x 2 +1 t ≤ 4t x −σ0 t σ0

(7.6.3)

and t ≥ 2. Let x = t. Then we see from (7.6.2) and (7.6.3) that |ζ(s)| ≤

1 2 + σ + 4t 1−σ0 . σ0 0 n t n≤t

(7.6.4)

By (7.6.1), we observe that t 1 du t 1−σ0 eα log t ≤ 1 + ≤ 1 + ≤ 1 + σ0 n σ0 1 − σ0 α 1 u n≤t and

2 t σ0

< 2,

4t 1−σ0 = 4eα .

Hence, we conclude from (7.6.4) that |ζ(s)| < u 1 log t where u 1 = u 1 (α) is a number depending only on α. Next we derive from Lemma 7.10 an estimate for ζ (s).

202

7 The Riemann Zeta Function and the Prime Number Theorem

Lemma 7.11 Let α ≥ 0. There exist positive numbers u 2 = u 2 (α) and t2 ≥ 2 depending only on α such that for σ > σ0 = 1 − we have

α with t ≥ t2 , log t

(7.6.5)

|ζ (s)| < u 2 (log t)2 .

Proof We may assume (7.6.5) with t2 sufficiently large number depending only on α. Let ρ = logα t and let s = σ + it with σ ≥ σ0 and t ≥ t2 . Then ζ (s) =

1 2πi

|z−s|=ρ

ζ(z) dz (z − s)2

by (2.3.2). Therefore, for |z − s| = ρ, we have |ζ (s)| ≤

1 2πρ M , M= 2π ρ2 ρ

(7.6.6)

where M = max |ζ(z)|. |z−s|=ρ

Further for estimating M, we observe in |z − s| = ρ that Re(z) ≥ Re(s) − ρ ≥ σ0 − ρ = 1 −

2α log t

and Im(z) ≥ Im(s) − ρ ≥ t2 − ρ ≥ t1 by taking t2 sufficiently large. Now we derive from Lemma 7.10 with α replaced by 2α that M ≤ u 3 log t with u 3 = u 3 (α). Hence, we conclude from (7.6.6) that |ζ (s)| ≤ u 3 (log t)2 . Lemma 7.12 Let 0 < δ < 1. Then there exists a number u 4 depending only on δ such that for σ ≥ δ and t ≥ 1, we have |ζ(s)| < u 4 t 1−δ . Proof We may assume that σ ≤ 2 and t exceeds a sufficiently large number depending on δ, otherwise the assertion follows immediately. We observe that the inequality (7.6.2) is valid for 2 ≥ σ ≥ σ0 for any σ0 > 0. Therefore, by putting σ0 = δ in (7.6.2), we get

7.6 Estimates for ζ(s) and ζ (s)

|ζ(s)| ≤

203

x 1 du 1 1 1 x 1−δ 1 + + + (2 + t) < 1 + + + (3 + t) δ . δ δ−1 δ δ δ t n t x x δx u δx 1 n≤x

By putting x = t, we have |ζ(s)| < 1 +

t 1−δ t 1−δ t 1−δ + + (3 + t) = 1 + t 1−δ 1−δ t δt

1 3+t 1 + + 1−δ t δt

< t 1−δ

since t ≥ 1.

4 1 + +2 1−δ δ

7.7 Introduction to the Prime Number Theorem In this section, we state the Prime Number Theorem proved by J. Hadamard and de la Vallée Poussin, independently, dating back to 1896. For convenience, we shall write PNT for the Prime Number Theorem. Here we also introduce some arithmetic functions and state PNT in terms of these functions. For x > 0, let 1, π(x) = p≤x

ϑ(x) =

log p

p≤x

and we recall that ψ(x) =

p

log p.

pm ≤x

Thus π(3/2) = 0, π(6) = 3, π(100) = 25, ϑ(10) = log 2 + log 3 + log 5 + log 7 and ψ(10) = 3 log 2 + 2 log 3 + log 5 + log 7. We have Theorem 7.13 (PNT) lim

x→∞

π(x) = 1. x/ log x

This is also written as π(x) ∼

x as x → ∞. log x

By definition of limit, PNT reads as: For > 0, there exists x0 = x0 () > 0 depending only on such that (1 − )

1 always contains a prime. For > 0, PNT implies that the interval (x, x + x) contains a prime whenever x exceeds a sufficiently large number depending only on , see Exercise 7.19. Now we give PNT in terms of ψ(x) and ϑ(x). We have the following. Theorem 7.14 The Prime Number Theorem is equivalent to each of the following: ψ(x) = 1. x→∞ x ϑ(x) (b) lim = 1. x→∞ x

(a) lim

The proof of Theorem 7.14 depends on the following lemmas which we shall prove first and then turn to the proof of Theorem 7.14. Lemma 7.15 For x ≥ 1, we have 0 ≤ ψ(x) − ϑ(x) ≤ 2x 1/2 (log x)2 . Proof We have ψ(x) =

log p +

log p + · · · +

log p =

pr ≤x

log p + . . .

pr ≤x

p2 ≤x

p≤x

and

1

log p = ϑ(x r ).

p≤x 1/r

Further

1

1

ϑ(x r ) = 0 if x r < 2, i.e. 1

ϑ(x r ) = 0 if r >

log x := r0 . log 2

Therefore 0 ≤ ψ(x) − ϑ(x) =

1

ϑ(x r ).

2≤r ≤r0 1

But, for r ≥ 2, we have ϑ(x r ) ≤ x 1/2 log x and hence 0 ≤ ψ(x) − ϑ(x) ≤ r0 x 1/2 log x ≤

x 1/2 (log x)2 < 2x 1/2 (log x)2 . log 2

7.7 Introduction to the Prime Number Theorem

205

For an integer x ≥ 1 and a prime p, we use in the proof of the next result ord p (x!) =

∞ x , pr r =1

where ord p (x!) denotes the highest power of p dividing x!. This is obtained by observing number of multiples of pr but not of pr +1 in [1, x] is equal to the that x x − . r pr pr +1 Lemma 7.16 For x ≥ 2, we have ψ(x) > C(x − 2) with C =

log 2 . 2

Proof For an improvement of the above inequality, see Exercise 7.15(a). Let

2n M= n

for n ≥ 1.

The proof depends on comparing an upper bound and lower bound for M. We have M=

(n + 1) . . . (2n) (2n)! ≥ 2n . = (n!)2 n!

For an upper bound of M, we write M=

pn p ,

p≤2n

where

2n [ log log p ]

np =

r =1

2n n − 2 . pr pr

We observe that

Therefore

and hence

2n n 2n n − 2 < − 2 − 1 = 2. pr pr pr pr

2n n − 2 ≤1 pr pr

206

7 The Riemann Zeta Function and the Prime Number Theorem

np ≤ Now log M =

n p log p ≤

p≤2n

log 2n . log p

log 2n log p = ψ(2n). log p p≤2n

Thus, by comparing the above upper and lower bounds for log M, we have ψ(2n) ≥ n log 2 for n ≥ 1. Further ψ(2n + 1) ≥ ψ(2n) ≥ n log 2 for n ≥ 1. Thus ψ(m) ≥

m−1 log 2 for m ≥ 2 2

and hence for x ≥ 2 ψ(x) ≥ ψ([x]) ≥

x −2 [x] − 1 log 2 > log 2. 2 2

Combining Lemmas 7.15 and 7.16, we have the following. Corollary 7.17 There exists an absolute positive constant x1 > 0 such that ϑ(x) > Further lim

x→∞

C x for x ≥ x1 . 2 ψ(x) = 1. ϑ(x)

Proof By Lemmas 7.15 and 7.16 ϑ(x) ≥ ψ(x) − 2x 1/2 (log x)2 ≥ C(x − 2) − 2x 1/2 (log x)2 . There exists sufficiently large absolute constant x2 > 0 such that 2x 1/2 (log x)2 < and hence ϑ(x) > C(x − 2) −

Cx for x ≥ x2 4

Cx Cx ≥ for x ≥ x2 . 4 2

7.7 Introduction to the Prime Number Theorem

207

Now by Lemma 7.15 ψ(x) 2x 1/2 (log x)2 4(log x)2 →0 −1≤ < ϑ(x) ϑ(x) C x 1/2

0≤

as x tends to ∞. Thus lim

x→∞

ψ(x) = 1. ϑ(x)

Proof of Theorem 7.14 By Corollary 7.17, we see that (a) holds if and only if (b) holds. Therefore, it suffices to show that the Prime Number Theorem is equivalent to (b). This amounts to proving lim

π(x)/ logx x ϑ(x)/x

x→∞

i.e. lim

x→∞

First we observe that ϑ(x) =

= 1,

π(x) log x = 1. ϑ(x)

log p ≤ π(x) log x

p≤x

implying 1≤

π(x) log x . ϑ(x)

Let δ > 0. Then ϑ(x) ≥

log p ≥ (1 − δ) log x π(x) − π(x 1−δ ) .

x 1−δ < p≤x

Therefore

π(x) log x 1 x 1−δ log x ≤ + ϑ(x) 1−δ ϑ(x)

(7.7.1)

by using π(x 1−δ ) ≤ x 1−δ . Let > 0 and choose δ = δ() > 0 such that 1 0 such that 2 log x < for x ≥ x3 . δ Cx 2 Hence 1≤

π(x) log x ≤ 1 + + = 1 + for x ≥ x3 . ϑ(x) 2 2

7.8 Equivalence of PNT and the Non-vanishing of ζ(s) on the Line σ = 1 The proof depends on the following result. Theorem 7.18 (Wiener–Ikehara) Let A(x) be non-negative and non-decreasing in [0, ∞). Let ∞ A(x)e−xs d x → f (s) in σ > 1, 0

where f (s) is analytic in σ ≥ 1 except at s = 1 where it has a simple pole with residue 1. Then A(x) = 1. lim x→∞ ex We shall prove Theorem 7.18 in the next section. In this section, we shall assume Theorem 7.18 to prove the following result. Theorem 7.19 Assume Theorem 7.18. Then the Prime Number Theorem is equivalent to Theorem 7.9. This is remarkable as PNT is an arithmetic statement, whereas Theorem 7.9 is an analytic statement. Proof First we assume PNT. Let ζ(1 + it0 ) = 0. Then t0 = 0 by Theorem 7.8 and lim

x→∞

ψ(x) =1 x

by Theorem 7.14. Then for > 0, there exists x0 = x0 () > 0 such that for x ≥ x0 , we have |ψ(x) − x| < x.

7.8 Equivalence of PNT and the Non-vanishing of ζ(s) on the Line σ = 1

We consider φ(s) = −

209

1 ζ (s) 1 − , σ > 0. s ζ(s) s−1

We observe, by Theorem 7.8 and Exercise 2.7(b), that φ(s) is analytic in σ > 0 and t > 0 except at the zeros of ζ(s) where it has simple poles. Let σ > 1. Further, by (7.4.3), we have

∞

φ(s) = 1

ψ(x) − x s+1

∞ 1

dx = xs

∞

1

ψ(x) − x d x. x s+1

Thus

x0

|φ(s)| ≤ 1

|ψ(x) − x| dx + x σ+1

∞ x0

|ψ(x) − x| dx < K + x σ+1

∞ 1

dx , =K+ xσ σ−1

where K is a constant. Therefore lim (σ − 1)φ(σ + it0 ) = 0.

σ→1+

Also φ(s) has a simple pole at 1 + it0 with residue r = 0. Therefore φ(σ + it0 ) =

r + A0 + A1 (σ − 1) + · · · σ−1

implying lim (σ − 1)φ(σ + it0 ) = r.

σ→1+

This is a contradiction. Now we assume Theorems 7.9 and 7.18 and prove PNT. Let A(x) = ψ(e x ), f (s) = −

ζ (s) . sζ(s)

We observe that A(x) ≥ 0 and non-decreasing in [0, ∞). By Theorems 7.8, 7.9 and Corollary 7.2, we see that f (s) is analytic in σ ≥ 1 except at s = 1 where it has a simple pole with residue 1, see Exercise 2.7(b). Putting e x = t, we have in σ > 1

∞

A(x)e 0

−xs

∞

dx =

ψ(e )e x

−xs

∞

dx =

0

1

ψ(t)dt = f (s) t s+1

by (7.4.3). Thus, all the assumptions of Theorem 7.18 are satisfied. Hence lim

x→∞

ψ(e x ) = 1, ex

210

7 The Riemann Zeta Function and the Prime Number Theorem

which is equivalent to lim

y→∞

ψ(y) = 1. y

Now the assertion follows from Theorem 7.14.

7.9 Lemmas for the Proof of the Wiener–Ikehara Theorem 7.18 We shall need the following results in the proof of Theorem 7.18. Lemma 7.20 We have

∞

sin2 v dv = π. v2

−∞

Proof Since the integrand is an even function, the integral is equal to 2 0

∞

sin2 v dv. v2

Integrating by parts and using sin 2v = 2 sin v cos v, this is equal to ∞ ∞ ∞ sin2 v sin 2v sin 2v 2 dv = 2 dv. +2 −v 0 v v 0 0 By writing 2v = w, the integral on the right-hand side is equal to

∞

2 0

sin w dw =2 w/2 2

Therefore, it suffices to show that

∞ 0

∞

0

sin w dw. w

π sin w = . w 2

For this, we consider the series s(x) =

sin 2x sin 3x sin x + + + ··· 1 2 3

and for n ≥ 1, we write sn (x) =

sin 2x sin nx sin x + + ··· + . 1 2 n

(7.9.1)

7.9 Lemmas for the Proof of the Wiener–Ikehara Theorem 7.18

Thus

x

sn (x) =

211

(cos t + cos 2t + · · · + cos nt)dt.

0

Now t 3t t 5t 3t = sin − sin + sin − sin 2 2 2 2 2 1 1 t − sin n − t + · · · + sin n + 2 2 t 1 t − sin . = sin n + 2 2

(cos t + cos 2t + · · · + cos nt)2 sin

Therefore

x

sn (x) = 0

sin(n + 21 )t x dt − . 2 sin 2t 2

(7.9.2)

We rewrite the integral on the right-hand side as

x

0

1 2 sin

t 2

−

x sin(n + 21 )t 1 1 sin n + t dt + dt. t 2 t 0

(7.9.3)

By writing (n + 21 )t = u, the second integral in (7.9.3) is equal to

(n+ 21 )x

0

sin u du = u/(n + 21 ) (n + 21 )

which tends to

∞ 0

(n+ 21 )x 0

sin u du, u

sin w dw w

as n → ∞. Further we show that the first integral in (7.9.3) tends to zero as n tends to infinity. Integrating by parts −

cos(n + 21 )t n+

1 2

1 2 sin

t 2

1 − t

x

x

+ 0

cos(n + 21 )t n+

0

1 2

1 1 sin 2

1 and this tends to zero as n → ∞. Here we observe that 2 sin t − 2 tends to zero. Hence, we derive from (7.9.1)–(7.9.3) that

∞

s(x) = lim sn (x) = n→∞

0

1 t

sin w x dw − . w 2

Further, by putting x = π on both sides, we see from (7.9.1) that

t 2

1 − t

dt

tends to zero as t

212

7 The Riemann Zeta Function and the Prime Number Theorem

∞

0= 0

which implies that

∞ 0

sin w π dw − w 2

sin w π dw = . w 2

Lemma 7.21 (Riemann–Lebesgue) Let ψ(θ) be bounded in (a, b) and assume that b ψ(θ)dθ exists. Then a

b

lim

m→∞

and

b

lim

m→∞

ψ(θ) cos mθ dθ = 0

a

ψ(θ) sin mθ dθ = 0.

a

Proof Let a = x0 < x1 < · · · < xn−1 < b = xn be a suitable partition of (a, b). For 1 ≤ r ≤ n, let Ur and L r be the upper and the lower bounds, respectively, of ψ(θ) in (xr −1 , xr ). Further, we write Sn = U1 (x1 − x0 ) + U2 (x2 − x1 ) + · · · + Un (xn − xn−1 ) and sn = L 1 (x1 − x0 ) + L 2 (x2 − x1 ) + · · · + L n (xn − xn−1 ).

b

Let > 0. Since we have

ψ(θ)dθ exists, there exists n 0 = n 0 () > 0 such that for n ≥ n 0

a

Sn − sn < . We assume that n ≥ n 0 and n 0 is sufficiently large. Let K be an upper bound of |ψ(θ)| in (a, b). For 1 ≤ r ≤ n, we write ψ(θ) = ψ(xr −1 ) + Wr (θ) with θ ∈ (xr −1 , xr ), where |Wr (θ)| ≤ Ur − L r . Next we observe that for 1 ≤ r ≤ n

7.9 Lemmas for the Proof of the Wiener–Ikehara Theorem 7.18

xr

ψ(θ) cos mθ dθ = ψ(xr −1 )

xr −1

xr

213

xr

cos mθ dθ +

xr −1

Wr (θ) cos mθ dθ.

xr −1

Therefore

b

ψ(θ) cos mθ dθ =

a

n

ψ(θ) cos mθ dθ

xr −1

r =1

=

xr

n

ψ(xr −1 )

xr

cos mθ dθ +

xr −1

r =1

n r =1

xr

Wr (θ) cos mθ dθ,

xr −1

where the absolute value of the first sum on the right-hand side is at most n r =1

|ψ(xr −1 )|

xr xr −1

cos mθ dθ ,

which is less than or equal to 2nmK and this tends to zero as m tends to ∞. The absolute value of the second sum on the right-hand side is less than or equal to n

(Ur − L r )(xr − xr −1 ) = Sn − sn < .

r =1

Hence

b

lim

m→∞ a

Similarly

lim

m→∞ a

b

ψ(θ) cos mθ dθ = 0.

