Complex Variables and Numerical Methods Gujarat Technological University 2017 [2 ed.] 9789352604821, 9352604822

Table of contents :
Title
1 Complex Numbers
2 Analytic Functions
3 Complex Integration
4 Power Series
5 Applications of Contour Integration
6 Conformal Mapping and Its Applications
7 Interpolation
8 Numerical Integration
9 Solutions of a System of Linear Equations
10 Roots of Algebraic and Transcendental Equations
11 Eigenvalues by Power and Jacobi Methods
12 Numerical Solution of Ordinary Differential Equations
Index

Citation preview

Complex Variables and Numerical Methods Gujarat Technological University 2017

Second Edition

About the Authors Ravish R Singh is presently Academic Advisor at Thakur Educational Trust, Mumbai. He obtained a BE degree from University of Mumbai in 1991, an MTech degree from IIT Bombay in 2001, and a PhD degree from Faculty of Technology, University of Mumbai, in 2013. He has published several books with McGraw Hill Education (India) Private Limited on varied subjects like Engineering Mathematics (I and II), Applied Mathematics, Electrical Engineering, Electrical and Electronics Engineering, etc., for all-India curricula as well as regional curricula of some universities like Gujarat Technological University, Mumbai University, Pune University, Jawaharlal Nehru Technological University, Anna University, Uttarakhand Technical University, and Dr A P J Abdul Kalam Technical University (formerly known as UPTU). Dr Singh is a member of IEEE, ISTE, and IETE, and has published research papers in national and international journals. His fields of interest include Circuits, Signals and Systems, and Engineering Mathematics. Mukul Bhatt is presently Assistant Professor, Department of Humanities and Sciences, at Thakur College of Engineering and Technology, Mumbai. She obtained her MSc (Mathematics) from H N B Garhwal University in 1992. She has published several books with McGraw Hill Education (India) Private Limited on Engineering Mathematics (I and II) and Applied Mathematics for all-India curricula as well as regional curricula of some universities like Gujarat Technological University, Mumbai University, Pune University, Jawaharlal Nehru Technological University, Anna University, Uttarakhand Technical University, and Dr A P J Abdul Kalam Technical University (formerly known as UPTU). She has seventeen years of teaching experience at various levels in engineering colleges in Mumbai and her fields of interest include Integral Calculus, Complex Analysis, and Operation Research. She is a member of ISTE.

Complex Variables and Numerical Methods Gujarat Technological University 2017 Second Edition Ravish R Singh Academic Advisor Thakur Educational Trust Mumbai, Maharashtra Mukul Bhatt Assistant Professor Department of Humanities and Sciences Thakur College of Engineering and Technology Mumbai, Maharashtra

McGraw Hill Education (India) Private Limited CHENNAI McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai 600 116 Complex Variables and Numerical Methods (Gujarat Technological University 2017) Copyright © 2017, 2016 by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listing (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. ISBN 13: 978-93-5260-482-1 ISBN 10: 93-5260-482-2 Managing Director: Kaushik Bellani Director—Products (Higher Education & Professional): Vibha Mahajan Manager—Product Development: Koyel Ghosh Senior Specialist—Product Development: Piyali Chatterjee Head—Production (Higher Education & Professional): Satinder S Baveja Assistant Manager—Production: Anuj K Shriwastava Assistant General Manager—Product Management (Higher Education & Professional): Shalini Jha Product Manager—Product Management: Ritwick Dutta General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.

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Dedicated To Our Parents Late Shri Ramsagar Singh and Late Shrimati Premsheela Singh Ravish R Singh Late Shri Ved Prakash Sharma and Late Shrimati Vidyavati Hemdan Mukul Bhatt

Contents Preface Roadmap to the Syllabus 1. Complex Numbers Introduction 1.1 Complex Numbers 1.2 Geometrical Representation of Complex Numbers (Argand’s Diagram) 1.2 1.4 Algebra of Complex Numbers 1.2 1.5 Different Forms of Complex Numbers 1.3 1.6 Modulus and Argument (or Amplitude) of Complex Numbers 1.7 Properties of Complex Numbers 1.4 1.8 De Moivre’s Theorem 1.25 1.9 Applications of De Moivre’s Theorem 1.36 1.10 Circular and Hyperbolic Functions 1.58 1.11 Inverse Hyperbolic Functions 1.61 1.12 Separation into Real and Imaginary Parts 1.72 1.13 Logarithm of a Complex Number 1.88 Points to Remember 1.104

xi xv 1.1–1.107

1.1 1.2 1.3

2. Analytic Functions

1.4

2.1–2.88

2.1 Introduction 2.1 2.2 Complex Variable 2.1 2.3 Basic Definitions 2.2 2.4 Limits 2.8 2.5 Continuity 2.12 2.6 Differentiability 2.15 2.7 Analytic Functions 2.19 2.8 Cauchy–Riemann Equations in Polar Form 2.21 2.9 Harmonic Functions 2.46 2.10 Properties of Analytic Functions 2.46 2.11 Construction of Analytic Functions: Milne–Thomson Method 2.63 Points to Remember 2.88

viii

Contents

3. Complex Integration

3.1–3.58

3.1 Introduction 3.1 3.2 Some Basic Definitions 3.1 3.3 Line Integral 3.2 3.4 Simply Connected and Multiply Connected Regions 3.20 3.5 Cauchy’s Integral Theorem 3.20 3.6 Cauchy’s Integral Formula 3.31 3.7 Liouville Theorem 3.32 3.8 Maximum Modulus Theorem 3.32 Points to Remember 3.57 4. Power Series

4.1–4.117

4.1 Introduction 4.1 4.2 Sequences and Series 4.1 4.3 Power Series 4.2 4.4 Convergence of a Power Series 4.3 4.5 Taylor’s Series 4.8 4.6 Laurent’s Series 4.18 4.7 Singular Points 4.50 4.8 Residues 4.59 4.9 Cauchy’s Residue Theorem 4.75 4.10 Argument Theorem 4.105 4.11 Rouche’s Theorem 4.107 Points to Remember 4.115 5. Applications of Contour Integration

5.1–5.51

5.1 5.2

Introduction 5.1 Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q 5.1 5.3 Evaluation of Improper Real Integral of a Rational Function 5.22 5.4 Evaluation of Improper Real Integral of a Rational Function Including Trigonometric Functions 5.35 5.5 Evaluation of Improper Integral when Simple Poles Lie on the Real Axis 5.45 Points to Remember 5.51 6. Conformal Mapping and Its Applications 6.1 Introduction 6.1 6.2 Conformal Mapping 6.1 6.3 Some Standard Transformations 6.2 6.4 Some Special Transformations 6.29 6.5 Schwarz—Christoffel Transformation 6.45 6.6 Bilinear Transformation 6.49 Points to Remember 6.73

6.1–6.74

Contents

7. Interpolation

ix

7.1–7.88

7.1 Introduction 7.1 7.2 Finite Differences 7.2 7.3 Different Operators and their Relations 7.6 7.4 Interpolation 7.19 7.5 Newton’s Forward Interpolation Formula 7.19 7.6 Newton’s Backward Interpolation Formula 7.30 7.7 Central Difference Interpolation 7.39 7.8 Gauss’s Forward Interpolation Formula 7.40 7.9 Gauss’s Backward Interpolation Formula 7.44 7.10 Stirling’s Formula 7.48 7.11 Interpolation with Unequal Intervals 7.55 7.12 Lagrange’s Interpolation Formula 7.56 7.13 Divided Differences 7.70 7.14 Newton’s Divided Difference Formula 7.71 7.15 Inverse Interpolation 7.84 Points to Remember 7.86 8. Numerical Integration

8.1–8.41

8.1 Introduction 8.1 8.2 Newton–Cotes Quadrature Formula 8.1 8.3 Trapezoidal Rule 8.2 8.4 Simpson’s 1/3 Rule 8.9 8.5 Simpson’s 3/8 Rule 8.19 8.6 Gaussian Quadrature Formulae 8.31 Points to Remember 8.40 9. Solutions of a System of Linear Equations

9.1–9.62

9.1 Introduction 9.1 9.2 Solutions of a System of Linear Equations 9.2 9.3 Elementary Transformations 9.2 9.4 Numerical Methods for Solutions of a System of Linear Equations 9.3 9.5 Gauss Elimination Method 9.4 9.6 Gauss Elimination Method with Partial Pivoting 9.15 9.7 Gauss–Jordan Method 9.20 9.8 Gauss–Jacobi Method 9.31 9.9 Gauss–Siedel Method 9.37 Points to Remember 9.61 10. Roots of Algebraic and Transcendental Equations 10.1 10.2 10.3 10.4

Introduction 10.1 Bisection Method 10.2 Regula Falsi Method 10.15 Newton–Raphson Method 10.22

10.1–10.50

x

Contents

10.5 Secant Method 10.39 Points to Remember 10.49 11. Eigenvalues by Power and Jacobi Methods

11.1–11.24

11.1 Introduction 11.1 11.2 Eigenvalues and Eigenvectors 11.1 11.3 Nature of Eigenvalues of Special Types of Matrices 11.2 11.4 Relations between Eigenvalues and Eigenvectors 11.3 11.5 Power Method 11.3 11.6 Jacobi Method 11.15 Points to Remember 11.24 12. Numerical Solution of Ordinary Differential Equations

12.1–12.47

12.1 Introduction 12.1 12.2 Euler’s Method 12.2 12.3 Modified Euler’s Method 12.9 12.4 Runge–Kutta Methods 12.21 Points to Remember 12.46 Index

I.1–I.2

Preface Mathematics is a key area of study in any engineering course. A sound knowledge of this subject will help engineering students develop analytical skills, and thus enable them to solve numerical problems encountered in real life, as well as apply mathematical principles to physical problems, particularly in the field of engineering.

Users This book is designed for the 4th semester GTU Mechanical Engineering students pursuing the course Complex Variables and Numerical Methods (CODE 2141905). It covers the complete GTU syllabus for the course on Complex Variables and Numerical Methods for the mechanical engineering branch.

Objective The crisp and complete explanation of topics will help students easily understand the basic concepts. The tutorial approach (i.e., teach by example) followed in the text will enable students develop a logical perspective to solving problems.

Features Each topic has been explained from the examination point of view, wherein the theory is presented in an easy-to-understand student-friendly style. Full coverage of concepts is supported by numerous solved examples with varied complexity levels, which is aligned to the latest GTU syllabus. Fundamental and sequential explanation of topics is well aided by examples and exercises. The solutions of examples are set following a ‘tutorial’ approach, which will make it easy for students from any background to easily grasp the concepts. Exercises with answers immediately follow the solved examples enforcing a practice-based approach. We hope that the students will gain logical understanding from solved problems and then reiterate it through solving similar exercise problems themselves. The unique blend of theory and application caters to the requirements of both the students and the faculty. Solutions of GTU examination questions are incorporated within the text appropriately.

xii

Preface

Highlights ∑ Crisp content strictly as per the latest GTU syllabus of Complex Variables and Numerical Methods (Regulation 2014) ∑ Comprehensive coverage with lucid presentation style ∑ Each section concludes with an exercise to test understanding of topics ∑ Solutions of GTU examination papers from 2010 to 2015 present appropriately within the chapters ∑ Solution of 2016 GTU examination paper can be accessible through weblink. ∑ Rich exam-oriented pedagogy: ã Solved Examples within chapters: 473 ã Solved GTU questions tagged within chapters: 150 ã Unsolved Exercises: 474

Chapter Organization The content spans the following twelve chapters which wholly and sequentially cover each module of the syllabus. o Chapter 1 introduces Complex Numbers. o Chapter 2 discusses Analytic Functions. o Chapter 3 presents Complex Integration. o Chapter 4 covers Power Series. o Chapter 5 deals with Applications of Contour Integration. o Chapter 6 presents Conformal Mapping and its Applications. o Chapter 7 explains Interpolation. o Chapter 8 introduces Numerical Integration. o Chapter 9 discusses Solutions of a System of Linear Equations. o Chapter 10 deals with Roots of Algebraic and Transcendental Equations. o Chapter 11 covers Eigenvalues by Power and Jacobi Methods. o Chapter 12 explains Numerical Solution of Ordinary Differential Equations.

Acknowledgements We are grateful to the following reviewers who reviewed various chapters of the script and generously shared their valuable comments: Ramesh S Damor Jyotindra C Prajapati Vijay Solanki Prakash Vihol

L D College of Engineering, Ahmedabad Marwadi Education Foundation Group of Institutions, Rajkot Government Engineering College, Patan Government Engineering College, Rajkot

Preface

Manokamna Agarwal Som Sahni Urvi Trivedi

xiii

Silver Oak College of Engineering and Technology, Ahmedabad Babaria Institute of Technology, Vadodara SAL Institute of Technology and Engineering Research, Ahmedabad

We would also like to thank all the staff at McGraw Hill Education (India), especially Piyali Chatterjee, Anuj Kr. Shriwastava, Koyel Ghosh, Satinder Singh Baveja, and Vibha Mahajan for coordinating with us during the editorial, copyediting, and production stages of this book. Our acknowledgements would be incomplete without a mention of the contribution of all our family members. We extend a heartfelt thanks to them for always motivating and supporting us throughout the project. Constructive suggestions for the improvement of the book will always be welcome. Ravish R Singh Mukul Bhatt

Publisher’s Note Remember to write to us. We look forward to receiving your feedback, comments, and ideas to enhance the quality of this book. You can reach us at [email protected]. Please mention the title and authors’ name as the subject. In case you spot piracy of this book, please do let us know.

RoAdmAP to the SyllAbuS This text is useful for Complex Variables and Numerical Methods (CoDe 2141905) Module 1: Complex Numbers and Functions Exponential, Trigonometric, De Moivre’s Theorem, Roots of a Complex Number, Hyperbolic Functions and their Properties, Multi-valued Function and its Branches: Logarithmic Function and Complex Exponent Function Limit, Continuity and Differentiability of Complex Function, Analytic Functions, Cauchy-Riemann Equations, Necessary and Sufficient Condition for Analyticity, Properties of Analytic Functions, Laplace Equation, Harmonic Functions, Harmonic Conjugate Functions and their Engineering Applications. GO TO CHAPTER 1: Complex Numbers CHAPTER 2: Analytic Functions

Module 2: Complex Integration: Curves, Line Integral (Contour Integral) and its Properties, Cauchy-Goursat Theorem, Cauchy Integral Formula, Liouville Theorem (without proof), Maximum Modulus Theorems (without proof) GO TO CHAPTER 3: Complex Integration

Module 3: Power Series Convergence (Ordinary, Uniform, Absolute) of Power Series, Taylor and Laurent Theorems (without proof), Laurent Series Expansions, Zeros of Analytic Functions , Singularities of Analytic Functions and their Classification. Residues: Residue Theorem, Rouche’s Theorem (without proof) GO TO CHAPTER 4: Power Series

xvi

Roadmap to the Syllabus

Module 4: Applications of Contour Integration Evaluation of Various Types of Definite Real Integrals using Contour Integration Method. GO TO CHAPTER 5: Applications of Contour Integration

Module 5: Conformal Mapping and its Applications Conformal and Isogonal Mappings, Translation, Rotation and Magnification, Inversion, Mobius (Bilinear), Schwarz-Christoffel Transformations GO TO CHAPTER 6: Conformal Mapping and its Applications

Module 6: Interpolation Interpolation: Finite Differences, Forward, Backward and Central Operators, Interpolation by Polynomials: Newton’s forward, Backward Interpolation Formulae, Newton’s Divided Formula, Gauss and Stirling’s Central Difference Formulae and Lagrange’s Interpolation Formulae for Unequal Intervals GO TO CHAPTER 7: Interpolation

Module 7: Numerical Integration Newton-Cotes Formula, Trapezoidal and Simpson’s Formulae, Error Formulae, Gaussian Quadrature Formulae GO TO CHAPTER 8: Numerical Integration

Module 8: Solutions of a System of Linear Equations Gauss Elimination, Partial Pivoting, Gauss-Jacobi Method and Gauss-Seidel Method GO TO CHAPTER 9: Solutions of a System of Linear Equations

Roadmap to the Syllabus

xvii

Module 9: Roots of Algebraic and Transcendental Equations Bisection, False Position, Secant and Newton-Raphson Methods, Rate of Convergence GO TO CHAPTER 10: Roots of Algebraic and Transcendental Equations

Module 10: Eigenvalues by Power and Jacobi Methods GO TO CHAPTER 11: Eigenvalues by Power and Jacobi Methods

Module 11: Numerical Solution of Ordinary Differential Equations Euler and Runge-Kutta Methods GO TO CHAPTER 12: Numerical Solution of Ordinary Differential Equations

CHAPTER

1

Complex Numbers chapter outline 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13

1.1

Introduction Complex Numbers Geometrical Representation of Complex Numbers (Argand’s Diagram) Algebra of Complex Numbers Different Forms of Complex Numbers Modulus and Argument (or Amplitude) of Complex Numbers Properties of Complex Numbers De Moivre’s Theorem Applications of De Moivre’s Theorem Circular and Hyperbolic Functions Inverse Hyperbolic Functions Separation into Real and Imaginary Parts Logarithm of a Complex Number

IntroductIon

The complex numbers are an extension of the real numbers obtained by introducing an imaginary unit i, where i = -1 . The operations of addition, subtraction, multiplication, and division are applicable on complex numbers. A negative real number can be obtained by squaring a complex number. With a complex number, it is always possible to find solutions to polynomial equations of degree more than one. Complex numbers are used in many applications, such as control theory, signal analysis, quantum mechanics, relativity, etc.

1.2

Chapter 1

1.2

complex numbers

Complex Numbers

A complex number z is an ordered pair (x, y) of real numbers x and y. It is written as z = (x, y) or z = x + iy, where i =

-1 is known as the imaginary unit. Here, x is called

the real part of z and is written as “Re (z)”, and y is called the imaginary part of z and is written as “Im (z)”. If x = 0 and y π 0 then z = 0 + iy = iy which is purely imaginary. If x π 0 and y = 0 then z = x + i 0 = x which is real. Hence, z is purely imaginary, if its real part is zero, and is real if its imaginary part is zero. This shows that every real number can be written in the form of a complex number by taking its imaginary part as zero. Hence, the set of real numbers is contained in the set of complex numbers. Two complex numbers are equal if and only if their corresponding real and imaginary parts are equal. If z = x + iy is a complex number then its conjugate or complex conjugate is defined as z = x – iy. The even power of i is either 1 or –1 and odd power of i is either i or –i. i 2 = i.i = –1 i3 = i2.i = –i i4 = (i2)2 = (–1)2 = 1 i5 = i. i 4 = i, etc.

1.3

GeometrIcal representatIon of complex numbers (arGand’s dIaGram) y

Any complex number z = x + iy can be represented as a point P(x, y) in the xy-plane with reference to the rectangular x and y axes. The plot of a given complex number z = x + iy, as the point P(x, y) in the xy-plane is known as Argand’s diagram (Fig. 1.1). The x-axis is called the real axis, y-axis is called the imaginary axis, and the xy-plane is called the complex plane.

P(x, y)

x′

1.4 alGebra of complex numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers. (a) Addition:

z1 + z2 = (x1 + iy1) + (x2 + iy2) = (x1 + x2) + i (y1 + y2)

x

O

y′

fig. 1.1

1.5

(b) Subtraction:

Different Forms of Complex Numbers

z1 – z2 = (x1 + iy1) – (x2 + iy2) = (x1 – x2) + i (y1 – y2)

(c) Multiplication:

z1 z2 = (x1 + iy1) (x2 + iy2) [∵ i2 = –1]

= (x1 x2 – y1 y2) + i (x2 y1 + y2 x1) z1 x + iy1 = 1 z2 x2 + iy2 ( x + iy1 ) ( x2 - iy2 ) = 1 ◊ ( x2 + iy2 ) ( x2 - iy2 )

(d) Division:

=

1.5 1.5.1

1.3

x1 x2 + y1 y2 x22

+

y22

Êy x -x y ˆ + i Á 1 22 12 2 ˜ Ë x2 + y2 ¯

dIfferent forms of complex numbers cartesian or rectangular form

If x and y are real numbers then z = x + iy is called the Cartesian form of the complex number.

1.5.2

polar form

The complex number z = x + iy can be represented by the point P whose cartesian co-ordinates are (x, y) (Fig. 1.2). If polar coordinates of the point P are (r, q ) then x = r cos q and y = r sin q. Hence, polar form of z is

y P(r, q ) r x'

z = r cos q + ir sin q = r (cos q + i sin q )

q O

x

Polar form can also be written as r – q. y'

1.5.3

exponential form

fig. 1.2

Polar form of z is z = r (cos q + i sin q ) cos q + i sin q = eiq Hence, exponential form of z is

[Using Expansion]

z = reiq

note eiq = cos q + i sin q, e–iq = cos q – i sin q. 1 iq - iq Hence, cos q = (e + e ) 2

and

sin q =

1 iq (e - e - iq ) 2i

1.4

Chapter 1

1.6

modulus and arGument (or amplItude) of complex numbers

Complex Numbers

Let

z = x + iy = r (cos q + i sin q )

||

Here ‘r’ is called the modulus or absolute value of z and is denoted by z or mod (z) and q is called argument or amplitude of z and is denoted by arg (z) or amp (z). z = r = x 2 + y2

Hence,

Ê yˆ arg (z) = q = tan -1 Á ˜ Ë x¯

note The value of q lying between –p and p is called the principal value of argument. The argument of z is the value of q which lies in the quadrant of point (x, y).

1.7

propertIes of complex numbers

Let z = x + iy and

z = x – iy. 1 (a) Re (z) = x = ( z + z ) 2 1 (b) Im (z) = y = (z - z ) 2i (c) ( z1 + z2 ) = z1 + z2 (d) ( z1 z2 ) = z1 z2 Êz ˆ

z

(e) Á 1 ˜ = 1 z2 Ë z2 ¯ (f ) z z = z 2 = | z |2 (g)

|| |z1z2| = |z1| |z2| and

(h)

È∵| z | = | z | = x 2 + y 2 ˘ ÎÍ ˚˙

arg (z1z2) = arg (z1) + arg (z2)

z1 z1 = z2 z2 and

Êz ˆ arg Á 1 ˜ = arg (z1) – arg (z2) Ë z2 ¯

1.7

Properties of Complex Numbers

1.5

example 1 [Summer 2015]

Arg (z1, z2) = Arg (z1) + Arg (z2)? Justify. Solution Yes,

Arg (z1 z2) = Arg (z1) + Arg (z2) z1 = r1eiq1 , z2 = r2 eiq2

Let

z1 z2 = (r1 eiq1 ) (r2 eiq2 ) = r1 r2 ei (q1 + q2 ) \

Arg ( z1 z2 ) = q1 + q 2 = Arg ( z1 ) + Arg( z2 )

example 2 Find the principal argument Arg z when z = Solution z= = =

-2 1+ 3 i -2 1 + 3i

◊

(1 - 3 i) (1 - 3 i)

-2 + 2 3 i 1+ 3

1 3 +i 2 2 Ê 3ˆ ÁË ˜ 2 ¯ Arg z = tan -1 Ê 1ˆ ÁË - ˜¯ 2 =-

( )

= tan -1 - 3 =

2p 3

-2 1+ 3 i

.

[Winter 2014]

1.6

Chapter 1

Complex Numbers

example 3 Find the modulus and principal value of the arguments of the following complex numbers: (i) (4 + 2i ) -3 + 2i

(

)

(ii) 2 + 6 3i 5 + 3i Solution (i)

z = (4 + 2i ) ( -3 + 2i )

= ( -12 + 4 2i - 6i - 2 2 )

= - (12 + 2 2 ) + i ( 4 2 - 6 ) |z| =

(12 + 2 2 )2 + (4

2 - 6)

2

= 2 36 + 2 + 12 2 + 8 + 9 - 12 2 = 2 55 È 4 2 -6 ˘ arg( z ) = tan -1 Í ˙ ÍÎ - (12 + 2 2 ) ˙˚ Ê 3-2 2 ˆ = tan -1 Á ˜ Ë 6+ 2 ¯ Aliter

z = (4 + 2i ) ( -3 + 2i ) = z1 z2 , say r =|z| = z1 z2 = z1 z2 = 4 + 2i -3 + 2i = ( 16 + 4 ) ( 9 + 2 ) = 220 = 2 55 q = arg( z ) = arg( z1 z2 ) = arg( z1 ) + arg( z2 )

= arg(4 + 2i ) + arg ( -3 + 2i )

1.7

Properties of Complex Numbers

Ê 2ˆ Ê 2ˆ = tan -1 Á ˜ + tan -1 Á Ë 4¯ Ë -3 ˜¯ Ê 2ˆ Ê 1ˆ = tan -1 Á ˜ - tan -1 Á Ë 2¯ Ë 3 ˜¯ Ê 1 2 Á 2- 3 -1 = tan Á Á 1+ 1 ◊ 2 ÁË 2 3

ˆ ˜ ˜ ˜ ˜¯

Ê 3-2 2 ˆ = tan -1 Á Ë 6 + 2 ˜¯ z=

2 + 6 3i

5 + 3i Ê 2 + 6 3i ˆ Ê 5 - 3i ˆ =Á ˜ ˜Á Ë 5 + 3i ¯ Ë 5 - 3i ¯

(ii)

28 + 28 3i 28 = 1 + 3i =

| z | = (1)2 +

( 3)

2

=2

Ê 3ˆ p arg ( z ) = tan -1 Á = Ë 1 ˜¯ 3

example 4 Evaluate [(1+i)100 + (1–i)100]. Solution Let

1 + i = reiq r = 1+ i = 1+1 = 2 p Ê 1ˆ q = tan -1 Á ˜ = tan -1 1 = Ë 1¯ 4

\

1 + i = 2e

and

1 - i = 2e

i

p 4

-i

p 4

1.7

1.8

Chapter 1

(1 + i )

100

Complex Numbers

+ (1 - i )

100

p Ê i ˆ = Á 2e 4 ˜ Ë ¯

100

p Ê -i ˆ + Á 2e 4 ˜ Ë ¯

100

= 250 (e25p i + e -25p i ) = 250 (2 cos 25p ) = 251 (-1) = -251

example 5 Express in polar form: Ê 2 + iˆ (i) Á Ë 3 - i ˜¯

2

(ii) 1 + sin a + i cos a

Solution 2

4 + i 2 + 4i Ê 2 +iˆ = ÁË 3 - i ˜¯ 9 + i 2 - 6i 3 + 4i = 8 - 6i 3 + 4i 8 + 6i = ◊ 8 - 6i 8 + 6i i = 2 1 = 0+i◊ 2

(i)

0+i◊

Let

1 = r (cos q + i sin q ) 2 2

1 Ê 1ˆ r = 02 + Á ˜ = Ë 2¯ 2 Ê 1ˆ p Á ˜ q = tan -1 Á 2 ˜ = tan -1 • = Ë 0¯ 2

and Hence, the polar form is 2

1Ê p pˆ Ê 2 +iˆ ÁË 3 - i ˜¯ = 2 ÁË cos 2 + i sin 2 ˜¯

1.7

(ii)

Properties of Complex Numbers

1.9

Êp ˆ Êp ˆ 1 + sin a + i cos a = 1 + cos Á - a ˜ + i sin Á - a ˜ Ë2 ¯ Ë2 ¯ Êp aˆ Êp aˆ Êp aˆ = 2 cos2 Á - ˜ + 2i sin Á − ˜ cos Á - ˜ Ë 4 2¯ Ë 4 2¯ Ë 4 2¯ q q˘ È 2 q Í∵ 1 + cos q = 2 cos 2 , sin q = 2 sin 2 cos 2 ˙ Î ˚ Êp aˆ È Êp aˆ Ê p a ˆ˘ = 2 cos Á - ˜ Ícos Á - ˜ + i sin Á - ˜ ˙ Ë 4 2¯ Î Ë 4 2¯ Ë 4 2 ¯˚

Comparing with the polar form r(cos q + i sin q ), Êp aˆ r = 2 cos Á - ˜ Ë 4 2¯ q=

p a 4 2

Hence, the polar form is Êp aˆ È Êp aˆ Ê p a ˆ˘ 1 + sin a + i cos a = 2 cos Á - ˜ Ícos Á - ˜ + i sin Á - ˜ ˙ Ë 4 2¯ Î Ë 4 2¯ Ë 4 2 ¯˚

example 6 Find the value of

- 5 + 12i .

Solution Let

- 5 + 12i = x + iy –5 + 12i = (x + iy)2 –5 + 12i = (x2 – y2) + i (2xy)

Comparing real and imaginary parts, x2 – y2 = –5 and 2xy = 12 xy = 6 Putting y =

6 in Eq. (1), x x2 -

36 x2

= –5

... (1)

1.10

Chapter 1

Complex Numbers

x4 + 5x2 – 36 = 0 (x2 + 9) (x2 – 4) = 0 x2 = –9, x2 = 4 Since x is real,

x=±2

and

y=±3 -5 + 12i = ± 2 ± 3i

Hence,

example 7 If x and y are real, solve the equation

3 y + 4i iy = 0. ix + 1 3 x + y

Solution 3 y + 4i iy =0 ix + 1 3 x + y iy (3 x + y) - (3 y + 4i )(ix + 1) =0 (ix + 1)(3 x + y) (-3 y + 4 x ) + i (3 xy + y 2 - 3 xy - 4) = 0 + i 0 (-3 y + 4 x ) + i( y 2 - 4) = 0 + i 0 Comparing real and imaginary parts, –3y + 4x = 0 and

\

y2 – 4 = 0 y=±2 x=±

3 2

x=±

3 , 2

Hence, the solution is y=±2

example 8 Solve the equation z2 + (2i – 3) z + 5 – i = 0. Solution z2 + (2i – 3) z + 5 – i = 0

[Summer 2015]

1.7

Properties of Complex Numbers

1.11

The equation is quadratic in z. z=

- (2i - 3) ± (2i - 3)2 - 4 (5 - i ) 2

- (2i - 3) ± 4i 2 + 9 - 12i - 20 + 4i 2 - (2i - 3) ± -15 - 8i = 2 =

=

- (2i - 3) ± 16i 2 + 1 - 8i 2

- (2i - 3) ± (1 - 4i )2 2 - (2i - 3) ± (1 - 4i ) = 2 - 2i + 3 + 1 - 4i -2i + 3 - 1 + 4i = and 2 2 = 2 - 3i and 1 + i =

Hence, the solution is z = 2 – 3i

and z = 1 + i

example 9 Prove that Re (z) > 0 and |z – 1| < |z + 1| are equivalent, where z = x + iy. Solution z = x + iy Re (z) > 0 x>0 Now,

... (1)

|z – 1| < |z + 1| |x + iy – 1| < |x + iy + 1| ( x − 1) 2 + y 2 < ( x + 1) 2 + y 2 x2 + 1 – 2x + y2 < x2 + 1 + 2x + y2 –2x < 2x 0 < 4x 00

... (2)

1.12

Chapter 1

Complex Numbers

From Eqs (1) and (2), Re (z) > 0

and

|z – 1| < |z + 1| are equivalent.

example 10 If i + z = i - z , show that z is completely real. Solution Let

z = x + iy i+z = i-z i + x + iy = i - x - iy x + i( y + 1) = - x + i(1 - y) 2

x + i( y + 1) = - x + i(1 - y)

2

x 2 + ( y + 1)2 = (- x )2 + (1 - y)2 x 2 + y2 + 2 y + 1 = x 2 + 1 - 2 y + y2 4y = 0 y=0 Hence, z = x is completely real.

example 11 Find the locus of z if z - 5 - 6i = 4. Solution Let

z = x + iy z - 5 - 6i = 4 x + iy - 5 - 6i = 4 ( x - 5) + i( y - 6) = 4 ( x - 5)2 + ( y - 6)2 = 4

Squaring both the sides, (x – 5)2 + (y – 6)2 = 16 which represents a circle with the centre at (5, 6) and a radius of 4 units. Hence, the locus of z is the circle.

1.7

Properties of Complex Numbers

1.13

example 12 z - 2i is purely imaginary, prove that the locus of z in the Argand 2z - 1 diagram is a circle. Find the centre and radius. If

Solution Let

u=

z - 2i 2z - 1

u=

( x + iy) - 2i 2( x + iy) - 1

=

and z = x + iy

x + i( y - 2) (2 x - 1) + 2iy

È x + i( y - 2) ˘ È (2 x - 1) - 2iy ˘ =Í ˙Í ˙ Î (2 x - 1) + 2iy ˚ Î (2 x - 1) - 2iy ˚ [ x(2 x - 1) + 2 y( y - 2)] + i[( y - 2)(2 x - 1) - 2 xy] = (2 x - 1)2 + 4 y 2 If u is purely imaginary, Re(u) = 0 x(2 x - 1) + 2 y( y - 2) (2 x - 1)2 + 4 y 2

=0

2 x 2 - x + 2 y2 - 4 y = 0 x x 2 + y2 - - 2 y = 0 2 Ê1 ˆ which represents a circle with the centre at Á , 1˜ and a radius of Ë4 ¯

example 13 z+i and z = x + iy then show that z+2 (i) locus of (x, y) is a straight line, if u is real (ii) locus of (x, y) is a circle, if u is purely imaginary If u =

Find the centre and radius of the circle.

17 units. 4

1.14

Chapter 1

Complex Numbers

Solution u=

z+i z+2

u=

x + iy + i x + iy + 2

and

z = x + iy

È x + i ( y + 1) ˘ È ( x + 2 - iy) ˘ =Í ˙ ˙Í Î ( x + 2) + iy ˚ Î ( x + 2) - iy ˚ [ x ( x + 2) + y ( y + 1)] + i [( y + 1)( x + 2) - xy] = ( x + 2 )2 + y 2 x ( x + 2) + y ( y + 1) Re (u) = ( x + 2 )2 + y 2 ( y + 1)( x + 2) - xy Im (u) = ( x + 2 )2 + y 2 x + 2y + 2 = ( x + 2 )2 + y 2 (i) If u is real then Im (u) = 0 x + 2y + 2 ( x + 2 )2 + y 2

=0

x + 2y + 2 = 0 which represents a straight line. Hence, the locus of (x, y) is a straight line. (ii) If u is purely imaginary then Re (u) = 0 x ( x + 2) + y ( y + 1) ( x + 2 )2 + y 2

=0

x2 + y2 + 2x + y = 0 5 1ˆ Ê which represents a circle with the centre at Á -1, - ˜ and a radius of units. Ë 2 2¯ Hence, the locus of (x, y) is a circle.