ψ(θ) sin mθ dθ = 0.

The Riemann–Lebesgue lemma is valid even when ψ is not bounded, see [[33], p. 172]. Next we show that Theorem 7.18 follows from the following result. Theorem 7.22 Let B(x) = e−x A(x) and suppose that the assumptions of Theorem 7.18 are satisfied. Then we have lim

y→∞

v sin2 v B y− dv = π. λ v2 −∞ λy

As stated above, we formulate the following. Lemma 7.23 Theorem 7.22 implies Theorem 7.18.

214

7 The Riemann Zeta Function and the Prime Number Theorem

Proof We assume Theorem 7.22. It suffices to show (i) lim sup B(x) ≤ 1 and x→∞

(ii) lim inf B(x) ≥ 1, x→∞

since then 1 ≤ lim inf B(x) ≤ lim sup B(x) ≤ 1 x→∞

x→∞

implies that lim B(x) = 1.

x→∞

First, we give a proof of (i). Let a > 0, λ > 0 be given such that y > λa . Then λy > a and we have a v sin2 v B y− dv ≤ π lim sup λ v2 y→∞ −a by Theorem 7.22 and non-negativity of the integrand. Since A(u) = eu B(u) nondecreasing, we have a y− a v y− v e λ ≥B y− e λ for − a ≤ v ≤ a, B y− λ λ which implies that a v−a v a − 2a ≥B y− e λ ≥B y− e λ. B y− λ λ λ Thus

a

lim sup y→∞

i.e.

−a

a − 2a sin2 v e λ B y− dv ≤ π, λ v2

a − 2a a sin2 v e λ lim sup B y − dv ≤ π. 2 λ y→∞ −a v

Let a → ∞, λ → ∞ such that π lim sup B(y) ≤ π implying (i).

a λ

→ 0. Then, by Lemma 7.20, we have

y→∞

Next we prove (ii). By (i), we see that B(x) ≤ c for x ≥ 0 where c > 0 is a constant. Let a and λ be given as in (i) and y be sufficiently large. Then

λy

−∞

−a λy a v sin2 v v sin2 v v sin2 v v sin2 v B y− dv = B y− + B y− + B y− 2 2 2 λ v λ v λ v λ v2 −∞ a −a −a ∞ a 2 2 2 sin v sin v v sin v ≤c dv + c dv + B y− dv. v2 v2 λ v2 −∞ a −a

Thus, by Theorem 7.22, we have

7.9 Lemmas for the Proof of the Wiener–Ikehara Theorem 7.18

π≤c

−a −∞

sin2 v dv + c v2

∞

a

sin2 v dv + lim inf y→∞ v2

215 a

−a

v sin2 v B y− dv. λ v2 (7.9.4)

Now, for −a ≤ v ≤ a, we have v y− v a y+ a e λ ≤B y+ e λ, B y− λ λ and therefore a a+v a 2a v ≤B y+ e λ ≤B y+ eλ. B y− λ λ λ Thus, by Lemma 7.20, we have

a

lim inf y→∞

−a

v sin2 v a 2a e λ π. B y− dv ≤ lim inf B y + 2 y→∞ λ v λ

Let a → ∞, λ → ∞ with

a λ

→ 0. Then, by (7.9.4), we have π ≤ lim inf B(y)π, y→∞

which implies (ii).

7.10 Proof of Theorem 7.18 By Lemma 7.23, it suffices to give a proof of Theorem 7.22 which we now prove. For σ > 1, we have ∞ ∞ 1 −sx = A(x) e d x, e−(s−1)x d x. f (s) = s−1 0 0 Thus

1 = g(s) := f (s) − s−1

∞

e−(s−1)x (B(x) − 1)d x.

(7.10.1)

0

Let > 0 and λ > 0. By (7.10.1), we have

∞

g (t) := g(1 + + it) =

(B(x) − 1)e−(+it)x d x.

(7.10.2)

0

We observe from the assumptions of Theorem 7.22 that g(s) is analytic in σ ≥ 1. We evaluate the integral 1 2

2λ −2λ

|t| i yt e dt. g (t) 1 − 2λ

(7.10.3)

216

7 The Riemann Zeta Function and the Prime Number Theorem

By (7.10.2), the above integral (7.10.3) is equal to 1 2

∞ |t| i yt −(+it)x (B(x) − 1)e d x dt. 1− e 2λ −2λ 0 2λ

(7.10.4)

Now we show that 1 2

∞ |t| |B(x) − 1|e−x d x < ∞ 1− dt 2λ −2λ 0 2λ

(7.10.5)

so that the order of integration in (7.10.4) can be interchanged by the Fubini theorem, see Lemma 2.9. For s > 1 and x > 0, we have

∞

f (s) =

A(u)e

−us

∞

du ≥

0

A(u)e

−us

∞

du ≥ A(x)

x

e−us du =

x

A(x)e−sx . s

Thus A(x) ≤ c1 esx , where c1 ≥ 1 is a positive constant depending only on s. Thus, |B(x) − 1|e−x ≤ c1 e(s−1−)x + e−x . By taking s = 1 + 2 , we have x

|B(x) − 1|e−x ≤ 2c1 e− 2 . Thus, the left-hand side of (7.10.5) is at most ∞ x |t| |t| t 4c1 2λ 4c1 λ 2c1 2λ 1− 1− 1− dt dt = dt = < ∞. e− 2 d x = 2λ −2λ 2λ 0 2λ −2λ 0

c1

2λ

Therefore, the order of integration in (7.10.4) can be interchanged. Thus, (7.10.4) and hence (7.10.3) is equal to 1 2

∞

(B(x) − 1)e

−x

0

Now, integrating by parts, we have

|t| i(y−x)t 1− e dt d x. 2λ −2λ 2λ

(7.10.6)

7.10 Proof of Theorem 7.18

217

1 2λ 1 0 1 2λ |t| t t 1− 1+ 1− ei(y−x)t dt = ei(y−x)t dt + ei(y−x)t dt 2 −2λ 2λ 2 −2λ 2λ 2 0 2λ t 1 2λ 1− = (ei(y−x)t + e−i(y−x)t )dt 2 0 2λ 2λ t = 1− cos((y − x)t)dt 2λ 0 2λ t sin((y − x)t) 2λ 1 1 = 1− + sin((y − x)t)dt 2λ y−x 2λ y − x 0 0 =

1 1 sin2 (λ(y − x)) (− cos(2λ(y − x)) + 1) = , 2λ (y − x)2 λ(y − x)2

which we substitute in (7.10.6) to conclude that 1 2

∞ sin2 λ(y − x) |t| i yt e dt = g (t) 1 − (B(x) − 1)e−x d x. (7.10.7) 2λ λ(y − x)2 −2λ 0 2λ

Since g(s) is analytic in σ ≥ 1, we see that g (t) → g(1 + it) as → 0 uniformly in −2λ ≤ t ≤ 2λ. Therefore 1 →0 2

lim

|t| i yt 1 2λ |t| i yt e dt = e dt. (7.10.8) g (t) 1 − g(1 + it) 1 − 2λ 2 −2λ 2λ −2λ 2λ

By combining (7.10.8) and (7.10.7), we have 1 2

∞ ∞ sin2 λ(y − x) sin2 λ(y − x) |t| eiyt dt = lim g(1 + it) 1 − B(x)e−x d x − lim e−x dx 2 →0 →0 2λ λ(y − x) λ(y − x)2 −2λ 0 0 ∞ ∞ sin2 λ(y − x) sin2 λ(y − x) B(x) dx − dx = 2 λ(y − x) λ(y − x)2 0 0 2λ

since the integrands are non-negative and increasing as → 0, see [[30], Sect. 10.8]. |t| By applying Lemma 7.21 with a = −2λ, b = 2λ, ψ(t) = g(1 + it) 1 − 2λ with |t| ≤ 2λ, we have lim

y→∞ 0

∞

B(x)

sin2 λ(y − x) d x = lim y→∞ λ(y − x)2

0

∞

sin2 λ(y − x) d x. λ(y − x)2

Putting λ(y − x) = v, we get lim

y→∞

∞ λy sin2 v sin2 v v sin2 v B y− dv = lim dv = =π 2 y→∞ −∞ v 2 λ v2 −∞ −∞ v λy

by Lemma 7.20.

218

7 The Riemann Zeta Function and the Prime Number Theorem Im(z)

Fig. 7.1 Rectangular contour for Cauchy residue theorem

(−T, u)

(T, u)

Γ3

Γ4

Γ2

(0, 0) Γ1

(−T, 0)

(T, 0)

Re(z)

7.11 Analytic Continuation of ζ(s) in C and Functional Equation for ζ(s) We show that ζ(s) has analytic continuation in the whole complex plane and it satisfies a functional equation. The proof is due to Riemann and depends on the following functional equation for the Theta functions. Theorem 7.24 For x > 0, we have ∞

e−πn

2

∞ 1 −πn 2 /x =√ e . x n=−∞

x

n=−∞

Now we give lemmas for the proof of Theorem 7.24.

Lemma 7.25 We have e

∞

Proof We have −∞

−πu 2

=

∞

e−πx e−2πi xu d x. 2

−∞

e−πx d x = 1, see Exercise 6.17(b). Therefore, we may suppose 2

that u = 0. Further there is no loss of generality in assuming that u > 0. We take the following contour. Let T be sufficiently large and = 1 + 2 + 3 + 4 . By Theorem 2.4, we have (Fig. 7.1) 0= Further

e−πz dz = 2

e−πz dz + 2

1

2

e−πz dz + 2

e−πz dz + 2

3

e−πz dz. 2

4

7.11 Analytic Continuation of ζ(s) in C and Functional Equation for ζ(s)

e

−πz 2

1

3

e−πz d x = − 2

For z 2

T −T

dz =

T

219

e−πx d x, 2

−T

e−π(x+iu) d x = −eπu 2

2

T

e−πx e−2πi xu d x. 2

−T

e−πz = e−π(T +iv) = e−πT eπv e−2πivT , 2

2

2

2

which tends to zero as T → ∞. Further 2 e−πz dz → 0 as T → ∞. 2

Similarly

e−πz dz → 0 as T → ∞. 2

4

Hence, by letting T → ∞, we have 0=

∞ −∞

e−πx d x − eπu 2

2

∞

e−πx e−2πi xu d x. 2

−∞

Since the first integral on the right-hand side is equal to 1 as already mentioned in the beginning of the proof of this lemma, we have eπu

2

∞

e−πx e−2πi xu d x = 1 2

−∞

implying

∞

−∞

e−πx e−2πi xu d x = e−πu . 2

2

Let F be a real-valued function such that F is continuous and periodic with period 1. Further assume that F is of bounded variation. Then F can be expanded as a Fourier series ∞ F(x) = am e2πimx , m=−∞

where

am = 0

see [[33], Sect. 9.2].

1

F(x) e−2πimx d x,

220

7 The Riemann Zeta Function and the Prime Number Theorem

For u > 0, we restrict to ∞

F(x) =

f (x + n),

(7.11.1)

n=−∞

where

f (x) = e−πx

2 −2

u

(7.11.2)

as it will suffice for our application. Its Fourier transform is

fˆ(x) =

∞

−∞

f (v)e2πivx dv.

(7.11.3)

We observe that ∞

F(x + 1) =

∞

f (x + 1 + n) =

n=−∞

f (x + n) = F(x).

n=−∞

Thus, F(x) is periodic with period 1 and we consider F(x) in 0 ≤ x ≤ 1. Further F(x) is of bounded variation and continuous in [0, 1]. Therefore, F has Fourier series expansion in [0, 1]. Lemma 7.26 We have

∞

∞

f (n) =

n=−∞

fˆ(n).

(7.11.4)

n=−∞

Proof As stated above F(x) =

∞

am e2πimx , 0 ≤ x ≤ 1

(7.11.5)

m=−∞

where am =

1

F(x)e−2πimx d x =

0

1

∞

f (x + n)e−2πimx d x =

0 n=−∞

∞

1

f (x + n)e−2πimx d x

n=−∞ 0

by (7.11.1). Since the terms of the series in the middle integral are non-negative, we see that the interchange of integration and summation is justified. By writing x + n = y, we have am =

∞ n=−∞ n

n+1

f (y) e−2πimy dy =

∞ −∞

f (y)e−2πimy dy = fˆ(−m).

Hence, by putting x = 0 in (7.11.1) and (7.11.5), we obtain

7.11 Analytic Continuation of ζ(s) in C and Functional Equation for ζ(s) ∞

∞

f (m) = F(0) =

m=−∞

am =

m=−∞

∞

∞

fˆ(−m) =

m=−∞

221

fˆ(m).

m=−∞

Proof of Theorem 7.24 Let u > 0 and ∞

G(u) =

e−πn

2 2

u

.

n=−∞

We recall that

f (x) = e−πx

and thus

∞

2 −2

u

f (−n) = G(u −1 ).

(7.11.6)

n=−∞

We show that

G(u) = u −1 G(u −1 ),

√ and then the assertion follows by taking u = x. By Lemma 7.25 we replace u by nu and we get ∞ ∞ 2 G(u) = e−πx e−2πinux d x. n=−∞ −∞

By putting ux = v, we see from (7.11.3) and (7.11.2) that G(u) = u −1

∞

∞

n=−∞ −∞

v 2

e−π( u ) e−2πinv dv = u −1

∞

fˆ(−n) = u −1

n=−∞

∞

f (−n) = u −1 G(u −1 )

n=−∞

by (7.11.4) and (7.11.6). Now we are ready to prove the functional equation for ζ(s).

Theorem 7.27 The function ζ(s) has analytic continuation in C except at s = 1 where it has a simple pole with residue 1. Further it satisfies the functional equation ζ(s) = 2s π s−1 sin

πs (1 − s)ζ(1 − s). 2

Proof For n ≥ 1, we consider

∞

x 2 s−1 e−n

0

Putting n 2 πx = y, the integral equals

1

2

πx

d x.

222

7 The Riemann Zeta Function and the Prime Number Theorem

∞

x 2 s−1 e−n 1

2

πx

1 s 2 (n π) 2

dx =

0

∞

y 2 s−1 e−y dy = 1

0

(s/2) in σ > 0. n s π s/2

Thus, in σ > 1 ∞ ( 2s )ζ(s) = π s/2 n=1

∞

x 2 s−1 e−n 1

2

πx

∞ ∞

dx =

0

0

x 2 s−1 e−n 1

2

πx

dx

n=1

since the series in the above integral is absolutely convergent on every closed interval. We put for x > 0 ∞ 2 θ(x) = e−n πx . n=1

Then

∞

e−n

2

πx

= 2θ(x) + 1

n=−∞

and

∞

e

1 + 1. = 2θ x

−n 2 π/x

n=−∞

Now we derive from Theorem 7.24 that 1 2θ(x) + 1 = √ x and therefore

1 θ(x) = √ θ x

1 2θ +1 , x

1 1 1 + √ − . x 2 2 x

Then in σ > 1 ( 2s )ζ(s) = π s/2

∞

1

x 2 s−1 θ(x)d x +

1

1

1

x 2 s−1 0

1 √ θ x

1 1 1 + √ − d x. x 2 2 x

We consider the second integral. We see that it is equal to

1

1

3

x 2 s− 2 θ

0

By putting x =

1 u

1 1 dx + . x s(s − 1)

in the integral, we see that the above expression is equal to

∞ 0

u − 2 s+ 2 −2 θ(u)du + 1

3

1 . s(s − 1)

7.11 Analytic Continuation of ζ(s) in C and Functional Equation for ζ(s)

223

Therefore, in σ > 1 ( 2s )ζ(s) = π s/2

∞

(x 2 s−1 + x − 2 s− 2 ) θ(x)d x + 1

1

1

1

1 . s(s − 1)

(7.11.7)

We denote by H (s) the right-hand side of (7.11.7). Since the series for θ(x) converges uniformly on closed intervals in [1, ∞), we derive that the integral in H (s) is analytic in C and hence H (s) is analytic in C except at s = 0, 1 where it has simple poles. We define in C π s/2 H (s) , (7.11.8) ζ(s) = ( 2s ) which agrees with the earlier definition of ζ(s) in σ > 1. We observe that ζ(s) is analytic in C except possibly at s = 0 and s = 1. But the simple pole of H (s) at s = 0 cancels with the simple pole of ( 2s ) at s = 0. Therefore, ζ(s) has pole only at s = 1 with residue lim (s − 1)H (s)

s→1

π 1/2 π s/2 =1 s = ( 2 ) ( 21 )

by Corollary 6.22(iii). Thus, ζ(s) given above is the required analytic continuation in C. Next, we turn to proving the functional equation for ζ(s). Since 1 1 1 1 1 1 (1 − s) − 1 = − s − , − (1 − s) − = s − 1 2 2 2 2 2 2 and

1 1 = , s(s − 1) (1 − s)(1 − s − 1)

we see that H (s) given by the right-hand side of (7.11.7) is invariant under the transformation s → 1 − s and hence we have H (s) = H (1 − s), i.e. π

− 2s

s 2

ζ(s) = π

Therefore 1

ζ(s) = π s− 2

− 1−s 2

1 1 − s ζ(1 − s). 2 2

( 21 − 21 s) ζ(1 − s). ( 2s )

But, by Theorem 6.21(c) and (b), we have

(7.11.9)

224

7 The Riemann Zeta Function and the Prime Number Theorem

( 21 − 21 s)(1 − 21 s) ( 21 − 21 s) 2s π 1/2 (1 − s) πs = (1 − s). = = 2s π −1/2 sin s ( 2 ) π/ sin πs 2 ( 2s )(1 − 21 s) 2

Hence ζ(s) = 2s π s−1 sin

πs (1 − s) ζ(1 − s). 2

Now we derive from the functional equation for ζ(s) information on the distribution of zeros of ζ(s). For studying the zeros of ζ(s), it is convenient to consider the function 1 (7.11.10) ξ(s) = s(s − 1) H (s). 2 Since H (s) is analytic in C except at s = 0, 1 where it has simple poles, we see that ξ(s) is entire. Further, by (7.11.8), we have ξ(s) =

s 1 ζ(s). s(s − 1)π −s/2 2 2

(7.11.11)

Since ( 2s ) has simple poles at s = −2m with integer m ≥ 0 and the pole at s = 0 cancels with the factor s, we see from the functional equation for ζ(s) that ζ(s) has simple zeros at s = −2, −4, −6 . . . . These are called the trivial zeros of ζ(s) and all other zeros of ζ(s) are called the non-trivial zeros of ζ(s). Now we formulate a result listing the properties of zeros of ξ(s) and hence of ζ(s). Lemma 7.28 The function ξ(s) satisfies the following properties: (i) ξ(s) = ξ(1 − s). (ii) The zeros of ξ(s) are symmetric around the lines σ = 21 and t = 0. Further ξ(s) is real on the line σ = 21 . (iii) ξ(0) = ξ(1) = 21 . (iv) All the zeros s = σ + it of ξ(s) satisfy 0 ≤ σ ≤ 1. (v) The non-trivial zeros of ζ(s) and ξ(s) are identical. (vi) log M(r ) ∼ 21 r log r as r → ∞ where M(r ) = max |ξ(s)|. In particular, ξ(s) |s|=r

is an entire function of order 1. 1 (vii) = ∞ where the sum is taken over all non-trivial zeros of ζ(s). In |ρ| ρ particular, ζ(s) has infinitely many non-trivial zeros. Proof (i) By (7.11.10) and (7.11.9), we have ξ(1 − s) =

1 1 (1 − s)(1 − s − 1)H (1 − s) = s(s − 1)H (s) = ξ(s). 2 2

(ii) By putting s = 21 + it in (i), we get ξ( 21 + it) = ξ( 21 − it). Further it is clear from the definition of ξ(s) that ξ(s) = ξ(s). This implies the first part of (ii). Since

7.11 Analytic Continuation of ζ(s) in C and Functional Equation for ζ(s)

ξ

1 + it 2

=ξ

1 − it 2

=ξ

1 + it 2

225

1 = ξ + it , 2

we conclude that ξ( 21 + it) is real. (iii) By (i) and (7.11.11), we have

s 1 1 1 1 s(s − 1)π −1/2 ζ(s) = lim ((s − 1)ζ(s)) π −1/2 = 2 2 s→1 2 2 s→1 2

ξ(0) = ξ(1) = lim

by Corollary 6.22(iii) and ζ(s) has a simple pole with residue 1 at s = 1. (iv) Assume that s = σ0 + it0 satisfies ξ(s0 ) = 0. Let σ0 > 1. Then ( s20 )ζ(s0 ) = 0 by (7.11.11). This is not possible since ζ(s) = 0 in σ > 1 by Corollary 7.2 and (s) has no zero by Corollary 6.22(i). Let σ0 < 0. Then ξ(1 − s0 ) = 0 by (i) and 1 − σ0 > 1. This is again not possible as above. Therefore, σ0 ≥ 0 and hence 0 ≤ σ0 ≤ 1. (v) Let s0 be a non-trivial zero of ζ(s). Then s0 is not a pole of ( 2s ) unless s0 = 0 s = 1. Therefore, ξ(s0 ) = 0 by (7.11.11). in which case lim s s→0 2 Let ξ(s0 ) = 0. Then s0 ∈ / {0, 1} by (iii). Since (s) has no zero by Corollary 6.27, we see that ζ(s0 ) = 0 by (7.11.11) again. (vi) Denote by u 1 , u 2 , u 3 and u 4 positive constants. For sufficiently large r , we see from (7.11.11) that M(r ) ≥ ξ(r ) ≥ π − 2 r

1

1 r 2

≥ e 2 r log r −u 1 r 1

(7.11.12)

by Theorem 6.25 with m = 1 and Lemma 6.26. Let r ≥ r0 where r0 is sufficiently large and let |s| = r . By Lemma 7.12 with δ = 21 and |ζ(s)| ≤ ζ(2) for σ ≥ 2, we derive that 1

|ζ(s)| < u 2 |s| 2 for σ ≥

1 . 2

Now we see from (7.11.11), Theorem 6.25 with m = 1 and Lemma 6.26 that |ξ(s)| < e 2 s| log |s|+u 3 |s| for σ ≥ 1

1 . 2

(7.11.13)

Also we derive from (i) that |ξ(s)| = |ξ(1 − s)| < e 2 |1−s| log |1−s|+u 3 |1−s| for σ < 1

1 . 2

(7.11.14)

By combining (7.11.13) and (7.11.14), we get M(r ) ≤ e 2 r log r +u 4 r . 1

(7.11.15)

226

7 The Riemann Zeta Function and the Prime Number Theorem

Now the assertion follows from (7.11.12) and (7.11.15). (vii) We assume that 1 < ∞, |ρ| ρ

(7.11.16)

where the sum is taken over all the non-trivial zeros of ζ(s) and we shall arrive at a contradiction. By (iii), ρ = 0 and we observe from (v) that the sum is taken over all the zeros of ξ(s). Let τ be the exponent of convergence for the sequence {|ρ|} of zeros of ξ(s) and k be the rank of ξ(s). By (vi) and Lemma 6.11, we observe that τ ≤ 1 and k ≤ τ ≤ k + 1. Then k ∈ {0, 1}. Further (6.5.5) holds by the Hadamard factorisation Theorem 6.12 with k ∈ {0, 1} and h ≤ 1. Then we derive from (6.5.5) and (7.11.16) that log M(r ) = O(r ) which contradicts (7.11.12) and the assertion follows. Finally, we conclude the following result on the zeros of ζ(s). Theorem 7.29 We have (i) All the non-trivial zeros s = σ + it of ζ(s) satisfy 0 ≤ σ ≤ 1 and they are symmetric around the lines σ = 21 and σ = 0. (ii) ζ(s) has no zero in 0 ≤ s ≤ 1. Proof (i) It follows immediately by combining Lemma 7.28(v), (iv) and (ii). (ii) Let 0 ≤ s ≤ 1 such that ζ(s) = 0. If s = 0, then ξ(s) = 0 by (7.11.11) contradicting Lemma 7.28(iii). Thus s = 0 and also s = 1 since ζ(s) has a pole at s = 1. We show that ζ(s) < 0 for 0 < s < 1 and then the assertion follows. In σ > 1, we have (1 − 21−s )ζ(s) = (1 − 21−s )

∞ ∞ ∞ 1 1 1 = − 2 s s n n (2n)s n=1 n=1 n=1

= (1s − 2s ) + (3−s − 4−s ) + (5−s − 6−s ) + · · · (7.11.17) and the ordering of the terms in the series is justified since it is absolutely convergent. We shall show that series on the right-hand side converges uniformly on compact subsets of σ > 0. Then, it is analytic in σ > 0 by Sect. 2.3 (iv). Also (1 − 21−s )ζ(s) is analytic in σ > 0 since the simple pole of ζ(s) at s = 1 cancels with the zero of (1 − 21−s ). Now we derive from Sect. 2.3 (ii) that (1 − 21−s )ζ(s) = (1−s − 2−s ) + (3−s − 4−s ) + . . . is valid in σ > 0. Therefore (1 − 21−s )ζ(s) > 0 for 0 < s < 1 implying

7.11 Analytic Continuation of ζ(s) in C and Functional Equation for ζ(s)

227

ζ(s) < 0 for 0 < s < 1. Let K be a compact subset of σ > 0. Then there exist δ > 0 and θ > 0 such that σ ≥ δ and |s| ≤ θ whenever s = σ + it ∈ K . Then (2n − 1)−s − (2n)−s = s and

∞ n=1

2n 2n−1

2n d x dx θ ≤θ ≤ s+1 1+δ x (2n − 1)1+δ 2n−1 x

1 < ∞. (2n − 1)1+δ

Hence, the series given by (7.11.17) converges uniformly on compact subsets of σ > 0.

7.12 The Function μ(σ) Let a ≤ b be fixed real numbers, s = σ + it with a ≤ σ ≤ b, |t| ≥ 1. The constants implied by O depend only on a and b, and thus on σ if a = b. Let χ(s) = 2s π s−1 sin

πs (1 − s). 2

Then ζ(s) = χ(s)ζ(1 − s) for s ∈ C by Theorem 7.27. Further χ(s) =

2s−1 π s sec πs 2 (s)

(7.12.1)

(7.12.2)

by Theorem 6.21(b). By writing sec

πsi −1 πs −πsi =2 e 2 +e 2 , 2

πs π|t| 1 . sec = 2e− 2 1 + O 2 |t|

we have

(7.12.3)

Then we derive from (7.12.2), (7.12.3) and Corollary 6.27(ii) that |χ(s)| = 2σ−1 π σ 2e−

π|t| 2

1

(2π)− 2 e

π|t| 2

1 −σ 1 1 1 |t| 2 1+O = . |t| 2 −σ 1 + O |t| 2π |t|

228

7 The Riemann Zeta Function and the Prime Number Theorem

In particular, we have

|χ(s)| ∼

|t| 2π

21 −σ

.

(7.12.4)

Now we have Lemma 7.30 Let −∞ < σ < ∞ and |t| ≥ 1. Then there exists χ = χ(σ) depending only on σ such that ζ(σ + it) = Oσ (|t|χ ) . Proof We assume that the constants implied by O in the proof depend only on σ. Since ζ(s) = ζ(¯s ), we may suppose that t ≥ 1. Let σ > 1. Then there exists δ > 0 1 such that σ ≥ 1 + δ and then ζ(s) = O(1). Further ζ(s) = O(t 2 ) for 21 ≤ σ ≤ 1 1 by Lemmas 7.12, 7.10 and ζ(s) = O(t) for 0 ≤ σ ≤ 21 , ζ(s) = O(t 2 −σ ) for −∞ < σ < 0 by (7.12.4). This implies the assertion of the lemma. Definition For −∞ < σ < ∞, let μ(σ) be the greatest lower bound of all χ = χ(σ) given by Lemma 7.30. Then for > 0, we have ζ(σ + it) = O |t|μ(σ)+ , where the constant implied by O depends only on and σ. We see from Theorem 2.27 that μ is convex and continuous in (−∞, ∞). We have μ(σ) = 0 for σ > 1 and μ(σ) = 21 − σ for σ < 0 by (7.12.1) and (7.12.4). In fact μ(σ) = 0 for σ ≥ 1 and μ(σ) =

1 − σ for σ ≤ 0 2

by continuity. Let 1 > 0. Now we apply Theorem 2.27 with f (s) = ζ(s), a = 0, b = 1, k1 =

1 + 1 , k2 = 1 2

for 0 ≤ σ ≤ 1. We observe that the assumption (2.7.7) is satisfied by Lemma 7.30. Hence, we conclude that μ(σ) ≤

1 (1 − σ) + 1 for 0 ≤ σ ≤ 1. 2

Letting 1 tend to zero, we get μ(σ) ≤

1 (1 − σ) for 0 ≤ σ ≤ 1, 2

7.12 The Function μ(σ)

which implies μ

229

1 1 ≤ and hence 2 4 ζ

1 + it 2

1 = O |t| 4 + for > 0.

7.13 Main Conjectures in the Theory of the Riemann Zeta Function In this section, we consider only the non-trivial zeros of ζ(s). We have shown that ζ(s) has infinitely many zeros. A famous conjecture of Riemann states as follows. The Riemann hypothesis: All the non-trivial zeros of ζ(s) lie on the line σ = 21 . Hardy proved in 1914 that ζ(s) has infinitely many zeros on the line σ = 21 . In fact, Selberg proved that a positive proportion of zeros of ζ(s) lie on the line σ = 21 and X. Gourdon showed that 41.28 percent zeros of ζ(s) lie on the line σ = 21 . It has also been confirmed by X. Gourdon that the first 1013 zeros of ζ(s) lie on the line σ = 21 but the Riemann hypothesis remains unproved. A weaker conjecture than the Riemann hypothesis, but still unproved, is the following. Lindelöf hypothesis: Let θ > 0. Then ζ

1 + it 2

= O |t|θ ,

(7.13.1)

where the constant implied by the O symbol depends only on θ. 13 Hardy proved (7.13.1) with θ = 16 + and Bourgain with 84 + for > 0. For 0 ≤ σ0 ≤ 1 and T > 0, we denote by N (σ0 , T ) the number of zeros, counted with multiplicity, of ζ(s) in the rectangle σ0 ≤ σ ≤ 1, 0 ≤ t ≤ T and we write N (T ) for N (0, T ). Thus N (T ) is the number of zeros of ζ(s), counted with multiplicity, in the rectangle 0 ≤ σ ≤ 1, 0 ≤ t ≤ T. For a given T , we know that N (T ) is finite and further N (T ) = 1. lim T →∞ T log T We have the following conjecture on the density of zeros of ζ(s). Density hypothesis: For 0 ≤ σ ≤ 1 and > 0, we have N (σ, T ) = O T 2−2σ+ . Here the constant implied by the symbol O depends only on σ and . 25 ≤ Ingham proved density hypothesis when 56 ≤ σ ≤ 1 and Bourgain when 32 σ ≤ 1. It is known that the Riemann hypothesis implies the Lindelöf hypothesis and the Lindelöf hypothesis implies the density hypothesis. Further Ingham showed that

230

7 The Riemann Zeta Function and the Prime Number Theorem

the density hypothesis implies 1 + . pn+1 − pn = O pn2 This is not a detailed survey on the above famous conjectures but we hope that it will help the readers to gather it from the literature. We conclude this section by proving that the Riemann hypothesis implies the Lindelöf hypothesis. The proof depends on the Borel–Carathéodry Lemma 5.5 and the Hadamard three-circle theorem Corollary 2.26. Theorem 7.31 The Riemann hypothesis implies the Lindelöf hypothesis. Proof By the Riemann hypothesis, since ζ(s) has no zero in σ > 21 , we derive from Theorem 2.28 that there exists a branch of logarithm analytic in σ > 21 except at s = 1 such that it is given by Theorem 7.7 in σ > 1. Let > 0 such that σ > 21 + . Further we suppose that t > 0 since ζ(s) is symmetric around the x-axis and also t exceeds a sufficiently large number t0 depending only on such that 1 < for t ≥ t0 . 2 log log t Let δ=

1 log log t

σ>

1 δ + . 2 2

and then

Let C1 , C2 , C3 and C4 be circles each with centres at s0 = δ −1 + it and passing through the points 1 + δ + it, σ + it, 21 + δ + it and 21 + 2δ + it, respectively, of circles C1 , C2 , C3 and C4 . Then we observe that the radii r1 , r2 , r3 and r4 of circles C1 , C2 , C3 and C4 are equal to δ −1 − 1 − δ, δ −1 − σ, δ −1 − 21 − δ and δ −1 − 21 − 2δ , respectively. For |s − s0 | < r4 , we have Re (log ζ(s)) = log |ζ(s)| log t by Lemma 7.12. Then we conclude from the Borel and Carthéodory Lemma 5.5 with f (s) = log ζ(s), r = r3 and R = r4 that M3 ≤ αδ −1 log t,

(7.13.2)

where α is a constant and Mi = max | f (s)| with 1 ≤ i ≤ 3. By the Hadamard threes∈Ci

circle theorem Corollary 2.26, we have

7.13 Main Conjectures in the Theory of the Riemann Zeta Function

M2 ≤ M11−a M3a ,

231

(7.13.3)

where log 1 + log(r2 /r1 ) = a= log(r3 /r1 ) log 1 +

log 1 + = r3 −r1 log 1 + r1

r2 −r1 r1

1+δ−σ δ −1 −1−δ 1/2 δ −1 −1−δ

= 2(1 − σ) + O(δ).