1.7

Properties of Complex Numbers

1.15

example 14 Find z if arg ( z + 1) =

2p p and arg ( z - 1) = . 3 6

Solution z = x + iy p arg (z + 1) = 6 p arg (x +iy + 1) = 6 p arg[( x + 1) + iy] = 6 Ê y ˆ p tan -1 Á = Ë x + 1˜¯ 6

Let

y 1 Êpˆ = tan Á ˜ = Ë ¯ x +1 6 3 x - y 3 = -1

Also,

arg ( z - 1) = arg ( x + iy - 1) = arg [( x - 1) + iy] = Ê y ˆ = tan -1 Á Ë x - 1˜¯

2p 3 2p 3 2p 3 2p 3

y Ê 2p ˆ = tan Á ˜ = - 3 Ë 3 ¯ x -1 x 3+y= 3 Solving Eqs (1) and (2), 1 3 , y= 2 2 1 3 z = +i 2 2 x=

Hence,

...(1)

...(2)

1.16

Chapter 1

Complex Numbers

example 15 If the sum and product of two numbers are real, show that the two numbers must be either real or conjugate. Solution Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers. Let z1 + z2 = a, where a is real. (x1 + iy1) + (x2 + iy2) = a + i · 0 (x1 + x2) + i (y1 + y2) = a + i · 0 Comparing real and imaginary parts, x1 + x2 = a y1 + y2 = 0

… (1) … (2)

Let z1 z2 = b, where b is real. (x1 + iy1) (x2 + iy2) = b + i · 0 (x1x2 – y1y2) + i (x2 y1 + x1 y2) = b + i · 0 Comparing real and imaginary parts, x1x2 – y1 y2 = b x2 y1 + x1 y2 = 0

… (3) … (4)

Substituting y2 = – y1 from Eq. (2) in Eq. (4), x2 y1 – x1 y1 = 0 y1(x2 – x1) = 0 y1 = 0 or x2 – x1 = 0, x1 = x2 If y1 = 0 then y2 = 0 z1 = x1 and z2 = x2 If x1 = x2 then z1 = x1 + iy1 and z2 = x1 – iy1 Hence, z1 and z2 both are either real or conjugate.

example 16 Prove that | exp (–2z) | < 1 if and only if Re z > 0. Solution z = x + iy Re( z ) = x e -2 z = e -2( x + iy ) = e -2 x e-2iy = e -2 x (cos 2 y - i sin 2 y)

[Winter 2012]

1.7

Properties of Complex Numbers

1.17

e -2 z = e- 4 x cos2 2 y + e - 4 x sin 2 2 y = e- 4 x = e- 2 x

... (1)

If Re z > 0, i.e., x > 0, e2x > e0 e2x > 1 e–2x < 1 e -2 z < 1

[From Eq. (1)]

exp (-2 z ) < 1

Conversely, if

e -2( x + iy ) < 1 e -2 x e -2iy < 1 È∵ reiq = r ˘ Î ˚

e -2 x < 1 1 e2 x

1 2 x log e > log 1 2x > 0 x>0 Hence, exp (-2z ) < 1 if and only if Re z > 0.

example 17 Show that

z - 1 £ arg ( z ) . z

Solution Let

z = reiq, where z = r and arg (z) = q

||

z reiq -1 = -1 z r = eiq – 1 = cos q + i sin q – 1

| | | | = |cos q – 1+ i sin q |

1.18

Chapter 1

Complex Numbers

= -2 sin 2

q q q + i 2 sin cos 2 2 2

= 2 sin

q q q - sin + i cos 2 2 2

= 2 sin

q 2

= 2 sin

q 2

£2

sin 2

q q + cos2 2 2

q 2

È sinq ˘ ÍÎ∵ q £ 1˙˚

£q £ arg( z )

example 18 Êp aˆ If sin a = i tan q, prove that cos q + i sin q = tan Á + ˜ . Ë 4 2¯ Solution i tan q = sin a i sin q sin a = cos q 1 Applying componendo and dividendo, cos q + i sin q 1 + sin a = cos q - i sin q 1 - sin a Êp ˆ 1 + cos Á - a ˜ iq Ë ¯ 2 e = e - iq Êp ˆ 1 - cos Á - a ˜ Ë2 ¯

e2iq =

Êp aˆ 2 cos2 Á - ˜ Ë 4 2¯ Êp aˆ 2 sin 2 Á - ˜ Ë 4 2¯

È Ê p a ˆ˘ (eiq )2 = Ícot Á - ˜ ˙ Î Ë 4 2 ¯˚

2

1.7

Properties of Complex Numbers

1.19

Êp aˆ eiq = cot Á - ˜ Ë 4 2¯ Èp Ê p a ˆ ˘ = tan Í - Á - ˜ ˙ Î2 Ë 4 2 ¯˚ Êp aˆ = tan Á + ˜ Ë 4 2¯ Êp aˆ cos q + i sin q = tan Á + ˜ Ë 4 2¯

example 19 Prove that tan -1 z =

i Ê i + zˆ log Á . Ë i - z ˜¯ 2

[Winter 2013]

Solution Let

tan–1 z = q z = tanq 1 Ê eiq - e - iq ˆ = Á iq i Ë e + e - iq ˜¯ Ê eiq - e - iq ˆ = -i Á iq Ë e + e - iq ˜¯ z -eiq + e - iq = iq i e + e - iq

Applying componendo and dividendo, i + z 2e - iq = i - z 2eiq i+z e -2iq = i-z Ê i + zˆ -2iq = log Á Ë i - z ˜¯ q=tan -1 z =

1 Ê i + zˆ log Á Ë i - z ˜¯ 2i

i Ê i + zˆ log Á Ë i - z ˜¯ 2

1.20

Chapter 1

Complex Numbers

example 20 iq

Prove that (1 - e )

-

1 2

+ (1 - e

- iq

)

-

1 2

1

qˆ2 Ê = Á 1 + cosec ˜ . Ë 2¯

Solution (1 - eiq )

-

1 2

+ (1 - e - iq )

-

1 2

= (1 - cos q - i sin q )

q q qˆ Ê = Á 2 sin 2 - i 2 sin cos ˜ Ë 2 2 2¯ qˆ Ê = Á 2 sin ˜ Ë 2¯ qˆ Ê = Á 2 sin ˜ Ë 2¯

qˆ Ê = Á 2 sin ˜ Ë 2¯

qˆ Ê = Á 2 sin ˜ Ë 2¯ qˆ Ê = Á 2 sin ˜ Ë 2¯ qˆ Ê = Á 2 sin ˜ Ë 2¯ qˆ Ê = Á 2 sin ˜ Ë 2¯

-

1 2

-

1 2

+ (1 - cos q + i sin q )

q q qˆ Ê + Á 2 sin 2 + i 2 sin cos ˜ Ë 2 2 2¯

-

-

1 2

1 2

-

1 2

1 1 È - ˘ ÍÊ sin q - i cos q ˆ 2 + Ê sin q + i cos q ˆ 2 ˙ ÁË ÍÁË 2 2 ˜¯ 2 2 ¯˜ ˙˚ Î

-

1 2

1 1 È - ˘ Êq qˆ¸ 2 ˙ Êq qˆ¸ 2 Ï Êq qˆ ÍÏ Ê q q ˆ ÍÌcos ÁË 2 - 2 ˜¯ - i sin ÁË 2 - 2 ˜¯ ˝ + Ìcos ÁË 2 - 2 ˜¯ + i sin ÁË 2 - 2 ˜¯ ˝ ˙ Ó ˛ ˙ ˛ ÍÓ Î ˚

-

1 2

-

1 2

-

1 2

-

1 2

-

1 2

1 1 È - ˘ ÍÏÔ - i ÊÁ p - q ˆ˜ ¸Ô 2 ÏÔ i ÊÁ p - q ˆ˜ ¸Ô 2 ˙ Í Ìe Ë 2 2 ¯ ˝ + Ìe Ë 2 2 ¯ ˝ ˙ ÍÔ ÔÓ Ô˛ ˙ Ô˛ ÍÓ ˙ Î ˚ Êp qˆ È i ÊÁ p - q ˆ˜ -iÁ - ˜ ˘ Íe Ë 4 4 ¯ + e Ë 4 4 ¯ ˙ Í ˙ Î ˚

È Ê p q ˆ˘ Í2 cos ÁË - ˜¯ ˙ 4 4 ˚ Î 1

È q ˆ ˘2 2Êp Í4 cos ÁË - ˜¯ ˙ 4 4 ˚ Î 1

È Ï Ê p q ˆ ¸˘ 2 Í2 Ì1 + cos Á - ˜ ˝˙ Ë 2 2 ¯ ˛˙˚ ÍÎ Ó

1

q ˘2 È Í 1 + sin 2 ˙ =Í ˙ Í si n q ˙ ÍÎ 2 ˙˚ 1

q ˘2 È = Ícosec + 1˙ 2 ˚ Î

È 2 q˘ Í∵ 1 + cos q = 2 cos 2 ˙ Î ˚

1.7

Properties of Complex Numbers

example 21 If a = i + 1, b = 1 – i and tanf =

1 then prove that x +1

( x + a )n - ( x + b )n = sin nf cosec nf. a -b

Solution a = i + 1,

b = 1 – i,

tanf =

1 x +1

cot f = x + 1, x = cot f – 1 ( x + a )n - ( x + b )n (cot f - 1 + i + 1)n - (cot f - 1 + 1 - i )n = a -b i +1-1+ i n

= =

Ê cos f ˆ Ê cos f ˆ ÁË sin f + i˜¯ - ÁË sin f - i˜¯

n

2i (cos f + i sin f )n - (cos f - i sin f )n 2i sin n f (eif )n - (e- if )n

=

2i sin n f einf - e- inf

= =

2i sin n f 2i sin nf 2i sin n f

= sin nf cosec n f

example 22 If (a1 + ib1) (a2 + ib2) . . . (an + ibn) = A + iB, prove that 2 2 2 2 2 2 2 2 (i) (a1 + b1 )(a2 + b2 ) (an + bn ) = A + B

Êb ˆ Êb ˆ Êb ˆ Ê Bˆ (ii) tan -1 Á 1 ˜ + tan -1 Á 2 ˜ +  + tan -1 Á n ˜ = tan -1 Á ˜ . Ë A¯ Ë a1 ¯ Ë a2 ¯ Ë an ¯

1.21

1.22

Chapter 1

Complex Numbers

Solution (a1 + ib1) (a2 + ib2) . . . (an + ibn) = A + iB (i)

… (1)

Taking the modulus of Eq. (1),

|( a1 + ib1)(a2 + ib2) L (an + ibn)| = |A + iB | |a1 + ib1| |a2 + ib2| L |an + ibn| = | A + iB | a12 + b12 a22 + b22  an2 + bn2 =

A2 + B 2

Squaring both the sides, (a12 + b12 )(a22 + b22 )  (an2 + bn2 ) = A2 + B2 (ii) Taking the argument of Eq. (1), arg [(a1 + ib1) (a2 + ib2) L (an + ibn)] = arg (A + iB) arg (a1 + ib1) + arg (a2 + ib2) + L + arg (an + ibn) = arg (A + iB) Êb ˆ Êb ˆ Êb ˆ Ê Bˆ tan -1 Á 1 ˜ + tan -1 Á 2 ˜ +  + tan -1 Á n ˜ = tan -1 Á ˜ Ë A¯ Ë a1 ¯ Ë a2 ¯ Ë an ¯

example 23 If x + iy = 3 a + ib , prove that

a b + = 4( x 2 - y 2 ). x y

Solution x + iy = 3 a + ib 1

(a + ib) 3 = x + iy a + ib = ( x + iy)3 = x 3 + i 3 y3 + 3 x 2 iy + 3 xi 2 y 2 = (xx 3 - 3 xy 2 ) + i(3 x 2 y - y3 ) Comparing real and imaginary parts,

Hence,

a = x3 – 3xy2

and

b = 3x2y – y3

a = x 2 - 3 y2 x

and

b = 3 x 2 - y2 y

a b + = 4( x 2 - y 2 ) x y

[∵ i 3 = -i ]

1.7

Properties of Complex Numbers

example 24 If (1 + cos q + i sin q )(1 + cos 2q + i sin 2q ) = u + iv, prove that 2 2 2 (i) u + v = 16 cos

q cos2 q 2

(ii)

3q v = tan . 2 u

Solution u + iv = (1 + cos q + i sin q)(1 + cos 2q + i sin 2q)

q q qˆ Ê = Á 2 cos2 + i 2 sin cos ˜ (2 cos2 q + i 2 sin q cos q ) Ë 2 2 2¯ qÊ q qˆ = 2 cos Á cos + i sin ˜ 2 cos q (cos q + i sin q ) 2¯ 2Ë 2 iq

q = 4 cos cos q ◊ e 2 ◊ eiq 2 q = 4 cos cos q ◊ e 2 iq = re , say where

i 3q 2

q r = u + iv = 4 cos cos q 2 q u2 + v 2 = 4 cos cos q 2 q u2 + v 2 = 16 cos2 cos2 q 2

and

f = arg(u + iv) =

3q 2

Ê v ˆ 3q tan -1 Á ˜ = Ë u¯ 2 v Ê 3q ˆ = tan Á ˜ Ë 2¯ u

example 25 Êpˆ Êpˆ If xr = cos Á r ˜ + i sin Á r ˜ , show that lim x1 x2 x3  xn = -1. Ë2 ¯ Ë2 ¯ nÆ•

1.23

1.24

Chapter 1

Complex Numbers

Solution Êpˆ Êpˆ xr = cos Á r ˜ + i sin Á r ˜ Ë2 ¯ Ë2 ¯ ip

= e2

r

ip e2

lim x1 x2 x3  xn = lim

n →•

ip

ip

ip

2 e2

3 e2

n e2

n Æ•

= lim e



Ê1 1 1 1ˆ ip Á + + + + ˜ Ë 2 2 2 23 2n ¯

n Æ•

lim x1 x2 x3  xn = lim

n Æ•

È Ê 1ˆ n ˘ ip Í1- Á ˜ ˙ Ë 2 ¯ ˙˚ e ÍÎ

[Using the sum of GP]

nƕ

= lim e e ip

Ê 1ˆ - ip Á ˜ Ë 2n ¯

n Æ•

=e e ip

-

ip 2•

= (cos p + i sin p ) e

-

ip •

= ( -1 + i ◊ 0 ) e 0 Hence,

lim x1 x2 x3 ... xn = -1 n Æ•

exercIse 1.1 1. Find the modulus and principal value of the argument: (i) - 3 - i 1 + 2i 1 - 3i 1 + 2i (iii) 1 - (1 - i)2 (ii)

1+ i 1- i (v) tan a – i

(iv)

-5p 1 3p È , (iii) 1, 0 Í Ans. : (i) 2, 6 (ii) 4 2 Í p Í (v) sec a , a ÍÎ 2

(iv) 1,

p˘ 4˙ ˙ ˙ ˙˚

1.8

De Moivre’s Theorem

1.25

2. Express in polar form: (i)

3 -i 1+ i 1- i

(ii) (iii)

2 + 6 3i 5 + 3i È p pˆ p p p p ˆ˘ Ê Ê Í Ans.: (i) 2 ÁË cos - i sin ˜¯ (ii) cos - i sin (iii) 2 ÁË cos - i sin ˜¯ ˙ 6 6 2 2 3 3 ˚ Î

3. Find the value of

3 - 4i. [ans. : 2 – i or –2 + i]

4. Find the locus of z, if

z -1 is purely imaginary. z +i [ans. : circle x2 + y2 – x – y = 0]

5. Find the locus of z if (i)

z -1 =1 z +1

Ê z - 1ˆ p (ii) arg ÁË ˜= z + 1¯ 4 ÈÎ Ans.: (i) x = 0 (ii) circle x 2 + y 2 - 2y - 1 = 0 ˘˚

6. Find two numbers whose sum is 4 and product is 8. [ans. : 2 ± 2i] 7. If x + iy =

2

2 2

a + ib , prove that (x + y ) = a + b

8. If |z| = 1, z ≠ 1, prove that

2.

z -1 is purely imaginary. z +1

9. If a = ei2a, b = ei2b, c = ei2g then prove that 10. If (a + i b) =

2

ab c + = 2 cos(a + b - g ). c ab

1 , prove that (a2 + b 2) (a2 + b2) = 1. a + ib

11. Find the value of x2 – 6x + 13, when x = 3 + 2i

1.8

de moIvre’s theorem

statement For any real number n, one of the values of (cos q + i sin q )n is cos nq + i sin nq. Hence, (cos q + i sin q )n = cos nq + i sin nq

1.26

Chapter 1

Complex Numbers

Proof Case I: n is a positive integer Let z1 = r1 (cos q1 + i sin q1), z2 = r2 (cos q2 + i sin q2), …, zn = rn (cos qn + i sin qn). z1 z2 = r1 (cos q1 + i sin q1) r2 (cos q2 + i sin q2) = r1 r2 [(cos q1 cos q2 – sin q1 sin q2) + i(sin q1 cos q2 + cos q1 sin q2)] = r1 r2 [cos (q1 + q2) + i sin (q1 + q2)] Similarly, z1 z2… zn = r1 (cos q1 + i sin q1) r2 (cos q2 + i sin q2)… rn (cos qn + i sin qn) = (r1 r2… rn) (cos q1 + i sin q1) (cos q2 + i sin q2)… (cos qn + i sin qn) = (r1 r2… rn)[cos (q1 + q2 + … + qn) + i sin (q1 + q2 + … + qn)]

… (1.1)

If z1 = z2 = … = zn = z = r (cos q + i sin q) then Eq. (1.1) reduces to zn = rn (cos nq + i sin nq) rn(cos q + i sin q )n = rn (cos nq + i sin nq) (cos q + i sin q)n = (cos nq + i sin nq), where n is a positive integer. Case II: n is a negative integer Let n = –m, where m is a positive integer. (cos q + i sin q)n = (cos q + i sin q)–m 1

=

(cos q + i sin q )m 1 = (cos mq + i sin mq ) (cos mq - i sin mq ) = (cos mq + i sin mq )(cos mq - i sin mq ) = cos mq - i sin mq

[ Using Case I]

= cos( - m)q + i sin(-m m)q [∵ cos(-q ) = cos q ,sin(-q ) = - sin q ] = cos nq + i sin nq , where n is a negetive integer. Case III: n is a rational number Let n =

p , where p and q are integers and q π 0. q q

Ê q qˆ Ê qˆ Ê qˆ Consider Á cos + i sin ˜ = cos Á q ◊ ˜ + i sin Á q ◊ ˜ q q q Ë ¯ Ë ¯ Ë q¯ = cos q + i sin q

[ Ussing Case I and Case II]

1.8

1.27

De Moivre’s Theorem

1

q q cos + i sin = (cos q + i sin q ) q q q

Hence,

p

p

Ê q qˆ q ÁË cos q + i sin q ˜¯ = (cos q + i sin q )

p

Ê qˆ Ê qˆ cos Á p ◊ ˜ + i sin Á p ◊ ˜ = (cos q + i sin q ) q Ë q¯ Ë q¯

[ Using Case I and Case II]

cos n q + i sin nq = (cos q + i sin q )n (cos q + i sin q)n = (cos nq + i sin nq), where n is a rational number. Hence,

(cos q + i sin q)n = cos nq + i sin nq for any real number n.

example 1 n

Ê np ˆ Ê np ˆ Ê 1 + sin a + i cos a ˆ Prove that Á = cos Á - na ˜ + i sin Á - na ˜ . ˜ Ë ¯ Ë ¯ 2 2 Ë 1 + sin a - i cos a ¯ Solution ˆ˘ Êp ˆ Êp È n Í 1 + cos ÁË 2 - a ˜¯ + i sin ÁË 2 - a ˜¯ ˙ Ê 1 + sin a + i cos a ˆ ˙ ÁË ˜ =Í Í 1 + sin a - i cos a ¯ Êp ˆ Êp ˆ˙ Í 1 + cos Á - a ˜ - i sin Á - a ˜ ˙ Ë2 ¯ Ë2 ¯˚ Î

n

aˆ Êp aˆ Êp aˆ˘ È 2Êp Í 2 cos ÁË 4 - 2 ˜¯ + 2i sin ÁË 4 - 2 ˜¯ cos ÁË 4 - 2 ˜¯ ˙ ˙ =Í Í aˆ Êp aˆ Êp aˆ˙ 2Êp Í 2 cos Á - ˜ - 2i sin Á - ˜ cos Á - ˜ ˙ Ë 4 2¯ Ë 4 2¯˚ Ë 4 2¯ Î Êp aˆ Êp aˆ˘ È Í cos ÁË 4 - 2 ˜¯ + i sin ÁË 4 - 2 ˜¯ ˙ ˙ =Í Í Êp aˆ Êp aˆ˙ Í cos Á - ˜ - i sin Á - ˜ ˙ Ë 4 2¯ Ë 4 2¯˚ Î È i ÊÁ p - a ˆ˜ Í e Ë 4 2¯ =Í Í - i ÊÁ p - a ˆ˜ ÍÎ e Ë 4 2 ¯

˘ ˙ ˙ ˙ ˙˚

È 2i ÊÁ p - a ˆ˜ ˘ = ÍÎe Ë 4 2 ¯ ˙˚

n

n

n

n

1.28

Chapter 1

Complex Numbers

=e

Ê np ˆ iÁ - na ˜ Ë 2 ¯

Ê np ˆ Ê np ˆ = cos Á - na ˜ + i sin Á - na ˜ Ë 2 ¯ Ë 2 ¯

example 2 If x +

1 = 2 cos q , prove that x

(i) x n +

1 x

n

= 2 cos n q

(ii)

x2n + 1 x

2 n -1

+x

=

cos n q cos( n - 1)q [Summer 2015]

Solution x + x -1 = 2 cos q = eiq + e - iq \ (i)

x=e xn +

1 x

n

[ Using Euler’s formula]

iq

( )

= ei q

n

+

1

(e ) iq

n

= einq + e- inq = 2 cos nq

(e ) = +x (e ) iq

x2n + 1 (ii)

x 2 n -1

iq

2n

2 n -1

+1 + eiq

Ê ei 2 nq + 1 ˆ Ê e - inq ˆ = Á i (2 n -1)q ˜Á ˜ + eiq ¯ Ë e - inq ¯ Ëe =

einq + e - inq

ei ( n -1)q + e- i ( n -1)q 2 cos nq = 2 cos( n - 1)q cos nq = cos( n - 1)q

1.8

De Moivre’s Theorem

example 3 If x +

1 1 1 = 2 cos q , y + = 2 cos f , z + = 2 cos y then prove that x y z

(i) xyz + (ii)

xm yn

1 = 2 cos(q + f + y ) xyz yn

+

xm

= 2 cos(mq - nf ) 1

m n (iii) x y +

m n

= 2 cos(mq + nf )

x y Solution x+

1 = 2 cosq x

x + x -1 = eiq + e - iq \x = e

(i)

xyz +

[Using Euler’s formula]

iq

1 1 = eiq eif eiy + iq if iy xyz e e e = ei (q +f +y ) + e - i (q +f +y ) = 2 cos(q + f + y )

(ii)

xm yn

+

yn xm

=

(eiq )m (eif )n

+

(eif )n (eiq )m

= ei ( mq - nf ) + e - i ( mq - nf ) = 2 cos( mq - nf ) (iii)

x m yn +

1 m n

x y

= (eiq )m (eif )n +

1 iq m

(e ) (eif )n

= ei ( mq + nf ) + e - i ( mq + nf ) = 2 cos(mq + nf )

1.29

1.30

Chapter 1

Complex Numbers

example 4 If x -

1 1 1 = 2i sin q , y - = 2i sin f , z - = 2i sin y, prove that x y z

(i) xyz + m

x

n

y

(ii)

1 = 2 cos (p + q + y ) xyz Ê q fˆ = 2 cos Á - ˜ . Ë m n¯ x

n

y

+m

Solution x-

x-

1 1 1 = 2i sin q , y - = 2i sin f , z - = 2i sin y x y z

1 = 2i sin q x

x - x -1 = eiq - e - iq \ x=e

y = eif, z = eiy

Similarly, (i)

[ Using Euler’s formula ]

iq

xzy +

1 1 = eiq eif eiy + iq if iy xyz e e e = ei (q +f +y ) + e - i (q +f +y ) = 2 cos(q + f + y )

(ii)

m

x

n

y

n

+m

y x

=

1

1

(eiq ) m

(eif ) n

1 (eif ) n

=e

Ê q fˆ iÁ - ˜ Ë m n¯

+

1

(eiq ) m +e

Ê q fˆ -i Á - ˜ Ë m n¯

Ê q fˆ = 2 cos Á - ˜ Ë m n¯

1.8

De Moivre’s Theorem

1.31

example 5 Prove that (4n)th power of 1 + 7i is equal to (–4)n, where n is a (2 - i )2 positive integer. Solution 1 + 7i ( 2 - i )2

=

1 + 7i

4 + i 2 - 4i 1 + 7i = 4 - 1 - 4i Ê 1 + 7i ˆ Ê 3 + 4i ˆ =Á Ë 3 - 4i ˜¯ ËÁ 3 + 4i ˜¯

3 + 4i + 21i + 28i 2 9 + 16 3 + 25i - 28 = 25 = -1 + i –1 + i = r (cos q + i sin q ) =

Let where

r = | - 1 + i | = (-1)2 + 12 = 2 3p Ê 1ˆ q = tan -1 Á ˜ = tan -1 (-1) = Ë -1¯ 4 [∵ the point (–1,1) lies in the second quadrant]

\

Now,

3p 3p ˆ Ê -1 + i = 2 Á cos + i sin ˜ Ë 4 4¯ È 1 + 7i ˘ Í 2˙ Î (2 - i ) ˚

4n

= (-1 + i )4 n 4n

È Ê 3p 3p ˆ ˘ = Í 2 Á cos + i sin ˜ ˙ 4 4 ¯˚ Î Ë 4n È Ê 3p ˆ Ê 3p ˆ ˘ = ( 2 ) Ícos Á 4 n ˜ + i sin Á 4 n ˜ ˙ Ë Ë ¯ 4 ¯˚ 4 Î = (2)2 n (cos 3n p + i sin 3n p ) = (4)n ÈÎ(-1)3n + 0 ˘˚ = ÈÎ4(-1)3 ˘˚ = (-4)n

n

1.32

Chapter 1

Complex Numbers

example 6 If a and b are the roots of the equation x2 – 2x + 2 = 0, find the value of a n + b n. Hence, deduce that a 8 + b 8 = 32. [Summer 2015] Solution x2 – 2x + 2 = 0 2± 4-8 x= 2 = 1 ± -1 = 1± i Since a and b are the roots of the equation, a = 1 + i, b = 1 – i 1 + i = r(cos q + i sin q)

Let

r = 1 + i = 1+1 = 2 and

p Ê 1ˆ q = tan -1 Á ˜ = tan -1 (1) = Ë 1¯ 4

\

p pˆ Ê a = 1 + i = 2 Á cos + i sin ˜ = 2 e 4 Ë 4 4¯

and

p pˆ Ê b = 1 - i = 2 Á cos - i sin ˜ = 2 e 4 Ë 4 4¯

ip

ip

n

ip ˆ ip Ê Ê - ˆ 4 a +b = Ë 2 e ¯ +Ë 2 e 4 ¯ n

n

= ( 2)

n

(e

inp 4

+e

n

np ˆ Ê = 2 2 Á 2 cos ˜ Ë 4 ¯ n

= 22

+1

Ê np ˆ cos Á ˜ Ë 4 ¯

Putting n = 8, 8p 4 5 = 2 cos 2p

a 8 + b 8 = 25 cos

= 25 = 32

-

inp 4

)

n

1.8

De Moivre’s Theorem

1.33

example 7 If a and b are the roots of the equation x2 – 2x + 4 = 0 then show that Ê np ˆ a n + b n = 2 n+1 cos Á ˜ and, hence, find the value of a 15 + b 15. Ë 3¯ Solution x2 – 2x + 4 = 0 2 ± 4 - 16 x= 2 = 1 ± -3 = 1± i 3 a and b are the roots of the equation. \ a=1+i 3,b=1–i 3 Let

1 + i 3 = r (cos q + i sinq )

where

r = 1 + i 3 = 1+ 3 = 4 = 2

and

q = arg (1 + i 3 ) = tan -1 ( 3 ) =

\

p pˆ Ê a = 1 + i 3 = 2 Á cos + i sin ˜ = 2e 3 Ë 3 3¯

and

b = 1 - i 3 = 2 Á cos - i sin ˜ = 2e Ë 3 3¯

p 3 ia

Ê

n

n

n

( ) + (2e ) ) = 2 (e + e

a +b =

ip 2e 3 n

-

inp 3

-

Ê np ˆ = 2 n +1 cos Á ˜ Ë 3 ¯ Putting n = 15, Ê 15p ˆ a 15 + b 15 = 216 cos Á Ë 3 ˜¯ = 216 (-1) = -216

ip 3

inp 3

Ê np ˆ = 2 n ◊ 2 cos Á ˜ Ë 3 ¯

= 216 cos 5p

pˆ

p

n

-

ip 3

1.34

Chapter 1

Complex Numbers

example 8 If a and b are roots of z2 sin2 q – z sin 2q + 1 = 0, prove that a n + b n = 2 cos nq cosecn q , where n is a positive integer. Solution z2 sin2 q – z sin 2q + 1 = 0 z= =

sin 2q ± sin 2 2q - 4 sin 2 q 2 sin 2 q 2 sin q cos q ± 4 sin 2 q cos2 q - 4 sin 2 q 2 sin 2 q

cos q ± cos2 q - 1 sin q cos q ± i sin q = sin q =

= e ± iq cosec q a and b are the roots of the equation. \ a = e iqcosec q,

b = e–iq cosec q

a n + b n = (eiq cosec q )n + (e–iq cosec q )n = (cosec q )n (einq + e–inq ) = (cosec q )n (2 cos nq ) = 2 cos nq cosecn q

example 9 If z = –1 + i 3 and n is an integer then prove that z2n + 2nzn + 22n = 0, if n is not a multiple of 3. Solution z = –1 + i 3 Let

–1 + i 3 = r (cos q + i sin q ) 2

where

r = -1 + i 3 = (-1)2 + ( 3 ) = 4 = 2

and

Ê 3ˆ 2p -1 q = tan -1 Á ˜ = tan ( - 3 ) = 3 1 Ë ¯ [∵ the point ( - 1, 3) lies in the second quadrant]

1.8

De Moivre’s Theorem

2p 2p ˆ Ê z = - 1 + i 3 = 2 Á cos + i sin ˜ = 2e Ë 3 3 ¯

\

zn Consider

2n

2n

+

=

zn

Ê 2ip ˆ n Á 2e 3 ˜ ÁË ˜¯ 2n

=e

2 inp 3

+e

-

+

1.35

2 ip 3

2n Ê 2ip ˆ Á 2e 3 ˜ ÁË ˜¯

n

2 inp 3

2 np 3 Let n = 3k + 1 where k is an integer then = 2 cos

zn 2

n

2n

+

z

= 2 cos

n

2p (3k + 1) 3

2p ˆ Ê = 2 cos Á 2kp + Ë 3 ˜¯ 2p 3 Ê 1ˆ = 2Á- ˜ Ë 2¯

= 2 cos

[∵ cos(2 kp + q ) = cos q ]

= -1 zn 2n

+

2n

= -1

zn

z2n + 2n zn + 22n = 0, if n is not a multiple of 3.

exercIse 1.2 1. If z =

1 2

+

i 2

then by using De Moivre’s theorem, simplify (z)10 + (z )10 ,

where z is the complex conjugate of z. [ans.: 0] 2.

If n is a positive integer, show that (a + ib)n + (a – ib)n = 2 rn cos nq, Ê bˆ where r2 = a2 + b2 and q = tan-1 Á ˜ ◊ Hence, or otherwise, deduce that Ë a¯ 8

8

(1 + i 3 ) + (1 - i 3 )

= -28 .

1.36

Chapter 1

3. If x +

Complex Numbers

1 1 1 r r = 2 cos q , y + = 2 cos f, prove that x y + r r = 2 cos(rq + rf) . x y x y

4. If 2 cos q = x +

1 cos n q x 2n + 1 , prove that 2 n -1 . = x cos( n - 1)q x +x

5. If sin a + sin b = 0 = cos a + cos b, prove that (i) cos 2a + cos 2b = 2 cos (p + a + b ) (ii) sin 2a + sin 2b = 2 sin (p + a + b ) 2 6. If a and b are the roots of the equation x - 2 3x + 4 = 0 then prove that 3 3 a + b = 0.

7. If a , b are the roots of the equation x2 – 2x + 2 = 0, prove that a n + b n n = 2 ◊ 2 2 cos np ◊ 4 Hence, show that a 8 + b 8 = 32. 8. If a, b are the roots of the equation x2 – 3x + 1 = 0, prove that a n + b n np = 2 cos . Hence, show that a 12 + b 12 = 2. 6

1.9 applIcatIons of de moIvre’s theorem 1.9.1

roots of an algebraic equations

De Moivre’s theorem can be used to find the roots of an algebraic equation. Let the equation be zn = cos q + i sin q 1

z = (cos q + i sin q ) n 1

= [cos(2 kp + q ) + i sin(2 kp + q )] n Ê 2 kp + q ˆ Ê 2 kp + q ˆ = cos Á + i sin Á ˜ Ë Ë n ¯ n ˜¯ Putting k = 0, 1, 2, . . . n – 1, all n roots of the equation are obtained.

notes (i) Complex roots always occur in conjugate pairs if all the coefficients of the equation including constant terms are real. (ii) The product of all the roots of the equation is known as the continued product.

1.9 Applications of De Moivre’s Theorem

1.37

example 1 Find all the values of the following: 1

2

(i) (-1) 5

(ii) (1 - i ) 3

(iii)

3

1+ i

+

3

2

1- i

.