Hence a ≤ 1 − 2 + O(δ)

(7.13.4)

since σ > 21 + . If s ∈ C1 , we observe that Re(s) ≥ 1 + δ and then we derive that M1 ≤

∞ ∞ 1 du 1 ≤ = , 1+δ 1+δ n u δ 1 n=2

(7.13.5)

where δ is a constant. Now, by (7.13.3), (7.13.2), (7.13.5) and (7.13.4), we conclude that log |ζ(σ + it)| ≤ | log ζ(σ + it)| ≤

1−a a α 1 log log t log t < log t. δ δ (log t)

Then we get that |ζ(σ + it)| < t for every σ > which implies

1 , 2

(7.13.6)

1 ζ 2 + it < t

by continuity.

7.14 Exercises 7.1 Show that the Möbius function μ(n) is multiplicative and satisfies d|n

μ(d) =

1 if n = 1 0 if otherwise.

(Hint: For positive integers a and b with (a, b) = 1, we have μ(ab) = 1 if and only if either μ(a) = 1, μ(b) = 1 or μ(a) = −1, μ(b) = −1. Let n =

232

7 The Riemann Zeta Function and the Prime Number Theorem

P1a1 · · · Pkak where P1 < P2 < · · · < Pk are prime numbers. Then observe that

μ(d) =

d|n

k i=1

μ(Pi ) +

μ(Pi P j ) + · · · + μ(P1 · · · Pk ).

i< j

) 7.2 Show that d(n) is multiplicative and d(n) = O (n ) for > 0. (Hint: For showing that d(n) is multiplicative, list all the divisors of n. For ap + 1 where n = pa p .) showing d(n) = O(n ), estimate the product p 1 p p≤2

7.3 Show that σa (n) is multiplicative. 7.4 For n ≥ 3, show that ω(n) = O

log n log log n

,

where ω(n) denotes the number of distinct prime factors of n. (Hint: Estimate the number of prime factors of n which are greater than log n.) 7.5 Let π(x, z) denote the number of positive integers n ≤ x coprime to all prime numbers p ≤ z. Then show that 1 1− + O 2z . π(x, z) = x p p≤z (Hint: By Exercise 6.1, write π(x, z) = 7.6 For σ > 1, prove that (i) ζ 2 (s) = (ii)

n≤x d|(n,Pz )

μ(d) where Pz =

p.)

p≤z

∞ d(n)

ζ 4 (s) = ζ(2s)

n=1 ∞ n=1 ∞

ns d 2 (n) ns

2ω(n) ζ 2 (s) = (iii) . ζ(2s) ns n=1 (Hint: Apply Theorem 7.1.) 7.7 For k ≥ 2, an integer ν is called k-powerful if p k |ν whenever p|ν for all primes. Let G(k) = ν ∈ N | ν is k-powerful

7.14 Exercises

233

and

1 if n ∈ G(k) 0 if otherwise.

f k (n) = Then show that

∞ p −2s ζ(2s)ζ(3s) f 2 (n) 1+ = in σ > 1. = s −s n 1− p ζ(6s) p n=1 Further give the Euler product for

∞ f 3 (n) in σ > 1. ns n=1

(Hint: Apply Theorem 7.1.) 7.8 (a) (The Möbius inversion formula) If f is an arithmetic function and g(n) = f (d) for n ≥ 1, then show that d|n

f (n) =

μ(d)g

n

d|n

(b) If h(n) =

μ(d) f

n d

d|n

d

for n ≥ 1.

for n ≥ 1, then show that

f (n) =

h(d) for n ≥ 1.

d|n

(c) Prove that (n) = −

μ(d) log d.

d|n

7.9 Let φ(n) be the Euler φ-function which counts the number of integers 1 ≤ m ≤ n with (m, n) = 1. Prove that 1 φ(n) = n 1− p p|n and derive that φ(n) multiplicative. is (Hint: Write n = 1 and apply Exercise 7.8(a).) d|n

1≤a≤n (a,n)=d

7.10 For an integer k ≥ 1, show that 1 φ(k) = log x n k n≤x

(n,k)=1

234

7 The Riemann Zeta Function and the Prime Number Theorem

as x → ∞. 1 = (Hint: Write μ(d).) n n≤x n≤x d|(n,k) (n,k)=1

7.11 Show that c1 n 2 ≤ φ(n)σ1 (n) ≤ c2 n 2 , where c1 and c2 are positive constants. pa p +1 − 1 .) (Hint: Use Exercise 7.9 and σ1 (n) = p−1 p|n 7.12 As x → ∞, prove that (n) = log x + O(1), (a) n n≤x log p (b) = log x + O(1), p p≤x 1 1 (c) where B is a constant. = log log x + B + O p log x p≤x (Hint: (a) Show that m log m + O(m) = log m! =

m (n) + O(m). n n≤m

log n if (c) Apply the Abel summation formula with λn = n for n ≥ 2, an = n 1 .) n is prime and 0 otherwise and f (n) = log n 7.13 Let {ak }∞ k=0 be a monotone decreasing sequence of real numbers whose limit is zero and {bk }∞ k=0 be a sequence of complex numbers such that its partial sums ∞ are bounded. Then show that ak bk converges. (Hint: Let Bn =

n

k=0

bk and Sn =

k=0

Sn − Sm = an (Bn − Bm ) +

n

n

ak bk . For n > m ≥ 0, write by (7.4.1)

k=0

(Bk − Bm )(ak − ak+1 ).)

k=m+1

7.14 Find absolute constants c1 > 0 and c2 > 0 such that c1 for n ≥ 2. 7.15 For n ≥ 1, prove that

n n < π(n) < c2 log n log n

7.14 Exercises

235

(a) ψ(n) ≥ (n − 2) log 2 and (b) ϑ(n) ≤ 2n log 2. (Hint: (a) Find a lower bound for the least common multiple of 1, 2, . . . , 2n + 1 1 by considering the integral x n (1 − x)n d x. 0

(b) The proof is by induction on n. We may assume that n = 2m + 1 is prime. Then show that ϑ(2m + 1) − ϑ(m + 1) ≤ 2m log 2.) ψ(x) 7.16 Apply (7.4.3) to show that if lim exists, then it has to be equal to 1. x→∞ x 7.17 Show that for infinitely many n, we have d(n) > 2(1−)(log n)/(log log n) . (Hint: Estimate when n is of the form

p. It is known that for > 0, there

p≤x

exists N0 = N0 () such that

d(n) < 2(1+)(log n)/(log log n) for n ≥ N0 .) 7.18 Show that PNT implies that lim

n→∞

pn = 1. n log n

7.19 For > 0, show that there exists a prime in the interval (x, x + x) whenever x exceeds a sufficiently large number depending only on . 7.20 Use 5 + 8 cos θ + 4 cos 2θ + cos 3θ ≥ 0 in place of 3 + 4 cos θ cos 2θ ≥ 0 to give a proof of Theorem 7.9. ∞ 7.21 Show that the series n −1−it is divergent. n=1

7.22 Let |t| ≥ 2. Assume that ζ(s) has no zero in σ > 1 −

1 . Then show that log |t|

ζ (s) 1 = O(log2 |t|) in σ > 1 − . ζ(s) 2 log |t| (Hint: Apply Lemma 5.5 as in the proof of Theorem 7.31.) 7.23 Assume the Riemann hypothesis. Let σ0 > 21 and > 0. Then ζ(s) = O(|t| ), uniformly in σ ≥ σ0 .

ζ (s) = O(|t| ) ζ(s)

236

7 The Riemann Zeta Function and the Prime Number Theorem

7.24 For a > 0, the Hurwitz Zeta function in σ > 1 is defined as ζ(s, a) =

∞ n=0

1 . (n + a)s

Show that ζ(s, a) is analytic in σ > 1. Further prove that (s)ζ(s, a) = 0

∞

x s−1 e−ax d x in s > 1. 1 − e−x

(Hint: By change of variable x = (n + a)t with n ≥ 0 in the above integral, we have ∞ e−nt e−at t s−1 dt. (n + a)−s (s) = 0

)

Chapter 8

The Prime Number Theorem with an Error Term

8.1 Introduction In this chapter, we prove in Sect. 8.5 the following. Theorem 8.1 For x ≥ 2, we have x dt 1/10 π(x) = + O xe−C(log x) , 2 log t where C > 0 is an absolute constant. The second term on the right-hand side is called an error term. Integrating by parts, we see that x x x dt = +O , log x (log x)2 2 log t x x and thus Theorem 8.1 implies PNT. Rewriting PNT as π(x) = +o , log x log x x dt it is clear from Theorem 8.1 that 2 log t is a better approximation than logx x to π(x). We have given a proof of PNT in the previous chapter by a method different from that of J. Hadamard and de la Vallée Poussin. The purpose of this chapter is to explain this basic and fundamental method of Hadamard and de la Vallée Poussin. This is done by proving Theorem 8.1, as in Siegel [27]. With a view to keep the details at a simpler level, we have not aimed to give a sharper error term. We refer the readers 1 of log x. For error term to Ingham [18] for error term with exponent 21 in place of 10 1 1 where exponent 10 is replaced by exponent smaller than 2 , the method of Vinogradov on estimating exponential sums is also required. For example, Korobov proved with exponent 35 + for > 0. Further the Riemann hypothesis implies

© Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9_8

237

238

8 The Prime Number Theorem with an Error Term

x

π(x) = 2

dt + O x 1/2 log x , log t

see Ingham [[18], p. 84]. We proved in the previous chapter that ζ(s) has no zero on the line σ = 1. We shall combine in Sect. 8.2 its proof with the estimates from Sect. 7.6 for obtaining a positive lower bound for the absolute value of ζ(1 + it) and this leads to a zero-free region for ζ(s) asymptotically close to the line σ = 1. We considered the function ψ(x) for a proof of PNT in the previous chapter. Analogously, we introduce the function x

ψ1 (x) = 0

to ψ1 (x) by

ψ(u)du and show in Sect. 8.3 that it suffices to prove an approximation

1 2 x 2

given by ψ1 (x) =

1 2 1/10 x + O x 2 e−C1 (log x) 2

with some positive constant C1 for the proof of Theorem 8.1. Further we give an integral representation for ψ1 (x) in Sect. 8.4. Then we are ready with all the ingredients required for the proof of Theorem 8.1 and we complete its proof in Sect. 8.5. The crucial idea of shifting the line of integration for the proof of Theorem 8.1 is explained while estimating (8.5.7). We denote by c1 , c2 , . . . positive constants in this chapter. Unless otherwise mentioned, we understand that they are absolute constants. 1 . We refer to [11, 16, 17, 27] for the topics in this chapter and for further Let δ = 10 studies and related topics.

8.2 Positive Lower Bound for |ζ(1 + i t)| and Zero-Free Region for ζ(s) We begin this section with a proof for the following positive lower bound for the absolute value of ζ(1 + it). Lemma 8.2 There exists c1 such that |ζ(1 + it)| > c1 (log t)−7 for t ≥ 2. Proof We may assume that t exceeds a sufficiently large constant, otherwise the assertion follows from Theorem 7.9. Let σ > 1. Then we have |ζ 3 (σ)ζ 4 (σ + it)ζ(σ + 2it)| ≥ 1 as in the proof of Theorem 7.9. Therefore, we derive from Lemma 7.10 with α = 0 that

8.2 Positive Lower Bound for |ζ(1 + it)| and Zero-Free Region for ζ(s)

|ζ(σ + it)| ≥ |ζ(σ)|−3/4 |ζ(σ + 2it)|−1/4 ≥ c2 (σ − 1)3/4 (log t)−1/4

239

(8.2.1)

since ζ(s) has a simple pole at s = 1. Further we observe that

σ

ζ(1 + it) − ζ(σ + it) = −

ζ (u + it)du.

(8.2.2)

1

Therefore, by (8.2.1) and Lemma 7.11 with α = 1, we get

|ζ(1 + it)| ≥ |ζ(σ + it)| −

σ

1

ζ (u + it)du

≥ c2 (σ − 1)3/4 (log t)−1/4 − c3 (σ − 1)(log t)2 ,

where c3 = u 2 (1) in Lemma 7.11. We take σ−1=

c2 2c3

4

(log t)−9 .

Then we get |ζ(1 + it)| ≥

1 c24 (log t)−7 . 16 c33

Now we derive from Lemma 8.2 a zero-free region for ζ(s) given by the following result. Lemma 8.3 Let t ≥ 2. There exist positive constants c4 and c5 such that for σ >1− we have

c4 , (log t)9

(8.2.3)

|ζ(σ + it)| ≥ c5 (log t)−7 .

In particular, we see from the above lemma that ζ(s) = 0 whenever σ satisfies (8.2.3). Thus, Lemma 8.3 gives zero-free region (8.2.3) for ζ(s). Proof We may assume that t exceeds a sufficiently large positive constant. Then σ > 1 − log1 t . Now, by applying Lemmas 8.2 and 7.11 with α = 1 to (8.2.2), we derive |ζ(σ + it)| ≥ |ζ(1 + it)| − c3 |σ − 1| log2 t ≥ c1 (log t)−7 − c3 |σ − 1| log2 t. (8.2.4) We may assume that |σ − 1| >

c1 (log t)−9 2c3 c1

otherwise the assertion follows from

(log t)−9 and the assertion follows from 2c3 (8.2.1). Therefore, we may suppose that σ − 1 < 0. Then σ < 1 − 2cc13 (log t)−9 which contradicts (8.2.3) if c4 = c1 /(2c3 ).

(8.2.4). If σ − 1 ≥ 0, then σ − 1 >

240

8 The Prime Number Theorem with an Error Term

Finally, by combining Lemmas 7.11 and 8.3, we derive the following estimate for the absolute value of ζ (s)/ζ(s). Corollary 8.4 Let |t| ≥ 2. There exist positive constants c6 and c7 such that for σ >1−

c6 , (log |t|)9

ζ (s)

9

ζ(s) ≤ c7 (log |t|) .

we have

Proof We may suppose that t > 0 since ζ(s) and ζ (s) are symmetric around the x-axis. Now the assertion follows by combining Lemma 7.11 with α = 1 and Lemma 8.3.

8.3 An Equivalent Version of Prime Number Theorem 8.1 with Error Term in Terms of ψ1 (x) We recall from Sect. 8.1 that ψ1 (x) =

x

ψ(u)du

for x ≥ 0.

0

Further we define

(x) =

∞ 1 , m m=1 p pm ≤x

which can be rewritten as

We approximate

(x) =

(n) . log n 2≤n≤x

(8.3.1)

(x) by π(x) as given in the following.

Lemma 8.5 For x ≥ 2, we have 1 (x) = π(x) + O(x 2 log x). Proof We have 0≤

(x) − π(x) ≤ 1 + ··· + 1, p≤x 1/2

p≤x 1/r

(8.3.2)

8.3 An Equivalent Version of Prime Number Theorem …

241

where either r = 1 in which case the right-hand side is zero or r is the least positive integer such that x 1/r ≤ 2. In the latter case, we have x 1/r ≤ 2 < x 1/(r −1) implying

log x log 2

≤r

0. Then x δ δ e−θ(log t) dt = O xe−θ(log x) , 2

where the constant implied by the symbol O depends only on θ. Proof We rewrite the integral on the left-hand side as

x

δ

e−θ(log t) dt =

2

x

δ

eθ(log t−(log t) )

2

dt . tθ

Since log t − (log t)δ is non-decreasing function of t, the integral is at most δ

eθ(log x−(log x) )

2

x

dt tθ

and the assertion follows immediately if θ = 1. If θ = 1, we integrate by parts to obtain x δ δ e−(log t) dt = xe−(log x) + O(1). 2

Now we shall show that it suffices to approximate ψ1 (x) by 21 x 2 for the proof of Theorem 8.1 as follows. Lemma 8.7 Assume that there exists c8 > 0 such that ψ1 (x) =

Then

x

π(x) = 2

where c9 = c8 /2.

1 2 δ x + O x 2 e−c8 (log x) . 2 dt δ + O xe−c9 (log x) , log t

(8.3.3)

(8.3.4)

242

8 The Prime Number Theorem with an Error Term

Proof Assume (8.3.3) with c8 > 0. We may assume that x exceeds a sufficiently large absolute constant otherwise (8.3.4) follows immediately. First, we prove that δ (8.3.5) ψ(x) = x + O xe−c9 (log x) . δ

Let h = xe−c9 (log x) . Then δ

h = x 2 e−c8 (log x) h −1 , 0 < h

0 and y > 0, we have 1 2πi

d+i∞ d−i∞

ys ds = s(s + 1)

0, if y ≤ 1 1 1 − y , if y ≥ 1.

√ Proof Let T ≥ 2. Draw a circle around the origin and with radius R = d 2 + T 2 > 2. This circle meets the line σ = d at d + i T and d − i T . Denote by 1 and 2 the part of the circle to the right and the left of the line σ = d, respectively (Figs. 8.1 and 8.2). We use the contour 1 when y ≤ 1 and 2 when y ≥ 1. By the Cauchy residue theorem 2.13, we have ⎧ 1 ys ⎪ ⎪ ds, if y ≤ 1 ⎪ ⎪ d+i T ⎨ 2πi 1 s(s + 1) 1 ys ds = ⎪ 2πi d−i T s(s + 1) ⎪ 1 1 ys ⎪ ⎪ + ds, if y ≥ 1. 1 − ⎩ y 2πi 2 s(s + 1) It suffices to show that both the integrals 1 2πi

1

1 ys ds and s(s + 1) 2πi

2

ys ds s(s + 1)

tend to zero as T → ∞. For s ∈ 1 , we have |y s | = y Re(s) ≤ 1 since y ≤ 1, Re(s) > 0 and for s ∈ 1 ∪ 2 , we have |s(s + 1)| = R|s + 1| ≥ R(|s| − 1) = R(R − 1) > since R > 2. Therefore

R2 , 2

244

8 The Prime Number Theorem with an Error Term

d + iT Γ1

0

Re(z)

d

d − iT Im(z) Fig. 8.1 Contour for Lemma 8.8 with y ≤ 1 Fig. 8.2 Contour for Lemma 8.8 with y ≥ 1

d + iT Γ2

Re(z)

−1 O d

d − iT Im(z)

8.4 Integral Representation for ψ1 (x)

1

2πi

1

245

1 2 ys 2 ds

≤ →0 2π R = s(s + 1) 2π R 2 R

as T tends to infinity and

1

2πi

2

2 yd ys 2y d ds

≤ →0 2π R = 2 s(s + 1) 2π R R

as T tends to infinity.

Now we derive from Theorem 7.7 and Lemma 8.8 the following integral representation for ψ1 (x). Lemma 8.9 For d > 1 and x > 1, we have 1 ψ1 (x) = − 2πi

d+i∞

d−i∞

x s+1 ζ (s) ds. s(s + 1) ζ(s)

(8.4.1)

Proof Let d > 1 and x > 1. By Theorem 7.7, we have 1 2πi

d+i∞ d−i∞

d+i∞ ∞ ζ (s) 1 xs xs (n) − ds = ds s(s + 1) ζ(s) 2πi d−i∞ s(s + 1) n=1 n s x s ∞ (n) d+i∞ n = ds 2πi s(s + 1) d−i∞ n=1

since the series converges uniformly on σ = d > 1. Further the above sum is equal to 1 (x − n)(n) x n≤x by Lemma 8.8. Therefore, we have −

1 2πi

d+i∞ d−i∞

x s+1 ζ (s) ds = (x − n)(n). s(s + 1) ζ(s) n≤x

(8.4.2)

Now we apply Abel’s summation formula from Sect. 7.4 with λn = n, an = (n), (n) = ψ(x) to the right-hand side of (8.4.2). We f (t) = x − t and A(x) = n≤x

derive from (7.4.2) that n≤x

(x − n)(n) = 1

x

ψ(u)du = ψ1 (x)

(8.4.3)

246

8 The Prime Number Theorem with an Error Term

since f (x) = 0 and (8.4.2) and (8.4.3).

1 0

ψ(u)du = 0. Now (8.4.1) follows immediately by combining

8.5 Proof of Theorem 8.1 By Lemma 8.7, it suffices to prove (8.3.3). We may assume that x ≥ x0 where x0 is sufficiently large absolute constant, otherwise the assertion follows immediately. Let t ∈ R and we write α(t) = 1 −

c10 c10 , β(t) = 1 + , 9 (log(|t| + 2)) (log(|t| + 2))9

(8.5.1)

where c10 = c6 /2 and c6 is the constant appearing in Corollary 8.4. We observe that α(−t) = α(t) and β(−t) = β(t) so that the curves are symmetric around x-axis. Let = {s | Re(s) > 1}. We observe that is convex region. For U ≥ 2, let L 1 , L 2 , L 3 and L 4 be given by Fig. 8.3 and L be the curve β(t) with parameter interval (−∞, ∞). Then (L 3 )∗ = {(β(t), t) | t ∈ [−U, U ]} and (L )∗ = {(β(t), t) | − ∞ < t < ∞} and = L 1 + (−L 2 ) + (−L 3 ) + (−L 4 ) is a closed path in . Further we see from x s−1 ζ (s) is analytic in . Therefore, we derive from TheoCorollary 7.2 that s(s + 1) ζ(s) rem 2.4 that 1 x s−1 ζ (s) ds = 0. 2πi s(s + 1) ζ(s)

Fig. 8.3 Shifting the line of integration to the left (I)

1 + iU

(β(U ), U )

L2

3 2

+ iU

0

L1

1

3 2

Re(z)

L3 1 − iU

(β(U ), −U )

L4

3 2

− iU

8.5 Proof of Theorem 8.1

247

Fig. 8.4 Shifting the line of integration to the left (II) (α(T ), T )

L2

L6 (β(T ), T ) L1 Re(z)

1

O

L3 (α(T ), −T )

L4

(β(T ), −T ) L5

Im(z)

Thus

x s−1 ζ (s) ds = s(s + 1) ζ(s) j=2 4

L 1

L j

x s−1 ζ (s) ds. s(s + 1) ζ(s)

(8.5.2)

Since |s(s + 1)| ≥ U 2 if s ∈ L 2 , we see from Corollary 8.4 that the absolute value of the integral along L 2 is O

x 1/2 (log U )9 U2

→0

as U tends to infinity. Further, since the absolute values of the integral along L 2 and L 4 are identical, we observe from (8.4.1) with d = 3/2 and (8.5.2) that ψ1 (x) 1 =− 2 x 2πi

L

x s−1 ζ (s) ds s(s + 1) ζ(s)

(8.5.3)

by letting U tend to infinity. Let T ≥ 2. We shall take T0 and T as functions of x0 and x, respectively, so that T ≥ T0 and T0 is sufficiently large. Let 1 be such that s ∈ 1 whenever Re(s) > c6 1− where c6 is the constant appearing in Corollary 8.4. We observe (log(|t| + 2))9 that 1 is simply connected. Let L 1 , L 2 , L 3 and L 4 be as in Fig. 8.4 and γ = L 1 + (−L 2 ) + (−L 3 ) + (−L 4 ).

248

8 The Prime Number Theorem with an Error Term

x s−1 ζ (s) is analytic in s(s + 1) ζ(s) 1 1 except at s = 1 where it has simple pole with residue − 2 , see Exercise 2.7 (b), and 1 ∈ / γ ∗ . Therefore, we derive from the Cauchy residue theorem 2.13 that

Then γ is a closed path in 1 . Further, by Corollary 8.4,

1 2πi Then 1 2πi

L1

γ

1 x s−1 ζ (s) ds = − . s(s + 1) ζ(s) 2

4 1 1 x s−1 ζ (s) x s−1 ζ (s) ds = ds − , s(s + 1) ζ(s) 2πi i=2 L i s(s + 1) ζ(s) 2

which we combine with (8.5.3) to derive 6 ψ1 (x) 1 x s−1 ζ (s) 1 ds + = − x2 2πi i=2 L i s(s + 1) ζ(s) 2

since L = L 1 + L 5 + L 6 . Let L = L 2 + · · · + L 6 . For 2 ≤ i ≤ 6, let Ji be the integral of along L i and put J =

6

x s−1 ζ (s) s(s + 1) ζ(s)

Ji . Then

j=2

ψ1 (x) 1 = −J + . 2 x 2

(8.5.4)

Now we estimate the integrals Ji with 2 ≤ i ≤ 6. By Corollary 8.4, we have ζ (s) = O (log |t|)9 if |t| ≥ 2, s ∈ 1 ζ(s) and

ζ (s) =O ζ(s)

1 |σ − 1|

if s ∈ L 1 ∪ L 3 , |t| < 2, since ζ (s)/ζ(s) has a simple pole at s = 1 with residue −1. Thus, we always have ζ (s) = O (log(|t| + 2)9 if s ∈ L . ζ(s) First, we estimate J5 and J6 . For s ∈ L 5 ∪ L 6 ∪ L 2 ∪ L 4 , c10

|x s−1 | = x σ−1 ≤ x (log T )9

8.5 Proof of Theorem 8.1

249

and thus c 10 J5 = J6 = O x (log T )9

∞ T

(log t)9 dt t2

c10 = O x (log T )9 T −1/2 .

c10 J2 = J4 = O x (log T )9 (log T )9 T −2 .

Further

(8.5.5)

(8.5.6)

Finally, we estimate J3 . Writing α = α(t), we see that for s ∈ L 3 |s(s + 1)| ≥ σ 2 + t 2 ≥ α2 + t 2 and we derive from (8.5.2) that |x s−1 | = x σ−1 = x

c

10 − (log(|t|+2)) 9

≤x

c

− (log11T )9

,

where c11 = c10 /2. Therefore c − 11 J3 = O x (log T )9

T 0

(log(t + 2))9 dt α2 + t 2

c − 11 = O x (log T )9 .

(8.5.7)

c − 11 The above estimate O x (log T )9 for J3 is a decreasing function of d((L 3 )∗ , {1}). It is obtained by shifting the line of integration to the left of the line σ = 1 by the Cauchy theorem. By combining (8.5.5), (8.5.6) and (8.5.7), we get c10 c 1 − 11 J = O x (log T )9 T − 2 + x (log T )9 . Let T be such that

which we rewrite as

c10

x (log T )9 T − 2 = x 1

1

c

− (log11T )9

,

c12

T 2 = x (log T )9 , c12 = c10 + c11 . By taking logarithm on both the sides, we have

Let T0 be given by

log x = (2c12 )−1 (log T )10 .

(8.5.8)

log x0 = (2c12 )−1 (log T0 )10 .

(8.5.9)

Then T0 is sufficiently large and T ≥ T0 , since x ≥ x0 with x0 is sufficiently large. Now

250

8 The Prime Number Theorem with an Error Term

δ −c log x J = O e 11 (log T )9 = O e−c13 (log x) , where c13 = c11 (2c12 )9/10 . Then we see from (8.5.4) that x2 δ + O x 2 e−c13 (log x) . 2

ψ1 (x) =

8.6 Exercises 8.1 Show that lim

n→∞

pn+1 − pn = 1. n

8.2 Let c > 0 and x > 0. For k ≥ 1, show that 1 2πi

c+i∞ c−i∞

xs ds = s(s + 1) · · · (s + k)

0, if x ≤ 1 1 1 1 − , if x ≥ 1. k! x

8.3 Let A > 0, B > 0, u > 0 and |t| ≥ 2. Assume that for σ >1− we have

u , (log |t|) A

ζ (s) = O (log(|t|)) B . ζ(s)

Then prove that

x

π(x) = 2

du 1/(A+1) + O xe−v(log x) , log u

where v and the constant implied by O symbol depend only on A, B and u. 8.4 Assume the Riemann hypothesis and let > 0. Then show that 3 ψ(x) = x + O x 4 + .

(8.6.1)

ζ (s) as given in Exercise 7.23. The error term in ζ(s) 1 (8.6.1) has been sharpened to O(x 2 (log x)2 ), see [[18], Theorem 30].) (Hint: Apply estimate for

Chapter 9

The Dirichlet Series and the Dirichlet Theorem on Primes in Arithmetic Progressions

9.1 Introduction In the last two chapters, we studied ζ(s) given by the series

∞ 1 in σ > 1. In this ns n=1

∞ an with an ∈ C and s ∈ C and these ns n=1 are known as the Dirichlet series. The numbers an are called the coefficients of the series. Several results analogous to those of power series stated in Sect. 2.1 and of the ∞ 1 series in Chap. 7 are also valid for the Dirichlet series and we shall prove them ns n=1 in Sects. 9.2 and 9.3. For example, we show that there exists σc ∈ C∞ such that the Dirichlet series converges in σ > σc and diverges in σ < σc . Further it is uniformly convergent on compact subsets of σ > σc . The line σ = σc is called the line of ∞ 1 convergence. For example, σc = 1 for . We also prove a theorem of Landau ns n=1 that a Dirichlet series with non-negative coefficients has a singularity at σ = σc . Further we apply this result in Sect. 9.4 to give another proof of Theorem 7.9 that ζ(s) has no zero on the line σ = 1. The proof depends on the Ramanujan identity given in Theorem 7.6. We recall that it has already been proved in Chap. 7 (see Theorems 7.9, 7.18 and 7.19) that non-vanishing of ζ(s) on the line σ = 1 is equivalent to the Prime Number Theorem. An important class of Dirichlet series is series given by the Dirichlet functions L(s, χ). For studying these functions, a knowledge of the Dirichlet characters is required and we give an account of the Dirichlet characters in Sects. 9.5 and 9.6. We prove in Sects. 9.7 and 9.8 that L-functions do not vanish at the point s = 1. This is crucial for the proof of the Dirichlet theorem that there are infinitely many primes in an arithmetic progression which we shall prove in Sect. 9.9. We refer to [5, 6, 11, 15, 16, 20, 31] for the topics in this chapter and for further studies and related topics.

chapter, we consider more general series

© Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9_9

251

252

9 The Dirichlet Series and the Dirichlet Theorem …

9.2 Convergence of the Dirichlet Series We recall that

∞ an n=1

ns

, an ∈ C, s ∈ C, s = σ + it

(9.2.1)

is called the Dirichlet series. We prove the following. Lemma 9.1 Assume that (9.2.1) converges at s0 = σ0 + it0 . Then (a) Let 0 < θ < π2 . Then (9.2.1) converges uniformly in 0 < | arg(s − s0 )| ≤ π2 − θ. (b) The series (9.2.1) converges in σ > σ0 and converges uniformly on compact subsets of σ > σ0 . Proof Assume that (9.2.1) converges at s0 = σ0 + it0 and let s = s0 + s . By writing an bn = s , we observe that n0 ∞ ∞ ∞ an bn = with bn finite. ns ns n=1 n=1 n=1

Thus, there is no loss of generality in assuming that s0 = 0. Let s = σ + it with σ > 0, > 0 and we put ∞ rn = aν . ν=n+1

Since

∞

aν is finite by assumption, there exists M0 depending only on such that

ν=1

|rn | < for n ≥ M0 and let M > M0 , N > M. Then, we have N N N 1 r M−1 an rn−1 − rn 1 rN + = = r − − . n s s s s s n n (n + 1) n M (N + 1)s n=M n=M n=M We observe that N N 1 1 1 1 ≤ rn − |r | − n (n + 1)s n s n=M (n + 1)s ns n=M n+1 N n+1 N d x dx |s| 0. Let K be a compact set in σ > 0. Then we see from Theorem 1.4 that there exist u > 0 and δ > 0 depending only on K such that |s| < u and σ > δ whenever s ∈ K . |s| u Therefore, < and we see from (9.2.2) that σ δ N a 2u n . < s n δ n=M Hence, (9.2.1) converges uniformly on K .

Let U = {σ | (9.2.1) converges} and V = {σ | (9.2.1) diverges} . By Lemma 9.1 (b), every element of V is less than all the elements of U . If U = ∅ and V = ∅, there exists a real number σc such that (9.2.1) converges for σ > σc and diverges for σ < σc . Further we put σc = ∞ if U = ∅ and σc = −∞ if V = ∅. Thus, there exists σc ∈ C∞ such that (9.2.1) converges for σ > σc and diverges for σ < σc . The series (9.2.1) may either converge or diverge when σ = σc . Definitions (i) σc is called the abscissa of convergence of (9.2.1). (ii) σ = σc is called the line of convergence of (9.2.1). (iii) σ > σc is called the half plane of convergence of (9.2.1). (iv) The series (9.2.1) converges absolutely if

254

9 The Dirichlet Series and the Dirichlet Theorem … ∞ ∞ |an | an < ∞. s= n nσ n=1 n=1

Then there exists σa ∈ C∞ such that (9.2.1) converges absolutely if σ > σa and diverges absolutely if σ < σa . Here σa = ∞ if the series (9.2.1) diverges absolutely everywhere and σa = −∞ if the series (9.2.1) converges absolutely everywhere. We call σa the abscissa of absolute convergence of (9.2.1). It is clear that if a Dirichlet series converges absolutely, then it converges. Therefore σc ≤ σa . On the other hand, we prove the following. Lemma 9.2 We have σa − σc ≤ 1 if σa < ∞.

(9.2.3)

|an | = 0. Thus, there exists a nσ < K for n ≥ 1. Now we consider

Proof Let > 0 and σc < σ < σc + ε. Then lim constant K such that

|an | nσ

n→∞

∞ ∞ |an | 1 ≤K < ∞. 1+σ+ 1+ n n n=1 n=1

Therefore σa ≤ 1 + σ + < 1 + σc + 2ε. This is true for every > 0. Hence, the assertion (9.2.3) holds.

∞

(−1)n−1 . ns n=1 This converges if and only if Re(s) > 0 and therefore σc = 0 by Lemma 9.1 (b). Further σa = 1 and hence σa − σc = 1. Next, we show that a Dirichlet series represents analytic function in its half plane of convergence. The inequality (9.2.3) is optimal. For this, we consider the example

Theorem 9.3 Let f (s) be given by the Dirichlet series f (k) (s) with k ≥ 1 are analytic in σ > σc given by f (k) (s) = (−1)k

∞ an n=1

ns

in σ > σc . Then

∞ an (log n)k in σ > σc . ns n=1

Proof Note that the series for f (k) (s) is obtained by term-wise differentiation. The assertion follows from Lemma 9.1 (b), (2.3.3) and Sect. 2.3 (iv).