2

Solution –1 = r (cos q + i sin q)

(i) Let

r = | –1| = 1

where

Ê 0ˆ q = tan -1 Á ˜ = tan -1 0 = p Ë -1¯ –1 = cos p + i sin p

and

[ the point (–1,0) lies in the second quadrant]

= cos (2kp + p) + i sin (2kp + p) 1

1

(-1) 5 = [cos(2 k + 1)p + i sin(2 k + 1)p ]5 = cos(2 k + 1)

p p + i sin (2 k + 1) 5 5 1

Putting k = 0, 1, 2, 3, 4, all five values of (-1) 5 are obtained. (ii) Let 1 – i = r (cos q + i sin q ) where and

r = 1 - i = 1 + (-1)2 = 2

p Ê -1ˆ q = tan -1 Á ˜ = tan -1 (-1) = Ë 1¯ 4 È Ê pˆ Ê p ˆ˘ 1 - i = 2 Ícos Á - ˜ + i sin Á - ˜ ˙ Ë ¯ Ë 4¯˚ 4 Î p pˆ Ê = 2 Á cos - i sin ˜ Ë 4 4¯ 2 (1 - i ) 3

2

È Ê p p ˆ˘3 = Í 2 Á cos - i sin ˜ ˙ 4 4¯˚ Î Ë 1 2Ê È 2p 2p ˆ ˘ 3 - i sin ˜ ˙ = Í( 2 ) Á cos Ë 4 4 ¯˚ Î 1

È Ê p p ˆ ˘3 = Í2 Á cos - i sin ˜ ˙ 2 2¯˚ Î Ë

[ the point (1, –1) lies in the fourth quadrant]

1.38

Chapter 1

Complex Numbers

1

È Ï Ê pˆ p ˆ ¸˘ 3 Ê = Í2 Ìcos Á 2kp + ˜ - i sin Á 2kp + ˜ ˝˙ Ë 2¯ 2 ¯ ˛˙˚ ÍÎ Ó Ë 1

p p˘ È = 2 3 Ícos(4 k + 1) - i sin (4 k + 1) ˙ [Using De Moivre’s theorem] 6 6˚ Î 2

Putting k = 0, 1, 2, all three values of (1 - i ) 3 are obtained. 1+ i

(iii) Let

2

= r (cos q + i sin q)

r=

1+ i 2

=

q = tan -1 (1) =

and

1+ i 2

= cos

1- i

and

2

1 1 + =1 2 2 p 4

p p + i sin 4 4

= cos

p p - i sin 4 4 1

3

1+ i 2

+3

1

1- i

p pˆ3 Ê p pˆ3 Ê = Á cos + i sin ˜ + Á cos - i sin ˜ Ë ¯ Ë 4 4 4 4¯ 2 1

È Ê pˆ p ˆ ˘3 Ê = Ícos Á 2 kp + ˜ + i sin Á 2 kp + ˜ ˙ Ë 4¯ 4¯˚ Î Ë 1

È Ê p ˆ ˘3 pˆ Ê + Ícos Á 2 kp + ˜ - i sin Á 2 kp + ˜ ˙ Ë 4¯˚ 4¯ Î Ë p p˘ È = Ícos(8k + 1) + i sin (8k + 1) ˙ 12 12 ˚ Î p p˘ È + Ícos(8k + 1) - i sin (8k + 1) ˙ 12 12 ˚ Î p = 2 cos(8k + 1) 12 Putting k = 0, 1, 2, all three values of

3

1+ i 2

+3

1- i 2

[Using De Moivre’s theorem]

are obtained.

1.9 Applications of De Moivre’s Theorem

example 2 3

Ê1 3ˆ 4 Find the continued product of all the values of Á + i . 2 ˜¯ Ë2 Solution 1 3 +i 2 2 = r (cos q + i sin q)

Let

z=

where

r=

and

Ê Á q = tan -1 Á Á ÁË

1 3 +i = 2 2

1 3 + =1 4 4

3ˆ 2 ˜ = tan -1 ( 3 ) = p ˜ 1 ˜ 3 2 ˜¯

1 3 p p +i = cos + i sin 2 2 3 3 3

3

Ê1 3ˆ4 Ê p pˆ4 Á 2 + i 2 ˜ = ÁË cos 3 + i sin 3 ˜¯ Ë ¯ 1

p pˆ4 Ê = Á cos3 ◊ + i sin 3 ◊ ˜ Ë 3 3¯ = (cos p

1 + i sin p ) 4 1

= [cos(2 kp + p ) + i sin (2 kp + p )]4 p p˘ È = Ícos(2 k + 1) + i sin (2 k + 1) ˙ 4 4˚ Î Putting

k = 0, 1, 2, 3, p

z0 = cos

i p p + i sin = e 4 4 4

z1 = cos

i 3p 3p + i sin =e 4 4

z2 = cos

5π i 5π 5π + i sin =e 4 4 4

3p 4

1.39

1.40

Chapter 1

Complex Numbers

i 7p 7p + i sin =e z3 = cos 4 4

7p 4

The continued product is z0 z1 z2 z3 =

ip e4

◊e

i

3p 4

◊e

i

5p 4

◊e

i

7p 4

Ê p 3p 5p 7p ˆ iÁ + + + ˜ Ë4 4 4 4 ¯

=e =e

i

16p 4

= ei 4 p = cos 4p + i sin 4p = 1+ i ◊0 3

Ê1 3ˆ4 Hence, the continued product of all the values of Á + i ˜ is 1. Ë2 2 ¯

example 3 Find all roots of the equation log z =

ip . 2

[Winter 2012]

Solution log z =

ip 2 ip

z=e2 p p + i sin 2 2 pˆ pˆ Ê Ê = cos Á 2 kp + ˜ + i sin Á 2 kp + ˜ Ë ¯ Ë 2 2¯ = cos

where

k = 0, 1, 2, 3, ...

example 4 Show that the nth roots of unity form a geometric progression with the 2π 2π ˆ Ê common ratio Á cos + i sin ˜ and show that the continued product Ë n n ¯ of all nth roots is (–1)n+1. [Summer 2015, 2014]

1.9 Applications of De Moivre’s Theorem

1.41

Solution To find nth roots of unity, consider the equation xn = 1 = cos 0 + i sin 0 = cos 2kp + i sin 2kp 1

x = (cos 2 k p + i sin 2 k p ) n = cos Putting

2kp 2kp + i sin n n

k = 0, 1, 2, . . . n – 1, all n roots are x0 = cos 0 + i sin 0 = 1 x1 = cos

2p 2p + i sin =e n n

2 ip n

2

4 ip n

Ê 2ip ˆ = Áe n ˜ = w2 ÁË ˜¯

6 ip n

Ê 2ip ˆ = Á e n ˜ = w3 ÁË ˜¯

4p 4p + i sin =e x2 = cos n n 6p 6p + i sin =e x3 = cos n n

= w , say

3

……………………………………………… ……………………………………………… xn -1

2(n - 1)p 2(n - 1)p = cos + i sin =e n n

2( n -1)ip n

Ê 2ip ˆ = Áe n ˜ ÁË ˜¯

n -1

= w n -1

Hence, 1, w, w2, w3 . . . , wn–1 represent n roots of unity which are in geometric progres2p 2p + i sin sion with the common ratio w = cos n n The continued product of roots is x0 ◊ x1 ◊ x2 ◊◊◊ xn -1 = 1◊ e =

2 ip n

◊e

4 ip n

◊e

6 ip n

◊◊◊ e

2( n -1)ip n

2 ip {1 + 2 + 3 +  + ( n - 1)} e n

=e

2 ip n ( n -1) ◊ n 2

[Using the sum of AP]

ip(n–1)

=e = einp ◊ e–ip = (cos np + i sin np ) (cos p – i sin p) = (–1)n (–1) = (– 1)n+1 Hence, the continued product of all nth roots of unity is (–1)n+1.

1.42

Chapter 1

Complex Numbers

example 5 Find and graph all sixth roots of unity in the complex plane. [Winter 2013] Solution Let

x6 = 1 = cos 0 + i sin 0 = cos 2kp + i sin 2kp 1

x = (cos 2kp + i sin 2kp ) 6 2 kp 2 kp = cos + i sin 6 6 kp kp = cos + i sin 3 3 Putting k = 0, 1, 2, 3, 4, 5, all the root are obtained (Fig. 1.3). x0 = cos 0 + i sin 0 = 1 x1 = cos

ip p p + i sin = e = w 3 3 3 2 ip 3

2p 2p + i sin =e 3 3 x3 = cos p + i sin p = -1

x2 = cos

x4 = cos

4p 4p + i sin =e 3 3

x5 = cos

5p 5p + i sin =e 3 3

4 ip 3

5ip 3

= w2

= w4 = w5

fig. 1.3

example 6 If w is a 7th root of unity then prove that S = 1 + w + w2m + w3m + w4m + w5m + w6m = 7 if m is a multiple of 7 and is 0 otherwise. Solution Putting n = 7 in Example 4, the 7th roots of unity are obtained as 1, w, w2, w3, w4, w5, w6. where

w = cos

i 2p 2p + i sin =e 7 7

2p 7

1.9 Applications of De Moivre’s Theorem

1.43

S = 1 + w m + w 2 m + w 3m +  + w 6 m = = =

1 [1 - (w m )7 ] 1- w m 1 - w 7m

[Using the Sum of GP]

1-wm 1 - (e2ip )m Ê 2ip ˆ 1- Áe Ë 7 ˜¯

m

Putting m = 7k, where k is an integer S = 1 + w 7k + (w 2)7k + (w 3)7k + L + (w 6)7k = 1 + (w 7)k + (w 7 )2k + (w 7 )3k + L + (w 7 )6k = 1 + (1)k + (1)2k + (1)3k + L + (1)6k =1+1+1+1+1+1+1 =7 Putting m = 7k + 1, where k is an integer

[∵ w 7 = cos 2p + i sin 2p = 1]

7 k +1

S=

1 - (e2ip )

2p ˆ Ê i Á 2 kp + ˜ 7 ¯

1- e Ë =

=

1 - (cos 2p + i sin 2p )7 k +1 È Ê 2p ˆ 2p ˆ ˘ Ê 1 - Ícos Á 2 kp + ˜¯ + i sin ÁË 2 kp + 7 ˜¯ ˙ Ë 7 Î ˚ 1 - (1 + i .0)7 k +1 2p 2p ˆ Ê + i sin ˜ 1 - Á cos Ë 7 7 ¯

1-1 1-w =0 =

example 7 If w is a complex cube root of unity, prove that (1 – w)6 = –27. Solution w is the root of the equation x3 = 1. \

w3= 1

1.44

Chapter 1

Complex Numbers

1 – w3= 0

... (1)

(1– w )(1 + w + w 2) = 0 Since w is a complex cube root of unity, w π1 1 + w + w2 = 0

\ Now,

… (2)

(1 – w)6 = [(1 – w)2]3 = [1 + w 2 – 2w]3 = (–w – 2w)3

[Using Eq. (2)]

3

= (–3w)

= –27 w 3 = –27

[Using Eq. (1)]

example 8 Solve the following equations: (i) x6 – i = 0 (ii) x9 + x5 – x4 – 1 = 0 (iii) x4 – x3 + x2 – x + 1 = 0 Solution (i)

x6 = i = cos

p p + i sin 2 2

pˆ pˆ Ê Ê = cos Á 2 kp + ˜ + i sin Á 2 kp + ˜ Ë Ë 2¯ 2¯ 1

p p ˘6 È x = Ícos(4 k + 1) + i sin (4 k + 1) ˙ 2 2˚ Î = cos(4 k + 1)

p p + i sin (4 k + 1) 12 12

[Using De Moivre’s theorem]

Putting k = 0, 1, 2, 3, 4, 5, all the 6 roots of the given equation are obtained. (ii)

x9 + x5 - x4 - 1 = 0 x 5 ( x 4 + 1) - 1( x 4 + 1) = 0 ( x 4 + 1)( x 5 - 1) = 0

1.45

1.9 Applications of De Moivre’s Theorem

The roots of x 9 + x 5 - x 4 - 1 = 0 are the roots of x4 + 1 = 0 and x5 – 1 = 0. x4 + 1 = 0 x 4 = -1 = cos p + i sin p = cos(2 k1p + p ) + i sin(2 k1p + p ) 1

x = [cos(2 k1 + 1)p + i sin(2 k1 + 1)p ] 4 = cos(2 k1 + 1) =e

i ( 2 k1 +1)

p p + i sin(2 k1 + 1) 4 4

p 4

Putting k1 = 0, 1, 2, 3, all the roots of x4 + 1 = 0 are obtained. Also, x5 – 1 = 0 x5 = 1 = cos 0 + i sin 0 = cos 2 k2p + i sin 2 k2p 1

x = (cos 2 k2p + i sin 2 k2p ) 5 = cos i

=e

2 k2 p 2k p + i sin 2 5 5

2 k2 p 5

Putting k2 = 0, 1, 2, 3, 4, all the roots of x5 – 1 = 0 are obtained. Hence, all the roots of the given equations are given by e where k1 = 0, 1, 2, 3 and k2 = 0, 1, 2, 3, 4. (iii)

x4 – x3 + x2 – x + 1 = 0 1 È1 - (- x )5 ˘˚ 1 - x + x2 - x3 + x 4 = Î 1 - (- x ) =

i ( 2 k1 +1)

p 4

i

and e

2 k2 p 5

,

[Using the sum of GP]

1 + x5 1+ x

( x 5 + 1) = ( x + 1)(1 - x + x 2 - x 3 + x 4 ) Thus, all the roots of x5 + 1 = 0 are the roots of x4 – x3 + x2 – x + 1 = 0 except x = – 1, which corresponds to x + 1 = 0.

1.46

Chapter 1

Complex Numbers

Consider x5 + 1 = 0 x 5 = -1 = cos p + i sin p = cos(2 kp + p ) + i sin (2 kp + p ) 1

x = [cos(2 k + 1)p + i sin (2 k + 1)p ]5 p p + i sin (2 k + 1) 5 5 5 Putting k = 0, 1, 2, 3, 4, all the 5 roots of x + 1 = 0 are p p x0 = cos + i sin 5 5 3p 3p x1 = cos + i sin 5 5 = cos(2 k + 1)

5p 5p + i sin = cos p + i sin p = -1 5 5 7p 7p x3 = cos + i sin 5 5 9p 9p x4 = cos + i sin 5 5 x2 = cos

Hence, x0, x1, x3, x4 are the roots of the given equation.

example 9 Prove that the roots of the equation x2 – 2ax cos q + a2 = 0 are also the roots of the equation x2n – 2anxn cos nq + a2n = 0. Solution x2 – 2ax cos q + a2 = 0 2 a cos q ± 4 a 2 cos2 q - 4 a 2 2 = a (cos q ± i sin q )

x=

= ae± iq The roots of the equation are ae± iq . Putting x = ae± iq in x2n – 2anxn cos nq + a2n, x2n – 2anxn cos nq + a2n = (ae ±iq)2n – 2an(ae± iq )n cos nq + a2n = a2n e±inq (e±inq – 2 cos nq + e∓ inq) = a2n e± inq (e± inq + e∓ inq – 2 cos nq )

... (1)

1.9 Applications of De Moivre’s Theorem

1.47

= a2n e±inq (2 cos nq – 2 cos nq ) =0 Hence, the roots of the equation x2 – 2ax cos q + a2 = 0 are also the roots of the equation x2n – 2anxn cos nq + a2n = 0.

example 10 If a, a 2, a 3, a 4 are roots of x5 – 1 = 0 then show that (1 – a ) (1 – a 2 ) (1 – a 3 ) (1 – a 4 ) = 5. Solution One root of x5 – 1 = 0 is obviously 1, the remaining roots are given as a, a2, a3, a4. x 5 - 1 = ( x - 1)( x - a )( x - a 2 )( x - a 3 )( x - a 4 ) ( x - 1)( x 4 + x 3 + x 2 + x + 1) = ( x - 1)( x - a )( x - a 2 )( x - a 3 )( x - a 4 ) [∵ xn – 1 = (x – 1) (xn–1 + xn–2 + xn–3 + L + 1)] x4 + x3 + x2 + x + 1 = (x – a) (x – a 2) (x – a 3) (x – a 4) Putting x = 1 on both the sides, 1 + 1 + 1+ 1+ 1 = (1 – a) (1 – a 2) (1 – a 3) (1 – a 4) (1 – a) (1 – a 2) (1 – a 3) (1 – a 4) = 5

example 11 If a, b, g, d are the roots of x4 + x3 + x2 + x + 1 = 0, find their values and show that (1 – a ) (1 – b ) (1 – g ) (1 – d ) = 5. Solution x 4 + x3 + x 2 + x + 1 = 0 (x - 1) (x 4 + x3 + x 2 + x + 1) = 0 x5 - 1 = 0

È∵ x n -1 = (x - 1) (x n -1 + x n - 2 +  + 1)˘ Î ˚

This shows that all the roots of x5 – 1 = 0, except x = 1, are the roots of x4 + x3 + x2 + x + 1 = 0. As solved in Example 8, all roots of x5 – 1 = 0 are x0 = 1

x1 = e

2 ip 5

x2 = e

4 ip 5

x3 = e

6 ip 5

x4 = e

8ip 5

Except x0 = 1 remaining roots x1, x2, x3, x4 are the roots of x4 + x3 + x2 + x + 1 = 0.

1.48

Chapter 1

Complex Numbers

a =e

2 ip 5

, b =e

4 ip 5

, g =e

6 ip 5

, d =e

8ip 5

Since a, b, g, d are the roots of x4 + x3 + x2 + x + 1 = 0, x4 + x3 + x2 + x + 1 = (x – a ) (x – b ) (x – g ) (x – d ) Putting x = 1 on both the sides, 1 + 1 + 1 + 1 + 1 = (1 - a ) (1 - b ) (1 - g ) (1 - d ) 5 = (1 - a ) (1 - b ) (1 - g ) (1 - d ) (1 - a ) (1 - b ) (1 - g ) (1 - d ) = 5

example 12 Find the cube roots of 1 – cos q – i sin q. Solution x 3 = 1 - cos q - i sin q q q q = 2 sin 2 - i 2 sin cos 2 2 2 = 2 sin

q È Êp qˆ Ê p q ˆ˘ Ícos ÁË - ˜¯ - i sin ÁË - ˜¯ ˙ 2Î 2 2 2 2 ˚

= 2 sin

Ï qÈ Ï Êp qˆ¸ Ê p q ˆ ¸˘ Ícos Ì2np + ÁË - ˜¯ ˝ - i sin Ì2 np + Á - ˜ ˝˙ Ë 2 2 ¯ ˛˙˚ 2 ÍÎ Ó 2 2 ˛ Ó 1

È qÏ Ê p qˆ p q ˆ ¸˘ 3 Ê x = Í2 sin Ìcos Á 2np + - ˜ - i sin Á 2np + - ˜ ˝˙ Ë 2Ó Ë 2 2¯ 2 2 ¯ ˛˙˚ ÍÎ 1

qˆ3 È 1Ê p -q ˆ 1Ê p -q ˆ˘ Ê = Á 2 sin ˜ Ícos Á 2 np + - i sin Á 2 np + ˙ ˜ Ë 2¯ Î 3Ë 2 ¯ 3Ë 2 ˜¯ ˚ 1

qˆ3 Ê x = Á 2 sin ˜ Ë 2¯

È Ï (4 n + 1)p - q ¸ Ï (4 n + 1)p - q ¸˘ ˝˙ ˝ - i sin Ì Ícos Ì 6 6 Ó ˛˚ ˛ Î Ó

Putting n = 0, 1, 2, cube roots of 1 – cos q – i sin q are obtained.

example 13 Show that the points representing the roots of the equation zn = i (z – 1)n on the Argand’s diagram are collinear.

1.9 Applications of De Moivre’s Theorem

Solution zn = i (z – 1)n n

Ê z ˆ ÁË z - 1˜¯ = i p p + i sin 2 2 pˆ pˆ Ê Ê = cos Á 2 kp + ˜ + i sin Á 2 kp + ˜ Ë Ë 2¯ 2¯ = cos

1

z p p ˘n È = Ícos(4 k + 1) + i sin (4 k + 1) ˙ z -1 Î 2 2˚ p p = cos (4 k + 1) + i sin (4 k + 1) 2n 2n =e

i (4 k +1)

p 2n

= eiq ,

where q = i (4 k + 1)

z = eiq z -1 z = zeiq - eiq z(1 - eiq ) = -eiq z=-

eiq

e

1 - eiq

-

iq 2

-

iq 2

◊ e

iq

e2 =e

-

iq 2

iq

-e2 iq

e2 =-2i sin

q 2

q q + i sin 2 2 = q 2i sin 2 1 q 1 = cot + 2i 2 2 cos

p 2n

1.49

1.50

Chapter 1

Complex Numbers

i q 1 = - cot + 2 2 2 1 i q p = - cot , where q = i(4 k + 1) 2 2 2 2n Putting k = 0, 1, 2, . . . ., n – 1, all roots of zn = i (z – 1)n are obtained. 1 The real parts of all the roots remain same as , which means the x-coordinate of 2 1 all the points represented by the roots on the Argand’s diagram is , i.e., constant. 2 1 Therefore, all the points lie on the line x = and, hence, are collinear. 2

exercIse 1.3 2

1. Find all the values of (1 - i) 3 . p 1 È ˘ - i(4 k +1) 3 6 Ans. : 2 e , k = 0, 1, 2˚˙ ÎÍ

1

2. Find the continued product of all the values of (1 + i)8 . [ans. : –(1 + i)] 3. Solve the equations: (i) x12 – 1 = 0

(ii) x5 +

(iv) (1 + x)6 + x6 = 0 (v) x14

(iii) x7 + x4 + x3 + 1 = 0 3 =i + 127x7 – 128 = 0.

È Ans. : ˘ Í ˙ Ê 12 k + 5 ˆ 1 iÁ p Ê1 Ê 3 iˆ 3ˆ Í ˙ Ë 30 ˜¯ 5 , k = 0, 1, 2, 3, 4 Í (i) ± 1, ± i, ± Á 2 ± i 2 ˜ , ± Á 2 ± 2 ˜ (ii) 2 e ˙ Ë ¯ Ë ¯ Í ˙ Í ˙ Ê 1 i ˆ p 1 3 1 Í(iii) - 1, ± i ± ,±Á (iv) - - i cot(2k + 1) , k = 0, 1, 2, 3, 4, 5˙ ˜ 2 2 2 12 Ë 2 Í ˙ 2¯ Í ˙ 2 k1p p ˙ Í(v) ei 7 , ei(2k2 +1) 7 , k = k = 0, 1, 2, 3, 4, 5, 6 Î ˚ 1 2

4. Prove that 1 + w + w2 = 0. 5. Prove that (i) a3n + b3n = 2, n is an integer (ii) aeax + bebx x È 3 3 ˘ x + cos x ˙ , where a, b are complex cube roots of = -e 2 Í 3 sin 2 2 ˚˙ ÎÍ unity.

1.51

1.9 Applications of De Moivre’s Theorem

6. Prove that the points representing the roots of the equation z3 = i (z – 1)3 on the Argand’s diagram are collinear. 7. If a, a2, a3, a4, a5, a6 are the roots of x7 – 1 = 0, prove that (1 – a) (1 – a2) (1 – a3) (1 – a4) (1 – a5) (1 – a6) = 7.

1.9.2

expansion of trigonometric functions

type I Expansion of sinnq, cosnq in terms of sin nq, cos nq, where n is a positive integer. Let x = cos q + i sin q = eiq 1 = cos q – i sin q = e–iq x Hence,

x+

Again,

1 = 2i sin q x xn = (cos q + i sin q )n = cos nq + i sin nq = einq 1 = 2 cos q and x

1 n

x–

= (cos q – i sin q )n = cos nq – i sin nq = e–inq

x xn +

1 n

= 2 cos nq

and

x

xn –

1 xn

= 2i sin nq n

To expand cosnq and sinnq , write cosnq =

1 1 Ê 1ˆ n x + ˜ and sin q = n Á Ë ¯ x (2i )n 2

n

1ˆ Ê ÁË x - x ˜¯ and

expand RHS using binomial expansion (x + a)n = xn + nC1 xn–1a + nC2 xn–2a2 + . . . + an

example 1 Prove that sin5q =

1 (sin 5q – 5 sin 3q + 10 sin q ). 16

Solution Let then

x = cos q + i sin q, 1 = cos q – i sin q x È1 Ê 1ˆ˘ sin 5 q = Í Á x - ˜ ˙ x¯˚ Î 2i Ë =

5

2 3 4 5 1 È 5 Ê 1ˆ Ê 1ˆ ˘ 4 Ê 1ˆ 3 Ê 1ˆ 2 Ê 1ˆ Í x + 5 ◊ x + 10 x + 10 x + 5 x + ÁË x ˜¯ ÁË x ˜¯ ÁË x ˜¯ ÁË x ˜¯ ÁË x ˜¯ ˙ (2i )5 ÎÍ ˚˙

[Using binomial expansion]

1.52

Chapter 1

Complex Numbers

=

1 Ê 5 10 5 1ˆ x - 5 x 3 + 10 x - + 3 - 5 ˜ 5 Á Ë x x 32i x ¯

=

1 ÈÊ 5 1 ˆ 1ˆ˘ Ê 3 1ˆ Ê Í x - 5 ˜¯ - 5 ÁË x - 3 ˜¯ + 10 ÁË x - ˜¯ ˙ 32i ÎÁË x ˚ x x

=

1 (2i sin 5q - 5 ◊ 2i sin 3q + 10 ◊ 2i sin q ) 32i

sin 5 q =

[∵ i 5 = i 4 ◊ i = i ] 1 È ˘ n Í∵ x - n = 2i sin nq ˙ x Î ˚

1 (sin 5q - 5 sin 3q + 10 sin q ) 16

example 2 Prove that cos6q + sin6q =

1 (3 cos 4q + 5). 8

Solution Let

x = cos q + i sin q ; then 6

È1 Ê 1ˆ˘ 1 cos6 q = Í Á x + ˜ ˙ = 6 Ë ¯ x 2 2 Î ˚

1 = cos q – i sin q x

1ˆ Ê ÁË x + x ˜¯

6

=

1 Ê 6 1 1 1 1 1 1ˆ x + 6 x 5 ◊ + 15 x 4 2 + 20 x 3 ◊ 3 + 15 x 2 ◊ 4 + 6 x ◊ 5 + 6 ˜ 6 Á Ë x 2 x x x x x ¯

=

˘ 1 ÈÊ 6 1 ˆ 1ˆ 1ˆ Ê Ê x + 6 ˜ + 6 Á x 4 + 4 ˜ + 15 Á x 2 + 2 ˜ + 20 ˙ 6 ÍÁ Ë ¯ Ë ¯ Ë ¯ 2 Î x x x ˚

= =

1 6

2 1

25

(2 cos6q + 6 ◊ 2 cos 4q + 15 ◊ 2 cos 2q + 20) (cos6q + 6 cos 4q + 15cos 2q + 10) 6

È1 Ê 1ˆ˘ sin 6 q = Í Á x - ˜ ˙ x¯˚ Î 2i Ë 2 3 1 È 6 5 Ê 1ˆ 4 Ê 1ˆ 3 Ê 1ˆ Í = + + + x 6 x 15 x 20 x Á ˜ Á ˜ Á ˜ Ë x¯ Ë x¯ Ë x¯ (2i )6 ÍÎ 4 5 6 Ê 1ˆ Ê 1ˆ Ê 1ˆ ˘ + 15 x 2 Á - ˜ + 6 x Á - ˜ + Á - ˜ ˙ Ë x¯ Ë x ¯ Ë x ¯ ˙˚

È n 1 ˘ Í∵ x + n = 2 cos nq ˙ x Î ˚ ...(1)

1.9 Applications of De Moivre’s Theorem

===-

˘ 1 ÈÊ 6 1 ˆ 1ˆ 1ˆ Ê Ê x + 6 ˜ - 6 Á x 4 + 4 ˜ + 15 Á x 2 + 2 ˜ - 20 ˙ 6 ÍÁ Ë Ë x ¯ x ¯ x ¯ 2 ÎË ˚ 1 26 1 25

1.53

[∵ i 6 = (i 2 )3 = -1]

(2 cos6q - 6 ◊ 2 cos 4q + 15 ◊ 2 cos 2q - 20) (cos6q - 6 cos 4q + 15cos 2q - 10)

...(2)

Adding Eqs (1) and (2), cos6q + sin6q =

1 25

(12 cos 4q + 20) =

1 (3cos 4q + 5) 8

example 3 Expand sin5q cos3q in a series of sines of multiples of q. Solution Let x = cos q + i sin q,

1 = cos q – i sin q x 5

È1 Ê 1ˆ˘ È1 Ê 1ˆ˘ sin 5 q cos3 q = Í Á x - ˜ ˙ Í Á x + ˜ ˙ Ë ¯ Ë 2 2 i x x¯˚ Î ˚ Î 2

3

3

3

=

1 Ê 1ˆ Ê 1ˆ 1 x- ˜ Áx- ˜ ◊ 3 5 Á x¯ Ë x ¯ (2) (2i ) Ë

=

1ˆ 1 Ê 1ˆ Ê x - ˜ Á x2 - 2 ˜ 8 Á Ë Ë ¯ x (2 i ) x ¯

=

1 Ê 2 1ˆÊ 3 1ˆ x - 2 + 2 ˜ Á x6 - 3x2 + 2 - 6 ˜ 8 Á Ë ¯ Ë (2 i ) x x x ¯

=

1 6 2 3 1ˆ Ê 8 4 6 2 4 ÁË x - 3 x + 3 - 4 - 2 x + 6 x - 2 + 6 + x - 3 + 4 - 8 ˜¯ x x x x x (2 i )

2

1ˆ Ê ÁË x + x ˜¯

3

[∵ i 5 = i 4 ◊ i = i ]

i

8

1 ÈÊ 8 1 ˆ 1ˆ 1ˆ 1 ˆ˘ Ê Ê Ê x - 8 ˜ - 2 Á x6 - 6 ˜ - 2 Á x 4 - 4 ˜ + 6 Á x2 - 2 ˜ ˙ 8 ÍÁ Ë ¯ Ë ¯ Ë ¯ Ë x x x x ¯˚ (2 i ) Î 1 = 8 (2i sin 8q - 2 ◊ 2i sin 6q - 2 ◊ 2i sin 4q + 6 ◊ 2i sin 2q ) (2 i ) 1 = 7 (sin 8q - 2 sin 6q - 2 sin 4q + 6 sin 2q ) 2 1 = (sin 8q - 2 sin 6q - 2 sin 4q + 6 sin 2q ) 128 =

1.54

Chapter 1

Complex Numbers

type II Expansion of sin nq, cos nq in powers of sin q, cos q By De Moivre’s theorem, cos nq + i sin nq = (cosq + i sinq)n = cosnq + nC1 cosn–1q · i sinq + nC2 cosn–2q (i sinq )2 + nC3 cosn–3q (i sinq )3 + L = (cosnq – nC2 cos n–2 q sin2q + L) + i (nC1 cos n–1 q sin q – nC3 cos n–3q sin 3q + L) Comparing real and imaginary parts, cos nq = cosn q – nC2 cosn–2q sin2 q + L sin nq = nC1 cosn–1q sin q – nC3 cosn–3 sin3 q + L

example 1 sin 7q = 7 - 56 sin 2 q + 112 sin 4 q - 64 sin 6 q and state the sin q result which you have used. [Summer 2015] Prove that

Solution cos 7q + i sin 7q = (cos q + i sin q )7 = cos7 q + 7 cos6 q ◊ i sin q + 21 cos5 q (i sin q )2 + 35 cos4 q (i sin q )3 + 35 cos3 q (i sin q )4 + 21 cos2 q (i sin q )5 + 7 cos q (i sin q )6 + (i sin q )7 = (cos7 q - 21 cos5 q sin 2 q + 35 cos3 q sin 4 q - 7 cos q sin 6 q ) + i(7 cos6 q sin q - 35 cos4 q sin 3 q + 21 cos2 q sin 5 q - sin 7 q ) Comparing imaginary parts, sin 7q = 7 cos6 q sin q - 35 cos4 q sin3 q + 21 cos2 q sin 5 q - sin 7 q sin 7q = 7(1 - sin 2 q )3 - 35(1 - sin 2 q )2 sin 2 q + 21(1 - sin 2 q )sin 4 q - sin 6 q sin q = 7(1 - sin 6 q + 3 sin 4 q - 3 sin 2 q ) - 35(1 - 2 sin 2 q + sin 4 q )sin 2 q + 21(sin 4 q - sin 6 q ) - sin 6 q = 7 - 56 sin 2 q + 112 sin 4 q - 64 sin 6 q

1.9 Applications of De Moivre’s Theorem

1.55

example 2 Expand

sin 5q in powers of cos q only. sin q

Solution (cos 5q + i sin 5q) = (cos q + i sin q)5 5

4

[Using De Moivre’s theorem] 3

= cos q + 5 cos q · i sin q + 10 cos q (i sin q)2 + 10 cos2 q (i sin q)3 + 5 cos q (i sin q)4 + (i sin q)5 = (cos5 q – 10 cos3 q sin2 q + 5 cos q sin4 q) + i (5 cos4 q sin q – 10 cos2 q sin3 q + sin5 q) Comparing imaginary parts, sin 5q = 5 cos4 q sin q – 10 cos2 q sin3 q + sin5 q = sin q (5 cos4 q – 10 cos2 q sin2 q + sin4 q)

sin 5q = 5 cos4 q – 10 cos2 q (1 – cos2 q) + (1 – cos2 q)2 sin q = 5 cos4 q – 10 cos2 q + 10 cos4 q + 1 – 2 cos2 q + cos4 q = 16 cos4 q – 12 cos2 q + 1

example 3 Prove that tan 5q =

5 tan 4

5 tan q - 10 tan 3 q + tan 5 q 1 - 10 tan 2 q + 5 tan 4 q

and, hence, deduce that

p p - 10 tan 2 + 1 = 0. 10 10

Solution Comparing real and imaginary parts in Example 2, cos 5q = cos5 q – 10 cos3 q sin2 q + 5 cos q sin4 q sin 5q = 5 cos4 q sin q – 10 cos2 q sin3 q + sin5 q tan 5q = =

sin 5q cos 5q 5cos4 q sin q - 10 cos2 q sin 3 q + sin 5 q cos5 q - 10 cos3 q sin 2 q + 5cos q sin 4 q

1.56

Chapter 1

Complex Numbers

Dividing the numerator and denominator by cos5q, tan 5q = Putting q =

5 tan q - 10 tan 3 q + tan 5 q 1 - tan 2 q + 5 tan 4 q

p , 10 p p p + tan 5 5 tan - 10 tan 3 p 10 10 10 tan 5 ◊ = p p 10 + 5 tan 4 1 - tan 2 10 10

… (1)

p p p - 10 tan 3 + tan 5 10 10 10 = tan p = • = 1 p p 2 0 1 - tan 2 + 5 tan 4 10 10

5 tan

Hence, 1 - tan 2

p p + 5 tan 4 =0 10 10

example 4 If sin 6q = a cos5q sin q + b cos3q sin3q + c cosq sin5q, find values of a, b, c. Solution (cos 6q + i sin 6q) = (cos q + i sin q)6 = cos6q + 6 cos5q (i sin q ) + 15 cos4q (i sin q)2 + 20 cos3q (i sin q)3+ 15 cos2q (i sin q)4 + 6 cos q (i sin q)5 + (i sin q)6 = (cos6q – 15 cos4q sin2q + 15 cos2q sin4q – sin6q) + i (6 cos5q sin q – 20 cos3q sin3q + 6 cos q sin5q) Comparing imaginary parts,

Given

sin 6q = 6 cos5q sin q – 20 cos3q sin3q + 6 cos q sin5q

...(1)

sin 6q = a cos5q sin q + b cos3q sin3q + c cos q sin5q

...(2)

Comparing Eqs (1) and (2), a = 6, b = –20, c = 6

1.57

1.9 Applications of De Moivre’s Theorem

example 5 Prove that

1 + cos6 q = 16 cos4 q – 24 cos2q + 9. 1 + cos 2q

Solution Comparing real parts in Example 4, cos 6q = cos6 q – 15 cos4 q sin2 q + 15 cos2 q sin4 q – sin6 q = cos6 q – 15 cos4 q (1 – cos2 q) + 15 cos2 q (1 – cos2 q)2 – (1 – cos2 q)3 = cos6 q – 15 cos4 q + 15 cos6 q + 15 cos2 q (1 – 2 cos2 q + cos4 q) – (1 – 3 cos2 q + 3 cos4 q – cos6 q) = 32 cos6 q – 48 cos4 q + 18 cos2 q – 1 1 + cos6q 32 cos6 q - 48 cos4 q + 18 cos2 q = 1 + cos 2q 2 cos2 q = 16 cos4 q – 24 cos2 q + 9

example 6 Using De Moivre’s theorem, prove that 2 (1 + cos 8q ) = (x4 – 4x2 + 2)2, where x = 2 cos q. Solution 2 (1 + cos 8q) = 2 . 2 cos2 4q = (2 cos 4q)2 cos 4q + i sin 4q = (cos q + i sin q)4 = cos4 q + 4 cos3 q (i sin q ) + 6 cos2 q (i sin q)2 + 4 cos q (i sin q)3 + (i sin q)4 = (cos4 q – 6 cos2 q sin2 q + sin4 q) + i (4 cos3 q sin q – 4 cos q sin3 q) Comparing real parts, cos 4q = cos4 q – 6 cos2 q sin2 q + sin4 q = cos4 q – 6 cos2 q (1 – cos2 q) + (1 – cos2 q)2 = cos4 q – 6 cos2 q + 6 cos4 q + 1 – 2 cos2 q + cos4 q = 8 cos4 q – 8 cos2 q + 1 Substituting in Eq. (1), 2 (1 + cos 8q ) = (16 cos4 q – 16 cos2 q + 2)2 = (x4 – 4x2 + 2)2

where x = 2cos q

… (1)

1.58

Chapter 1

Complex Numbers

exercIse 1.4 1. Expand cos6q – sin6q in terms of cosine multiples of q. ans. :

1 (cos 6q + 15 cos 2q) 16

2. Expand cos8q in terms of cosine multiples of q. ans.

1 (cos 8q + 8 cos 6q + 28 cos 4q + 56 cos 2q + 35) 27

3. Prove that cos8q + sin8q = 4. Prove that cos4q sin3q = –

1 (cos 8q + 28 cos 4q + 35). 64 1 (sin 7q + sin 5q – 3 sin 3q – sin q). 64

5. Prove that –212cos6q sin7q = (sin13q – sin11q – 6 sin 9q + 6 sin 7q + 15 sin 5q – 15 sin 3q – 20 sin q). 6. Prove that –256 sin7 q cos2 q = cos 9q – 5 sin 7q + 8 sin 5q – 14 sin q. 7. Expand

sin7q in powers of sinq only. sin q ÈÎ Ans. : 7 − 56 sin2 q + 112 sin4 q − 64 sin6 q ˘˚

8. Prove that

sin 6q = 32 sin5q – 32 sin3q + 6 sin q. cos q

9. Prove that tan 7q =

7 tan q - 35 tan3 q + 21 tan5 q - tan7 q 1 - 21 tan2 q + 35 tan4 q - 7 tan6 q

and, hence, deduce 1 - 21tan2

p p p - 7 tan6 + 35 tan4 = 0. 14 14 14

10. Prove that 1 – cos 10q = 2 (16 sin5 q – 20 sin3 q + 5 sin q )2. 11. Prove that

1.10 1.10.1

1 + cos7q = (x3 – x2 – 2x + 1)2, where x = 2 cos q. 1 + cos q

cIrcular and hyperbolIc functIons circular function

From Euler’s formula, eiq = cos q + i sin q e = cos q – i sin q – iq

\

cos q =

eiq + e - iq eiq - e - iq , sin q = 2 2i

1.10

Circular and Hyperbolic Functions

1.59

If z = x + iy is a complex number then cos z =

eiz - e - iz eiz + e - iz , sin z = 2i 2

These are called circular functions of complex numbers.

1.10.2

hyperbolic function

If z is a complex number then the sine hyperbolic of z is denoted by sinh z and is given as e z - e- z sinh z = 2 and the cosine hyperbolic of z is denoted by cosh z and is given as e z + e- z 2 From these expressions, other hyperbolic functions can also be obtained using

cosh z =

tanh z =

sinh z 1 1 1 , cosech z = , sech z = . and coth z = cosh z sinh z cosh z tanh z

From the above definitions of sinh z, cosh z, tanh z, the following range of hyperbolic functions is obtained:

note

–•

0

•

sinh z

–•

0

•

cosh z

•

1

•

tanh z

–1

0

1

sinh (–z) = –sinh z, cosh (–z) = cosh z

1.10.3

relation between circular and hyperbolic functions

(i) sin iz = i sinh z Proof

z

and

sinh z = – i sin iz

By Euler’s formula, sin z =

eiz - e- iz 2i

sin iz =

ei z - e - i 2i

Replacing z by iz, 2

2

z

(e - z - e z ) = -i 2

È 1 ˘ Í∵ i = − i ˙ Î ˚

1.60

Chapter 1

Complex Numbers

=i

(e z - e - z ) 2

= i sinh z

1 sin iz i = –i sin iz

and

sinh z =

(ii) cos iz = cosh z Proof

By Euler’s formula, cos z =

eiz + e - iz 2

cos iz =

ei z + e - i 2

Replacing z by iz,

2

2

z

e- z + e z 2 = cosh z

=

(iii) tan iz = i tanh z and tanh z = – i tan iz Proof

sin iz cos iz i sinh z = cosh z = i tanh z

tan iz =

1 tan iz i = –i tan iz

tanh z =

1.10.4 A.

(i) (ii) (iii) B. (iv) (v)

formulae on hyperbolic functions cosh2 z – sinh2 z = 1 coth2 z – cosech2 z = 1 sech2 z + tanh2 z = 1 sinh 2z = 2 sinh z cosh z cosh 2z = cosh2 z + sinh2 z = 2 cosh2 z – 1 = 1 + 2 sinh2 z

1.11

Inverse Hyperbolic Functions

1.61

2 tanh z (vi) tanh 2z =

1 + tanh 2 z

C. (vii) sinh 3z = 3 sinh z + 4 sinh3 z (viii) cosh 3z = 4 cosh3 z – 3 cosh z (ix) tanh 3z =

3 tanh z + tanh 3 z

1 + 3 tanh 2 z (x) sinh (z1 ± z2) = sinh z1 cosh z2 ± cosh z1 sinh z2

D.

(xi) cosh (z1 ± z2) = cosh z1 cosh z2 ± sinh z1 sinh z2 tanh z1 ± tanh z2 (xii) tanh (z1 ± z2) = 1 ± tanh z tanh z 1

E.

z1 + z2 z -z cosh 1 2 2 2

(xiii) sinh z1 + sinh z2 = 2 sinh (xiv) sinh z1– sinh z2 = 2 cosh

2

z1 + z2 z -z sinh 1 2 2 2

(xv) cosh z1+ cosh z2 = 2 cosh

z1 + z2 z -z cosh 1 2 2 2

z1 + z2 z -z sinh 1 2 2 2 (xvii) 2 sinh z1cosh z2 = sinh (z1 + z2) + sinh (z1 – z2) (xvi) cosh z1– cosh z2 = 2 sinh

F.

(xviii) 2 cosh z1sinh z2 = sinh (z1+ z2) – sinh (z1 – z2) (xix) 2 cosh z1 cosh z2 = cosh (z1+ z2) + cosh (z1 – z2) (xx) 2 sinh z1 sinh z2 = cosh (z1 + z2) – cosh (z1– z2)

1.11

Inverse hyperbolIc functIons

If x = sinh u then u = sinh–1 x is called the sine hyperbolic inverse of x, where x is real. Similarly, cosh–1 x, tanh–1 x, coth–1 x, sech–1 x and cosech–1 x are defined. The inverse hyperbolic functions are many-valued functions but their principal values are only considered. If x is real, (i) sinh–1 x = log ( x + x 2 + 1) (ii) cosh–1 x = log ( x + x 2 - 1) (iii) tanh–1 x =

Ê 1+ x ˆ 1 log ÁË 1 - x ˜¯ 2

[Summer 2014]

1.62

Proof

Chapter 1

Complex Numbers

(i) Let sinh–1 x = y x = sinh y = 1

2x = e y – 2y

ey

y

e y - e- y 2 =

e2 y - 1 ey

e – 2x e – 1 = 0 This equation is quadratic in e y. ey =

2 x ± 4 x2 + 4 2

ey = x ± x 2 + 1

(

y = log x ± x 2 + 1

)

But x – x 2 + 1 < 0 and log of a negative number is not defined.

(

)

(

)

2 y = log x + x + 1

sinh–1 x = log x + x 2 + 1 (ii) Let

cosh–1 x = y x = cosh y = 2x = e y +

1 e

y

e y + e- y 2

=

e2 y + 1 ey

2xe y = e 2y + 1 e 2y – 2xe y + 1 = 0 ey =

2 x ± 4 x2 - 4 2

e y = x ± x2 - 1

( y = log ( x -

) - 1)

2 y = log x ± x - 1

Consider

x2

e y = x – x2 - 1 e- y =

1

.

x + x2 - 1

x - x2 - 1 x + x2 - 1

... (1.2) ... (1.3)

1.11

=

Inverse Hyperbolic Functions

1.63

x + x2 - 1 x2 - x2 + 1

= x + x2 - 1

(

- y = log x + x 2 - 1

)

(

)

(

)

y = - log x + x 2 - 1

... (1.4)

Equating Eqs (1.3) and (1.4),

)

(

log x - x 2 - 1 = - log x + x 2 - 1

... (1.5)

From Eqs (1.2) and (1.5),

( x = ± log ( x +

) + 1)

y = ± log x + x 2 - 1 cosh–1

x2

... (1.6)

)

(

È ˘ x = cosh α log x + x 2 + 1 ˚

)

(

= cosh ÈÎlog x + x 2 + 1 ˘˚ [∵ cosh (–z) = cosh z]

(

cosh–1 x = log x + x 2 + 1 (iii) Let

)

tanh–1 x = y x = tanh y x e y - e- y = 1 e y + e- y

Using componendo and dividendo, 1 + x e y + e- y + e y - e- y = 1 - x e y + e- y - e y + e- y =

2e y 2e - y

= e2 y 1+ x e2 y = 1- x Ê 1+ x ˆ 2 y = log Á Ë 1 - x ˜¯ y=

1 Ê 1+ x ˆ log Á Ë 1 - x ˜¯ 2

1.64

Chapter 1

Complex Numbers

tanh -1 x =

1 Ê 1+ x ˆ log Á Ë 1 - x ˜¯ 2

example 1 Prove that (cosh x + sinh x)n = cosh nx + sinh nx. Solution (cosh x + sinh x )n = (cos ix - i sin ix )n = cos(nix ) - i sin(nix ) [ Using De Moivre’s theorem ] = cosh nx - i ◊ i sinh nx = cosh nx + sinh nx

example 2 3

Ê 1 + tanh x ˆ Prove that Á = cosh 6x + sinh 6x. Ë 1 - tanh x ˜¯

Solution sinh x ˆ Ê 1+ Ê 1 + tanh x ˆ Á cosh x ˜ ÁË 1 - tanh x ˜¯ = Á sinh x ˜ Á 1˜ Ë cosh x ¯

3

3

Ê cosh x + sinh x ˆ =Á Ë cosh x - sinh x ˜¯ Ê cos ix - i sin ix ˆ =Á Ë cos ix + i sin ix ˜¯

3

3

ÈÊ cos ix - i sin ix ˆ Ê cos ix - i sin ix ˆ ˘ = ÍÁ ˜˙ ˜Á ÎË cos ix + i sin ix ¯ Ë cos ix - i sin ix ¯ ˚ È (cos ix - i sin ix )2 ˘ =Í ˙ Î cos2 ix + sin 2 ix ˚

3

3

= (cos ix - i sin ix )6 = cos 6ix - i sin 6ix = cosh 6 x - i ◊ i sinh 6 x = cosh 6 x + sinh 6 x

[Using De Moivre’s theorem]

1.11

Inverse Hyperbolic Functions

1.65

example 3 1

Prove that

= cosh 2 x.

1

11-

1 1 - cosh 2 x

Solution 1 1

11-

1 1 - cosh 2 x

1

=

[∵ cosh 2 x - sinh 2 x = 1]

1

1-

1 - sinh 2 x 1 = 1 11 + coseech 2 x 1-

1 =

[∵ 1 - coth 2 x = - cosech 2 x ]

1 1coth 2 x 1

=

1 - tanh 2 x 1

=

[∵ 1 - tanh 2 x = sech 2 x ]

2

sech x = cosh 2 x

example 4 Prove that tanh (log 3 ) = 0.5. Solution - e - log

3

+ e - log 1 33 = 1 3+ 3 3 -1 = 3 +1 = 0.5

3

tanh (log 3 ) =

e log

3

e log

3

1.66

Chapter 1

Complex Numbers

example 5 Solve the equation 17 cosh x + 18 sinh x = 1 for real values of x. Solution 17 cosh x + 18 sinh x = 1 Ê e x + e- x ˆ Ê e x - e- x ˆ + 18 Á =1 17 Á ˜ 2 ¯ 2 ˜¯ Ë Ë 35e x - e - x = 2 35e2 x - 1 = 2e x 35e2 x - 2e x - 1 = 0 This equation is quadratic in ex. 2 ± 4 + 140 2 ± 12 14 -10 = = , 70 70 70 70 For real values of x, ex should be positive. 14 1 \ ex = = 70 5 1 x = log = - log 5 5 ex =

example 6 Solve the equation 7 cosh x + 8 sinh x = 1 for real values of x. Solution 7 cosh x + 8 sinh x = 1 Ê e x - e- x ˆ Ê e x + e- x ˆ + 8Á 7Á ˜ ˜ =1 Ë Ë 2 ¯ 2 ¯ 1 15e x - x = 2 e 15e2 x - 2e x - 1 = 0 This equation is quadratic in ex. ex =

2 ± 4 + 60 2 ± 8 1 1 = = ,30 30 3 5

For real values of x, ex should be positive.

1.11

ex =

Inverse Hyperbolic Functions

1 3

x = log

1 = - log 3 3

example 7 1 4 If tanh x = , prove that sinh 2 x = 2 3 Solution sinh 2 x =

2 tanh x

1 - tanh 2 x 1 = 1 1− 4 4 = 3

example 8 Ê p xˆ Prove that sinh–1 (tan q) = log tan Á + ˜ . Ë 4 2¯ Solution

(

sinh -1 (tan q ) = log tan q + tan 2 q + 1 = log(tan q + sec q ) Ê sin q + 1ˆ = log Á Ë cos q ˜¯ È Êp ˆ ˘ Í cos ÁË 2 - q ˜¯ + 1 ˙ ˙ = log Í Í Êp ˆ ˙ Í sin ÁË 2 - q ˜¯ ˙ Î ˚

)

È ˘ Êp qˆ 2 cos2 Á - ˜ Í ˙ Ë 4 2¯ ˙ = log Í Í Êp qˆ Êp qˆ˙ Í 2 sin ÁË 4 - 2 ˜¯ cos ÁË 4 - 2 ˜¯ ˙ Î ˚

1.67

1.68

Chapter 1

Complex Numbers

È Ê p q ˆ˘ = log Ícot Á - ˜ ˙ Î Ë 4 2¯˚ È Ê p p q ˆ˘ = log Ítan Á - + ˜ ˙ Î Ë 2 4 2¯˚ È Ê p q ˆ˘ = log Ítan Á + ˜ ˙ Î Ë 4 2¯˚

example 9 Prove that Ê x ˆ -1 -1 (i) tanh x = sinh Á ˜. Ë 1 - x2 ¯ (ii) sinh -1 x = cosh -1

(

)

1 + x2 .

Solution Ê Ê x ˆ x x2 (i) sinh -1 Á = + log Á ˜ 1 - x2 Ë 1 - x2 ¯ Ë 1 - x2 1 x Ê = log + Á Ë 1 - x2 1 - x2 Ê x +1 ˆ = log Á ˜ Ë 1 - x2 ¯ Ê 1+ x 1+ x ˆ = log Á ˜ Ë 1- x 1+ x ¯ Ê 1+ x ˆ = log Á ˜ Ë 1- x ¯ 1 Ê 1+ xˆ = log Á Ë 1 - x ˜¯ 2 = tanh -1 x (ii) cosh -1

(

)

( = log (

1 + x 2 = log

= sinh

ˆ È ˘ + 1˜ Í∵ sinh -1 x = log x + x 2 + 1 ˙ Î ˚ ¯ ˆ ˜ ¯

1 + x2 + 1 + x2 - 1 1 + x2 + x

-1

x

)

)

(

)

(

)

È ˘ -1 2 Î∵ cosh x = log x + x - 1 ˚

1.11

Inverse Hyperbolic Functions

1.69

example 10 Prove that tanh–1 (sin q) = cosh–1 (sec q). Solution tanh -1 (sin q ) =

1 Ê 1 + sin q ˆ log Á Ë 1 - sin q ˜¯ 2

=

1 È (1 + sin q ) (1 + sin q ) ˘ log Í ˙ 2 Î (1 - sin q ) (1 + sin q ) ˚

=

1 Ê 1 + sin q ˆ log Á Ë cos q ˜¯ 2

=

1 ◊ 2 log (sec q + tan q ) 2

(

2

= log sec q + sec 2 q - 1

)

= cosh -1 (sec q )

(

)

È ˘ 2 -1 Î∵ cosh x = log x + x - 1 ˚

example 11 È1 + 1 + z 2 Prove that cosech–1 z = log Í z ÍÎ of z? Solution y = cosech–1z

Let

e2 y

cosech y = z 2 =z e y - e- y 1 2 ey - y = z e 2 y - e -1 = 0 z

˘ ˙ . Is it defined for all values ˙˚

1.70

Chapter 1

Complex Numbers

This equation is quadratic in ey. 2 4 ± 2 +4 z z ey = 2 1 + 1 + z2 z È1 + 1 + z2 y = log Í z Í Î

È∵ 1 - 1 + z 2 < 0 and e y cannot be negative˘ ˙˚ ÍÎ

=

˘ ˙ ˙ ˚

It is not defined for z < 0.

example 12 È Ê x - aˆ ˘ i Ê aˆ Prove that tan -1 Íi Á = - log Á ˜ . ˜ ˙ Ë x¯ 2 Î Ë x + a¯ ˚ Solution Let

È Ê x - aˆ˘ tan -1 Íi Á ˜˙ = q Î Ë x + a¯˚ eiq - e - iq Ê x - aˆ iÁ tan q = = Ë x + a ˜¯ i(eiq + e - iq ) x - a - eiq + e - iq = iq x+a e + e - iq

Applying componendo – dividendo, x + a + x - a eiq + e - iq - eiq + e - iq = x + a - x + a eiq + e - iq + eiq - e - iq x e- iq = = e -2iq a ei q a = e2iq x Ê aˆ 2iq = log Á ˜ Ë x¯ 1 i Ê aˆ Ê aˆ q = log Á ˜ = - log Á ˜ Ë x¯ Ë x¯ 2i 2 È Ê x - aˆ˘ i Ê aˆ tan -1 Íi Á ˜¯ ˙ = - 2 log ÁË x ˜¯ Ë x + a Î ˚

1.11

Inverse Hyperbolic Functions

1.71

exercIse 1.5 n

Ê 1 + tanh x ˆ = cosh 2nx + sinh 2nx. 1. Prove that Á Ë 1 - tanh x ˜¯ 2. Prove that cosec x + coth x = coth 1

3. Prove that

1

11-

x . 2

= - sinh2 x.

1 1 + sinh2 x

4. If 6 sinh x + 2 cosh x + 7 = 0, find tanh x. 3 15 ˘ È Í Ans. : 5 , - 17 ˙ Î ˚ 5. Find the value of tanh log

5. È Í Ans. : Î

6. If tanh x =

1 , find cosh 2x. 2

7. Prove that sin–1 x =

1 log(ix + 1 - x 2 ). i

2˘ 3 ˙˚

4 5˘ È Í Ans. : 3 , 3 ˙ Î ˚

[Summer 2013]

8. Prove that cos–1x = -i log(x ± 1 - x 2 ). 9. Prove that sin–1ix = 2np + i log(x + 1 + x 2 ). Êp xˆ 10. Prove that sinh–1(tan x) = log tan Á + ˜ . Ë 4 2¯ 11. Prove that (i) tanh–1(cos q ) = cosh–1(cosec q ). (ii) sinh–1(tan q ) = log (sec q + tan q ). ip Ê 3i ˆ 12. Prove that cosh-1 Á ˜ = log 2 + . Ë 4¯ 2 13. If sinh–1(x + iy) + sinh–1(x – iy) = sinh–1a, prove that 2 (x2 + y2) a2 – 2x2 + 2y2.

a2 + 1 =

1.72

1.12

Chapter 1

Complex Numbers

separatIon Into real and ImaGInary parts

To separate real and imaginary parts of a complex number, the following results are used: (i) sin (x ± iy) = sin x cos iy ± cos x sin iy = sin x cosh y ± icos x sinh y (ii) cos (x ± iy) = cos x cos iy ∓ sin x sin iy = cos x cosh y ∓ i sin x sinh y (iii) tan ( x ± iy) = =

2 sin ( x ± iy) cos( x ∓ iy) ◊ 2 cos( x ± iy) cos( x ∓ iy) sin 2 x ± sin 2iy sin 2 x ± i sinh 2 y = cos 2 x + cos 2iy cos 2 x + cosh 2 y

(iv) sinh (x ± iy) = sinh x cosh iy ± cosh x sinh iy = sinh x cos iiy ± cosh x (–i sin iiy) = sinh x cos (–y) ± cosh x [–i sin (–y)] = sinh x cos y ± i cosh x sin y (v) cosh (x ± iy) = cosh x cosh iy ± sinh x sinh iy = cosh x cos y ± i sinh x sin y 2 sinh ( x ± iy) cosh ( x ∓ iy) ◊ 2 cosh ( x ± iy) cosh ( x ∓ iy) sinh 2 x ± sinh 2iy sinh 2 x ± i sin 2 y = . = cosh 2 x + cosh 2iy cosh 2 x + cos 2 y

(vi) tanh ( x ± iy) =

example 1 Separate real and imaginary parts of (i) sinh z Ê 3i ˆ (ii) cos-1 Á ˜ (iii) cos–1 (ix) Ë 4¯ (iv) sin–1 (eiq)

(v) sin–1 (cosec q )

Solution (i)

sinh z = -i sin iz = -i sin i( x + iy) = -i sin(ix - y) = -i(sin ix cos y - cos ix sin y) = -i(i sinh x cos y - cosh x sin y) = -i 2 sinh x cos y + i cosh x sin y = sinh x cos y + i cosh x sin y

[Summer 2014]

1.12

(ii)

Let cos-1

Separation into Real and Imaginary Parts

1.73

3i = x + iy 4

3i = cos( x + iy) = cos x cosh y - i sin x sinh y 4 Comparing real and imaginary parts, cos x cosh y = 0

...(1)

3 4

...(2)

sin x sinh y = -

From Eq. (1),

cos x = 0 x=

Putting x =

[∵ cosh y π 0]

p 2

p in Eq. (2), 2 sin

p 3 sinh y = 2 4 3 sinh y = 4 Ê 3ˆ y = sinh -1 Á ˜ Ë 4¯ Ê3 ˆ 9 = log Á + + 1˜ Ë4 16 ¯ 3 5 Ê ˆ = log Á + ˜ Ë 4 4¯ = log 2 3i p = + i log 2 4 2

Hence,

cos-1

(iii) Let

cos-1 (ix ) = a + ib cos(a + ib ) = ix cos a cos ib - sin a sin ib = ix cos a cosh b - i sin a sinh b = 0 + ix

Comparing real and imaginary parts, cos a cosh b = 0,

… (1)

sin a sinh b = –x

… (2)

From Eq. (1), cos a = 0

[ cosh b π 0]

1.74

Chapter 1

Complex Numbers

a= Putting a =

p 2

p in Eq. (2), 2 sin

p sinh b = –x 2 sinh b = –x b = sinh–1 (–x) = –sinh–1(x)

(

= - log x + x 2 + 1 cos–1 (ix) =

Hence, (iv) Let sin–1 (eiq) = x + iy

)

(

p – i log x + x 2 + 1 2

)

eiq = sin (x + iy)

cos q + i sin q = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sinh y Comparing real and imaginary parts, cos q = sin x cosh y sin q = cos x sinh y Eliminating y from Eq. (1) and Eq. (2), cosh 2 y - sinh 2 y = 1=

cos2 q sin 2 x

-

sin 2 q cos2 x

cos2 q cos2 x - sin 2 q sin 2 x sin 2 x cos2 x

sin 2 x cos2 x = cos2 q cos2 x - (sin 2 q )(1 - cos2 x ) (1 - cos2 x ) cos2 x = (1 - sin 2 q ) cos2 x - sin 2 q + sin 2 q cos2 x cos4 x = sin 2 q cos2 x = sin q cos x = ± sin q x = cos-1 ( ± sin q ) From Eq. (2), sin 2 q = cos2 x sinh 2 y

… (1) … (2)

1.12

Separation into Real and Imaginary Parts

1.75

cos2 x = sin q ,

Putting

sin 2 q = sin q sinh 2 y sinh 2 y = sin q sinh y = ± sin q y = sinh -1 ( ± sin q ) = log ( ± sin q + sin q + 1 ) sin -1 (eiq ) = cos-1 ( ± sin q ) + i log ( ± sin q + sin q + 1 )

Hence,

(v) Let sin–1 (cosec q ) = x + iy cosec q = sin ( x + iy) cosec q = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sinh y Comparing real and imaginary parts, cosec q = sin x cosh y 0 = cos x sinh y From Eq. (2), cos x = 0, x = Putting x =

p 2

p in Eq. (1), 2 p cosh y 2 = cosh y y = cosh–1 (cosec q)

cosec q = sin

(

= log cosec q + cosec2q - 1 = log (cosec q + cot q ) Ê 1 + cos q ˆ = log Á Ë sin q ˜¯ q ˆ Ê 2 cos2 Á 2 ˜ = log Á q q˜ Á 2 sin cos ˜ Ë 2 2¯ qˆ Ê = log Á cot ˜ Ë 2¯ Hence,

sin -1 (cosecq ) =

p q + i log cot 2 2

)

… (1) … (2)

1.76

Chapter 1

Complex Numbers

example 2 [Winter 2012]

Expand cosh (z1 + z2). Solution cosh ( z1 + z2 ) = cos i( z1 + z2 ) = cos(iz1 + iz2 ) = cos iz1 cos iz2 - sin iz1 sin iz2

= cosh z1 cosh z2 - (i sinh z1 )(i sinh z2 ) = cosh z1 cosh z2 + sinh z1 sinh z2

example 3 If cos (u + iv) = x + iy, show that (i) (1 + x)2 + y2 = (cosh v + cos u)2 (ii) (1 – x)2 + y2 = (cosh v – cos u)2 Solution cos (u + iv) = x + iy cos u cos iv – sin u sin iv = x + iy cos u cosh v – i sin u sinh v = x + iy (i) Consider 1 + x + iy = 1 + cos u cosh v – i sin u sinh v Taking modulus on both the sides and squaring, (1 + x)2 + y2 = (1 + cos u cosh v)2 + sin2 u sinh2 v = 1 + 2 cos u cosh v + cos2 u cosh2 v + (1 – cos2 u) (cosh2 v – 1) = 1 + 2 cos u cosh v + cos2 u cosh2 v + cosh2 v – 1 – cos2 u cosh2 v + cos2 u = (cosh v + cos u)2. (ii) Consider 1 – x – iy = 1 – cos u cosh v + i sin u sinh v Taking modulus on both the sides and squaring, (1 – x)2 + y2 = (1–cos u cosh v)2+ sin2u sinh2 v = 1 – 2 cos u cosh v + cos2 u cosh2 v + (1 – cos2 u) (cosh2 v – 1) = (cosh v – cos u)2

example 4 If sinh (q + if) = eia, prove that (i) sinh4 q = cos2 a (ii) cos2 f = cos2 a

1.12

Separation into Real and Imaginary Parts

1.77

Solution sinh (q + if) = cos a + i sin a cos a + i sin a = sinh q cosh if + cosh q sinh if = sinh q cos i(if) + cosh q [–i sin i (if)] = sinh q cos (–f) + cosh q [–i sin (–f)] = sinh q cos f + i cosh q sin f Comparing real and imaginary parts, cos a = sinh q cos f

… (1)

sin a = cosh q sin f

… (2)

(i) Eliminating f from Eqs (1) and (2), cos2 f + sin 2 f =

cos2 a sinh 2 q

+

2

1=

cos a

+

sin 2 a cosh 2 q sin 2 a

sinh 2 q cosh 2 q sinh2 q cosh2 q = cos2 a cosh2 q + sin2 a sinh2 q sinh2 q (1 + sinh2 q) = cos2 a (1 + sinh2 q) + (1 – cos2 a) sinh2 q sinh2 q + sinh4 q = cos2 a + cos2 a sinh2 q + sinh2 q – cos2 a sinh2 q sinh4 q = cos2 a ... (3) (ii) From Eq. (1), cos2 a = sinh2 q cos2 f cos2 f =

cos2 a

sinh 2 q cos2 f = cos a

=

cos2 a cos a

[Using (3)]

example 5 If cos (q + if) = cos a + i sin a then prove that cos 2q + cosh 2f = 2. Solution cos(q + if) = cos a + i sin a cos a + i sin a = cos q cos if – sin q sin if = cos q cosh f – i sin q sinh f Comparing real and imaginary parts, cos a = cos q cosh f sin a = – sin q sinh f

...(1) ...(2)

1.78

Chapter 1

Complex Numbers

Squaring and adding Eqs (1) and(2), cos2 a + sin 2 a = cos2 q cosh 2 f + sin 2 q sinh 2 f (1 + cos 2q ) (1 + cosh 2f ) (1 - cos 2q ) (cosh 2f - 1) 1= + 2 2 2 2 4 = 1 + cosh 2f + cos 2q + cos 2q cosh 2f + cosh 2f - 1 - cos 2q cosh 2f + cos 2q = 2(cosh 2f + cos 2q ) \ cosh 2f + cos 2q = 2

example 6 If cos (q + if) = r(cos a + i sin a), prove that f=

1 È sin(q - a ) ˘ log Í ˙ 2 Î sin(q + a ) ˚

Solution cos(q +if) = r(cos a + isin a) r(cos a + i sin a) = cos q cos if – sin q sin if = cos q cosh f – i sin q sinh f Comparing real and imaginary parts, r cos a = cos q cosh f r sin a = – sin q sinh f Dividing Eq. (2) by Eq. (1), r sin a - sin q sinh f = r cos a cos q cosh f tan a = - tan q tanh f tanh f = - tan a cot q f = tanh -1 (- tan a cot q ) =

1 Ê 1 - tan a cot q ˆ log Á Ë 1 + tan a cot q ˜¯ 2

=

1 Ê cos a sin q - sin a cos q ˆ log Á Ë cos a sin q + sin a cos q ˜¯ 2

=

È sin(q - a ) ˘ 1 log Í ˙ 2 Î sin(q + a ) ˚

... (1) ... (2)

1.12

1.79

Separation into Real and Imaginary Parts

example 7 If sin (q + if ) = tan a + i sec a, show that cos2q cosh 2f = 3. Solution sin (q + if) = tan a + i sec a tan a + i sec a = sin q cos if + cos q sin if = sin q cosh f + i cos q sinh f Comparing real and imaginary parts, tan a = sin q cosh f sec a = cos q sinh f Eliminating a from Eqs (1) and (2), sec2 a – tan2 a = cos2 q sinh2 f – sin2 q cosh 2 f

… (1) … (2)

(1 + cos 2q ) (cosh 2f - 1) (1 - cos 2q ) (1 + cosh 2f ) 2 2 2 2 4 = cosh 2f – 1 + cos 2q cosh 2f – cos 2q – 1– cosh 2f + cos 2q + cos 2q cosh 2f = –2 + 2 cos 2q cosh 2f \ cos 2q cosh 2f = 3 1=

example 8 If sin (q + if) = R (cos a + i sin a ), prove that 1 (i) R2 = (cosh 2f – cos 2q ) 2 (ii) tan a = tanh f cot q. Solution sin (q + if) = R (cos a + i sin a) R (cos a + i sin a) = sin q cosh f + i cos q sinh f Comparing real and imaginary parts, R cos a = sin q cosh f R sin a = cos q sinh f Eliminating a from Eqs (1) and (2),

… (1) … (2)

R 2 (cos2 a + sin 2 a ) = sin 2 q cosh 2 f + cos2 q sinh 2 f (1 - cos 2q ) (1 + cosh 2f ) (1 + cos 2q ) (cosh 2f - 1) + R2 = 2 2 2 2 2 cosh 2f - 2 cos 2q = 4 1 = (cosh 2f - cos 2q ) 2

1.80

Chapter 1

Complex Numbers

Dividing Eq. (2) by Eq. (1), tan a =

cos q sinh f sin q cosh f

= cot q tanh f

example 9 pˆ Ê If u + iv = cosh Á a + i ˜ , find (u2 – v2). Ë 4¯ Solution pˆ Ê u + iv = cosh Á a + i ˜ Ë 4¯ pˆ Ê = cos Á ia - ˜ Ë 4¯ p p + sin ia sin 4 4 p p = cosh a cos + i sinh a sin 4 4 1 1 = cosh a + i sinh a 2 2

= cos ia cos

Comparing real and imaginary parts, cosh a u=

,

sinh a v=

2 u2 - v 2 =

2

1 1 (cosh 2 a - sinh 2 a ) = 2 2

example 10 Prove that 2e2x = cosh 2v – cos 2u, where ez = sin (u + iv) and z = x + iy. Solution ez = sin u cos iv + cos u sin iv ex+iy = sin u cosh v + cos u ( i sinh v) ex eiy = sin u cosh v + i cos u sinh v ex (cos y + i sin y) = sin u cosh v + i cos u sinh v

1.12

Separation into Real and Imaginary Parts

1.81

Comparing real and imaginary parts, ex cos y = sin u cosh v

… (1)

x

e sin y = cos u sinh v

… (2)

Squaring and adding Eqs (1) and (2), e2x (cos2 y + sin2 y) = sin2 u cosh2 v + cos2 u sinh2 v e2x = (1 – cos2 u) cosh2 v + cos2 u (cosh2 v – 1) = cosh2 v – cos2 u 1 + cosh 2v 1 + cos 2u 2 2 2e2x = cosh 2v – cos 2u =

example 11 If sin–1(a + ib ) = x + iy, prove that sin2 x and cosh2 y are the roots of the equation l2 – (a 2 + b 2 + 1) l + a 2 = 0. Solution sin-1 (a + ib) = x + iy a + ib = sin (x + iy) = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sinh y Comparing real and imaginary parts, a = sin x cosh y b = cos x sinh y Consider a 2 + b 2 + 1 = sin2 x cosh2 y + cos2 x sinh2 y + 1 = sin2 x cosh2 y + (1 – sin2 x) (cosh2 y – 1) + 1 = sin2 x cosh2 y + cosh2 y – 1 – sin2 x cosh2 y + sin2 x + 1 = cosh2 y + sin2 x Also, a2 = cosh2 y sin2 x Substituting (1 + a 2 + b 2) and a 2 in the given equation, l2 – (cosh2 y + sin2 x) l + cosh2 y sin2 x = 0 2 Comparing with x – (sum of roots) x + product of roots = 0, cosh2 y and sin2 x are the roots of the given equation.

example 12 Separate into real and imaginary parts: (i) tan (x + iy) (ii) tan–1 (x + iy)

(iii) tan–1 (eiq ).