9.2 Convergence of the Dirichlet Series

255

A power series must have a singularity on the circle of convergence. But the Dirichlet series need not have a singularity on its line of convergence. For this, we ∞ consider again the series (−1)n−1 n −s . This has σc = 0. Therefore, it is analytic n=1

in σ > 0 by Theorem 9.3. Further (1 − 2−s )ζ(s) =

∞ (−1)n−1 n=1

ns

in σ > 1

and the left-hand side is entire since the pole of ζ(s) at s = 1 cancels with the zeros of 1 − 21−s at s = 1. Therefore (1 − 21−s )ζ(s) =

∞ (−1)n−1 n=1

ns

in σ > 0

by Identity theorem for holomorphic functions, see Sect. 2.3(ii). Thus, the function representing the Dirichlet series has no singularity on its line of convergence. On the other hand, if the coefficients of the Dirichlet series are non-negative, we prove the following. Theorem 9.4 (Landau) Let an ≥ 0 for n ≥ 1 and f (s) be given by the Dirichlet ∞ an series with σc < ∞. Then f (s) is not analytic at s = σc . ns n=1 Proof We may assume that σc = 0. We suppose that f (s) is analytic at s = 0 and we shall arrive at a contradiction (Fig. 9.1). Since f (s) is analytic at s = 0, there exists > 0 such that f (s) is analytic in D(0, ). Let 0 < δ < such that δ 2 + 2δ < .

Fig. 9.1 Dirichlet series with non-negative coefficients and analytic at abscissa of convergence

(0,

2 + 2)

(0,

−

− δ 2 + 2δ

−δ

o

δ 2 + 2δ)

1

256

9 The Dirichlet Series and the Dirichlet Theorem …

Then f (s) is analytic in D(0,

√

δ 2 + 2δ) and

D(1, 1 + δ) ∩ {s ∈ C|Re(s) ≤ 0} ⊆ D(0, δ 2 + 2δ). Then f (s) is analytic in D(1, 1 + δ) since f (s) is analytic in Re(s) > 0. Thus ∞ (s − 1)ν

ν!

ν=0

f (ν) (1) < ∞ for some s < 0.

By Theorem 9.3, we have f (ν) (1) =

∞

an

n=1

(− log n)ν . n

Therefore ∞> =

∞ ∞ (s − 1)ν ν=0 ∞ n=1

=

ν! an n

∞ an n=1

n

∞ ν=0

an

n=1

∞ ∞ (− log n)ν (1 − s)ν (log n)ν = an n ν! n ν=0 n=1 ∞

an ((1 − s) log n)ν = e(1−s) log n n n n=1

n 1−s =

∞ an n=1

ns

.

This is not possible since σc = 0. The interchange of infinite sums above is justified since, because of an ≥ 0 and s < 0, the series are of positive terms.

9.3 Multiplication of Two Dirichlet Series ∞ ak

∞ bm Let s be given. Suppose that and are absolutely convergent. Therefore, s s k m m=1 k=1 its terms can be rearranged in any way. Thus

∞ ∞ ∞ ∞ ∞ ak bm −s a b (km) = cn n −s , = k m s s k m m=1 n=1 k=1 k=1 m=1 where cn =

km=n

ak bm .

9.3 Multiplication of Two Dirichlet Series

257

Therefore, a product of two absolutely convergent Dirichlet series is an absolutely convergent Dirichlet series in a half plane where both the series converge absolutely. Example 9.1 In σ > 1,

∞ ∞ ∞ 1 (m) log n = s s k m ns m=1 n=1 k=1 since

(m) = log n.

m|n

Theorem 9.5 (Uniqueness Theorem) Suppose that F(s) =

∞ an n=1

and G(s) =

in σ > σa

ns

∞ bn n=1

in σ > σa

ns

converge absolutely. Let {sk }∞ k=1 be a sequence of complex numbers such that Re(sk ) → ∞ as k → ∞ and F(sk ) = G(sk ) for k ≥ 1. Then an = bn for n ≥ 1. Proof The proof is by contradiction. We put H (s) = F(s) − G(s) Then H (s) =

and h n = an − bn . ∞ hn n=1

ns

.

We may assume that there exists at least one n such that h n = 0. Let N be the first positive integer such that h N = 0. Then H (s) =

∞ hN hn + . s N ns n=N +1

Therefore h N = H (s)N s − N s Let s = sk for k ≥ 1. Then H (sk ) = 0 and

∞ hn . ns n=N +1

258

9 The Dirichlet Series and the Dirichlet Theorem …

h N = −N sk

∞ hn . n sk n=N +1

|h N | ≤ N σk

∞ |h n | . n σk n=N +1

Therefore

(9.3.1)

Let σa < α < σa + 1 be fixed and k be sufficiently large integer such that σk > σa + 1. Then, since α > σa , we have ∞ |h n | 1. ζ(2s) ns n=1

(9.4.1)

Since each of ζ(s + ai) and ζ(s − ai) has a zero at s = 1 and ζ(s) has a simple pole at s = 1 by Theorem 7.8, we see that the numerator of the left-hand side of (9.4.1) is analytic in σ > 0. Further ζ(2s) is non-zero and analytic in σ > 21 by Corollary 7.2.

9.4 Another Proof of Theorem 7.9 That ζ(1 + it) = 0

259

Therefore, the left-hand side of (9.4.1) is analytic in σ > 21 . In fact, it is analytic in σ ≥ 21 except at s = 21 where it has removable singularity since ζ(s) has a pole at s = 1 by Theorem 7.8. On the other hand, the right-hand side of (9.4.1) is a Dirichlet series with nonnegative terms and we conclude from Theorem 9.4 that σc < 21 . Therefore (9.4.1) is valid at s = 21 . This is a contradiction since the right-hand side of (9.4.1) is ≥ 1 whereas the left-hand side tends to zero as s approaches 21 from right.

9.5 Characters of Finite Abelian Groups Let G be a finite abelian group of order h. Then the fundamental theorem on finitely generated abelian groups implies that G is a direct product of finitely many cyclic groups. Thus G = G1 × G2 × · · · × Gk , where G j with 1 ≤ j ≤ k is a cyclic group with generator A j and order r j . Then every element A ∈ G can be uniquely written as A = At11 · · · Atkk ,

0 ≤ tj < rj, 1 ≤ j ≤ k

(9.5.1)

and h = r1 r2 · · · rk . Let E denote the unit element of G and A−1 the inverse element of A ∈ G. Then a character χ on G is a function from G into the multiplicative group C∗ of non-zero elements of C such that χ(AB) = χ(A)χ(B) for A, B ∈ G. Thus, χ is a homomorphism of G into C∗ . The character χ E given by χ E (A) = 1 for every A ∈ G is called the Principal character of G. Then 1 = χ(E) = χ(Ah ) = (χ(A))h for A ∈ G. the set of all characters of G. This is an abelian group, with unit We denote by G element χ E , under product operation given by χ1 χ2 (A) = χ1 (A)χ2 (A) for A ∈ G. is is an abelian group with unit element χ E . Since every element of G Then G determined by its values at A1 , . . . , Ak and χ(A j ) is an r j th root of unity for 1 ≤ ≤ r1r2 · · · rk = h. For 1 ≤ j ≤ k and 0 ≤ t j < r j , let j ≤ k, we observe that |G|

260

9 The Dirichlet Series and the Dirichlet Theorem …

χ jt j (A j ) = e

2πit j rj

, χ jt j (Ai ) = 1 for i = j.

Then the characters χ jt j with 1 ≤ j ≤ k, 0 ≤ t j < r j are distinct and they are ≥ h and hence r1 · · · rk = h in number. Therefore |G| = h. |G| = |G|

(9.5.2)

given by For A ∈ G given by (9.5.1), we define the function ψ from G into G ψ(A) = χt11 χt22 · · · χtkk (A), where χ j = χ j1 for 1 ≤ j ≤ k. Then ψ is a homomorphism, injective and also onto are isomorphic groups. by (9.5.2). Hence G and G

9.5.1 Properties of Characters such that χ(A) = 1. (i) For A ∈ G with A = E, there exists χ ∈ G Proof Let A = E be given by (9.5.1). Then (t1 , . . . , tk ) = (0, . . . 0). There is no loss t 2πi 1 of generality in assuming that t1 > 0 and then χ1t1 (A) = e r1 = 1 since 0 < t1 < r1 . we have (ii) For χ ∈ G,

χ(A) =

A∈G

h, if χ = χ E 0, otherwise.

(9.5.3)

Proof Denote the sum in (9.5.3) by S and let B ∈ G. Then χ(AB) = χ(A) = S. χ(B)S = A∈G

A∈G

Then (χ(B) − 1)S = 0. If S = 0, then χ(B) = 1 for every B ∈ G. Therefore χ = χ E and S = h. (iii) Let A ∈ G. Then χ∈G

χ(A) =

h, if A = E 0, otherwise.

Then Proof Denote by T the sum in (9.5.4). Let χ1 ∈ G.

(9.5.4)

9.5 Characters of Finite Abelian Groups

χ1 (A)T =

261

χ1 χ(A) =

χ∈G

χ(A) = T,

χ∈G

which implies (χ1 (A) − 1)T = 0. If T = 0, then χ1 (A) = 1 for every χ1 ∈ G. Therefore A = E by (i) and T = h.

9.6 The Dirichlet Characters For m > 0, let (Z/mZ)∗ be the multiplicative group of reduced residue classes (mod m). We write G = (Z/mZ)∗ in this section. Thus |G| = φ(m). For an integer a, we denote by a¯ the residue class (mod m) containing a. Thus, 1¯ is the identity element of G. Definition 9.1 A Dirichlet character (mod m) is a function from Z into C such that χ(ab) = χ(a)χ(b) χ(a) = χ(b)

for a, b ∈ Z,

if a ≡ b

(mod m)

and χ(a) = 0

if and only if (a, m) > 1.

Thus, a Dirichlet character (mod m) is completely multiplicative, periodic with period m and vanishes at all integers a with (a, m) > 1. There is (1-1) correspondence between the set of all Dirichlet characters χ (mod m) and the characters f χ on G given by ¯ = χ(a) f χ (a)

for (a, m) = 1.

Therefore, there are φ(m) distinct Dirichlet characters (mod m) which we denote by χ1 , χ2 , . . . , χφ(m) where χ1 is the Principal Dirichlet character (mod m) given by χ1 (a) = 1 if (a, m) = 1 and 0 otherwise. Further, for (a, m) = 1, we have

φ(m), if a ≡ 1 (mod m) χ(a) = (9.6.1) 0, otherwise, χ( mod m)

where the sum is taken over all the distinct Dirichlet characters (mod m). This follows from (9.5.4) since the sum on the left-hand side in (9.6.1) is equal to the sum over all the characters of G at a. ¯ Similarly, for a Dirichlet character χ (mod m), we derive from (9.5.3) that

262

9 The Dirichlet Series and the Dirichlet Theorem …

χ(a) =

a( mod m)

φ(m), if χ = χ1 0, otherwise,

(9.6.2)

where the sum is taken over any reduced residue system (mod m). Example 9.2 The Dirichlet characters (mod 4). We have m = 4 and φ(m) = 2. Thus there are two distinct Dirichlet characters (mod 4) given by 123 4 n χ1 (n) 1 0 1 0 χ2 (n) 1 0 −1 0 Example 9.3 The Dirichlet characters (mod 5). Here m = 5 and φ(m) = 4. Thus, there are four distinct Dirichlet characters (mod 5). Every character (mod 5) satisfies χ(2)χ(3) = χ(6) = 1 and χ(4) = (χ(2))2 . Therefore, they are given by n χ1 (n) χ2 (n) χ3 (n) χ4 (n)

12 3 11 1 1 −1 −1 1 i −i 1 −i i

4 1 1 −1 −1

5 0 0 0 0

9.7 The Dirichlet L-Functions Let m > 0 be an integer and χ be a Dirichlet character (mod m) which we shall assume from now onwards in this chapter. Then the Dirichlet L-function is defined as ∞ χ(n) L(s, χ) = for s = σ + it. ns n=1 Thus L(s, χ) is a Dirichlet series. If m = 1, we observe that L(s, χ) = ζ(s). We prove the following. Theorem 9.6 L(s, χ) with χ = χ1 is analytic in σ > 0. Proof Let χ = χ1 and we write x = mq + r with 0 ≤ r < m. Then

χ(n) =

n≤x

m

2m

χ(n) +

n=1

n=m+1

mq+r

=

n=mq+1

χ(n)

χ(n) + · · · +

mq n=m(q−1)+1

mq+r

χ(n) +

n=mq+1

χ(n)

9.7 The Dirichlet L-Functions

263

by (9.6.2). Therefore mq+r χ(n) = χ(n) ≤ r < m. n≤x n=mq+1 By Lemma 9.1(b) and Theorem 9.3, it suffices to show that L(s, χ) converges for s > 0. Let s > 0 and M > 0 be an integer. We apply (7.4.2) with λn = n + M − 1 and an = χ(n) for n ≥ 1, f (t) = t −s and A(t) = χ(n). We have |A(t)| < m M≤n≤t

as above and x ∞ χ(n) A(x) m A(t) dt m m = ≤ + s dt + sm = s + s −→ 0 xs xs s s+1 s+1 n t t x M M M M≤n≤x as M tends to infinity. Hence L(s, χ) converges.

We observe in the proof of Theorem 9.6 that σc = 0. Therefore σa ≤ 1 by Lemma 9.2. In fact σa = 1 since ∞ |χ(n)| n=1

n

=

∞ 1 =∞ n n=1

(n,m)=1

by Exercise 7.10. We shall be brief in our details for the results whose proofs are similar to the one already proved for ζ(s) in Chap. 7. We show in the following result that L(s, χ) has the Euler product. Lemma 9.7 We have χ( p) −1 (a) L(s, χ) = 1− s in σ > 1 where the product converges absolutely p p and uniformly on compact subsets of σ > 1. (b) L(s, χ) = 0 in σ > 1. 1 1− . (c) L(s, χ1 ) has a simple pole at s = 1 with residue p p|m (d) log L(s, χ) = (e)

∞ (n) χ(n) n=2

log n n s

in σ > 1 where the logarithm has principal branch.

L(s, χ) ≥ 1 in σ > 1.

χ

(f) σc = 1 if χ = χ1 and σc = 0 otherwise. Further σa = 1. Proof (a) We apply Corollary 7.4 with f (n) = tive. Therefore, we have for σ > 1

χ(n) ns

which is completely multiplica-

264

9 The Dirichlet Series and the Dirichlet Theorem …

χ( p) χ( p 2 ) 1+ s + L(s, χ) = + ··· p p 2s p χ( p) −1 = 1− s p p and the product is absolutely convergent. The product is also uniformly convergent on compact subsets of σ > 1 by Theorem 6.7 (a). (b) The assertion follows by combining Lemma 9.7(a) with Corollary 6.6. (c) By Lemma 9.7(a) and Theorem 7.1, we have

χ1 ( p) ps ( p,m)=1 1 1− s = ζ(s) p p|m

L(s, χ1 ) =

1−

−1

=

1−

p

1 ps

−1 1 1− s p p|m (9.7.1)

for σ > 1. Now the assertion follows since ζ(s) has a simple pole at s = 1 with residue 1 by Theorem 7.8. (d) As in the proof of Theorem 7.7, we have

log L(s, χ) =

p

=

∞ χ( p k ) p

=

χ( p) − log 1 − s p

k=1

kp ks

∞ (n) χ(n) n=1

log n n s

in σ > 1,

(9.7.2)

where the logarithm has principal branch. (e) By (9.7.2), we have χ

∞ 1 log L(s, χ) = χ( p k ) in σ > 1, kp ks χ p k=1

where the sum is taken over all the Dirichlet characters (mod m). Then we derive from (9.6.1) that χ

log L(s, χ) = φ(m)

p pk ≡1( mod m)

∞ 1 in σ > 1. kp ks k=1

9.7 The Dirichlet L-Functions

265

By exponentiating both sides, we get

L(s, χ) ≥ 1 in σ > 1.

χ

(f) Since

∞ |χ(n)| n=1

and

nσ

∞ |χ(n)| n=1

nσ

≤

∞ 1 < ∞ in σ > 1 nσ n=1

∞ 1 = ∞ for σ = 1 = n n=1 (n,m)=1

by Exercise 7.10, we observe that σa = 1. Further σc = 1 if χ = χ1 by (9.7.1). Let χ = χ1 . Then σc ≤ 0 by Theorem 9.6. By Lemma 9.2 and σa = 1, we have σc ≥ σa − 1 = 0 and hence σc = 0.

9.8 Non-vanishing of L(1, χ) Theorem 9.8 L(1, χ) = 0. Proof Let χ = χ0 and assume that L(1, χ0 ) = 0. Then χ0 = χ1 by Lemma 9.7(c). First, we assume that χ0 is not real. Then χ0 is also a Dirichlet character (mod m) different from χ0 such that L(1, χ0 ) = 0. Therefore, we see from (9.7.1) that lim

s→1+

L(s, χ) = 0

χ

since ζ(s) has a simple pole at s = 1 by Theorem 7.8. Here the product is taken over all the Dirichlet characters (mod m). This contradicts Lemma 9.7(e). Thus, we may assume that χ0 is a real character. Therefore, χ0 assumes only the values 0, 1 and −1. We consider g(s) = Therefore in σ > 1, we have

ζ(s)L(s, χ0 ) in σ > 1. ζ(2s)

(9.8.1)

266

9 The Dirichlet Series and the Dirichlet Theorem …

χ0 ( p) χ0 ( p 2 ) g(s) = (1 + p ) 1 + + + ··· ps p 2s p

∞ b( p k ) = 1+ , p ks p k=1

s

where b( p k ) = χ0 ( p k ) + χ0 ( p k+1 ) = χ0 ( p k )(1 + χ0 ( p)) ≥ 0.

Let the function b be extended multiplicatively to all positive integers. Then b(1) = 1 and we have b( p k ) ≤ 2 for k ≥ 0. Then for > 0 and n ≥ n 0 () > 0, we have 0 ≤ b(n) ≤ 2ω(n) ≤ 2c1 log n/ log log n ≤ c2 n , where c1 and c2 are positive constants, see Exercise 7.4. Therefore, the Dirichlet ∞ b(n) series converges for s > 1. Now we derive from Theorem 9.3 that ns n=1 g(s) =

∞ b(n) n=1

ns

(9.8.2)

analytic in σ > 1 and thus σc ≤ 1 where σc is the abscissa of convergence of the above series (9.8.2). Since b(n) ≥ 0, we derive from Theorem 9.4 that g(s) is not analytic at s = σc . Since ζ(s) is analytic in σ > 0 except at s = 1 where it has a simple pole and L(1, χ0 ) = 0, we see that the numerator of g(s) in (9.8.1) is analytic in σ > 0 and further the denominator of ζ(2s) is analytic in σ > 21 where it has no zero. Therefore g(s) is analytic in σ > 21 . Thus σc ≤ 21 . In fact, σc < 21 since g(s) has a removable singularity at s = 21 as ζ(s) has a pole at s = 1. Therefore (9.8.2) is valid for s = 21 . Then 1 ≥ b(1) = 1, 0=g 2 which is a contradiction.

9.9 The Dirichlet Theorem on Primes in Arithmetic Progression Theorem 9.9 Let a and m > 0 be integers such that (a, m) = 1. Then there are infinitely many primes p ≡ a (mod m). It is clear that the assumption (a, m) = 1 is necessary, otherwise there is no prime p ≡ a (mod m).

9.9 The Dirichlet Theorem on Primes in Arithmetic Progression

Proof We show that

p≡a

p (mod m)

267

1 =∞ p

and the assertion follows immediately. For σ > 1, we derive from Lemma 9.7(d) that log L(s, χ) =

χ( p) ps

p

where R(s, χ) =

+ R(s, χ),

∞ χ( p k ) p

kp ks

k=2

and logarithm has principle value. Since (a, m) = 1, there exists an integer b such that ab ≡ 1(modm). Then (b, m) = 1 and therefore χ(b) = 0. Now

χ(b) log L(s, χ) =

χ

where R(s) =

1 χ( pb) + R(s), ps χ p

χ(b)R(s, χ)

χ

and the sum is taken over all the Dirichlet characters (mod m). By (9.6.1), the sum χ( pb) is zero unless pb ≡ 1 (mod m), i.e. p ≡ a (mod m), and φ(m) when χ

p ≡ a (mod m). Thus

χ(b) log L(s, χ) = φ(m)

χ

p≡a

(mod m)

1 + R(s). ps

(9.9.1)

Further |R(s)| ≤

χ

∞ 1 |R(s, χ)| ≤ φ(m) kσ p p k=2

= φ(m)

p

1 p σ ( p σ − 1)

1 ≤ 2φ(m)ζ(2). ≤ 2φ(m) p 2σ p Let s > 0 and s → 1+ . Then R(s) remains bounded whereas the left-hand side in (9.9.1) tend to infinity by Theorems 9.6 and 9.7(c). Therefore

268

9 The Dirichlet Series and the Dirichlet Theorem …

p≡a

as s → 1+ and hence

(mod m)

p≡a

p (mod m)

1 =∞ ps

1 = ∞. p

9.10 The Prime Number Theorem in an Arithmetic Progression For integers k and m > 0 with (k, m) = 1, let

π(x; m, k) =

1,

p≤x p≡k (mod m)

the number of primes p ≡ k (mod m). Thus π(x; 1, 0) = π(x). If the primes are approximately distributed equally evenly in all reduced classes (mod m), we expect π(x; m, k) =

x 1 φ(m) log x

as

x →∞

since a residue class (mod m) contains no prime if it is not reduced. The above asymptotic formula is proved if m is fixed and it is called the Prime Number Theorem in arithmetic progression. The proof runs quite parallel to the one given in Chap. 8, and therefore we shall not give its details. We refer to LeVeque [[20], Sect. 7.4] for its exposition. The error term in the above result is not uniform in m and this restricts the scope of its applications. It has been proved that the error term is uniform if m is less than any power of log x, see [[16], Ch 2] and the proof depends on delicate considerations regarding Siegel zero, see [[15], Ch 21]. Such versions of the Prime Number Theorem in arithmetic progression have several important applications.

9.11 Exercises 9.1 (a) Show that σc = ∞ for

∞ ∞ n! 1 and σ = −∞ for . c s s n n!n n=1 n=1

9.11 Exercises

269 ∞ an

(b) Find σc and σa for the series

n=1

ns

where

an = n − 2 , (−1)n n − 2 , log n, (log(n + 1))−1 . 1

9.2 Let F(s) =

1

∞ f (n) in σ > σa where f (n) is completely multiplicative. Then ns n=1 ∞ F (s) f (n)(n) in σ > σa . =− F(s) ns n=1

9.3 (a) Let f (s) =

∞ an n=1

ns

in σ > σa . Then show that

1 T →∞ 2T

T

lim

(Hint: | f (s)|2 = f (s) f (s) =

−T

| f (s)|2 dt =

n=1

∞ |an |2 n=1

∞ |an |2

n 2σ

+

n 2σ

.

(9.11.1)

am an n it .) m σ nσ m m n,m=n

(b) Calculate the left-hand side of (9.11.1) with f (s) = ζ(s), 9.4 Let χ be a Dirichlet character. Show that

1 , (ζ(s))2 . ζ(s)

3 L (σ, χ)L 4 (σ + it, χ)L(σ + 2it, χ2 ) ≥ 1 and derive that L(s, χ) does not vanish on the line σ = 1. (Hint: The proof is similar to that of Theorem 7.9. We may assume that χ = χ0 and use Re(3 + 4z + z2 ) ≥ 0 whenever |z| = 1.) 9.5 Show that 1 χ(k)π(x, χ), π(x; m, k) = φ(m) χ where the sum is taken over all Dirichlet characters modulo m and π(x, χ) = χ( p). p≤x

Chapter 10

The Baker Theorem

10.1 Introduction A complex number α is called algebraic number if there exists a non-zero polynomial f (X ) ∈ Q[X ] such that f (α) = 0 and algebraic integer if f (X ) ∈ Z[X ] such that f (X ) is monic and f (α) = 0. In fact α satisfies the unique polynomial P(X ) of minimal degree with relatively prime integer coefficients such that the leading coefficient of P is positive. The polynomial P is called the minimal polynomial of α. If P(X ) = a0 X n + a1 X n−1 + · · · + an , we define the height H (P) of P as H (P) = max{|a0 |, |a1 |, . . . , |an |}. Further the degree of α, denoted by deg(α), and the height of α, denoted by H (α), are defined as deg α = deg P, H (α) = H (P), where we write deg P for the degree of P. The denominator d(α) of an algebraic number is the least positive integer such that d(α)α is an algebraic integer. We observe that d(α) ≤ H (α). The number α is called transcendental if α is not algebraic and irrational if α ∈ / Q. For example, e and π are well-known examples of transcendental numbers. Let α be an algebraic number different from 0 and log α be an arbitrary but fixed branch of logarithm for α such that log α = 0. Then for irrational algebraic number β, Gel’fond and Schneider, independently, solved in 1934 Hilbert famous seventh problem that αβ = eβ log α is transcendental. This is known as the Gel’fond-Schneider theorem. There is no loss of generality in assuming that α ∈ / {0, 1} in place of α = 0 with log α = 0 in the Gel’fond-Schneider theorem, see Exercise 10.1. Let n ≥ 1 and α1 , . . . , αn be non-zero algebraic numbers such that log α1 , . . . , log αn are arbitrary but fixed branches of logarithms for α1 , . . . , αn . Then the Gel’fondSchneider theorem is equivalent to if log α1 and log α2 are linearly independent over © Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9_10

271

272

10 The Baker Theorem

Q, then log α1 and log α2 are linearly independent over the field of algebraic numbers. Baker [[5], Ch 2] in 1966–67 extended the Gel’fond-Schneider theorem as follows. Theorem 10.1 Let α1 , . . . , αn be non-zero algebraic numbers such that log α1 , . . . , log αn are linearly independent over Q. Then log α1 , . . . , log αn are linearly independent over the field of algebraic numbers. The proof of Theorem 10.1 depends on the Thue-Siegel lemma, which we shall prove in Sect. 10.2, and on the Cauchy residue theorem. We prove Theorem 10.1 in Sect. 10.3. Further we derive the following extension of the Gel’fond-Schneider theorem from Theorem 10.1 in Sect. 10.4. Corollary 10.2 Let α1 , . . . , αm be non-zero algebraic numbers different from 0, 1 and β1 , . . . , βm are algebraic numbers such that 1, β1 , . . . , βm are linearly indepenβ β dent over Q. Then α1 1 · · · αmm is transcendental. For algebraic numbers α1 , . . . , αn different from zero and algebraic numbers β1 , . . . , βn , we say that = β1 log α1 + · · · + βn log αn is a linear form in logarithms of algebraic numbers with algebraic coefficients. Baker [[5], Ch 2] proved that is transcendental if = 0. If log α1 , . . . , log αn are linearly independent over Q and β1 , . . . , βn are not all zero, the method of proof of Theorem 10.1 allows to give an explicit positive lower bound for || in terms of n, the degree [Q(α1 , . . . , αn , β1 , . . . , βn ) : Q], the heights of α1 , . . . , αn , β1 , . . . , βn and the choice of logarithms log α1 , . . . , log αn for α1 , . . . , αn . It is not always easy to check the linear independence of log α1 , . . . , log αn over Q. In fact, it is possible to give positive lower bound for || if = 0. Several sharpenings and extensions for lower bounds for || have been obtained and these constitute the Theory of linear forms in logarithms (more precisely Theory of linear forms in logarithms of algebraic numbers with algebraic coefficients) having important applications in several directions, see [5, 7–9, 26]. For example, we state the following result (without proof) from this theory. Let β1 , . . . , βn ∈ Z with absolute values not exceeding B, where B ≥ 2. Assume that the heights of α1 , . . . , αn do not exceed A1 , . . . , An where each A j ≥ 3. Put n log An and = logAn . Then Baker[[7], Ch 1] [Q(α1 , . . . , αn ) : Q] = d, = proved: If = 0, we have

j=1

|| > exp −(16nd)200n log log B .

(10.1.1)

We refer to [5–9, 26] for the topics in this chapter and for further studies and related topics.

10.2 The Thue-Siegel Lemma

273

10.2 The Thue-Siegel Lemma We prove the following version of the Thue-Siegel lemma given by Ramachandra [22]. Lemma 10.3 Assume that the coefficients of linear forms yk = ak1 x1 + ak2 x2 + · · · + akq xq

(10.2.1)

with 1 ≤ k ≤ p are algebraic integers in a field K of degree h over Q. Suppose that the maximum of absolute values of conjugates of ak j with 1 ≤ k ≤ p and 1 ≤ j ≤ q do not exceed A where A ≥ 1. Assume that 2q > ph(h + 1). Then there exist xi ∈ Z, not all zero, with ph(h+1)

|x j | < 1 + (2q A) 2q− ph(h+1) for 1 ≤ j ≤ q. Proof For an integer X ≥ 1, let I X be the set of all q-tuples (x1 , . . . , xq ) such that x j ∈ Z and |x j | ≤ X for 1 ≤ j ≤ q. Then |I X | = (2X + 1)q . Let Jq AX

⎧ ⎫ q ⎨ ⎬ = (y1 , y2 , . . . , y p ) yk = ak j x j with (x1 , . . . , xq ) ∈ I X . ⎩ ⎭ j=1

We observe that yk with 1 ≤ k ≤ p are algebraic integers in K such that the maximum of the absolute values of the conjugates of yk is at most q AX . If yk satisfies ykdk + a1 ykdk −1 + · · · + adk = 0 with ai ∈ Z, dk ≤ h, then |a j | ≤

dk (q AX ) j for 1 ≤ j ≤ dk j

since |a j | is absolute value of jth elementary symmetric function of yk and its conjugates. Therefore

274

10 The Baker Theorem

⎛

⎞ p h h 2 (q AX ) j + 1 ⎠ |Jq AX | ≤ ⎝h j j=1 ≤ h p 2hp (q AX )

ph(h+1) 2

h

h

j

j=1

+

p

1 2 j+1

since q ≥ 2, A ≥ 1, X ≥ 1. Further h

h j=1

j

+

1

p

2 j+1

≤

2h − 1 + h

1 2

hp

2

(4q AX )

.

ph(h+1) 2

.

Then there exist distinct q-tuples (x1 , . . . , xq ) and (x1 , . . . , xq ) in I X which map to the same p-tuple in Jq AX . Therefore x j = x j − x j with 1 ≤ j ≤ q are not all zero and (x1 , . . . , xq ) is a solution of (10.2.1). Further |x j | ≤ 2X for 1 ≤ j ≤ q. Let

ph(h+1)

λ = (2q A) 2q− ph(h+1) . We observe that λ > 1 since 2q > ph(h + 1). Let X be the integer satisfying λ − 1 ≤ 2X < λ + 1. Then (4q AX )

ph(h+1) 2

= (2q A) = λq−

Hence

ph(h+1) 2

ph(h+1) 2

(2X )

(2X )

ph(h+1) 2

ph(h+1) 2

< (2X + 1)q .

ph(h+1)

|x j | < 1 + (2q A) 2q− ph(h+1) .

10.3 Proof of the Baker Theorem 10.1

275

10.3 Proof of the Baker Theorem 10.1 Assume that log α1 , . . . , log αn are linearly independent over Q. We prove by contradiction. Suppose that there exist algebraic numbers β1 , . . . , βn , not all zero, such that β1 log α1 + · · · + βn log αn = 0. By permuting α1 , . . . , αn , if necessary, we may suppose that βn = 0. Further we put βj = −

β j βn

for 1 ≤ j ≤ n.

Then βn = −1 and β1 log α1 + · · · + βn−1 log αn−1 = log αn .

(10.3.1)

For λ1 , . . . , λn with 0 ≤ λi ≤ L, let γi = λi + λn βi for 1 ≤ i ≤ n. Then

λn γn−1 λn−1 βn−1 λn−1 λn γ β α1 1 · · · αn−1 = α1λ1 · · · αn−1 = α1λ1 · · · αn−1 αn α11 · · · αn−1

by (10.3.1). Step I. Construction of auxiliary function: We denote by u 0 , u 1 , . . . , u 11 effectively computable positive numbers (determined explicitly) depending only on n, α1 , . . . , αn , β1 , . . . , βn and the choice of branches of logarithms log α1 , . . . , log αn for α1 , . . . , αn . Let h ≥ u 0 , where u 0 is sufficiently large. We put K = Q(α1 , . . . , αn , β1 , . . . , βn ), [K : Q] = d, 1 b = , R = 3n 2 , n h 1 = h, h r = [h r −1 h b ] for 2 ≤ r ≤ R, L = [h 3−b ], k = [h 3+b ], kr −1 for 2 ≤ r ≤ R. k1 = k, kr = 2 Then kr ≥

k 3 R−1

for 1 ≤ r ≤ R. We consider the auxiliary function

(z 1 , . . . , z n−1 ) =

L λ1 =0

···

L λn =0

γ z

γ

n−1 p(λ1 , . . . , λn )α11 1 · · · αn−1

z n−1

,

(10.3.2)

276

10 The Baker Theorem

where p(λ1 , . . . , λn ) ∈ Z, not all zero, which we shall determine under the conditions m 1 ,...,m n−1 (r, . . . , r ) = 0 for 1 ≤ r ≤ h and m 1 + · · · + m n−1 ≤ k with m i ≥ 0

(10.3.3) where

m 1 ,...,m n−1 (z 1 , . . . , z n−1 ) =

∂ ∂z 1

m 1

∂ ··· ∂z m−1

m n−1 (z 1 , . . . , z n−1 ).