1.82

Chapter 1

Complex Numbers

Solution (i)

tan ( x + iy) =

sin ( x + iy) cos( x + iy)

=

2 sin ( x + iy) cos( x - iy) 2 cos( x + iy) cos( x - iy)

=

sin 2 x + sin 2iy cos 2 x + cos 2iy

sin 2 x + i sinh 2 y cos 2 x + cosh 2 y sin 2 x Real partt = cos 2 x + cosh 2 y =

Imaginary part = (ii) Let

sinh 2 y cos 2 x + cosh 2 y

tan -1 ( x + iy) = a + ib tan -1 ( x - iy) = a - ib

...(1) ...(2)

Adding Eqs (1) and (2), 2 a = tan -1 ( x + iy) + tan -1 ( x - iy) È x + iy + x - iy ˘ = tan -1 Í ˙ Î 1 - ( x + iy)( x - iy) ˚ ˆ Ê 1 2x a = tan -1 Á ˜ 2 2 2 Ë 1- x - y ¯ Subtracting Eq. (2) from Eq. (1), 2ib = tan -1 ( x + iy) - tan -1 ( x - iy) È x + iy - x + iy ˘ = tan -1 Í ˙ Î 1 + ( x + iy)( x - iy) ˚ 2iy tan 2ib = 1 + x 2 + y2 2iy i tanh 2b = 1 + x 2 + y2 ˆ Ê 2y 2b = tanh -1 Á 2 2˜ Ë 1+ x + y ¯ 2y È Í1 + 2 1 1 + + y2 x = log Í 2y Í 2 Í1 - 1 + x 2 + y2 Î

˘ ˙ ˙ ˙ ˙ ˚

1.12

=

b=

Separation into Real and Imaginary Parts

1.83

Ê 1 + x 2 + y2 + 2 y ˆ 1 log Á ˜ 2 Ë 1 + x 2 + y2 - 2 y ¯

È (1 + y)2 + x 2 ˘ 1 log Í 2 2˙ 4 ÍÎ (1 - y) + x ˙˚

È (1 + y)2 + x 2 ˘ ˆ i Ê 1 2x + tan -1 Á log Í ˜ 2 2˙ 2 Ë 1 - x 2 - y2 ¯ 4 ÍÎ (1 - y) + x ˙˚ tan–1 (eiq) = x + iy … (1) tan -1 ( x + iy) =

Hence, (iii) Let

tan–1 (e–iq) = x – iy Adding Eqs (1) and (2),

… (2)

Ê ei q + e - i q ˆ p 2 x = tan -1 (ei q ) + tan -1 (e - i q ) = tan -1 Á = tan -1 • = np + iq -iq ˜ 2 Ë 1- e ◊e ¯ np p + 2 4 Subtracting Eq. (2) from Eq. (1), x=

Ê eiq - e - iq ˆ 2iy = tan -1 (ei q ) - tan -1 (e - iq ) = tan -1 Á ˜ Ë 1 + eiq ◊ e - iq ¯ 2i sin q 2 i tanh 2 y = i sin q tan 2iy =

2 y = tanh -1 (sin q ) =

1 Ê 1 + sin q ˆ log Á Ë 1 - sin q ˜¯ 2

È Êp ˆ˘ Í 1 + cos ÁË 2 - q ˜¯ ˙ 1 ˙ = log Í Í 2 Êp ˆ˙ Í 1 - cos ÁË 2 - q ˜¯ ˙ Î ˚ È qˆ˘ 2Êp Í 2 cos ÁË 4 - 2 ˜¯ ˙ 1 ˙ = log Í Í qˆ ˙ 2 2Êp Í 2 sin ÁË 4 - 2 ˜¯ ˙ ˚ Î È Ê p q ˆ˘ 1 log Ícot Á - ˜ ˙ 2 Î Ë 4 2¯˚ Êp qˆ = log cot Á - ˜ Ë 4 2¯ =

2

1.84

Chapter 1

Complex Numbers

Êp qˆ = - log tan Á - ˜ Ë 4 2¯ 1 Êp qˆ y = - log tan Á - ˜ Ë 4 2¯ 2 1ˆ p i Ê Êp qˆ Hence, tan -1 (eiq ) = Á n + ˜ - log tan Á - ˜ Ë Ë 4 2¯ 2¯ 2 2

example 13 2x Êp ˆ 2 2 = 1. If tan Á + ia ˜ = x + iy, prove that x + y + Ë6 ¯ 3 Solution Êp ˆ tan Á + ia ˜ = x + iy Ë6 ¯ Êp ˆ -1 ÁË 6 + ia ˜¯ = tan ( x + iy)

... (1)

Êp ˆ -1 ÁË 6 - ia ˜¯ = tan ( x - iy)

... (2)

Adding Eqs (1) and (2), 2p = tan -1 ( x + iy) + tan -1 ( x - iy) 6 È x + iy + x - iy ˘ = tan -1 Í ˙ Î 1 - ( x + iy)( x - iy) ˚ 2x Êpˆ tan Á ˜ = Ë 3 ¯ 1 - x 2 - y2 2x 3= 1 - x 2 - y2 1 - x 2 - y2 = x 2 + y2 +

2x 3

2x

=1

3

1.12

Separation into Real and Imaginary Parts

1.85

example 14 If tan (x + iy) = i, where x and y are real then show that x is indeterminate and y is infinite. Solution tan (x + iy) = i x + iy = tan–1 (i) x – iy = tan–1 (–i) Adding Eqs (1) and (2), 2x = tan–1 i + tan–1 (–i) È i + ( -i ) ˘ = tan -1 Í ˙ Î 1 - i ( -i ) ˚ Ê i-i ˆ = tan -1 Á Ë 1 + i 2 ˜¯ 0 0 = indeterminate Subtracting Eq. (2) from Eq. (1), 2iy = tan–1 i – tan–1 (–i) -1 È i - ( -i ) ˘ = tan Í ˙ Î 1 + i( -i ) ˚ = tan–1 i tan 2iy = i i tanh 2y = i tanh 2y = 1 1 Ê 1 + 1ˆ 2y = tanh–1 (1) = log Á Ë 1 - 1˜¯ 2 = tan–1

y=

1 log • = • 4

example 15 2x Êp ˆ If cot Á + ia ˜ = x + iy, prove that x 2 + y 2 + = 1. Ë3 ¯ 3 Solution Êp ˆ cot Á + ia ˜ = x + iy Ë3 ¯ 1 Êp ˆ tan Á + ia ˜ = Ë3 ¯ x + iy

… (1) … (2)

1.86

Chapter 1

Complex Numbers

Ê 1 ˆ p + ia = tan -1 Á 3 Ë x + iy ˜¯ Ê 1 ˆ p - ia = tan -1 Á 3 Ë x - iy ˜¯ ˆ Êp ˆ Êp -1 Ê 1 ˆ -1 Ê 1 ˆ ÁË 3 + ia ˜¯ + ÁË 3 - ia ˜¯ = tan ÁË x + iy ˜¯ + tan ÁË x - iy ˜¯ 1 ˘ È 1 Í x + iy + x - iy ˙ 2p ˙ = tan -1 Í 3 Í1 - 1 . 1 ˙ ÍÎ x + iy x - iy ˙˚ ˆ Ê 2x = tan -1 Á 2 ˜ Ë x + y2 - 1¯ 2x Ê 2p ˆ tan Á ˜ = 2 Ë 3 ¯ x + y2 - 1 2x - 3= 2 x + y2 - 1 2x x 2 + y2 + =1 3

exercIse 1.6 1. Separate into real and imaginary parts: (i) cot (x + iy) (iv) (sin q + i cos q)

(ii) sec (x + iy)

(iii) cosec (x + iy)

i.

2 (cos x cosh y - i sin x sinh y ) ˘ sin 2 x - i sinh 2y È (ii) Í Ans. : (i) cosh 2y - cos 2 x ˙ cos 2 x + cosh 2y Í ˙ p Í ˙ q2 (sin x cosh y - i cos x sinh y ) (iii) (iv)) e 2 Í ˙ cosh 2y - cos 2 x Î ˚

2. If sin (a + ib) = x + iy, prove that (i) x2 sech2 b + y2 cosech2 b = 1

(ii)

x2 cosec2 a – y2 sec2 a = 1.

3. If sin (q + if) = p (cos a + i sin a), prove that p2 = tan a = tanh f cot q.

1 (cosh 2f – cos 2q), 2

4. If cosh (q + if) = eia, prove that sin2 a = sin4f = sinh4 q.

1.12

Separation into Real and Imaginary Parts

1.87

5. If cos (x + iy) cos (u + iv) = 1, where x, y, u, v are real then prove that tanh2 v cosh2 y = sin2 x. Ê u + iv ˆ sin u + i sinh v = 6. Prove that tan Á . Ë 2 ˜¯ cos u + cosh v 7. If tan (x + iy) = eiq, prove that (i) q =

np p + 2 4

(ii) y =

Êp qˆ 1 log tan Á + ˜ . Ë 4 2¯ 2

8. If tan (q + if) = tan a + i sec a then prove that (i) e2f = cot

α 2

(ii) 2q = np +

π +a. 2

9. Prove that all solutions of the equation sin z = 2i cos z are given by z=

np i + log 3. 2 2

2x Êp ˆ 10. If cot Á + ia ˜ = x + iy, prove that x2 + y2 = 1. Ë6 ¯ 3 2y ip ˆ Ê 2 2 = 1. 11. If tanh Á a + ˜ = x + iy, prove that x + y + Ë 6¯ 3 12. Separate real and imaginary parts of cos–1 (eiq). È Ans.: sin-1 sin q + i log ( 1 + sin q - sin q )˘ Î ˚

(

)

13. Prove that sin–1 (ix) = i log x + x 2 + 1 + 2np . 14. Separate into real and imaginary parts: (i) cos–1 (i)

-1 Ê 5i ˆ (ii) cos Á ˜ Ë 12 ¯

Ê 3i ˆ (iii) sin-1 Á ˜ Ë 4¯

(iv) sinh–1 (ix)

(v) tanh–1 (i).

(

)

p p 2 ˘ È Í Ans. : (i) 2 + i log 2 - 1 (ii) 2 + i log 3 ˙ Í ˙ ip ˙ ip Í (v) (iii) i log 2 (iv) cosh-1 x + ÍÎ 2 4 ˙˚ 15. Prove that sin–1 (cosec q) =

q p + i log cot . 2 2

16. If log sin (x + iy) = a + ib then prove that a = tan b = cot x tanh y.

1 Ê cosh 2y - cos 2 x ˆ log Á ˜¯ , Ë 2 2

1.88

1.13

Chapter 1

Complex Numbers

loGarIthm of a complex number

If z and w are two complex numbers and z = ew then w = log z is called the logarithm of the complex number z. Let z = x + iy = reiq Ê yˆ where r = z = x 2 + y 2 and q = arg (z) = tan–1 Á ˜ Ë x¯ log z = log (r eiq) = log r + log eiq = log r + iq log e = log r + iq 2 2 -1 Ê y ˆ = log x + y + i tan Á x ˜ Ë ¯

log ( x + iy) =

Hence,

(

)

1 Ê yˆ log x 2 + y 2 + i tan -1 Á x ˜ Ë ¯ 2

This is called the principal value of log (x + iy). The general value of log (x + iy) is given as Log (x + iy) = log r + i (2np + q) È 1 Ê yˆ˘ log x 2 + y 2 + i Í2 np + tan -1 Á ˜ ˙ Ë x¯˚ 2 Î = 2 np i + log( x + iy)

=

(

)

The general value of log (x + iy) is denoted by Log (x + iy), beginning with a capital L to distinguish it from its principal value which is denoted by log (x + iy).

example 1 Find the value of (i) log–3 (-2)

(ii) log (1 + i).

Solution 1 Ê 0ˆ log 4 + i tan -1 Á ˜ Ë -2 ¯ log 2 + ip loge (-2) 2 = = (i) log -3 (-2) = loge (-3) 1 log 3 + ip -1 Ê 0 ˆ log 9 + i tan Á ˜ Ë -3 ¯ 2 (ii) log (1 + i ) =

1 ip Ê 1ˆ 1 log 2 + i tan -1 Á ˜ = log 2 + Ë 1¯ 2 2 4

1.13

Logarithm of a Complex Number

example 2 Prove that logi i =

4n + 1 , where n, m are integers. 4m + 1

Solution logi i =

log i log i

1 pˆ Ê log 1 + i Á 2np + ˜ Ë 2 2¯ = 1 pˆ Ê log 1 + i Á 2mp + ˜ Ë 2 2¯ = =

i ( 4n + 1) p

i ( 4 m + 1) p 4n + 1 4m + 1

example 3 Show that

(

Solution (i)

sin log i i = sin(i log i ) Ê ip ˆ = sin Á i ˜ Ë 2¯ Ê -p ˆ = sin Á Ë 2 ˜¯ = -1

(ii)

)

(i ) sin log i i = -1

cos log i i = cos(i log i )  iπ  = cos  i   2  −π  = cos   2  =0

(

)

(ii ) cos log i i = 0

1.89

1.90

Chapter 1

Complex Numbers

example 4 Show that cos(i z ) = cos(iz ) for all z.

[Winter 2012]

Solution cos(i z ) = cosh z =

e z + e- z 2

=

e z + e- z 2

= cosh z = cos(iz )

example 5 Show that sin (iz ) = sin(i z ) if and only if z = np i (n Œ ) . [Winter 2014] Solution Let

sin (i z ) = sin (i z ) ¤ i sinh z = i sinh z Ê e z - e- z ˆ Ê e z - e- z ˆ ¤ i Á = iÁ ˜ ˜¯ Ë Ë 2 ¯ 2 Ê e z - e- z ˆ Ê e z - e- z ˆ = i ¤ -i Á ˜¯ ÁË ˜¯ Ë 2 2 ¤ - e z + e- z = e z - e- z ¤ 2e z = 2e - z ¤ e2 z = 1 g1 ¤ 2 z Log e = Log È Ê 0ˆ˘ ¤ 2 z = log 1 + 0 + i Í2 np + tan -1 Á ˜ ˙ Ë 1¯˚ Î ¤ 2 z = log 1 + 2 np i ¤ 2 z = 2 npii ¤ z = np i, n Œ 

1.13

Logarithm of a Complex Number

1.91

example 6 Prove that log (1 + cos 2q + i sin 2q ) = log (2 cos q ) + iq. Solution log (1 + cos 2q + i sin 2q ) = =

1 Ê sin 2q ˆ 2 log ÈÎ(1 + cos 2q ) + sin 2 2q ˘˚ + i tan -1 Á Ë 1 + cos 2q ˜¯ 2 1 Ê 2 sin q cos q ˆ log(4 cos4 q + 4 sin 2 q cos2 q ) + i tan -1 Á Ë 2 cos2 q ˜¯ 2

1 log ÈÎ4 cos2 q (cos2 q + sin 2 q )˘˚ + i tan -1 (tan q ) 2 1 = log (2 cos q )2 + iq 2 = log (2 cos q ) + iq =

example 7 Simplify log (eia + eib). Solution log (eia + eib) = log [cos a + i sin a) + (cos b + i sin b)] = log [(cos a + cos b) + i (sin a + sin b)] È Êa +bˆ Êa -bˆ Êa +bˆ Êa - b ˆ˘ = log Í2 cos Á ˜¯ cos ÁË 2 ˜¯ + 2i sin ÁË 2 ˜¯ cos ÁË 2 ˜¯ ˙ Ë 2 Î ˚ È Êa -bˆ Ï Êa +bˆ Ê a + b ˆ ¸˘ = log Í2 cos Á + i sin Á ˝˙ Ìcos Á ˜ ˜ Ë ¯ Ë ¯ Ë 2 ˜¯ ˛˙˚ 2 2 ÍÎ Ó È Êa +bˆ È Êa + b ˆ˘ Êa - b ˆ˘ = log Í2 cos Á + log Ícos Á + i sin Á ˙ ˜ ˜ Ë 2 ˜¯ ˙˚ Ë 2 ¯˚ Î Ë 2 ¯ Î Êa + bˆ ˜¯

iÁ È Êa - b ˆ˘ = log Í2 cos Á + log e Ë 2 ˙ ˜ Ë 2 ¯˚ Î È Êa - b ˆ˘ Êa + b ˆ +i = log Í2 cos Á Ë 2 ˜¯ ˙˚ ÁË 2 ˜¯ Î

[∵ log e = 1]

1.92

Chapter 1

Complex Numbers

example 8 Ê x - iˆ Prove that i log Á = p – 2 tan–1x. Ë x + i ˜¯ Solution Ê x - iˆ i log Á = i [ log( x - i ) - log( x + i )] Ë x + i ˜¯ È1 Ê 1ˆ 1 Ê 1ˆ˘ = i Í log( x 2 + 1) + i tan -1 Á - ˜ - log ( x 2 + 1) - tan -1 Á ˜ ˙ Ë ¯ Ë x¯˚ x 2 Î2 = i ÈÎi (- cot -1 x - cot -1 x )˘˚ = 2 cot -1 x Êp ˆ = 2 Á - tan -1 x ˜ Ë2 ¯ = p - 2 tan -1 x

example 9 Express log[sin( x + iy)] in the form of a + ib. Solution log[sin( x + iy)] = log(sin x cos iy + cos x sin iy) = log(sin x cosh y + i cos x sinh y) =

Ê cos x sinh y ˆ 1 log(sin 2 x cosh 2 y + cos2 x sinh 2 y) + i tan -1 Á 2 Ë sin x cosh y ˜¯

1 log[sin 2 x cosh 2 y + (1 - sin 2 x )(cosh 2 y - 1)] + i tan -1 (cot x ◊ tanh y) 2 1 = log(sin 2 x cosh 2 y + cosh 2 y - 1 - sin 2 x cosh 2 y + sin 2 x ) + i tan -1 (cot x tanh y) 2 1 = log(sinh 2 y + sin 2 x ) + i tan -1 (cot x tanh y) 2

=

example 10 È sin( x + iy) ˘ Prove that log Í = 2i tan–1 (cot x tanh y). ˙ Î sin( x - iy) ˚

1.13

Logarithm of a Complex Number

1.93

Solution È sin ( x + iy) ˘ log Í ˙ = log [sin (x + iy)] – log [sin (x – iy)] Î sin ( x - iy) ˚ = log (sin x cosh y + i cos x sinh y) – log (sin x cosh y – i cos x sinh y) =

Ê cos x sinh y ˆ 1 log (sin 2 x cosh 2 y + cos2 x sinh 2 y) + i tan -1 Á 2 Ë sin x cosh y ˜¯ -

Ê - cos x sinh y ˆ 1 log (sin 2 x cosh 2 y + cos2 x sinh 2 y) - i tan -1 Á 2 Ë sin x cosh y ˜¯

= i tan–1 (cot x tanh y) + i tan–1 (cot x tanh y) = 2i tan–1 (cot x tanh y)

example 11 2 2 È Ê a + ib ˆ ˘ a - b Prove that cos Íi log Á = . Ë a - ib ˜¯ ˙˚ a 2 + b2 Î Solution

È Ê a + ib ˆ ˘ cos Íi log Á = cos[i log (a + ib) - i log (a - ib)] Ë a - ib ˜¯ ˙˚ Î È1 b 1 Ê bˆ˘ = cos i Í log (a 2 + b2 ) + i tan -1 - log (a 2 + b2 ) - i tan -1 Á - ˜ ˙ Ë a¯˚ a 2 Î2 b bˆ Ê = cos i Á i tan -1 + i tan -1 ˜ Ë a a¯ bˆ Ê = cos Á -2 tan -1 ˜ Ë a¯ Ê = cos Á 2 tan -1 Ë

bˆ a ˜¯

= cos 2q , where tan q = =

1 - tan 2 q 1 + tan 2 q

b2 a2 = b2 1+ 2 a 1-

b a

1.94

Chapter 1

Complex Numbers

=

a 2 - b2 a 2 + b2

example 12 Find the value of (–i)i. Solution Let x + iy = (-i )i Taking logarithm on both the sides, log ( x + iy) = i log (-i ) ip ˆ Ê1 = i Á log 1 - ˜ Ë2 2¯ == x + iy =

i 2p 2

p 2 p e2 p

(-i )i = e 2

Hence,

example 13 Find the principal value of i(1 – i). Solution Let x + iy = i (1- i ) Taking logarithm on both the sides, log( x + iy) = log i (1- i ) = (1 - i ) log i È1 Ê 1ˆ ˘ = (1 - i ) Í log 1 + i tan -1 Á ˜ ˙ Ë 0¯ ˚ Î2 = (1 - i ) ÈÎ0 + i tan -1 (•)˘˚ Ê ip ˆ = (1 - i ) Á ˜ Ë 2¯ =

ip i 2p 2 2

[Winter 2013]

1.13

Logarithm of a Complex Number

1.95

ip p + 2 2 p ip = + 2 2 =

x + iy = e

Ê p ip ˆ + ËÁ 2 2 ¯˜ p

ip

= e2 e 2 p 2

p pˆ Ê = e Á cos + i sin ˜ Ë 2 2¯ p

= e 2 (0 + i ) =

p ie 2 p

Hence, the principal value of i (1- i ) is ie 2 .

example 14 For the principal branch, show that Log (i3) π 3 Log i. [Summer 2013] Solution For the principal branch, Log i 3 = log i 3 = log (-i ) 1 Ê -1ˆ log(1 + 0) + i tan -1 Á ˜ Ë 0¯ 2 ip =2 =

3 Log i = 3 log i È1 Ê 1ˆ ˘ = 3 Í log(1 + 0) + i tan -1 Á ˜ ˙ Ë 0¯ ˚ 2 Î Ê ip ˆ = 3Á ˜ Ë 2¯ From Eqs (1) and (2), Log(i3) π 3 Log i

...(1) [ For the principal branch ]

...(2)

1.96

Chapter 1

Complex Numbers

example 15 Separate the real and imaginary parts of (i) log1–i (1 + i) (ii) (1 + i)i (iii) (–i)–(1–i) Solution (i) Let x + iy = log1–i (1 + i) = log (1 + i ) log (1 - i )

=

1 log 2 + i tan -1 1 2

1 log 2 + i tan -1 (-1) 2 1 ip log 2 + 2 4 = 1 ip log 2 2 4 ip ˆ Ê ip ˆ Ê ÁË log 2 + 2 ˜¯ ÁË log 2 + 2 ˜¯ = ip ˆ Ê ip ˆ Ê ÁË log 2 - 2 ˜¯ ÁË log 2 + 2 ˜¯ p2 + ip log 2 4 = p2 (log 2)2 + 4 Comparing real and imaginary parts, (log 2)2 -

p2 4 , x= p2 2 (log 2) + 4 (log 2)2 -

y=

(ii) Let x + iy = (1 + i)i Taking logarithm on both the sides, log (x + iy) = i log (1 + i) Ê1 ˆ = i Á log 2 + i tan -1 1˜ Ë2 ¯ =

i p log 2 2 4

p log 2 (log 2)2 +

p2 4

1.13

i

log 2 -

x + iy = e 2

e

= ei log =e

-

p 4

2

e

Logarithm of a Complex Number

p 4

-

p 4

ÈÎcos(log 2 ) + i sin (log 2 )˘˚

Comparing real and imaginary parts, -

x=e -

y=e (iii) Let

p 4

cos(log 2 )

p 4

sin (log 2 )

x + iy = (-i )- (1- i ) log( x + iy) = -(1 - i ) log(-i ) ip ˆ Ê1 = -(1 - i ) Á log 1 - ˜ Ë2 2¯ Ê ip ˆ = -(1 - i ) Á - ˜ Ë 2¯ ip 2 p -i 2 2 p p =i + 2 2 =

p

x + iy = e 2

+i

p

= e2 e

p 2

i

p 2

p

p pˆ Ê = e 2 Á cos + i sin ˜ Ë 2 2¯ =

p ie 2

Comparing real and imaginary parts, x=0 y=

p e2

1.97

1.98

Chapter 1

Complex Numbers

example 16 Find the real and imaginary parts of tanh–1 (x + iy). Solution tanh -1 ( x + iy) =

È 1 + ( x + iy) ˘ 1 log Í ˙ 2 Î 1 - ( x + iy) ˚

1 [log{(1 + x ) + iy} - log{(1 - x ) - iy}] 2 1 È1 y ˆ˘ Ê y ˆ 1 Ê = Í log{(1 + x )2 + y 2 } + i tan -1 Á - log{(1 - x )2 + y 2 } - i tan -1 Á Ë 1 + x ˜¯ 2 Ë 1 - x ˜¯ ˙˚ 2 Î2 =

=

ÏÔ (1 + x )2 + y 2 ¸Ô˘ Ï -1 Ê y ˆ 1 È1 Ê y ˆ¸ og Ì + tan -1 Á Í lo ˝˙ + i Ìtan Á ˝ ˜ 2 2 Ë 1+ x ¯ Ë 1 - x ˜¯ ˛ 2 ÍÎ 2 ÓÔ (1 - x ) + y ˛Ô˙˚ Ó

È Ï ¸˘ y y Í + Ô Ô˙ 2 2 1 1 ÔÏ (1 + x ) + y Ô¸ -1 Ô 1 + x 1 - x Ô˙ = Í log Ì + i tan Ì ˝ ˝ 2 2 2 Í2 Ê y ˆ Ê y ˆ Ô˙ ÔÓ (1 - x ) + y Ô˛ Ô 1 Í ÔÓ ÁË 1 + x ˜¯ ÁË 1 - x ˜¯ Ô˛˙˚ Î ÏÔ (1 + x )2 + y 2 ¸Ô ˆ˘ Ê 1 È1 -1 y - xy + y + xy = Í log Ì ˝ + i tan Á ˜˙ 2 2 2 2 2 ÍÎ 2 Ë 1 - x - y ¯ ˚˙ ÓÔ (1 - x ) + y ˛Ô =

ÏÔ (1 + x )2 + y 2 ¸Ô 1 ˆ Ê 1 2y -1 log Ì ˝ + i tan Á ˜ 2 2 2 2 4 Ë 1- x - y ¯ ÓÔ (1 - x ) + y ˛Ô 2

Hence, Re{tanh -1 ( x + iy)} =

and

2 2 1 ÔÏ (1 + x ) + y Ô¸ log Ì 2 2˝ 4 ÔÓ (1 - x ) + y Ô˛

Im{tanh -1 ( x + iy)} =

ˆ Ê 1 2y tan -1 Á ˜ 2 2 2 Ë 1- x - y ¯

example 17 If ia+ib = a + ib, prove that a 2 + b 2 = e–(4k + 1)pb. Solution ia + ib = a + ib Taking logarithm on both the sides, (a + ib) log i = log (a + ib)

1.13

or

Logarithm of a Complex Number

1.99

Log (a + ib ) = (a + ib ) (2 kp i + log i ) ip ˆ Ê = (a + ib ) Á 2 kp i + ˜ Ë 2¯ = i (4 k + 1) a + ib = e

where

i ( 4 k +1)

pb pa - (4 k + 1) 2 2

pa pb - ( 4 k +1) 2 e 2

r = a + ib = e a2 + b 2 = e

- ( 4 k +1)

- ( 4 k +1)

=e

- ( 4 k +1)

pb pa i ( 4 k +1) 2 e 2

= reiq , sayy

pb 2

pb 2

a 2 + b 2 = e - ( 4 k +1)pb

example 18 By considering only the principle value, express (1 + i 3 )1+i form of (a + ib). Solution Let

a + ib = (1 + i 3 )1+ i

3

3

Taking logarithm on both the sides, log (a + ib) = (1 + i 3 ) log (1 + i 3 ) È1 ˘ log (a + ib) = (1 + i 3 ) Í log (1 + 3) + i tan -1 3 ˙ Î2 ˚ ip ˆ Ê1 = (1 + i 3 ) Á log 4 + ˜ Ë2 3¯ ip ˆ Ê = (1 + i 3 ) Á log 2 + ˜ Ë 3¯ = log 2 -

pˆ p 3 Ê + i Á 3 log 2 + ˜ Ë 3¯ 3

Ê pˆ p 3ˆ Ê Á log 2 ˜ i Á 3 log 2 + 3 ˜¯ 3 ¯ Ë

a + ib = eË

Ê p 3ˆ Á log 2 ˜ 3 ¯

= eË

e

È Ê pˆ p ˆ˘ Ê Ícos ÁË 3 log 2 + ˜¯ + i sin ÁË 3 log 2 + ˜¯ ˙ 3 3 ˚ Î

in the

1.100

Chapter 1

Complex Numbers

Comparing real and imaginary parts, a=e b=e

Ê p 3ˆ ÁË log 2 ˜ 3 ¯

Ê p 3ˆ ÁË log 2 ˜ 3 ¯

pˆ Ê cos Á 3 log 2 + ˜ Ë 3¯ pˆ Ê sin Á 3 log 2 + ˜ Ë 3¯

example 19 Separate the real and imaginary parts of ( i ) i . Solution a + ib = i

Let

Taking logarithm on both the sides, log (a + ib) = log i 1 = log i 2 1 ip = ◊ 2 2 iπ

a + ib = e 4 iπ

i =e4

Hence,

( i)

Let

i

= x + iy

Taking logarithm on both the sides, i log i = log ( x + iy) ip e4

ip

log e 4 = log ( x + iy)

p p ˆ ip Ê log ( x + iy) = Á cos + i sin ˜ Ë 4 4¯ 4 Ê 1 i ˆ ip =Á + ˜ Ë 2 2¯ 4 ip p = 4 2 4 2

1.13

-p

x + iy =

ip

e4 2 e4 2

=e

-

p

Logarithm of a Complex Number

p p ˆ Ê + i sin ÁË cos ˜ 4 2 4 2¯

4 2

Comparing real and imaginary parts, p 4 2

x=e

p

cos

4 2 p -

y=e

4 2

p

sin

4 2

example 20 Ê

1ˆ

p -ÁË 2 m + 2 ˜¯ p Prove that i = cos q + i sin q, where q = (4 n + 1) e . 2 Solution ii

Let

ii = x + iy

Taking logarithm on both the sides, i Log i = Log (x + iy) i (2 mp i + log i ) = Log ( x + iy) ip ˆ Ê Log (x + iy) = i Á 2 mp i + ˜ Ë 2¯ ( x + iy) = e i =e i

1ˆ Ê - Á 2m+ ˜ p Ë 2¯ 1ˆ Ê - Á 2m+ ˜ p Ë 2¯

i

Now,

i i = cos q + i sin q Ê

=i

-Á e Ë

1ˆ 2m+ ˜ p 2¯

Taking logarithm on both the sides, Log (cos q + i sin q ) = e =e

1ˆ Ê -Á 2 m + ˜ p Ë 2¯ 1ˆ Ê -Á 2 m + ˜ p Ë 2¯

Logi ip ˆ Ê ÁË 2 np i + 2 ˜¯ Ê

1ˆ

p - Á 2 m + 2 ˜¯ p = i(4 n + 1) e Ë = if , say 2

1.101

1.102

Chapter 1

Complex Numbers

cos q + i sin q = eif

\

eiq = eif Comparing both the sides, 1ˆ

Ê

p - Á 2 m + 2 ¯˜ p q = f = (4 n + 1) .e Ë 2

example 21 Prove that ii is wholly real and find its principal value. Also show that the values of ii form a geometric progression. [Summer 2014] Solution Let

x + iy = i i log( x + iy) = log i i log( x + iy) = i log i 1ˆ Ê1 = i Á log 1 + i tan -1 ˜ Ë2 0¯ = i(0 + i tan -1 •) Ê ip ˆ = iÁ ˜ Ë 2¯ =x + iy = e

p 2

-

p 2

Hence, ii is wholly real and its principal value is e Let

-

p 2

z = x + iy = i i Log( x + iy) = Log i i = i Log i = i(2 np i + log i ) ip ˆ Ê = i Á 2 np i + ˜ Ë 2¯ = -2 np -

p 2

◊

1.13

x + iy = e =e

Logarithm of a Complex Number

-

p 2 e -2 np

-

p 2

1.103

(e -2p )n

\ z = ka n , where k = e

-

p 2

, a = e -2p

Putting n = 0, 1, 2, 3, 4. … z0 = k , z1 = ka , z2 = ka 2 , z3 = ka 3 ,… Hence, the values of ii form a geometric progression with a common ratio a = e -2p ◊

exercIse 1.7 1. Find the general value of (i) log (–i) (v) cos (log ii)

(ii) log ( 3 - i)

(iii) log2 5

(iv) sin (log ii)

È ˘ pˆ pˆ Ê Ê (ii) log 2 + i Á 2p n - ˜ Í Ans. : (i) i ÁË 2p n - ˜¯ ˙ Ë 2 6¯ Í ˙ 2 2 2 2 Í(iii) [(log 5 log 2 + 4p mn) + i (n log 2 - m log 5)2p ] / [(log 2) + 4p m ]˙ Í(iv) - 1 ˙ (v) 0 Î ˚

2. Find the general value of log (1 + i) + log (1 – i). [ans. : log 2] 3. Find the general value of log (1 + i 3) + log(1 - i 3) ◊ [ans. : 2log 2] –i

4. Prove that sin loge (i ) = 1. 5. Show that log (–log i) = log

p ip - . 2 2

6. Prove that log (1 + i tan a) = log sec a + ia. q ˆ iq Ê 1 ˆ Ê1 = log Á sec ˜ - . 7. Prove that log Á Ë 1 + eiq ˜¯ Ë2 2¯ 2 8. Separate i(1–i) into real and imaginary parts. pˆ Ê È 2 np + ˜ ˘ ËÁ 2¯ Í Ans.: ie ˙ ÍÎ ˙˚

1.104

Chapter 1

Complex Numbers

9. Find the principal value of (x + iy)i and show that it is entirely real if 1 log(x 2 + y 2 ) is a multiple of p. 2 1 È ˘ 2 2 ÍHint : put 2 log(x + y ) = np ˙ Î ˚ y   tan −1 x  [coslog(x 2 + y 2 ) + i sinlog(x 2 + y 2 )  Ans.: e

10. If ia+ib = a + ib, prove that a2 + b2 = e–(4n+1)pb.

points to remember Algebra of Complex Numbers (i) Addition: z1 + z2 = (x1 + x2) + i (y1 + y2) (ii) Subtraction: z1 – z2 = (x1 – x2) + i (y1 – y2) (iii) Multiplication: z1 z2 = (x1 x2 – y1y2) + i (x2y1 + y2x1) z1 Ê x1 x2 + y1 y2 ˆ Ê y1 x2 - x1 y2 ˆ = +i z2 ÁË x22 + y22 ˜¯ ÁË x22 + y22 ˜¯

(iv) Division:

Different Forms of Complex Numbers (i) Cartesian or Rectangular Form: z = x + iy (ii) Polar Form: z = r (cos q + i sinq ) which is also denoted as r – q (iii) Exponential Form: z = reiq

Modulus and Argument (or Amplitude) of Complex Numbers 2 2 Modulus: | z | = r = x + y -1 Ê y ˆ Argument (or Amplitude): arg( z ) = q = tan Á ˜ Ë x¯

Properties of Complex Numbers (i) Re (z) = x =

1 1 (z + z ), Im (z) = y = (z – z ) 2 2i

(ii) ( z1 + z2 ) = z1 + z2 (iii)

( z1 z2 ) =

z1 z2

Points to Remember

Êz ˆ z (iv) Á 1 ˜ = 1 Ë z2 ¯ z2 (v) z z = z 2 = z (vi)

| | | |2 z1 z2 = | z1 | | z2 |

(vii) arg (z1 z2) = arg (z1) + arg (z2) (viii)

z1 z1 = z2 z2

Êz ˆ (ix) arg Á 1 ˜ = arg (z1) – arg (z2) Ë z2 ¯

De Moivre’s Theorem (cosq + i sin q )n = cos nq + i sin nq, where n is any real number.