Here we remark that 0,0,...,0 (z 1 , . . . , z n−1 ) = (z 1 , . . . , z n−1 ). The Equations (10.3.3) are satisfied if and only if L λ1 =0

···

L

n−1 p(λ1 , . . . , λn )α1λ1 r · · · αnλn r γ1m 1 · · · γn−1 =0

m

λn =0

for 1 ≤ r ≤ h and m 1 + · · · + m n−1 ≤ k with m i ≥ 0. These are linear equations in p(λ1 , . . . , λn ) with coefficients in K . Denoting by D the least common multiple of denominators of α1 , . . . , αn , β1 , . . . , βn , we multiply each of the above equations on both sides by D Lh+k so that we may assume that the coefficients of p(λ1 , . . . , λn ) are algebraic integers in K . The number of variables p(λ1 , . . . , λn ) in these equations is equal to (L + 1)n whereas the number of equations is ≤ h(k + 1)n−1 . Further we check that (L + 1)n > d(d + 1)h(k + 1)n−1 if u 0 is sufficiently large. Now we derive from Lemma 10.3 with q = (L + 1)n , p ≤ h(k + 1)n−1 and h = d that there exist integers p(λ1 , . . . , λn ), not all zero, satisfying (10.3.3) and (10.3.4) | p(λ1 , . . . , λn )| ≤ u 1Lh+k . Step II. Induction: We prove the following. Lemma 10.4 For 1 ≤ ν ≤ R, we have m 1 ,...,m n−1 (r, . . . , r ) = 0

(10.3.5)

for 1 ≤ r ≤ h ν and m 1 + · · · + m n−1 ≤ kν with m i ≥ 0. Proof If ν = 1, the assertion follows by (10.3.3), since h 1 = h and k1 = k. We assume that the assertion holds for ν − 1 with 2 ≤ ν ≤ R and we prove for ν. The proof is by contradiction. We suppose that there exist integers with 1 ≤ ≤ h ν and m 1 , . . . , m n−1 with m 1 + m 2 + · · · + m n−1 ≤ kν , m i ≥ 0 such that m 1 ,...,m n−1 (, . . . , ) = 0. In fact h ν−1 < ≤ h ν by induction hypothesis.

(10.3.6)

10.3 Proof of the Baker Theorem 10.1

277

Let f (z) = m 1 ,...,m n−1 (z, . . . , z). Then we see from (10.3.2) that f (z) = (log α1 )m 1 · · · (log αn−1 )m n−1

L λ1 =0

···

L

n−1 p(λ1 , . . . , λn )α1λ1 z · · · αnλn z γ1m 1 · · · γn−1 .

m

λn =0

(10.3.7) For 0 ≤ m ≤ kν−1 − kν and 1 ≤ r ≤ h ν−1 , we have f (m) (r ) =

j1 ,..., jn−1 j1 +···+ jn−1 =m

m! m + j ,...,m n−1 + jn−1 (r, . . . , r ). j1 ! · · · jn−1 ! 1 1

Since m 1 + j1 + · · · + m n−1 + jn−1 = m 1 + · · · + m n−1 + m ≤ kν + kν−1 − kν = kν−1 , we have

f (m) (r ) = 0 for 1 ≤ r ≤ h ν−1 , 0 ≤ m ≤ kν−1 − kν

by induction hypothesis. Therefore, we derive from the Cauchy residue theorem 2.13 that 1 f (z) F() dz = f (), (10.3.8) 2πi :|z|=5h ν (z − ) F(z) where

We have

and

F(z) = ((z − 1) · · · (z − h ν−1 ))kν−1 −kν +1 . ν−1 −kν +1)h ν−1 |F()| ≤ h (k ν

|F(z)| ≥ (4h ν )(kν−1 −kν +1)h ν−1 for |z| = 5h ν .

Therefore, since kν−1 − kν + 1 = kν−1 − we have

Further

kν−1 kν−1 +1> , 2 2

F() < 2−h ν−1 kν−1 . max |z|=5h ν F(z)

278

10 The Baker Theorem

max | f (z)| ≤ u 2Lh ν +k

|z|=5h ν

by (10.3.7) and (10.3.4). By using the above estimates in (10.3.8), we have | f ()| ≤

1 2π5h ν −h ν−1 kν−1 Lh ν +k 2 u2 ≤ 2−h ν−1 kν−1 u 3Lh ν +k . 2π 4h ν

(10.3.9)

Next, we find a lower bound for the absolute value of f (). By (10.3.7), (10.3.6) and (10.3.4), we have (log α1 )−m 1 · · · (log αn−1 )−m n−1 f () =

L λ1 =0

···

L

n−1 p(λ1 , . . . , λn )α1λ1 · · · αnλn γ1m 1 · · · γn−1

m

λn =0

(10.3.10) is a non-zero algebraic number in a field of degree d and the maximum of absolute values of its conjugates does not exceed u 4Lh ν +k and its denominator is also at most u 4Lh ν +k . Now, by using that the absolute value of norm of a non-zero algebraic integer is at least 1, we get that the absolute value of the left-hand side of (10.3.9) is at least u 5−(Lh ν +k) and hence | f ()| ≥ u 6−(Lh ν +k) (10.3.11) since (log α1 )−m 1 · · · (log αn−1 )−m n−1 ≥ u −k 7 . By combining (10.3.9) and (10.3.11), we have 2−h ν−1 kν−1 > (u 3 u 6 )−(Lh ν +k) . Therefore h ν−1 k < u 7 Lh ν ≤ u 7 h 3−b h b h ν−1 = u 7 h 3 h ν−1 . Then

1 3+b < u7h3 h 2

which is not possible if u 0 is sufficiently large. Step III An application of Theorem 2.32: Let g(z) = 0,...,0 (z, . . . , z) =

L λ1 =0

···

L

p(λ1 , . . . , λn )α1λ1 z · · · αnλn z

λn =0

by (10.3.2). For 1 ≤ r ≤ h R and 0 ≤ m ≤ k R , we have g (m) (r ) =

j1 ,..., jn−1 j1 +···+ jn−1 =m

m! j ,..., j (r, . . . , r ) = 0 j1 ! · · · jn−1 ! 1 n−1

10.3 Proof of the Baker Theorem 10.1

279

by Lemma 10.4 since j1 + · · · + jn−1 = m ≤ k R . We observe that g(z) is not identically zero since p(λ1 , . . . , λn ) are not all zero and log α1 , . . . , log αn are linearly independent over Q. Denote by N R the number of zeros of g(z) counted with multiplicity in D(0, h R ). Then N R ≥ h R k R ≥ u 8 h R k.

(10.3.12)

On the other hand, we conclude from Theorem 2.32 with F(z) = g(z), n = (L + 1)n , ≤ u 9 L and R = h R that N R ≤ u 10 ((L + 1)n + Lh R ) < 2u 10 Lh R since h R ≥ u 11 h 1+(R−1)b = u 11 h 1+(3n

2

−1) n1

(10.3.13)

> h 3n > (L + 1)n .

By combining (10.3.12) and (10.3.13), we get u 8 k < 2u 10 L . Then u 8 h 2b < 4u 10 which is not possible if u 0 is sufficiently large and the proof of Theorem 10.1 is complete.

10.4 Proof of Corollary 10.2 β

The proof is by induction on m. Let m = 1. Assume that α1 1 = α2 algebraic. Then β1 log α1 − log α2 = 0 for a suitable choice of branch of logarithm for α2 . Therefore log α1 and log α2 are linearly independent over Q. This contradicts Theorem 10.1 with n = 2. Let m ≥ 2. We assume that the assertion is valid for all positive integers n with n < m and we prove for m. Assume that β

βm = αm+1 α1 1 · · · αm

(10.4.1)

algebraic and 1, β1 , β1 , . . . , βm are linearly independent over Q. Let αm+1 = 1. Then βm = 0 otherwise the assertion follows by induction hypothesis. Then γm−1 γ = αm , α11 · · · αm−1 β

where γ j = − βmj for 1 ≤ j < m and 1, γ1 , . . . , γm−1 are linearly independent over Q. This is not possible by induction hypothesis. Hence we may assume that αm+1 = 1. We put βm+1 = −1 and we rewrite (10.4.1) as β

β

βm m+1 α1 1 · · · αm αm+1 = 1.

(10.4.2)

By Theorem 10.1, we derive that there exist ρ1 , . . . , ρm+1 ∈ Q, not all zero, such that

280

10 The Baker Theorem ρ

ρ

m+1 α11 · · · αm+1 = 1.

(10.4.3)

We may assume that ρm+1 = 0 by permuting the suffixes if necessary. Then we derive from (10.4.2) and (10.4.3) that δm = 1, α1δ1 · · · αm

where δ j = ρm+1 β j − βm+1 ρ j (1 ≤ j ≤ m) are linearly independent over Q since 1, andβ1 , . . . , βm are linearly independent over Q. Then δm = 0. Further

m−1 = αm where j = − α11 · · · αm−1

δj for 1 ≤ j < m δm

such that 1, 1 , . . . , m−1 are linearly independent over Q. This contradicts induction hypothesis.

10.5 An Application of (10.1.1) For an integer x with |x| > 1, let P(x) be the greatest prime factor of x and we write P(±1) = 1. We derive from (10.1.1) the following result which extends a result of Stφrmer [30]. Theorem 10.5 For integers x > 0 and k > 0, we have lim P(x(x + k)) = ∞ effectively.

x→∞

By effectively, we mean that for > 0 there exists x0 which can be determined explicitly in terms of k and such that P(x(x + k)) > for x ≥ x0 . Proof Let > 0 and P(x(x + k)) ≤ . It suffices to show that x is bounded by an effectively computable number depending only on k and . We may assume that x > k 2 . We write x = P1A1 . . . PnAn , x + k = P1B1 . . . PnBn , where A1 , . . . , An , B1 , . . . , Bn are non-negative integers and P1 , . . . , Pn are prime numbers ≤ . We have

0 < log

x +k x

k k 1 = log 1 + < < x− 2 . x x

Therefore 0 < (B1 − A1 ) log P1 + · · · + (Bn − An ) log Pn < x − 2 . 1

(10.5.1)

10.5 An Application of (10.1.1)

For 1 ≤ i ≤ n, we observe that 2 Ai ≤ PiAi ≤ x. Therefore Ai ≤ Similarly Bi < 2 log(x + k) < 4 log x for 1 ≤ i ≤ n. Then

281

log x < 2 log x. log 2

|Bi − Ai | < 4 log x for 1 ≤ i ≤ n. Now we apply (10.1.1) with n ≤ , d = 1, Ai ≤ and B = 4 log x for deriving that (10.5.2) (B1 − A1 ) log P1 + · · · + (Bn − An ) log Pn > (log x)−u 12 , where u 12 = u 12 ( ) and u 13 = u 13 ( ) are effectively computable. By (10.5.1) and (10.5.2), we have x ≤ (log x)2u 12 and hence x ≤ u 13 .

10.6 Exercises 10.1 Show that the assumption α = 0 with log α = 0 can be replaced by α ∈ / {0, 1} in the Gel’fond-Schneider theorem. 10.2 Assume (10.1.1). Then show that there exists an effectively computable number C depending only on n and d such that |α1b1 · · · αnbn − 1| > exp(−C log log B) whenever α1b1 · · · αnbn = 1. 10.3 For a positive integer x > 1, derive from (10.1.1) that P(x(x + 1))≥ c log log x where c is an effectively computable absolute constant. pn be the nth convergent 10.4 Let α be an algebraic number of degree ≥ 2 and qn in the continued fraction expansion (see [17], Ch X) of α. Then derive from (10.1.1) that lim P( pn qn ) = ∞ effectively. n→∞

1 pn < 2 for n ≥ 1.) (Hint: Use α − qn qn

References

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Index

A Abel summation formula, 196, 198 Abscissa of convergence, 253 Algebraically independent, 64 Algebraic integer, 271 Algebraic number, 271 Analytic continuation, 187, 198, 218 Analytic functions, 9 Analytic homeomorphism, 73 Argument principle, 46 Automorphism, 73 of unit open disc, 73, 76, 79 of upper half plane, 74, 81

B Baker, 64 Baker theorem, 271, 272 Bernoulli numbers, 167, 183 Bernoulli polynomials, 167 Beta function, 180 Bijection, 2 Blaschke product, 182 Borel and Carathéodory lemma, 119, 230

C Casorati - Weierstrass theorem, 32, 129 Cauchy - Goursat theorem, 26 Cauchy integral formula, 25 for open convex sets, 28 for open sets, 39 Cauchy residue theorem, 42 Cauchy - Riemann equations, 99 Cauchy theorem for open convex sets, 29 for open sets, 41

for simply connected regions, 34 Chains and cycles, 12 Characters, 259 Dirichlet character, 261 Principal Dirichlet character, 261 Closed polygon, 33 Completely multiplicative, 189, 191 Complex integral, 8 Component, 2 Conformally equivalent, 73 Conformal mapping, 73 Connected, 1 simply connected region, 1, 14, 18, 33, 42, 61, 85, 105 Convex function, 23 Convex set, 21 Curve, 9

D Density hypothesis, 229, 230 Dirichlet L-function, 262 Dirichlet problem, 112 Dirichlet series, 251, 252 Dirichlet theorem, 266 Duplication formula for Gamma function, 158

E Effectively computable, 275 End point of a curve, 9 Entire function, 9 of finite order, 144 Equicontinuous, 88 Equivalence relation in homotopic paths, 13 Euclid, 188

© Springer Nature Singapore Pte Ltd. 2020 T. N. Shorey, Complex Analysis with Applications to Number Theory, Infosys Science Foundation Series, https://doi.org/10.1007/978-981-15-9097-9

285

286 Euler constant, 134 Euler formula for gamma function, 161 Euler–Maclaurin–Jacobi sum formula, 169 Euler product for Dirichlet L-function, 263 for the Riemann Zeta function, 189 Exponent of convergence for the sequence, 144 Extended complex plane, 4

F Factorisation theorem for an arbitrary region, 148 Fubini theorem, 39, 40 Functional equation for (z), 158 for ζ (s), 218 for the theta functions, 218 Fundamental theorem of algebra, 30 of arithmetic, 189, 191 of finite abelian groups, 259

G Gamma function, 157 Gel’fond, 64 Gel’fond-Schneider theorem, 271, 272 Gourdon, 229 Great Picard theorem, 118

H Hadamard, 203, 237 Hadamard factorisation theorem, 145 Hadamard three-circle theorem, 55, 230 Half plane of convergence, 253 Harmonic conjugate, 103 Harmonic function, 101 Height, 271 Holomorphic functions, 9 Homeomorphism, 2 Homotopic paths, 13 Hurwitz theorem, 87 Hurwitz Zeta function, 236

I Identity theorem for holomorphic functions, 31 Infinite product, 134 absolutely convergent, 136 uniformly convergent, 137

Index Ingham, 229, 237, 238 Initial point of a curve, 9 Integral representation for (z), 163 Inverse function theorem, 50

J Jensen formula, 45, 61

K Korobov, 237

L Landau theorem, 132, 255 Laplace equation, 101 Laurent series, 31 LeVeque, 268 Lindelöf hypothesis, 229, 230 Linear fractional transformation, 98 Line of convergence, 253 Liouville theorem for entire functions, 30 for harmonic functions, 116 Little Picard theorem, 117 Logarithmic branch, 58, 59 analytic, 58–60 principal, 22 principal value at a point, 58

M Maximum modulus principle, 49, 50, 52 Maximum principle for harmonic functions, 109 Mean Value Property (MVP), 108 Meromorphic functions, 32, 151 Method of Vinogradov, 237 Minimal polynomial, 271 Minimum principle for harmonic functions, 110 Mittag-Leffler theorem, 152 Montel, 88 Morera theorem, 31, 41 Multiplicative, 189 Multiplicative group of reduced residue classes, 261

N Non-trivial zeros of ζ (s), 224 Non-vanishing of ζ (1 + it), 199

Index of L(1, χ), 265 Normal family, 88 Null-homotopic, 13, 25

O Open mapping theorem for analytic functions, 48, 50, 91, 151

P Parameter interval, 9 Paths, 9 closed, 9 homotopic, 12 polygonal, 3, 4 Phragmen–Lindelöf method, 52 Point at infinity, 4 Poisson kernel, 112 Polygon path, 3, 33 Power series, 29, 30 circle of convergence, 29 disc of convergence, 29 radius of convergence, 29 Preserve angle, 73 Prime Number Theorem, 203, 208 Primitive function, 26, 44 with error term, 237 Principal part, 32 Product formula for (z), 158 Product formula for sin π z, 155 R Ramanujan identity, 193, 251, 258 Rank of a function, 144 Region, 1 unbounded, 10 Riemann, 188, 218, 229 Riemann hypothesis, 187, 229, 250 Riemann integral, 9 Riemann–Lebesgue lemma, 212 Riemann mapping theorem, 1, 73, 74, 85, 107 Riemann sphere, 1, 4, 8 Riemann Zeta function, 187, 229 Rotation, 73 Rouché theorem, 51

287 S Schottky lemma, 117 Schwarz lemma, 76 Shifting the line of integration, 238 Siegel, 237 Siegel zero, 268 Singularity essential, 32 isolated, 32 pole, 32 order, 32 simple, 32 removable, 32 Spherical coordinates, 6 Star shaped, 21 Stereographic projection, 5 Stirling formula, 171 Stφrmer, 280

T Theory of linear forms in logarithms, 272 Thue-Siegel lemma, 272 Tijdeman, 63 Transcendental numbers, 64, 271

U Uniformly bounded family, 88 Uniformly continuous, 88

W Weierstrass elementary factors, 140 Weierstrass elliptic function, 185 Weierstrass factorisation theorem, 133, 142, 154 Wiener–Ikehara theorem, 187 Winding number, 10

Z Zero, 30 order, 30 simple, 30 Zero-free region for ζ (s), 239