Circular and Hyperbolic Functions iz - iz eiz + e - iz (i) sin z = e - e , cos z = 2 2i

(ii) sinh z =

e z + e- z e z - e- z , cosh z = 2 2

Relation between Circular and Hyperbolic Functions (i) sin iz = i sinh z, sinh z = –i sin iz (ii) cos iz = cosh z (iii) tan iz = i tanh z, tanh z = –i tan iz

Formulae on Hyperbolic Functions (i) cosh2 z – sinh2 z = 1 (ii) coth2 z – cosech2 z = 1 (iii) sech2 z + tanh2 z = 1 (iv) sinh 2z = 2 sinh z cosh z (v) cosh 2z = cosh2 z + sinh2 z = 2 cosh2 z – 1 = 1 + 2 sinh2 z 2 tanh z (vi) tanh 2z =

1 + tanh 2 z

(vii) sinh 3z = 3 sinh z + 4 sinh3 z (viii) cosh 3z = 4 cosh3 z – 3 cosh z

1.105

1.106

Chapter 1

(ix) tanh 3z =

Complex Numbers

3 tanh z + tanh3 z

1 + 3 tanh 2 z (x) sinh (z1 ± z2) = sinh z1 cosh z2 ± cosh z1 sinh z2

(xi) cosh (z1 ± z2) = cosh z1 cosh z2 ± sinh z1 sinh z2 (xii) tanh (z1 ± z2) =

tanh z1 ± tanh z2 1 ± tanh z1tanh z2

z1 + z2 z -z cosh 1 2 2 2 z1 + z2 z1 - z2 sinh z1 – sinh z2 = 2 cosh sinh 2 2 z1 + z2 z -z cosh 1 2 cosh z1 + cosh z2 = 2 cosh 2 2 z +z z -z cosh z1 – cosh z2 = 2 sinh 1 2 sinh 1 2 2 2 2 sinh z1 cosh z2 = sinh (z1 + z2) + sinh (z1 – z2)

(xiii) sinh z1 + sinh z2 = 2 sinh (xiv) (xv) (xvi) (xvii)

(xviii) 2 cosh z1 sinh z2 = sinh (z1 + z2) – sinh (z1 – z2) (xix) 2 cosh z1 cosh z2 = cosh (z1 + z2) + cosh (z1 – z2) (xx) 2 sinh z1 sinh z2 = cosh (z1 + z2) – cosh (z1 – z2)

Inverse Hyperbolic Functions (i) sinh–1 x = log (x + x 2 + 1) (ii) cosh–1 x = log (x + x 2 - 1) (iii) tanh–1 x =

1 Ê 1+ x ˆ log Á Ë 1 - x ˜¯ 2

Separation into Real and Imaginary Parts (i) sin (x ± iy) = sin x cosh y ± icos x sinh y (ii) cos (x ± iy) = cos x cosh y ± i sin x sinh y (iii) tan (x ± iy) = sin 2 x ± i sinh 2 y cos 2 x + cosh 2 y (iv) sinh (x ± iy) = sinh x cos y ± i cosh x sin y (v) cosh (x ± iy) = cosh x cos y ± i sinh x sin y

Points to Remember

(vi) tanh (x ± iy) =

sinh 2 x ± i sin 2 y cosh 2 x + cos 2 y

Logarithm of a Complex Number Principal Value: log (x + iy) =

1 Ê yˆ log (x 2 + y 2 ) + i tan -1 Á ˜ Ë x¯ 2

= log r + iq General Value: Log (x + iy) =

È 1 Ê yˆ˘ log (x 2 + y 2 ) + i Í2np + tan -1 Á ˜ ˙ Ë x¯˚ 2 Î

= log r + i (2np + q )

1.107

2 Analytic Functions CHAPTER

chapter outline 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11

2.1

Introduction Complex Variable Basic Definitions Limits Continuity Differentiability Analytic Functions Cauchy–Riemann Equations in Polar Form Harmonic Functions Properties of Analytic Functions Construction of Analytic Functions: Milne–Thomson Method

IntroductIon

The theory of functions of a complex variable plays a very important role in solving a large number of problems in the field of engineering and science. The functions of a complex variable are particularly concerned with the analytic function of a complex variable. Because the separate real and imaginary parts of any analytic function must satisfy Laplace’s equation, complex analysis is widely applicable to two-dimensional problems in Physics.

2.2

complex VarIable

Any variable of the form z = x + iy is known as a complex variable, where x and y are real and i = -1. Functions of a complex Variable If z = x + iy be a complex variable then w = f (z) = u + iv is known as a function of the complex variable z and is denoted by w = f (z).

2.2

Chapter 2 Analytic Functions

2.3

basIc deFInItIons

Distance: z - z0 represents the distance between two points z and z0. Circle: z - z0 = r represents a circle with centre at the point z0 and radius r. Interior of a circle: z - z0 < r represents the interior of the circle. Exterior of a circle: z - z0 > r represents the exterior of the circle. Annulus: The region between two concentric circles of radii r1 and r2 (r2 > r1) and centre at z0 is known as the annulus region and is represented as r1 < z – z0 < r2

|

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Neighbourhood: The set of all points for which z - z0 < r is known as the neighbourhood of z0. Boundary point: A point which does not lie in the interior or exterior of a region is known as a boundary point. Open set: A set that does not contain its boundary points is known as an open set. Closed set: A set that contains all its boundary points is known as a closed set. Connected set: If any two points of the set can be joined by a polygonal line such that all the points of the line also belong to the set then the set is known as a connected set. Domain: A set which is open and connected is known as a domain. Bounded region: A region which can be enclosed in a circle of finite radius is known as a bounded region. Compact region: A region that is closed and bounded is known as a compact region.

example 1 Classify and sketch the following regions: (i) |z – 4| > 3 (ii) 1 < |2z – 6| < 2 (iii) (iv) –p < Re z < p (v) |z| < 1 and |z| > 2 y (vi) Re(iz + 2) > 0

|z – 1| + |z + 1| £ 2

Solution (i)

z-4 >3 ( x - 4) + iy > 3 2

( x - 4) + iy > 9 ( x - 4) + y > 9 which represents the exterior of the circle with its centre at (4, 0) and a radius of 3 (Fig. 2.1). Hence, the region is open and unbounded. 2

2

O

x (4,0)

Fig. 2.1

2.3  Basic Definitions        2.3

y

(ii) 1 < 2 z - 6 < 2 1 2 1 2 1 4 1 4

< z-3 2

x2 + y2 < 1 and x2 + y2 > 4 which represents the interior of the circle with its centre at (0, 0) and a radius of 1, and the exterior of the circle with its centre at (0, 0) and a radius of 2 (Fig. 2.5). Hence, the region is not connected. (vi)

Re (iz + 2) > 0 Re (ix - y + 2) > 0 (2 - y) > 0 2>y y 1 p (ii) 0 £ arg z £ [Winter 2014] 4 Solution Let z = x + iy (i) Im z > 1 y >1 Yes, it is a domain (Fig. 2.7). (ii)

p 4 y Ê ˆ p 0 £ tan -1 Á ˜ £ Ë x¯ 4 0 £ arg z £

Fig. 2.7

y p £ tan x 4 y 0 £ £1 x

tan 0 £

Yes, it is a domain (Fig. 2.8).

Fig. 2.8

example 3 Sketch the set S = (z: –1 < Im (z) < 2}. Is it connected? Solution Let

z = x + iy -1 < Im( z ) < 2 -1 < y < 2

Since the lines y = 2 and y = –1 bounding the points of the set are not included in the set, the set is open (Fig. 2.9). Also, each pair of points within the set can be joined by a polygonal line. Hence, the set S is connected.

Fig. 2.9

2.6

Chapter 2 Analytic Functions

example 4 Sketch the region |z| £ 1. Is it a domain?

[Summer 2013]

Solution Let

z = x + iy z £1 x + iy £ 1 2

x + iy £ 1 x 2 + y2 £ 1 which represents the interior of the circle with its centre at (0, 0) and a radius of 1 (Fig. 2.10). Since, the region z £ 1 represents the boundary and the interior of the circle, it is closed.

||

Fig. 2.10

Hence, the given region is not a domain.

example 5 Sketch the region |z – 2 + i| £ 1. Is it a domain?

[Winter 2013]

Solution Let

z = x + iy z-2+i £1 ( x + iy) - 2 + i £ 1 ( x - 2) + i( y + 1) £ 1 2

( x - 2) + i( y + 1) £ 1 ( x - 2)2 + ( y + 1)2 £ 1 which represents the interior of the circle with its centre at (2, –1) and a radius of 1 Fig. 2.11 (Fig. 2.11). Since the region z – 2 + i £ 1 represents the boundary and the interior of the circle, it is closed. Hence, the given region is not a domain.

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example 6 Is the set |z – 1 + 2i| £ 2 a domain? Solution Let

z = x + iy

[Summer 2014]

2.3  Basic Definitions        2.7

z - 1 + 2i £ 2 x + iy - 1 + 2i £ 2 ( x - 1) + i( y + 2) £ 2 2

( x - 1) + i( y + 2) £ 4 ( x - 1)2 + ( y + 2)2 £ 4 which represents a circle with its center at (1, –2) and a radius of 2 (Fig. 2.12). Since the set z – 1 + 2i £ 2 represents the boundary and the interior of the circle, it is closed.

|

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Hence, the given region is not a domain.

Fig. 2.12

exercIse 2.1 I.

classify and sketch the following regions:

1. Re z ≥ 4

ÎÈans.: Closed, unbounded˚˘

2. 0 < 2z – 1 £ 2

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3. z £ z + 1

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ÎÈans.: Bounded, neither open nor closed˚˘ ÎÈans.: Closed, unbounded˚˘

4. z – 1 + 3i £ 1

|

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ÈÎans.: Closed, unbounded˘˚ II.

sketch the following sets and determine which are domains:

1. z – 4 ≥ z

| |

|

|

2. 2z + 3 > 4 3. z – 2 + i £ 1

|

|

ÈÎans.: No ˘˚ ÈÎans.: Yes˘˚ ÈÎans.: No ˘˚

4. Im z = 1 ÎÈans.: No ˚˘

2.8

Chapter 2 Analytic Functions

2.4

lImIts

Let w = f (z) be a single-valued function. The limit of f (z) is w0 = f (z0) as z approaches z0 if for each positive number Œ there exists a positive number d such that f ( z ) - w0 < Œ whenever z - z0 < d Symbolically, it is represented as lim f ( z ) = w0 z Æ z0

example 1

||

Using the definition of limit, show that if f (z) = iz in the open disk z < 1 then lim f ( z ) = i [Winter 2014] z Æ1 Solution Let

f(z) = iz, z0 = 1,

w0 = i

f ( z ) - w0 < Œ,

where Œ> 0

iz - i < Œ i z -1 < Œ z -1 < Œ

ÎÈ∵ i = 1˚˘

||

This condition is satisfied by the points in the open disk z < 1. Assuming Π= d, the following conclusion can be made: For every Π> 0, there exists d > 0 such that f (z) Рi < Πwhenever z Р1 < d Hence, lim f ( z ) = i.

|

|

z Æ1

example 2 iz i = . z Æ1 3 3

Using the definition of limit, show that lim Solution f (z) = Let

iz i , z0 = 1, w0 = 3 3

f ( z ) - w0 < Œ, iz i - 0

|

|

2.4

Limits

i z -1

0, there exists d > 0 such that iz i - 0, d2 > 0 such that x < d1 , y - 3 < d 2 z - z0 < d1 + d 2 = d \ z - 3i < d , where d = d1 + d 2 Now,

f ( z ) = 3 x + iy 2 = 3 x + i [( y - 3) + 3]

2

= 3 x + i ÈÎ( y - 3)2 + 9 + 6 ( y - 3)˘˚ f ( z ) - w0 = ÈÎ3 x + i( y - 3)2 + 9i + 6i( y - 3)˘˚ - 9i = 3 x + i ( y - 3)2 + 6i ( y - 3) 2

£ 3x + y - 3 + 6 y - 3 < 3 d1 + d 22 + 6d 2

2.9

2.10

Chapter 2 Analytic Functions

Assuming 3d1 + d 22 + 6d 2 = Œ, f ( z ) - w0 < Œ f ( z ) - 9i < Œ Thus, for every Œ > 0, there exists d > 0 such that f ( z ) - 9i < Œ whenever z - 3i < d lim f ( z ) = 9i

Hence, or

z Æ 3i

lim (3 x + i y 2 ) = 9 i

z Æ 3i

example 4 Show that the limit of the function does not exist. f (z) = z 2 =0

for z π i for z = i

Solution lim f ( z ) = lim z 2 z Æi

z Æi

= i2 = -1 f (i) = 0

Given

Since lim f ( z ) π f (i ), the limit does not exist. z Æi

example 5 Show that lim

zÆ0

z does not exist. z

Solution f (z) =

Let

z z

z zÆ0 z x + iy = lim z Æ 0 x - iy

lim f ( z ) = lim

zÆ0

2.4

Limits

2.11

The necessary condition for lim f ( z ) to exist is that f (z) should approach the same zÆ0

value along all the paths leading to z = 0. Let the path be y = mx. lim

zÆ0

x + iy x + imx = lim x Æ 0 x - iy x - imx 1 + im = lim x Æ 0 1 - im 1 + im = 1 - im

Since the limit value depends on m, it takes different values along different paths. Hence, the limit does not exist.

example 6 Show that lim

zÆ0

3 x 2 + y2 ( x + y )2

does not exist.

Solution The necessary condition for lim f ( z ) to exist is that f (z) should approach the same zÆ0

value along all the paths leading to z = 0. Let the path be y = mx. lim

zÆ0

3 x 2 + y2 ( x + y )2

= lim

x Æ0

= lim

x Æ0

=

3x 2 + m2 x 2 ( x + mx )2 3 + m2 (1 + m)2

3 + m2 (1 + m)2

Since the limit value depends on m, it takes different values along different paths. Hence, the limit does not exist.

2.12

2.5

Chapter 2 Analytic Functions

contInuIty

f (z) is said to be continuous at z = z0 if lim f ( z ) exists. z Æ z0

example 1 Determine whether the function f (z) is continuous at the origin. f (z) =

( x + y )2

, x 2 + y2 =0 ,

zπ0 z=0

Solution lim f ( z ) = lim

zÆ0

zÆ0

( x + y )2 x 2 + y2

Let z Æ 0 along the line y = mx. lim f ( z ) = lim

zÆ0

x Æ0

= lim

x Æ0

=

( x + mx )2 x 2 + m2 x 2 (1 + m)2 1 + m2

(1 + m)

2

1 + m2

Since the limit depends on m, it takes different values along different paths. Thus, the limit does not exist. Hence, f (z) is not continuous at the origin.

example 2 Discuss the continuity of f (z) at the origin. z f (z) = , z π 0 z = 0, z = 0 Solution z z x - iy = lim z Æ 0 x + iy

lim f ( z ) = lim

zÆ0

zÆ0

2.5

Continuity

2.13

Let z Æ 0 along the line y = mx. x - i mx lim f ( z ) = lim zÆ0 x Æ 0 x + i mx 1- i m = lim x Æ0 1 + i m 1- i m = 1+ i m Since the limit depends on m, it takes different values along different paths. Thus, the limit does not exist. Hence, f (z) is not continuous at the origin.

example 3 Show that f (z) =

Re ( z 2 ) , z

= 0, is continuous at z = 0.

zπ0 z=0

Solution lim f ( z ) = lim

zÆ0

zÆ0

Re( z 2 ) z

= lim

Re ÈÎ( x + iy)2 ˘˚

= lim

Re ÈÎ( x 2 - y 2 ) + i 2 xy ˘˚

zÆ0

x + iy

zÆ0

= lim

zÆ0

x 2 + y2 x 2 - y2 x 2 + y2

Let z Æ 0 along the line y = mx. lim f ( z ) = lim

zÆ0

x Æ0

= lim

x Æ0

= lim

x Æ0

x 2 - m2 x 2 x 2 + m2 x 2 x 2 (1 - m 2 ) x 1 + m2 x(1 - m 2 ) 1 + m2

=0 Since, the limit does not depend on m, it is independent of path.

2.14

Chapter 2 Analytic Functions

Hence, the limit exists. Also, lim f ( z ) = f (0) = 0 zÆ0

Hence, f (z) is continuous at z = 0.

example 4 Determine whether f(z) is continuous at z = i. Redefine if necessary to make it continuous. f ( z ) = z 2 + iz + 2, = i,

zπi z=i

Solution

lim f ( z ) = lim ( z 2 + iz + 2 ) z Æi

z Æi

= i2 + i2 + 2 = 2i 2 + 2 =0 But f (i) = i [Given] lim f ( z ) π f (i ) \ z Æi

Hence, f (z) is not continuous at z = i. To make it continuous, it can be redefined as f ( z ) = z 2 + iz + 2, = 0,

zπi z=i

example 5 Determine whether the function f (z) is continuous at z = –i. z 2 + 3iz - 2 , z+i =5 ,

f (z) =

z π -i z = -i

Can the function be redefined to make it continuous at z = –i? Solution z 2 + 3iz - 2 È0 ˘ form ˙ Í z Æ- i z+i Î0 ˚ 2 z + 3i = lim [Using L’ Hospital’s rule] z Æ- i 1 = -2i + 3i =i

lim f ( z ) = lim

z Æ- i

2.6

Differentiability

But f (–i) = 5 [Given] \ lim f ( z ) π f (-i ) z Æ- i

Hence, f (z) is not continuous at z = –i. To make it continuous, it can be redefined as z 2 + 3iz - 2 , z π -i z+i =i , z = -i

f (z) =

2.6

dIFFerentIabIlIty

A single-valued function f (z) is said to be differentiable at a point z0 if f ( z0 + Dz ) - f ( z0 ) f ′( z0 ) = lim exists. Dz Æ 0 Dz note

Derivative of f(z) at a point z0 can also be written as f ′( z0 ) = lim

z Æ z0

f ( z ) - f ( z0 ) z - z0

example 1 Find the derivative of f ( z ) =

1+ z by using the definition at z = 2. 1- z

Solution f (z) =

1+ z 1- z

For any z, f ¢( z ) = lim

Dz Æ 0

f ( z + Dz ) - f ( z ) Dz

At z = 2, f (2 + Dz ) - f (2) Dz 1 + (2 + Dz ) 1 + 2 1 - (2 + Dz ) 1 - 2 = lim Dz Æ 0 Dz 3 + Dz +3 = lim -1 - Dz Dz Æ 0 Dz 3 + Dz - 3 - 3Dz = lim Dz Æ 0 -(1 + Dz ) Dz

f ¢(2) = lim

Dz Æ 0

2.15

2.16

Chapter 2 Analytic Functions

-2 Dz -(1 + Dz ) Dz 2 = lim Dz Æ 0 1 + Dz =2 = lim

Dz Æ 0

example 2 Show that f (z) = i z3 is differentiable for all z. Solution f (z) = iz3 For any z, f ¢( z ) = lim

Dz Æ 0

= lim

Dz Æ 0

= lim

f ( z + Dz ) - f ( z ) Dz i( z + Dz )3 - iz 3 Dz i ÈÎ z 3 + ( Dz )3 + 3z 2 Dz + 3z( Dz )2 ˘˚ - iz 3 Dz

Dz Æ 0

= lim i ÈÎ( Dz ) + 3z + 3z Dz ˘˚ Dz Æ 0 2

2

= 3i z 2 f¢(z) exists for all values of z. Hence, f (z) is differentiable.

example 3 Show that f (z)= z Im (z) is differentiable only at z = 0 and f¢(0) = 0. Solution f ( z ) = z Im ( z ) = ( x + iy) Im ( x + iy) = ( x + iy) y For any z, f ( z + Dz ) - f ( z ) Dz ( z + Dz ) Im ( z + Dz ) - z Im ( z ) = lim Dz Æ 0 Dz ( x + iy + Dx + i Dy)( y + Dy) - ( x + iy) y = lim Dz Æ 0 Dx + i Dy ( x + iy) y + ( x + iy) Dy + ( Dx + i Dy)( y + Dy) - ( x + iy) y = lim Dz Æ 0 Dx + i Dy

f ¢( z ) = lim

Dz Æ 0

2.6

= lim

Dz Æ 0

Differentiability

2.17

( x + iy)Dy + ( Dx + i Dy)( y + Dy) Dx + i Dy

For f¢(z) to exist, Dz may approach zero along any path. (i) Let the path be parallel to the x-axis, i.e., Dy = 0, y Dx f ¢( z ) = lim Dx Æ 0 Dx =y (ii) Let the path be parallel to the y-axis, i.e., Dx = 0, ( x + iy) Dy + i Dy ( y + Dy) f ¢( z ) = lim Dy Æ 0 i Dy x + iy + i( y + Dy) = lim Dy Æ 0 i x + 2iy = i = -ix + 2 y Since f ¢(z) is different along different paths, f (z) is not differentiable. At z = 0, x = 0, y = 0 \ f ¢(z) = 0 at z = 0 along all paths. Hence, f (z) is differentiable only at z = 0.

example 4 Show that f (z) = |z|2 is continuous but not differentiable, at each point in the plane. [Winter 2014] Solution Let

z = x + iy f (z) = z 2 = x2 + y2 2 z , being a polynomial function, is continuous everywhere in the z-plane. For any z, f ( z + Dz ) - f ( z ) f ¢( z ) = lim Dz Æ 0 Dz

||

||

2

= lim

Dz Æ 0

z + Dz - z

2

Dz

= lim

( x + Dx )2 + ( y + Dy)2 - ( x 2 + y 2 ) Dx + i Dy

= lim

x 2 + ( Dx )2 + 2 x Dx + y 2 + ( Dy)2 + 2 y Dy - x 2 - y 2 Dx + i Dy

= lim

( Dx )2 + 2 x Dx + ( Dy)2 + 2 y Dy Dx + i Dy

Dz Æ 0

Dz Æ 0

Dz Æ 0

2.18

Chapter 2 Analytic Functions

For f¢(z) to exist, Dz may approach zero along any path. (i) Let the path be parallel to the x-axis, i.e., Dy = 0, ( Dx )2 + 2 x Dx f ¢( z ) = lim Dx Æ 0 Dx = lim ( Dx + 2 x ) Dx Æ 0

= 2x (ii) Let the path be parallel to the y-axis, i.e., Dx = 0, ( Dy)2 + 2 y Dy f ¢( z ) = lim Dy Æ 0 i Dy Dy + 2 y = lim Dy Æ 0 i 2y = i = -2iy Since f¢(z) is different along different paths, f (z) is not differentiable.

exercIse 2.2 limits 1. Use the definition of limit to prove that (i)

lim zÆ0

z2 =0 z

(ii)

lim z Æi

z -i 1 = 2 z + 1 2i

2. Show that the limit of the function f(z) as z tends to zero does not exist. (i)

Ê zˆ f (z) = Á ˜ Ë z¯

2

Ê 3. Prove that lim Á z + zÆi Ë

(ii)

f (z) = (z )2 , = 0,

zπ0 z=0

1ˆ ˜ = 0. z¯

continuity 1. Show that the function f(z) is continuous of z = i. z2 + 1 , z +i = 0,

f (z) =

zπi z=i

2. Discuss the continuity of the function f(z) = (x + y2) + i xy in the complex plane. ÈÎans.: Continuous everwhere ˘˚

2.7 Analytic Functions

2.19

3. Show that the function f(z) is not continuous at z = i. f (z) = z 2 , = 0,

zπi z =i

4. Determine whether the function f(z) is continuous at the origin. If not then redefine it to make it continuous. z Re(z) , |z| = 2,

f (z) =

zπ0 z=0 ÈÎans.: Not continuous, define f (0) = 0 ˘˚

differentiability 1. Show that the functions — z , Re(z) and Im(z) are not differentiable in the complex plane. 2. Find the derivative of f (z) =

2z - i at z = —i using definition. z + zi 1+ i ˘ È Í ans.: 2 ˙ Î ˚

3. Show that the function f(z) = z2 — z is not differentiable in the complex plane. 4. Determine whether the function f(z) is differentiable at the origin. x 3 y(y - ix) , x6 + y 2 = 0,

f (z) =

zπ0 z=0

ÎÈans.: No ˚˘

2.7 analytIc FunctIons A function f(z) is said to be an analytic function if it is defined and differentiable at each point of a region R. This function is also known as a regular or holomorphic function. entire Function A function f(z) is said to be an entire function if it is analytic everywhere in the finite plane (complex plane). An entire function is not analytic at z = •, e.g., cos z, sin z, ez and polynomial functions, etc.

2.20

2.7.1

Chapter 2 Analytic Functions

necessary conditions for f(z) to be analytic (cauchy—riemann equations)

The necessary conditions for a function f (z) = u + iv to be an analytic function at all the points in a region R are (i)

∂u ∂v = ∂x ∂y

(ii)

∂u ∂v =∂y ∂x

Proof Let f (z) = u + iv be an analytic function and Du, Dv be the increments in u and v corresponding to increments Dx, Dy in x and y respectively. f ( z + Dz ) - f ( z ) Dz Æ 0 Dz [(u + Du) - i(v + Dv)] - (u + iv) = lim Dz Æ 0 Dz Du + i Dv … (2.1) = lim Dz Æ 0 Dz Since f (z) is differentiable at each point of the region R, Dz may approach zero along any path. Consider two paths as follows: f ′( z ) = lim

(i) parallel to x-axis

Dy = 0 Dz = Dx + i Dy = Dx

Substituting in Eq. (2.1), Du + i Dv Dx Æ 0 Dx Du Dv = lim + i lim Dx Æ 0 Dx Dx Æ 0 Dx ∂v ∂u f ′( z ) = +i ∂x ∂x

f ′( z ) = lim

(ii) parallel to y-axis

…(2.2)

Dx = 0 Dz = Dx + i Dy = i Dy

Substituting in Eq. (2.1), Du + i Dv Dy Æ 0 i Dy Du Dv = lim + lim Dy Æ 0 i Dy Dy Æ 0 Dy ∂u ∂v + f ′ ( z ) = -i ∂y ∂y f ′( z ) = lim

…(2.3)

2.8

Cauchy—Riemann Equations in Polar Form

2.21

Since f (z) is differentiable, f ¢(z) must be unique. Equating Eqs (2.2) and (2.3), ∂u ∂v ∂v ∂u +i = -i ∂x ∂x ∂y ∂y Comparing real and imaginary parts, ∂u ∂v ∂u ∂v = , =∂x ∂y ∂y ∂x These equations are known as Cauchy–Riemann (C–R) equations.

2.7.2  Sufficient Conditions for f(z) to be analytic For a function f (z) = u + iv, if the partial derivatives in the region R and

2.8

∂u ∂u ∂v ∂v , , and are continuous ∂x ∂y ∂x ∂y

∂u ∂v ∂u ∂v = , =then the function of f (z) is analytic. ∂x ∂y ∂y ∂x

cauchy—rIemann equatIons In polar Form

If f (z) = u + iv is an analytic function where u and v are functions of r ,q and z = reiq then ∂u 1 ∂v = , ∂r r ∂q Proof

∂u ∂v = -r ∂q ∂r

f ( z ) = u + iv, z = reiq u + iv = f (reiq )

...(2.4)

Differentiating Eq. (2.4) partially w.r.t. r, ∂u ∂v +i = f ¢(reiq ) ◊ eiq ∂r ∂r = f ¢( z ) ◊ eiq f ¢( z ) =

∂v ˆ 1 Ê ∂u +i ˜ iq Á Ë ∂ ∂r ¯ r e

...(2.5)

Differentiating Eq. (2.4) partially w.r.t. q, ∂u ∂v +i ∂q ∂q

= f ¢(reiq ) ◊ ireiq = f ¢( z ) ◊ ireiq

f ¢( z ) =

∂v ˆ 1 Ê ∂u +i ˜ iq Á Ë ∂ q ∂ q¯ ire

Since f (z) is differentiable, f ¢(z) must be unique.

...(2.6)

2.22

Chapter 2 Analytic Functions

Equating Eqs (2.5) and (2.6), ∂v ˆ 1 Ê ∂u 1 Ê ∂u ∂v ˆ + i ˜ = iq Á +i ˜ iq Á Ë ¯ Ë ∂ ∂ q q¯ r r ∂ ∂ e ire u 1 ∂v i ∂u ∂u ∂v + +i =r ∂q r ∂q ∂r ∂r Comparing real and imaginary parts, ∂u 1 ∂v = , ∂r r ∂q ∂u 1 ∂v = , ∂r r ∂q

∂v 1 ∂u =∂r r ∂q ∂u ∂v = -r ∂q ∂r

example 1 Find the real and imaginary part of f (z) = z2 + 3z. Solution f ( z ) = z 2 + 3z = ( x + iy)2 + 3( x + iy) = ( x 2 + i 2 y 2 + 2ixy) + 3( x + iy) = ( x 2 - y 2 + 3xx ) + i(2 xy + 3 y) Real part = x2 – y2 + 3x Imaginary part = 2xy + 3y

example 2 Find the real and imaginary parts of f ( z ) = 2 z 3 - 3z. Solution f ( z ) = 2 z 3 - 3z = 2( x + iy)3 - 3( x + iy) = 2[ x 3 + i 3 y3 + 3 x 2 ◊ iy + 3 x(iy)2 ] - 3( x + iy) = 2[ x 3 - iy3 + i ◊ 3 x 2 y - 3 xy 2 ] - 3( x + iy) = (2 x 3 - 6 xy 2 - 3 x ) + i(6 x 2 y - 2 y3 - 3 y) Real part = 2 x 3 - 6 xy 2 - 3 x Imaginary part = 6 x 2 y - 2 y3 - 3 y

[Summer 2013]

2.8

Cauchy—Riemann Equations in Polar Form

2.23

example 3 Write the function f ( z ) = z +

1 in f (z) = u(r, q) + iv (r, q) form. z [Winter 2013]

Solution f (z) = z +

1 z

Let z = reiq f ( z ) = reiq +

1

reiq 1 = reiq + e - iq r 1 = r (cos q + i sin q ) + (cos q - i sin q ) r 1ˆ 1ˆ Ê Ê = Á r + ˜ cos q + i Á r - ˜ sin q Ë ¯ Ë r r¯

example 4 Determine whether the function 2 xy + i( x 2 - y 2 ) is analytic or not. Solution Let

f ( z ) = 2 xy + i( x 2 - y 2 ) u + iv = 2 xy + i( x 2 - y 2 )

Comparing real and imaginary parts, u = 2 xy, ∂u = 2 y, ∂x ∂u = 2 x, ∂y ∂u ∂v \ π , ∂x ∂y

v = x 2 - y2 ∂v = 2x ∂x ∂v = -2 y ∂y ∂v ∂u π∂x ∂y

C–R equations are not satisfied. Hence, f (z) is not analytic.

2.24

Chapter 2 Analytic Functions

example 5 Show that f (z) = z3 is analytic everywhere. Solution f (z) = z3 u + iv = ( x + iy)3 = x 3 + (iy)3 + 3 x 2 iy + 3 x (iy)2 = x 3 - 3 xy 2 + i(3 x 2 y - y3 ) Comparing real and imaginary parts, u = x 3 - 3 xy 2 , ∂u = 3 x 2 - 3 y2 , ∂x ∂u = -6 xy, ∂y ∂u ∂v \ = , ∂x ∂y

v = 3 x 2 y - y3 ∂v = 6 xy ∂x ∂v = 3 x 2 - 3 y2 ∂y ∂u ∂v =∂y ∂x

C–R equations are satisfied. Hence, z3 is analytic everywhere.

example 6 Test the analyticity of the function w = sin z. Solution w = sin z u + iv = sin( x + iy) = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sin nh y Comparing real and imaginary parts, u = sin x cosh y, ∂u = cos x cosh y, ∂x ∂u = sin x sinh y, ∂y ∂u ∂v \ = , ∂x ∂y C–R equations are satisfied. Hence, sin z is analytic.

v = cos x sinh y ∂v = - sin x sinh y ∂x ∂v = cos x cosh y ∂y ∂u ∂v =∂y ∂x

[Summer 2014]

2.8

Cauchy—Riemann Equations in Polar Form

2.25

example 7 Are |z|, Re (z), Im(z) analytic? Give reasons. Solution z = x + iy

Let

z = x 2 + y2 Re( z ) = x Im( z ) = y Since all the above functions are real, v = 0. ∂v ∂v \ = 0, = 0 for all functions. ∂x ∂y ∂u ∂u and are not zero. Thus, C–R equations are not satisfied for all the ∂x ∂y functions. Hence, z , Re(z), Im(z) are not analytic. But

||

example 8 State the basic difference between the limit of a function of a real variable and that of a complex variable. Solution In real calculus, x approaches x0 only along the line called real line; whereas in case of the complex variable z, z approaches z0 from any direction in the z-plane.

example 9 Show that the function f ( z ) = z is nowhere differentiable. [Summer 2014] Solution f (z) = z u + iv = x - iy Comparing real and imaginary parts, u = x, v = -y ∂u ∂v = 1, =0 ∂x ∂x ∂u ∂v = 0, = -1 ∂y ∂y ∂u ∂v \ π ∂y ∂y

2.26

Chapter 2 Analytic Functions

C–R equations are not satisfied. Hence, f (z) is nowhere differentiable.

example 10 Discuss the differentiability of f (z) = x2 + iy2.

[Summer 2015]

Solution f ( z ) = x 2 + iy 2 u + iv = x 2 + iy 2 Comparing real and imaginary parts, u = x2 , ∂u = 2 x, ∂x ∂u = 0, ∂y

v = y2 ∂v =0 ∂x ∂v = 2y ∂y

∂u ∂v except at the origin, i.e., at x = 0, y = 0. π ∂x ∂y Since C–R equations are satisfied only at the origin, f (z) is differentiable only at z = 0.

example 11 Show that f (z) = z|z| is not analytic anywhere. Solution f (z) = z | z | u + iv = ( x + iy) x 2 + y 2 = x x 2 + y 2 + iy x 2 + y 2 Comparing real and imaginary parts, u = x x 2 + y2 , ∂u 1 = x 2 + y2 + x ◊ ◊ 2 x, 2 ∂x 2 x + y2 =

2 x 2 + y2 x +y 2

2

v = y x 2 + y2 ∂v 1 = y◊ ◊ 2x 2 ∂x 2 x + y2 =

xy x + y2 2

2.8

Cauchy—Riemann Equations in Polar Form

∂u 1 = x◊ ◊ 2 y, 2 ∂y 2 x + y2 = \

∂v 1 = x 2 + y2 + y ◊ ◊ 2y 2 ∂y 2 x + y2

xy

=

x 2 + y2

∂u ∂v π , ∂x ∂y

x 2 + 2 y2 x 2 + y2

∂u ∂v π∂y ∂x

C–R equations are not satisfied. Hence, f (z) is not analytic anywhere.

example 12 Find the analytic region of f ( z ) = ( x - y)2 + 2i( x + y). Solution f ( z ) = ( x - y ) 2 + 2i ( x + y ) u + iv = ( x - y)2 + 2i( x + y) Comparing real and imaginary parts, u = ( x - y )2 , ∂u = 2( x - y), ∂x ∂u = -2( x - y), ∂y

v = 2( x + y ) ∂v =2 ∂x ∂v =2 ∂y

All the partial derivatives are continuous. For f(z) to be analytic, u and v should satisfy C–R equations. ∂u ∂v = , ∂x ∂y 2( x - y) = 2, x - y = 1,

2.27

∂u ∂v =∂y ∂x - 2( x - y) = -2 x- y =1

f (z) is analytic at all points satisfying x – y = 1. Hence, the analytic region is the line x – y = 1.

2.28

Chapter 2 Analytic Functions

example 13 [Winter 2014]

Examine the analyticity of sinh z. Solution Let

w = sinh z u + iv = -i sin i z = -i sin i ( x + iy) = -i sin (ix - y) = -i(sin ix cos y - cos ix sin y) = -i (i sinh x cos y - cosh x sin y) = sinh x cos y + i cosh x sin y

Comparing real and imaginary parts, u = sinh x cos y, ∂u = cosh x cos y, ∂x ∂u = - sinh x sin y, ∂y ∂u ∂v \ = , ∂x ∂y

v = cosh x sin y ∂v = sinh x sin y ∂x ∂v = cosh x cos y ∂y ∂u ∂v =∂y ∂x

C–R equations are satisfied. Hence, sinh z is analytic.

example 14 Show that w = ez is analytic everywhere in the complex plane. Hence, dw find . dz Solution w = ez u + iv = e x + iy = e x eiy = e x (cos y + i sin y) = e x cos y + ie x sin y

2.8

Cauchy—Riemann Equations in Polar Form

2.29

Comparing real and imaginary parts, u = e x cos y, ∂u = e x cos y, ∂x ∂u = -e x sin y, ∂y ∂u ∂v \ = , ∂x ∂y

v = e x sin y ∂v = e x sin y ∂x ∂v = e x cos y ∂y ∂u ∂v =∂y ∂x

C–R equations are satisfied. Also ex, cos y and sin y are continuous for all values of x and y. Hence, ez is analytic everywhere in the complex plane. Since w = u + iv is analytic everywhere, dw ∂u ∂v = +i dz ∂x ∂x = e x cos y + ie x sin y = e x (cos y + i sin y) = e x eiy = e x + iy = ez

example 15 Prove that cosh z is an analytic function and, hence, find its derivative. Solution f ( z ) = cosh z = cos iz u + iv = cos i( x + iy) = cos(ix - y) = cos ix cos y + sin ix sin y = cosh x cos y + i sinh x sin y Comparing real and imaginary parts, u = cosh x cos y, v = sinh x sin y ∂u ∂v = sinh x cos y, = cosh x sin y ∂x ∂x ∂v ∂u = sinh x cos y = - cosh x sin y, ∂y ∂y ∂u ∂v ∂u ∂v \ = , =∂x ∂y ∂y ∂x

2.30

Chapter 2 Analytic Functions

C–R equations are satisfied. Hence, cosh z is an analytic function. ∂u ∂v +i ∂x ∂x = sinh x cos y + i cosh x sin y = -i sin ix cos y + i cos ix sin y = -i(sin ix cos y - cos ix sin y) = -i sin(ix - y)

f ′( z ) =

= -i sin(ix + i 2 y) = -i sin i( x + iy) = -i 2 sinh( x + iy) = sinh z

example 16 Find the constants a, b, c if f ( z ) = x + ay + i(bx + cy) is analytic. Solution f ( z ) = ( x + ay) + i(bx + cy) u + iv = ( x + ay) + i(bx + cy) Comparing real and imaginary parts, u = x + ay, ∂u = 1, ∂x ∂u = a, ∂y

v = bx + cy ∂v =b ∂x ∂v =c ∂y

Since f (z) is analytic, C–R equations are satisfied. ∂u ∂v = , ∂x ∂y

∂u ∂v =∂y ∂x

1 = c, c =1

a = -b

example 17 Find the constants a, b, c, d so that the function f ( z ) = x 2 + axy + by 2 + i(cx 2 + dxy + y 2 ) is analytic.

2.8

Cauchy—Riemann Equations in Polar Form

2.31

Solution f ( z ) = x 2 + axy + by 2 + i(cx 2 + dxy + y 2 ) u + iv = x 2 + axy + by 2 + i(cx 2 + dxy + y 2 ) Comparing real and imaginary parts, u = x 2 + axy + by 2 , v = cx 2 + dxy + y 2 ∂u ∂v = 2 x + ay, = 2cx + dy ∂x ∂x ∂u ∂v = ax + 2by, = dx + 2 y ∂y ∂y Since f(z) is analytic, C–R equations are satisfied. ∂u ∂v ∂u ∂v = , =∂x ∂y ∂y ∂x 2 x + ay = dx + 2 y … (1), ax + 2by = -(2cx + dy) … (2) Comparing coefficients of x and y in Eqs (1) and (2), d = 2, a=2 a = -2c, 2b = - d c = -1,

b = -1

example 18 Find the values of a and b such that the function f ( z ) = x 2 + ay 2 - 2 xy + i(bx 2 - y 2 + 2 xy) is analytic. Also, find f¢(z). Solution f ( z ) = ( x 2 + ay 2 - 2 xy) + i(bx 2 - y 2 + 2 xy) u + iv = ( x 2 + ay 2 - 2 xy) + i(bx 2 - y 2 + 2 xy) Comparing real and imaginary parts, u = x 2 + ay 2 - 2 xy, v = bx 2 - y 2 + 2 xy ∂u ∂v = 2 x - 2 y, = 2bx + 2 y ∂x ∂x ∂u ∂v = 2 ay - 2 x, = -2 y + 2 x ∂y ∂y Since f(z) is analytic, C–R equations are satisfied. ∂u ∂v ∂u ∂v = , =∂x ∂y ∂y ∂x 2 x - 2 y = -2 y + 2 x … (1), 2 ay - 2 x = -2bx - 2 y … (2)

2.32

Chapter 2 Analytic Functions

Comparing coefficients of x and y in Eqs (1) and (2), 2 a = -2, a = -1,

- 2 = -2b b =1

∂u ∂v +i ∂x ∂x = (2 x - 2 y) + i(2bx + 2 y) = (2 x - 2 y ) + i (2 x + 2 y )

f ′( z ) =

= 2( x + i y + ix + iy) = 2[ x(1 + i ) + iy(1 + i )] = 2(1 + i )( x + iy) = 2(1 + i )z 2

[∵ b = 1] [∵ i 2 = -1]

example 19 If u + iv is analytic, show that v – iu and –v + iu are also analytic. Solution Since u + iv is analytic, C–R equations are satisfied. ∂u ∂v = ∂x ∂y ∂u ∂v =∂y ∂x

… (2 )

(i) C–R equations for v – iu are ∂v ∂(-u) ∂u = =- , ∂x ∂y ∂y

∂v ∂(-u) ∂u == ∂y ∂x ∂x

Equations (1) and (2) are satisfied. Hence, v – iu is analytic. (ii) C–R equations for –v + iu are ∂(- v) ∂u = , ∂x ∂y ∂v ∂u = , ∂x ∂y

… (1)

∂(- v) ∂u =∂y ∂x ∂v ∂u = ∂y ∂x

Equations (1) and (2) are satisfied. Hence, –v + iu is also analytic.

2.8

2.33

Cauchy—Riemann Equations in Polar Form

example 20 Show that an analytic function with constant imaginary part is constant. Solution f (z) = u + iv is an analytic function. ∂u ∂v ∂u ∂v = , =∂x ∂y ∂y ∂x Imaginary part is constant. v=c ∂v ∂v = 0, =0 ∂x ∂y But Now,

Hence,

∂u ∂v = =0 ∂x ∂y ∂u ∂v +i ∂x ∂x = 0 + i ◊0 =0 f (z) = constant

f ′( z ) =

example 21 If f ( z ) and f ( z ) are both analytic then show that f (z) is constant. Solution Let

f ( z ) = u + iv f ( z ) = u - iv

f (z) is analytic. ∂u ∂v = ∂x ∂y ∂u ∂v =∂y ∂x

… (1) … (2 )

f ( z ) is analytic. ∂u ∂(- v) = ∂x ∂y ∂v =∂y ∂u ∂(- v) =∂y ∂x ∂v = ∂x

… (3)

… ( 4)

2.34

Chapter 2 Analytic Functions

Adding Eqs (1) and (3), ∂u 2 =0 ∂x ∂u =0 ∂x Subtracting Eq. (2) from Eq. (4), ∂v 0=2 ∂x ∂v =0 ∂x ∂u ∂v f ′( z ) = +i ∂x ∂x = 0 + i ◊0 =0 Hence, f (z) = constant

example 22 If f(z) is an analytic function with constant modulus, prove that f(z) is constant. Solution f ( z ) = constant u2 + v 2 = constant u2 + v 2 = constant Differentiating Eq. (1) partially w.r.t x, 2u

… (1)

∂u ∂v + 2v =0 ∂x ∂x ∂u ∂v u +v =0 ∂x ∂x

…(2)

Differentiating Eq. (1) partially w.r.t y, ∂u ∂v 2u + 2v =0 ∂y ∂y ∂u ∂v u +v =0 ∂y ∂y

…(3)

Since f (z) = u + iv is analytic, C–R equations are satisfied. ∂u ∂v = , ∂x ∂y

∂u ∂v =∂y ∂x

2.8

2.35

Cauchy—Riemann Equations in Polar Form

Substituting in Eq. (3), ∂u Ê ∂v ˆ uÁ- ˜ + v =0 Ë ∂x ¯ ∂x ∂u ∂v -u =0 ∂x ∂x Multiplying Eq. (2) by u and Eq. (4) by v and adding,

…(4)

v

∂u =0 ∂x ∂u =0 ∂x Multiplying Eq. (2) by v and Eq. (4) by u and subtracting, (u2 + v 2 )

∂v =0 ∂x ∂v =0 ∂x ∂u ∂v f ′( z ) = +i ∂x ∂x = 0 + i ◊0 =0 f (z) = constant

(v 2 + u2 )

Hence,

example 23 Show that every analytic function w = u + iv can be expressed as a function of z alone, not as a function of z–. Solution

z = x + iy, z = x - iy 1 1 x = ( z + z ), y = (z - z ) 2 2i – \ u and v can be considered as functions of z and z – (Fig. 2.13). Hence, w also depends on z and z . w = u + iv ∂w ∂u ∂v = +i ∂z ∂z ∂z Ê ∂u ∂x ∂u ∂y ˆ Ê ∂v ∂x ∂ v ∂ y ˆ =Á ◊ + ◊ + ◊ +i ◊ Ë ∂x ∂z ∂y ∂z ˜¯ ÁË ∂x ∂z ∂y ∂z ˜¯ È ∂u 1 ∂u Ê 1 ˆ ˘ È ∂v 1 ∂v Ê 1 ˆ ˘ = Í ◊ + Á- ˜˙ +iÍ ◊ + Á- ˜˙ Î ∂x 2 ∂y Ë 2i ¯ ˚ Î ∂x 2 ∂y Ë 2i ¯ ˚

v

u

y

x

y

x

z

z

Fig. 2.13

2.36

Chapter 2 Analytic Functions

=

1 ∂u i ∂u i ∂v 1 ∂v + + 2 ∂x 2 ∂y 2 ∂x 2 ∂y

È 1 ∂u i ∂u i ∂u 1 ∂u ˘ [ Using C- R equations ] =Í + ˙ Î 2 ∂x 2 ∂y 2 ∂y 2 ∂x ˚ =0 – \ w does not depend on z . Hence, w can be expressed as a function of z alone, not as a function of z–.

example 24 If w = f (z) is analytic, prove that

∂w dw ∂w , where z = x + iy. = -i = ∂y dz ∂x

Solution Let

w = u + iv

Since w is an analytic function, ∂u ∂v ∂u ∂v = , =∂x ∂y ∂y ∂x w = f ( z ) = u + iv dw ∂u ∂v = f ′( z ) = +i dz ∂x ∂x =

∂v Ê ∂u ˆ +i ∂y ÁË ∂y ˜¯

[ Using C- R equations ]

Ê 1 ∂v ∂u ˆ = iÁ Ë i ∂y ∂y ˜¯ Ê ∂v ∂ u ˆ = i Á -i - ˜ Ë ∂y ∂y ¯ Ê ∂u ∂v ˆ = -i Á +i ˜ ∂y ¯ Ë ∂y ∂ (u + iv) ∂y ∂w = -i ∂y dw ∂u ∂v = f ¢( z ) = +i dz ∂x ∂x ∂ = (u + iv) ∂x ∂w = ∂x = -i

Again,

... (1)

... (2)

2.8

Cauchy—Riemann Equations in Polar Form

2.37

From Eqs (1) and (2), dw ∂w ∂w = = -i dz ∂x ∂y

example 25 Show that f ( z ) = xy is not differentiable at origin although C–R equations are satisfied. [Summer 2015] Solution f ( z) =

xy

u + iv =

xy

Comparing real and imaginary parts, u=

At the origin,

At the origin,

xy ,

v=0

∂u u( x + Dx, y) - u( x, y) = lim ∂x Dx Æ0 Dx ∂u u( Dx, 0) - u(0, 0) = lim ∂x Dx Æ0 Dx 0-0 = lim Dx Æ 0 Dx =0 ∂u u( x, y + Dy) - u( x, y) = lim Dy ∂y DyÆ0 ∂u u(0, Dy) - u(0, 0) = lim ∂y DyÆ0 Dy 0-0 = lim Dy Æ 0 Dy =0 ∵ v=0 ∂v = 0, ∂x ∂u ∂v \ = , ∂x ∂y

∂v =0 ∂y ∂u ∂v =∂y ∂x

Hence, C–R equations are satisfied at the origin.

2.38

Now,

Chapter 2 Analytic Functions

f ′(0) = lim

zÆ0

= lim

f ( z ) - f (0 ) z-0 xy - 0

z Æ 0 x + iy Let z Æ 0 along the line y = mx.

f ′(0) = lim

x Æ0

= lim

x Æ0

= lim

x Æ0

x.mx x + imx x m x(1 + im) m 1 + im

Since limit depends on m (i.e., on the path), it is not unique. Therefore, f¢(0) does not exist. Hence, f (z) is not analytic at the origin.

example 26 -v For what values of z does the function w defined by z = e (cos u + i sin u ) cease to be analytic? Solution

z = e - v (cos u + i sin u) = e - v eiu 2

= ei v eiu i iv + u =e( ) = eiw log z = log eiw = iw 1 w = log z i = -i log z dw i =dz z w ceases to be analytic, where

dw dw does not exist. i.e., Æ •. dz dz

2.8

At z = 0

Cauchy—Riemann Equations in Polar Form

dw Æ •. dz

Hence, w is not analytic at z = 0.

example 27 Prove that f ( z ) =

x 3 (1 + i ) - y3 (1 - i ) x 2 + y2

= 0,

, zπ0 z=0

satisfy C–R equations at the origin but f¢(0) does not exist. Solution f (z) = and

x 3 (1 + i ) - y3 (1 - i ) x 2 + y2

, zπ0

f (0 ) = 0 Ê x 3 - y3 ˆ Ê x 3 + y3 ˆ u + iv = Á 2 ˜ +iÁ ˜ Ë x + y2 ¯ Ë x 2 + y2 ¯

Comparing real and imaginary parts, u=

x 3 - y3 x 2 + y2

,

v=

x 3 + y3 x 2 + y2

∂u u( x + Dx, y) - u( x, y) = lim ∂x Dx Æ0 Dx At the origin,

At the origin,

u( Dx, 0) - u(0, 0) ∂u = lim ∂x Dx Æ0 Dx Dx - 0 = lim Dx Æ 0 Dx =1 u( x, y + Dy) - u( x, y) ∂u = lim ∂y DyÆ0 Dy ∂u u(0, Dy) - u(0, 0) = lim ∂y DyÆ0 Dy -( Dy) - 0 = lim Dy Æ 0 Dy = -1 ∂v v( x + Dx, y) - v( x, y) = lim ∂x Dx Æ0 Dx

2.39

2.40

Chapter 2 Analytic Functions

At the origin,

At the origin,

\

v( Dx, 0) - v(0, 0) ∂v = lim ∂x Dx Æ0 Dx Dx - 0 = lim Dx Æ 0 Dx =1 ∂v v( x, y + Dy) - v( x, y) = lim D y Æ 0 ∂y Dy ∂v v(0, Dy) - v(0, 0) = lim ∂y DyÆ0 Dy Dy = lim x Æ 0 Dy =1 ∂u ∂v = , ∂x ∂y

∂u ∂v =∂y ∂x

Hence, C–R equations are satisfied at the origin. Now,

f ′(0) = lim

f ( z ) - f (0 ) z-0

zÆ0

( x - y 3 ) + i( x 3 + y 3 ) -0 ( x 2 + y2 ) = lim zÆ0 x + iy Let z Æ 0 along the line y = mx. 3

f ′(0) = lim

zÆ0

= lim

x Æ0

( x 3 - m 3 x 3 ) + i( x 3 + m 3 x 3 ) ( x 2 + m 2 x 2 )( x + imx ) (1 - m3 ) + i(1 + m3 ) (1 + m 2 )(1 + im)

Since limit depends on m (i.e., on the path), it is not unique. Therefore, f¢(0) does not exist. Hence, f (z) is not analytic at the origin.

example 28 z2 , zπ0 z = 0, z=0

Show that the function f ( z ) =

2.8

Cauchy—Riemann Equations in Polar Form

2.41

satisfies the Cauchy–Riemann equations at the origin, but f¢(0) fails to exist. [Winter 2012] Solution f (z) = u + iv = = = =

z2 , zπ0 z ( x - iy)2 ( x - iy) ¥ ( x + iy) ( x - iy) ( x 2 - y 2 - 2ixy)( x - iy) x 2 + y2 ( x 3 - xy 2 - 2 xy 2 ) - i( x 2 y - y3 + 2 x 2 y) x 2 + y2 Ê y3 - 3 x 2 y ˆ + i Á 2 ˜ ( x 2 + y2 ) Ë x + y2 ¯

( x 3 - 3 xy 2 )

Comparing real and imaginary parts, u=

x 3 - 3 xy 2 x 2 + y2

,

v=

y3 - 3 x 2 y x 2 + y2

Also, f (z) = 0 when z = 0, i.e., u = 0, v = 0 at the origin. ∂u u( x + Dx, y) - u( x, y) = lim ∂x Dx Æ0 Dx At the origin,

At the origin,

u( Dx, 0) - u(0, 0) ∂u = lim Dx ∂x Dx Æ0 Dx - 0 = lim Dx Æ 0 Dx =1 ∂u u( x, y + Dy) - u( x, y) = lim ∂y DyÆ0 Dy u(0, Dy) - u(0, 0) ∂u = lim Dy ∂y DyÆ0 0-0 = lim Dy Æ 0 Dy =0 v( x + Dx, y) - v( x, y) ∂v = lim ∂x Dx Æ0 Dx

2.42

Chapter 2 Analytic Functions

v( Dx, 0) - v(0, 0) ∂v = lim Dx ∂x Dx Æ0 0-0 = lim Dx Æ 0 Dx =0 v( x, y + Dy) - v( x, y) ∂v = lim ∂y DyÆ0 Dy

At the origin,

v(0, Dy) - v(0, 0) ∂v = lim D y Æ 0 Dy ∂y Dy - 0 = lim Dy Æ 0 Dy =1

At the origin,

\

∂u ∂v = = 1, ∂x ∂y

∂u ∂v ==0 ∂y ∂x

Hence, C–R equations are satisfied at the origin. f ( z ) - f (0 ) Now, f ¢(0) = lim zÆ0 z-0 ( x 3 - 3 xy 2 ) + i( y3 - 3 x 2 y) x 2 + y2 = lim zÆ0 x + iy Let z Æ 0 along the line y = mx. f ¢(0) = lim

x Æ0

= lim

x Æ0

=

( x 3 - 3m 2 x 3 ) + i(m3 x 3 - 3mx 3 ) ( x 2 + m 2 x 2 )( x + imx ) (1 - 3m 2 ) + i(m3 - 3m) (1 + m 2 )(1 + im)

(1 - 3m 2 ) + i(m3 - 3m) (1 + m 2 )(1 + im)

Since the limit depends on m (i.e., on the path), it is not unique. Hence, f¢(0) does not exist.

example 29 Find p such that the function f ( z ) = r 2 cos 2q + ir 2 sin pq is analytic. Solution f ( z ) = r 2 cos 2q + ir 2 sin pq u + iv = r 2 cos 2q + ir 2 sin pq

2.8

Cauchy—Riemann Equations in Polar Form

2.43

Comparing real and imaginary parts, u = r 2 cos 2q , v = r 2 sin pq Since f (z) is analytic, C–R equations are satisfied. ∂u 1 ∂v = , ∂r r ∂q From the second equation,

∂u ∂v = -r ∂q ∂r

r 2 (-2 sin 2q ) = -r (2r sin pq ) sin 2q = sin pq On comparing, p=2

example 30 Show that f ( z ) =

1 is analytic everywhere except at z = 0 and find z

f¢(z). Solution Let

z = reiq 1 f (z) = z 1 u + iv = iq re 1 = e - iq r 1 = (cos q - i sin q ) r

1 1 u = cos q , v = - sin q r r ∂u 1 ∂v 1 = - 2 cos q , = sin q ∂r ∂r r 2 r ∂u 1 ∂v 1 = - sin q , = - cos q ∂q r ∂q r ∂u 1 ∂v ∂u ∂v \ = , = -r ∂r r ∂q ∂q ∂r C–R equations are satisfied. All the derivatives are continuous everywhere except at r = 0 (i.e., z = 0). Hence, f (z) is analytic everywhere except at z = 0. Now, f (z) = u + iv

2.44

Chapter 2 Analytic Functions

Differentiating w.r.t. r, ∂z ∂u ∂v = +i ∂r ∂r ∂r 1 1 f ′( z ) ◊ eiq = - 2 cos q + i 2 sin q r r (cos q - i sin q ) =r2 f ′( z ) ◊

f ′( z ) =

-eiq

r 2 eiq 1 = - 2 2iq r e 1 = - iq 2 (re ) 1 =- 2 z

example 31 Show that f (z) = log z is analytic everywhere except at the origin and find its derivative. Solution Let

z = reiq f ( z ) = log z u + iv = log(reiq ) = log r + log eiq = log r + iq

Comparing real and imaginary parts, u = log r , v =q ∂u 1 ∂v = , =0 ∂r r ∂r ∂u ∂v = 0, =1 ∂q ∂q ∂u 1 ∂v ∂u ∂v \ = , = -r ∂r ∂r r ∂q ∂q C–R equations are satisfied.

2.8

Cauchy—Riemann Equations in Polar Form

2.45

∂u 1 = is not continuous at r = 0 (i.e., z = 0). ∂r r Hence, log z is analytic everywhere except at the origin (z = 0). Now, f (z) = u + iv But

Differentiating w.r.t. r, ∂z ∂u ∂v = +i ∂r ∂r ∂r 1 f ′( z ) ◊ eiq = + i(0) r f ′( z ) ◊

f ′( z ) =

1

reiq 1 = z

exercIse 2.3 1. Show that an analytic function with a constant real part is constant. 2. Determine which of the following functions are analytic. (i) xy + iy

(ii) ex(cos y – i sin y)

(iii) z2 + z

– (v) z + 2z

(iv) z3

[ans.: (i) not analytic (ii) not analytic (iii) analytic (iv) analytic (v) not analytic] 3. Find the constants a, b, c, d, and e if the function f (z) = (ax 4 + bx 2 y 2 + cy 4 + dx 2 - 2y 2 ) + i(4 x 3 y - exy 3 + 4 xy ) is analytic. [ans.: a = 1, b = -6, c = 1, d = 2, e = 4 ] 4. Determine p such that the function Ê px ˆ 1 f (z) = log(x 2 + y 2 ) + tan-1 Á ˜ is analytic. Ë y ¯ 2 [ans.: p = –1] 5. For what values of z does the function cease to be analytic? (i)

z2 - 4 z2 + 1

(ii)

z z2 - 1

(iii) tan2 z

(2n - 1)p È ˘ , n = 1, 2, ...˙ Í ans. : (i) z = ±i (ii) z = ±1 (iii) z = 2 Î ˚

2.46

Chapter 2 Analytic Functions

xy 2 (x + iy ) ,z π 0 x2 + y 4 = 0, z=0

6. Show that f (z) =

is not analytic at the origin although C–R equations are satisfied at the origin.

2.9

harmonIc FunctIons

A real function f of two variables x and y is said to be a harmonic function in a region R if it has continuous second-order partial derivatives and satisfies the Laplace equation ∂2f ∂x

2

+

∂2f ∂y 2

=0

—2f = 0, where — 2 =

2.10

∂2 ∂x 2

+

∂2 ∂y 2

propertIes oF analytIc FunctIons

(1) If f (z) = u + iv is an analytic function, u and v are harmonic functions. Proof f (z) = u + iv is an analytic function. ∂u ∂v = ∂x ∂y ∂u ∂v =∂y ∂x

 (2.7)  (2.8)

Differentiating Eq. (2.7) w.r.t. x and Eq. (2.8) w.r.t. y, ∂2 u ∂x 2 ∂2 u

=

∂2 v ∂x ∂y

=-

 (2.9)

∂2 v ∂y ∂x

 (2.10)

∂y Adding Eqs (2.9) and (2.10), 2

∂2 u ∂x

2

+

∂2 u ∂y

2

=

∂2 v ∂2 v ∂x ∂y ∂y ∂x

=0 Hence, u is a harmonic function.

È ∂2 v ∂2 v = Í∵ ÍÎ ∂x ∂y ∂y ∂x

˘ ˙ ˙˚

2.10

2.47

Properties of Analytic Functions

Differentiating Eq. (2.7) w.r.t. y and Eq. (2.8) w.r.t. x, ∂2 u ∂2 v = ∂y ∂x ∂y 2

 (2.11)

∂2 u ∂2 v =- 2 ∂x ∂y ∂x

 (2.12)

Subtracting Eq. (2.12) from Eq. (2.11), ∂2 u ∂2 u ∂2 v ∂2 v = + 2 ∂y ∂x ∂x ∂y ∂y 2 ∂x ∂2 v ∂x 2

+

∂2 v ∂y 2

=0

È ∂2 u ∂2 u ˘ = Í∵ ˙ ÍÎ ∂y ∂x ∂x ∂y ˙˚

Hence, v is a harmonic function. note u and v of an analytic function u + iv are called conjugate harmonic functions of each other. (2) If f (z) = u + iv is an analytic function then the family of curves u(x, y) = c1 and the family of curves v(x, y) = c2 cut orthogonally. Proof f (z) = u + iv is an analytic function.

and

∂u ∂v = ∂x ∂y

 (2.13)

∂u ∂v =∂y ∂x

 (2.14)

u(x, y) = c1 (Fig. 2.14) ∂u dx ∂u dy ◊ + ◊ =0 ∂x dx ∂y dx

y

x

∂ u ∂ u dy + ◊ =0 ∂x ∂ y dx

x

Ê ∂u ˆ ÁË ∂x ˜¯

dy == m1 , say dx Ê ∂u ˆ ÁË ∂y ˜¯ which represents the slope of the family of curves u(x, y) = c1 v(x, y) = c2 ∂v dx ∂v dy ◊ + ◊ =0 ∂x dx ∂y dx

u

Fig. 2.14

2.48

Chapter 2 Analytic Functions

∂v ∂v dy + ◊ =0 ∂x ∂y dx Ê ∂v ˆ ÁË ∂x ˜¯

dy == m2 , say dx Ê ∂v ˆ ÁË ∂y ˜¯ which represents slope of family of curves v(x, y) = c2 È Ê ∂u ˆ ˘ È Ê ∂v ˆ ˘ Í Á ˜ ˙Í Á ˜ ˙ Ë ∂x ¯ ˙ Í Ë ∂x ¯ ˙ m1m2 = ÍÍ Ê ∂u ˆ ˙ Í Ê ∂v ˆ ˙ Í Á ˜ ˙Í Á ˜ ˙ ÍÎ Ë ∂y ¯ ˙˚ ÍÎ Ë ∂y ¯ ˙˚ È Ê ∂u ˆ ˘ È Ê ∂u ˆ ˘ ÍÁ ˜ ˙ Í -Á ˜ ˙ Ë ∂x ¯ ˙ Í Ë ∂y ¯ ˙ =Í Í Ê ∂u ˆ ˙ Í Ê ∂u ˆ ˙ ÍÁ ˜ ˙ Í Á ˜ ˙ ÍÎ Ë ∂y ¯ ˙˚ ÍÎ Ë ∂x ¯ ˙˚

[ Using Eqs (2.7) and (2.8)]

= -1 Hence, both the families of curves cut orthogonally. note In polar form, the condition of orthogonality for the families of curves u(r, q) = c1 and v(r, q) = c2 is Ê dq ˆ Ê dq ˆ ◊Ár ˜ = -1 ÁË r dr ˜¯ Ë dr ¯ v = c u=c 1

2

example 1 Show that u = 2x – x3 + 3xy2 is harmonic. Solution u = 2 x - x 3 + 3 xy 2 ∂u ∂u = 2 - 3 x 2 + 3 y2 , = 6 xy ∂x ∂y ∂2 u ∂x 2 ∂2 u ∂x

2

+

∂2 u ∂y 2

= -6 x = -6 x + 6 x =0

Hence, u is harmonic.

∂2 u ∂y 2

= 6x

2.10

Properties of Analytic Functions

2.49

example 2 Verify whether the function u = x 3 - 3 xy 2 + 3 x 2 - 3 y 2 + 1 is harmonic. Solution u = x 3 - 3 xy 2 + 3 x 2 - 3 y 2 + 1 ∂u = 3 x 2 - 3 y 2 + 6 x, ∂x ∂2 u ∂x 2 ∂2 u ∂x 2

+

∂2 u ∂y 2

= 6 x + 6,

∂u = -6 xy - 6 y ∂y ∂2 u ∂y 2

= -6 x - 6

= (6 x + 6) + (-6 x - 6)

=0 Hence, u is harmonic.

example 3 If f (z) = ez then show that u and v are harmonic functions. Solution f (z) = ez u + iv = e x + iy = e x eiy = e x (cos y + i sin y) Comparing real and imaginary parts, =

u = e x cos y,

v = e x sin y

∂u = e x cos y, ∂x

∂v = e x sin y ∂x

∂2 u

∂x

2

∂2 u

∂2 v

2

∂ u

= -e x cos y,

2

+

= e x sin y

∂x ∂v = e x cos y ∂y

∂y ∂ u

∂2 v

∂x ∂u = -e x sin y, ∂y 2

2

= e x cos y,

∂y

2

= e x cos y + (- e x cos y) =0

2

∂y

2

= -e x sin y

2.50

Chapter 2 Analytic Functions

∂2 v ∂x 2

+

∂2 v ∂y 2

= e x sin y + (-e x sin y)

=0 Hence, u and v are harmonic functions.

example 4 Prove that u = e- y cos x and v = e - x sin y the satisfy the Laplace equation, but u + iv is not an analytic function of z. Solution u = e - y cos x , ∂u = -e - y sin x, ∂x ∂2 u

∂x 2

+

∂2 v

= -e- y cos x,

∂x 2 ∂u = -e - y cos x, ∂y

= e - x sin y ∂x 2 ∂v = e - x cos y ∂y

∂2 u

= e - y cos x,

∂2 v

∂y 2 ∂2 u

v = e - x sin y ∂v = -e - x sin y ∂x

∂2 u ∂y 2

∂y 2

= -e - x sin y

= -e - y cos x + e - y cos x =0

∂ v ∂x

2

∂ v 2

2

+

∂y 2

= e - x sin y - e - x sin y =0

Hence, u and v satisfy the Laplace equation. ∂u ∂v π , ∂x ∂y

∂u ∂v π∂y ∂x

C–R equations are not satisfied. Hence, u + iv is not an analytic function.

example 5 Prove that u = x 2 - y 2 and v = regular.

y x + y2 2

are harmonic but u + iv is not

2.10

Properties of Analytic Functions

Solution u = x 2 - y2 , ∂u = 2 x, ∂x ∂2 u ∂x 2

∂y 2

x + y2 2

∂v y 2 xy = (2 x ) = 2 ∂ x ( x 2 + y 2 )2 ( x + y 2 )2 È ( x 2 + y 2 )2 - x ◊ 2( x 2 + y 2 )2 ◊ 2 x ˘ = 2 y Í ˙ ∂x 2 ( x 2 + y 2 )4 ÍÎ ˙˚ 2 2 2 y( y - 3 x ) = ( x 2 + y 2 )3 ∂2 v

= 2,

∂u = -2 y ∂y ∂2 u

y

v=-

È ( x 2 + y 2 ) - y (2 y ) ˘ ∂v = -Í ˙ 2 2 2 ∂y ˙˚ ÎÍ ( x + y )

= -2

x 2 - y2

== ∂2 v ∂y 2

= = =

∂2 u ∂x 2

+

∂2 u ∂y 2

( x 2 + y 2 )2 y2 - x 2

( x 2 + y 2 )2 2 y( x 2 + y 2 )2 - ( y 2 - x 2 ) ◊ 2( x 2 + y 2 ) ◊ 2 y ( x 2 + y 2 )4 2 y( x 2 + y 2 - 2 y 2 + 2 x 2 ) ( x 2 + y 2 )3 2 y(3 x 2 - y 2 ) ( x 2 + y 2 )3

= 2-2 =0

∂2 v ∂x

2

-

∂2 v ∂y

2

=

2 y( y 2 - 3 x 2 ) (x + y ) 2

2 3

+

2 y(3 x 2 - y 2 ) ( x 2 + y 2 )3

=0 Hence, u and v are harmonic. ∂u ∂v π , ∂x ∂y C–R equations are not satisfied. Hence, u + iv is not regular.

∂u ∂v π∂y ∂x

2.51

2.52

Chapter 2 Analytic Functions

example 6 If u(x, y) and v(x, y) are harmonic functions in a region R, prove that the Ê ∂u ∂v ˆ Ê ∂u ∂v ˆ function Á - ˜ + i Á + ˜ is an analytic function of z = x + iy. Ë ∂y ∂x ¯ Ë ∂x ∂y ¯ Solution Ê ∂u ∂v ˆ Ê ∂u ∂ v ˆ U + iV = Á - ˜ + i Á + ˜ Ë ∂y ∂x ¯ Ë ∂x ∂ y ¯ Comparing real and imaginary parts, ∂u ∂v ∂u ∂v U= V= - , + ∂y ∂x ∂x ∂y Let

∂U ∂2 u ∂2 v = , ∂x ∂x ∂y ∂x 2

∂V ∂ 2 u ∂ 2 v = + ∂ x ∂ x 2 ∂x ∂y

∂U ∂ 2 u ∂2 v , = ∂y ∂y 2 ∂y ∂x

∂V ∂2 u ∂2 v = + 2 ∂y ∂y ∂x ∂y

=

∂2 u ∂y 2

-

∂2 v , ∂x ∂y

=

∂2 u ∂2 v + 2 ∂x ∂y ∂y

Since u and v are harmonic functions, ∂2 u ∂x 2

+

∂2 u ∂y 2 ∂2 u ∂x 2

\ and

∂2 v

= 0, =-

∂x 2 ∂2 u ∂y 2

,

+

∂2 v ∂y 2 ∂2 v ∂x 2

=0 =-

∂2 v ∂y 2

∂U ∂2 u ∂2 v =- 2 ∂y ∂x ∂y ∂x ∂V ∂2 u ∂2 v = ∂y ∂x ∂y ∂x 2 ∂U ∂V = , ∂x ∂y

∂V ∂U =∂y ∂x

Hence, U + iV, i.e., the given function, is analytic.

example 7 Verify that the families of curves u = c1 and v = c2 cut orthogonally when u + iv = z3.

2.10

Properties of Analytic Functions

2.53

Solution u + iv = z 3 = ( x + iy)3 = x 3 + i 3 y3 + 3 x 2 ◊ iy + 3 x ◊ i 2 y 2 = ( x 3 - 3 xy 2 ) + i(3 x 2 y - y3 ) Comparing real and imaginary parts, =

v = 3 x 2 y - y3

u = x 3 - 3 xy 2 , Slope of families of curves u = c1 is

Ê ∂u ˆ ÁË ∂x ˜¯ dy =dx Ê ∂u ˆ ÁË ∂y ˜¯ ==

3 x 2 - 3 y2 -6 xy

x 2 - y2 = m1 , say 2 xy

Slope of families of curves v = c2 is Ê ∂v ˆ ÁË ∂x ˜¯

dy =dx Ê ∂v ˆ ÁË ∂y ˜¯ =-

6 xy

3 x - 3 y2 2 xy =- 2 = m2 , say x - y2 2

Ê x 2 - y2 ˆ Ê 2 xy ˆ - 2 m1m2 = Á ˜ ˜ Á Ë 2 xy ¯ Ë x - y 2 ¯ = -1 Hence, families of curves u = c1 and v = c2 cut orthogonally.

example 8 Verify that the families of curves u = c1 and v = c2 cut orthogonally, 2 when w = ez .

2.54

Chapter 2 Analytic Functions

Solution w = ez

2 2

u + iv = e( x + iy ) = ex

2

- y2 + 2 ixy

= ex

2

- y2 i 2 xy

= ex

2

e

-y

2

(cos 2 xy + i sin 2xy)

Comparing real and imaginary parts, u = ex

2

- y2

v = ex

cos 2 xy,

2

- y2

sin 2 xy

Slope of families of curves u = c1 is Ê ∂u ˆ ÁË ∂x ˜¯ dy =dx Ê ∂u ˆ ÁË ∂y ˜¯ =-

ex ex

2

2

- y2

-y

2

◊ 2 x cos 2 xy + e x

2

(-2 y) cos 2 xy + e x

- y2

(-2 y sin 2 xy)

2

2

-y

(-2 x sin 2 xy)

x cos 2 xy - y sin 2 xy y cos 2 xy + x sin 2 xy = m1 , say =

Slope of families of curves v = c2 is Ê ∂v ˆ ÁË ∂x ˜¯ dy =dx Ê ∂v ˆ ÁË ∂y ˜¯ =-

ex

2

- y2

x 2 - y2

◊ 2 x sin 2 xy + e x

e (-2 y)sin 2 xy + e x sin 2 xy + y cos 2 xy =- y sin 2 xy + x cos 2 xy = m2 , say

2

- y2

(2 y cos 2 xy)

x 2 - y2

(2 x cos 2 xy)

Ê x cos 2 xy - y sin 2 xy ˆ Ê x sin 2 xy + y cos 2 xy ˆ m1 m2 = Á Ë x sin 2 xy + y cos 2 xy ˜¯ ÁË x cos 2 xy - y sin 2 xy ˜¯ = -1 Hence, families of curves u = c1 and v = c2 cut orthogonally.

2.10

Properties of Analytic Functions

2.55

example 9 Show that the families of the curves r n = a sec nq and r n = b cosec nq cut orthogonally. Solution Let u = r n - a sec nq = 0

 (1)

and v = r - b cosec nq = 0

 (2 )

n

Differentiating Eq. (1) w.r.t. r, nr n -1 - a(sec nq tan nq ) ◊ n

dq =0 dr dq r n -1 = dr a sec nq tan nq

r

dq rn = dr a sec nq tan nq a sec nq = a sec nq tan nq = cot nq = m1 , say

Differentiating Eq. (2) w.r.t. r, nr n -1 - b(-cosec nq cot nq )n

dq =0 dr dq -r n -1 = dr b cosec nq cot nq

r

dq rn =dr b cosec nq cot nq =-

rn

r n cot nq 1 =cot nq = m2 , say 1 ˆ Ê m1 m2 = cot nq Á Ë cot nq ˜¯ = -1 Hence, families of the given curves cut orthogonally.

2.56

Chapter 2 Analytic Functions

example 10 ∂2

∂2

∂2 . Show that 2 + 2 = 4 ∂z ∂ z ∂x ∂y Solution Let f (z) be a function of x and y (Fig. 2.15). where z = x + iy z = x - iy 1 \ x = (z + z ) 2 1 y = (z - z ) 2i ∂f ∂f ∂x ∂f ∂y = ◊ + ◊ ∂z ∂x ∂z ∂y ∂z

2

2

=

∂f Ê 1 ˆ ∂f Ê 1 ˆ + ∂x ÁË 2 ˜¯ ∂y ÁË 2i ˜¯

=

1 Ê ∂f 1 ∂f ˆ + 2 ÁË ∂x i ∂y ˜¯

∂ ∂ 1 ∂ = + ∂z ∂x i ∂y ∂f ∂ f ∂ x ∂ f ∂y = ◊ + ◊ ∂ z ∂x ∂ z ∂y ∂ z =

∂f Ê 1 ˆ ∂f Ê 1 ˆ + ∂x ÁË 2 ˜¯ ∂y ÁË 2i ˜¯

=

1 Ê ∂f 1 ∂f ˆ 2 ÁË ∂x i ∂y ˜¯

∂ 1 ∂ ∂ = ∂ z ∂x i ∂y

From Eqs (1) and (2), 2

∂ Ê ∂ ˆ Ê ∂ 1 ∂ ˆÊ ∂ 1 ∂ ˆ 2 = + ∂z ÁË ∂z ˜¯ ÁË ∂x i ∂y ˜¯ ÁË ∂x i ∂y ˜¯ 4

∂2 ∂2 ∂2 = 2 + 2 ∂z ∂ z ∂x ∂y

This is also known as the complex form of the Laplace equation.

f

y

x z

Fig. 2.15

(1)

(2)

2.10

Properties of Analytic Functions

2.57

example 11 If f (z) = u + iv is analytic in the domain D then prove that Ê ∂2 ∂2 ˆ 2 2 Re f ( z ) = 2 f ¢( z ) . + Á 2 2˜ ∂y ¯ Ë ∂x

[Summer 2015]

Solution Let f (z) = u + iv Since f (z) is analytic, u and v are harmonic functions. ∂2 u ∂x 2 ∂2 v ∂x

2

+ +

∂2 u ∂y 2 ∂2 v ∂y 2

=0

...(1)

=0

...(2)

f ¢( z ) =

∂u ∂v +i ∂x ∂x 2

Ê ∂u ˆ Ê ∂v ˆ f ¢( z ) = Á ˜ + Á ˜ Ë ∂x ¯ Ë ∂x ¯

2

Ê ∂2 Ê ∂2 ∂2 ˆ 2 ∂2 ˆ 2 Á 2 + 2 ˜ Re f ( z ) = Á 2 + 2 ˜ u ∂y ¯ ∂y ¯ Ë ∂x Ë ∂x = =

∂2 ∂x 2

u2 +

∂2 ∂y 2

u2

∂ Ê ∂u ˆ ∂ Ê ∂u ˆ 2u Á 2u ˜ + ∂x Ë ∂x ¯ ∂y ÁË ∂yy ˜¯

2 È ∂ 2 u Ê ∂u ˆ 2 ∂ 2 u Ê ∂u ˆ ˘ = 2 Íu 2 + Á ˜ + u 2 + Á ˜ ˙ Ë ∂x ¯ ∂y Ë ∂y ¯ ˙˚ ÍÎ ∂x È Ê ∂ 2 u ∂ 2 u ˆ Ê ∂u ˆ 2 Ê ∂u ˆ 2 ˘ = 2 Íu Á 2 + 2 ˜ + Á ˜ + Á ˜ ˙ ∂y ¯ Ë ∂x ¯ Ë ∂y ¯ ˙˚ ÍÎ Ë ∂x

ÈÊ ∂u ˆ 2 Ê ∂u ˆ 2 ˘ = 2 ÍÁ ˜ + Á ˜ ˙ ÍÎË ∂x ¯ Ë ∂y ¯ ˙˚ = 2 f ¢( z )

2

[ Using Eq. (1)]

2.58

Chapter 2 Analytic Functions

example 12 If f (z) is an analytic function of z, prove that Ê ∂2 ∂2 ˆ 2 2 Á 2 + 2 ˜ f ( z ) = 4 f ′( z ) . ∂y ¯ Ë ∂x Solution Let f (z) = u + iv Since f (z) is analytic, u and v are harmonic functions. ∂2 u ∂x 2 ∂2 v ∂x 2

+ +

∂2 u ∂y 2 ∂2 v ∂y 2

=0

 (1)

=0

 (2 )

f ′( z ) =

∂u ∂v +i ∂x ∂x 2

Ê ∂u ˆ Ê ∂v ˆ f ′( z ) = Á ˜ + Á ˜ Ë ∂x ¯ Ë ∂x ¯ 2

Ê ∂u ˆ Ê ∂v ˆ 2 f ′( z ) = Á ˜ + Á ˜ Ë ∂x ¯ Ë ∂x ¯

2

2

 (3)

Ê ∂2 Ê ∂2 ∂2 ˆ 2 ∂2 ˆ 2 2 Á 2 + 2 ˜ f ( z ) = Á 2 + 2 ˜ (u + v ) ∂y ¯ ∂y ¯ Ë ∂x Ë ∂x = =

∂2 ∂x 2

(u2 + v 2 ) +

∂2 ∂y 2

(u

2

+ v2

)

∂v ˆ ∂ Ê ∂u ∂ Ê ∂u ∂v ˆ + 2 v ˜ + Á 2u 2u + 2v ˜ Á ∂x ¯ ∂y Ë ∂y ∂x Ë ∂x ∂y ¯

2 2 2 ÈÏ 2 2 2 Ô ∂ u Ê ∂u ˆ ¸Ô ÏÔ ∂ v Ê ∂v ˆ ¸Ô ÏÔ ∂ u Ê ∂u ˆ ¸Ô = 2 ÍÌu 2 + Á ˜ ˝ + Ìv 2 + Á ˜ ˝ + Ìu 2 + Á ˜ ˝ Ë ∂x ¯ Ô Ô ∂y Ë ∂y ¯ Ô Ë ∂x ¯ Ô Ô ∂x Í Ô ∂x ˛ Ó ˛ Ó ˛ ÎÓ

ÏÔ ∂2 v Ê ∂v ˆ 2 ¸Ô˘ + Ì v 2 + Á ˜ ˝˙ ÔÓ ∂y Ë ∂y ¯ Ô˛˙˚ È Ê ∂ 2 u ∂ 2 u ˆ Ê ∂u ˆ 2 Ê ∂v ˆ 2 Ê ∂u ˆ Ê ∂v ˆ = 2 Íu Á 2 + 2 ˜ + Á ˜ + Á ˜ + Á ˜ + Á ˜ Í Ë ∂x ∂y ¯ Ë ∂x ¯ Ë ∂x ¯ Ë ∂y ¯ Ë ∂y ¯ Î Ê ∂2 v ∂2 v ˆ ˘ +v Á 2 + 2 ˜ ˙ ∂y ¯ ˙˚ Ë ∂x 2

2

2.10

Properties of Analytic Functions

ÈÊ ∂u ˆ 2 Ê ∂v ˆ 2 Ê ∂v ˆ 2 Ê ∂u ˆ 2 ˘ = 2 ÍÁ ˜ + Á ˜ + Á - ˜ + Á ˜ ˙ ÍÎË ∂x ¯ Ë ∂x ¯ Ë ∂x ¯ Ë ∂x ¯ ˙˚ ÈÊ ∂u ˆ 2 Ê ∂v ˆ 2 ˘ 4 = ÍÁ ˜ + Á ˜ ˙ ÍÎË ∂x ¯ Ë ∂x ¯ ˙˚ = 4 f ′( z )

2.59

È Using equations (1) and (2 ) ˘ Í ˙ ÍÎand C- R equations ˙˚

2

Aliter Ê ∂2 ∂2 ˆ ∂2 È 2 + = 4 f ( z ) f ( z ) ◊ f ( z )˘˚ Á 2 ˜ ∂z ∂ z Î ∂y 2 ¯ Ë ∂x =4

∂2 [ f (z) ◊ f (zz )] ∂z ∂ z

∂ È∂ ˘ f ( z ) ◊ f ( z )}˙ { Í ∂z Î ∂ z ˚ ∂ È ∂ ˘ f ( z) f ( z )˙ =4 ∂z ÍÎ ∂z ˚ =4

=4

ÈÎ∵ f ( z ) iss a function of z alone ˘˚

∂ [ f ( z ) ◊ f ′( z )] ∂z

= 4 f ′( z )

∂ f ( z) ∂z

[∵ f ′( z ) isa function of z alone ]

= 4 f ′( z ) f ′( z ) = 4 f ′( z ) f ′( z ) = 4 f ′( z )

2

example 13 Ê ∂2 ∂2 ˆ If f (z) is an analytic function of z, prove that Á 2 + 2 ˜ log f ( z ) = 0. ∂y ¯ Ë ∂x Solution Let

f ( z ) = u + iv f ( z ) = u2 + v 2 2

f ( z ) = u2 + v 2

2.60

Chapter 2 Analytic Functions

Since f (z) is an analytic function, u and v are harmonic functions. ∂2 u ∂x

2

∂2 v ∂x 2

+ +

∂2 u ∂y 2 ∂2 v ∂y 2

(

log f ( z ) = log f ( z )

=0

 (1)

=0

 (2 )

1 2 2

)

1 2 log f ( z ) 2 1 = log(u2 + v 2 ) 2 2 2 Ê ∂ Ê ∂2 ∂ ˆ ∂2 ˆ È 1 ˘ + = + log f ( z ) log(u2 + v 2 )˙ Á 2 Á 2 2 ˜ Í2 2˜ ∂y ¯ ∂y ¯ Î Ë ∂x Ë ∂x ˚ =

= ∂2 ∂x

2

log(u2 + v 2 ) =

˘ ∂2 1 È ∂2 2 2 2 2 Í 2 log(u + v ) + 2 log(u + v )˙ 2 ÍÎ ∂x ∂y ˙˚ ∂ È 1 Í ∂x Î u 2 + v 2

=2

∂v ˆ ˘ Ê ∂u ÁË 2u ∂x + 2v ∂x ˜¯ ˙ ˚

∂ È u ux + v v x ˘ ∂x ÍÎ u2 + v 2 ˙˚

ÈÏ ∂ ∂ 2 ¸ 2 2 2 ˘ Í Ì x (u ux + v v x )˝ (u + v ) - (u ux + v v x ) ∂x (u + v ) ˙ ∂ ˛ ˙ = 2ÍÓ Í ˙ (u2 + v 2 )2 Í ˙ Î ˚ 2 È(uuxx + vv xx + ux2 + v x2 )(u2 + v 2 ) = 2 2 2 Î (u + v ) -(u ux + v vx )(2u ux + 2 v vx )˘˚ =

2

È(u3uxx + u2 v v xx + u2 ux2 + u2 v x2 + uv 2 uxx + v3 vxx (u + v ) Î 2

2 2

+ v 2 ux2 + v 2 v x2 ) - (2u2 ux2 + 2uvux v x + 2uvux v x + 2v 2 v x 2 )˘˚ =

2

Èu3uxx + v3 v xx + uv 2 uxx + u2 v v xx - u2 ux2 + u2 v x2 (u + v ) Î 2

2 2

+ v 2 ux2 - v 2 vx2 - 4uvux v x ˚˘

(3)

2.10

2.61

Properties of Analytic Functions

Similarly, ∂2 2 Èu3uyy + v3 v yy + uv 2 uyy + u2 vv yy - u2 uy2 + u2 v y2 log(u2 + v 2 ) = 2 2 2 2 Î ∂y (u + v ) + v 2 uy2 - v 2 v y2 - 4 uvuy v y ˘˚

(4)

Adding Eqs (3) and (4), Ê ∂2 ∂2 ˆ 2 2 2 Èu3 (uxx + uyy ) + v3 (v xx + v yy ) + uv 2 (uxx + uyy ) + Á 2 ˜ log(u + v ) = 2 ∂y 2 ¯ (u + v 2 )2 Î Ë ∂x + u2 v(v xx + v yy ) - u2 (ux2 + uy2 ) + u2 (v x2 + v y2 ) + v 2 (ux2 + uy2 ) - v 2 (v x2 + v y2 ) - 4uv(ux v x + uy v y )˘˚ =

2

Èu3 (0) + v3 (0) + uv 2 (0) + u2 v(0) - u2 (ux2 + uy2 ) (u + v 2 )2 Î 2

+ u2 (uy2 + ux2 ) + v 2 (ux2 + uy2 ) - v 2 (ux2 + uy2 ) - 4uv(-ux uy + uy vx )˘˚ È Using equationns (1) ˘ Í ˙ Îand (2) and C- R equations˚ =0 Ê ∂2 ∂2 ˆ Hence, Á 2 + 2 ˜ log f ( z ) = 0 ∂y ¯ Ë ∂x Aliter Ê ∂2 ∂2 ˆ ∂2 È 1 2˘ Á 2 + 2 ˜ log f ( z ) = 4 ∂z ∂z Í 2 log f ( z ) ˙ ∂y ¯ Ë ∂x Î ˚

{

}

=2

∂2 È g f (z) ◊ f (z) ˘ log ˚ ∂z ∂ z Î

=2

∂2 Èlog f ( z ) + log f ( z )˘˚ ∂z ∂ z Î

=2

˘ ∂ È ∂ ∂ log f ( z ) + log f ( z )˙ Í ∂z Î ∂ z ∂z ˚

˘ ∂ È 1 f ′( z )˙ [∵log f (z) isa function of z alone ] Í0 + ∂z Î f (z ) ˚ =0 [∵ f ( z ) and f ′( z ) are functionsof z alone ] =2

2.62

Chapter 2 Analytic Functions

example 14 If f (z) is an analytic function of z in any domain, prove that Ê ∂2 ∂2 ˆ p 2 p-2 2 . Á 2 + 2 ˜ f ( z ) = p f ′( z ) f ( z ) ∂y ¯ Ë ∂x Solution p

Ê ∂2 ∂2 ˆ ∂2 È 2 p + = f ( z ) 4 f ( z ) ˘˙ 2 Á 2 2˜ Í Î ˚ ∂z ∂ z ∂y ¯ Ë ∂x

p

∂2 È =4 f ( z ) ◊ f (zz )˘˚ 2 ∂z ∂ z Î

p˘ ∂ È ∂ È∵ f ( z ) = f ( z )˘ f ( z ) ◊ f ( z )} 2 ˙ { Í Î ˚ ∂z Î ∂ z ˚ p p ˘ p ∂ È -1 =4 Í{ f ( z )} 2 ◊ { f ( z )} 2 f ′( z )˙ [∵ f ( z ) isa function of z alone ] 2 ∂z Î ˚

=4

= 4◊

p p p ∂ -1 f ( z )] 2 f ′( z ) ◊ [ f ( z )] 2 [ 2 ∂z

= 4◊

p p p p -1 -1 f ( z )] 2 f ′ ( z ) ◊ [ f ( z )] 2 f ′( z ) [ 2 2 p

= p 2 [ f ( z ) ◊ f ( z )] 2 = p2 ÈÎ f ( z ) ◊

2

[ f ′(( z ) ◊ f ′(z)]

p-2 ˘ f (z) 2

˚

2 = p2 ÈÍ f ( z ) ˘˙ Î ˚

= p 2 f ′( z )

-1

p-2 2

f (z)

◊ È f ′ ( z ) ◊ f ′( z )˘ Î ˚

È f ′( z ) 2 ˘ ÍÎ ˙˚ p-2

exercIse 2.4 -1 Ê y ˆ 1. Prove that tan ÁË x ˜¯ is harmonic.

2. Prove that log x 2 + y 2 is harmonic.

[∵ f ( z ) isa function of z alone ]

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.63

3. If f(z) is analytic, prove that Ê ∂2 ∂2 ˆ 2 2 (i) Á 2 + 2 ˜ Re f (z) = 2 f ¢(z) ∂y ¯ Ë ∂x Ê ∂2 ∂2 ˆ 2 2 (ii) Á 2 + 2 ˜ Im f (z) = 2 f ¢(z) ∂y ¯ Ë ∂x 2

2

¸ 2 ¸ Ï∂ Ï∂ (iii) Ì f (z) ˝ + Ì f (z) ˝ = f ¢(z) ˛ Ó ∂y Ó ∂x ˛

{

4. Prove that log (x - 1)2 + (y - 2)2

} is harmonic in every region which does

not include the point (1, 2).

2.11

constructIon oF analytIc FunctIons: mIlne—thomson method

case (i) If the real part u of f (z) is given ∂u ∂u and . ∂x ∂y ∂u ∂v +i Step 2: Find f ′( z ) = ∂x ∂x ∂u ∂u = -i ∂x ∂y Step 1: Find

[ Using C- R equations ]

Step 3: Put x = z and y = 0 in f ¢(z). Step 4: Integrate f ¢(z) to obtain f (z). case (ii)

If the imaginary part v of f (z) is given

∂v ∂v and . ∂x ∂y ∂u ∂v Step 2: Find f ′( z ) = +i ∂x ∂x ∂v ∂v = +i [ Using C- R equations ] ∂y ∂x Step 3: Put x = z and y = 0 in f ′( z ). Step 1: Find

Step 4: Integrate f¢(z) to obtain f(z).

example 1 Find the analytic function f (z) = u + iv if u = x3 – 3xy. [Summer 2014]

2.64

Chapter 2 Analytic Functions

Solution u = x 3 - 3 xy ∂u = 3 x 2 - 3 y, ∂x ∂u ∂v +i ∂x ∂x ∂u ∂u = -i ∂x ∂y

∂u = -3 x ∂y

f ¢( z ) =

[ Using C- R equations ]

= (3 x 2 - 3 y) - i(-3 x ) = 3 x 2 - 3 y + 3ix Putting x = z, y = 0, f¢(z) = 3z2 + 3iz Integrating w.r.t. z, f ( z ) = Ú (3z 2 + 3iz ) dz = z3 + i

3z 2 +c 2

example 2 Construct the analytic function whose real part is u = x 3 - 3 xy 2 + 3 x 2 - 3 y 2 + 1. Solution u = x 3 - 3 xy 2 + 3 x 2 - 3 y 2 ∂u = 3 x 2 - 3 y 2 + 6 x, ∂x ∂u ∂v f ′( z ) = +i ∂x ∂x ∂u ∂u = -i ∂x ∂y

+1 ∂u = -6 xy - 6 y ∂y

[ Using C- R equations ]

= (3 x 2 - 3 y 2 + 6 x ) - i(-6 xy - 6y)

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.65

Putting x = z, y = 0, f ′( z ) = (3z 2 + 6 z ) + i(0) Integrating w.r.t. z, f ( z ) = Ú (3z 2 + 6 z )dz Ê z3 ˆ Ê z2 ˆ = 3Á ˜ + 6 Á ˜ + c Ë 3¯ Ë 2¯ = z 3 + 3z 2 + c

example 3 Find the analytic function w = u + iv whose imaginary part is given by v = ex (x sin y + y cos y). Solution v = e x ( x sin y + y cos y) ∂v = e x ( x sin y + y cos y) + e x (sin y) ∂x ∂v = e x ( x cos y + cos y - y sin y) ∂y ∂u ∂v f ′( z ) = +i ∂x ∂x ∂v ∂v = +i [ Using C- R equatio ons] ∂y ∂x = e x ( x cos y + cos y - y sin y) + i e x ( x sin y + y cos y + sin y) Putting x = z, y = 0, f ′( z ) = e z ( z cos 0 + cos 0 - 0) + i e z ( z sin 0 + 0 + sin 0) = e z ( z + 1) Integrating w.r.t. z, f ( z ) = Ú ( z + 1)e z dz = ( z + 1) e z - (1) e z + c = ze z + c

2.66

Chapter 2 Analytic Functions

example 4 Find the analytic function f (z) = u + iv, whose real part u is Solution u=

x x + y2 2

∂u (1)( x 2 + y 2 ) - x(2 x ) y2 - x 2 = = ∂x ( x 2 + y 2 )2 ( x 2 + y 2 )2 ∂u x 2 xy =- 2 ◊ 2y = - 2 ∂y ( x + y 2 )2 ( x + y 2 )2 ∂u ∂v +i f ′( z ) = ∂x ∂x ∂u ∂u = -i [ Using C-- R equations ] ∂x ∂y =

È 2 xy ˘ - i Í- 2 2 2˙ (x + y ) ÍÎ ( x + y ) ˙˚ y2 - x 2 2

2 2

Putting x = z, y = 0, f ′( z ) =

- z2

z4 1 =- 2 z

+ i (0 )

Integrating w.r.t. z, f ( z) = Ú -

1 z2

dz

Ê z -1 ˆ = -Á ˜ +c Ë -1 ¯ =

1 +c z

x x + y2 2

.

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.67

example 5 Determine the analytic function whose real part is

sin 2 x . cosh 2 y - cos 2 x

Solution u= ∂u = ∂x = =

sin 2 x cosh 2 y - cos 2 x 2 cos 2 x(cosh 2 y - cos 2 x ) - sin 2 x(2 sin 2 x ) (cosh 2 y - cos 2 x )2 2 cos 2 x cosh 2 y - 2 cos2 2 x - 2 sin 2 2 x (cosh 2 y - cos 2 x )2 2 cos 2 x cosh 2 y - 2 (cosh 2 y - cos 2 x )2

È ˘ ∂u 1 = sin 2 x Í◊ (2 sinh 2 y)˙ 2 ∂y ÍÎ (cosh 2 y - cos 2 x)) ˙˚ =-

2 sin 2 x sinh 2 y (cosh 2 y - cos 2 x )2

∂u ∂v +i ∂x ∂x ∂u ∂u = -i ∂x ∂y

f ′( z ) =

=

[ Using C- R equations ]

2 cos 2 x cosh 2 y - 2 (cosh 2 y - cos 2 x )

2

+i

2 sin 2 x sinh 2 y (cosh 2 y - cos 2 x )2

Putting x = z, y = 0, f ′( z ) =

2 cos 2 z - 2

(1 - cos 2 z )2 2 =1 - cos 2 z 2 =2 sin 2 z = -cosec 2 z

Integrating w.r.t. z, f ( z ) = Ú -cosec 2 z dz = cot z + c

[∵ cosh 0 = 1,sinh 0 = 0]

2.68

Chapter 2 Analytic Functions

example 6 Find the analytic function f (z) = u + iv, where v =

2 sin x sinh y . cos 2 x + cosh 2 y

Solution 2 sin x sinh y cos 2 x + cosh 2 y

v=

È cos x(cos 2 x + cosh 2 y) - sin x(-2 sin 2 x ) ˘ ∂v = 2 sinh y Í ˙ ∂x (cos 2 x + cosh 2 y)2 ÍÎ ˙˚ È cosh y(cos 2 x + cosh 2 y) - sinh y(2 sinh 2 y) ˘ ∂v = 2 sin x Í ˙ ∂y (cos 2 x + cosh 2 y)2 ÍÎ ˙˚ ∂u ∂v +i ∂x ∂x ∂v ∂v = +i ∂y ∂x

f ′( z ) =

[ Using C- R equations ]

È cosh y(cos 2 x + cosh 2 y) - 2 sinh y sinh 2 y ˘ = 2 sin x Í ˙ (cos 2 x + cosh 2 y)2 ÍÎ ˙˚ È cos x(cos 2 x + cosh 2 y) + 2 sin x sin 2 x ˘ + i 2 sinh y Í ˙ (cos 2 x + cosh 2 y)2 ÍÎ ˙˚ Putting x = z, y = 0, È cosh 0(cos 2 z + cosh 0) - 0 ˘ f ′( z ) = 2 sin z Í ˙ + i (0 ) (cos 2 z + cosh 0)2 Î ˚ =

2 sin z(cos 2 z + 1)

(cos 2 z + 1)2 2 sin z = cos 2 z + 1 2 sin z = 2 cos2 z = tan z sec z Integrating w.r.t. z, f ( z ) = Ú tan z sec z dz + c = sec z + c

[∵ sinh 0 = 0]

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.69

example 7 Find the analytic function u + iv, if u = ( x - y)( x 2 + 4 xy + y 2 ) . Also, find the conjugate harmonic function v. Solution u = ( x - y)( x 2 + 4 xy + y 2 ) = x 3 + 4 x 2 y + xy 2 - x 2 y - 4 xy 2 - y3 = x 3 + 3 x 2 y - 3 xy 2 - y3 ∂u = 3 x 2 + 6 xy - 3 y 2 ∂x ∂u = 3 x 2 - 6 xy - 3 y 2 ∂y ∂u ∂v +i ∂x ∂x ∂u ∂u = -i ∂y ∂x

f ′( z ) =

[ Using C- R equations ]

= (3 x 2 + 6 xy - 3 y 2 ) - i(3 x 2 - 6 xy - 3 y) Putting x = z, y = 0, f ′( z ) = 3z 2 - i(3z 2 ) = 3(1 - i )z 2 Integrating w.r.t. z, f ( z ) = Ú 3(1 - i )z 2 dz = (1 - i )z 3 + c u + iv = (1 - i )( x + iy)3 + c = (1 - i ) ÈÎ x 3 + i 3 y3 + 3 x 2 (iy) + 3 x(iy)2 ˘˚ + c = (1 - i ) ÈÎ x 3 - iy3 + 3ix 2 y - 3 xy 2 ˘˚ + c = x 3 - iy3 + 3ix 2 y - 3 xy 2 - ix 3 + i 2 y3 - 3i 2 x 2 y + 3ixy 2 + c = ( x 3 - y3 + 3 x 2 y - 3 xy 2 ) + i(- y3 + 3 x 2 y - x 3 + 3 xy 2 ) + c Comparing imaginary parts, v = - x 3 - y3 + 3 x 2 y + 3 xy 2 + c

2.70

Chapter 2 Analytic Functions

example 8 1 log( x 2 + y 2 ) is harmonic. Determine its analytic 2 function. Find also its conjugate. Show that u =

Solution 1 log( x 2 + y 2 ) 2 1 ∂u 1 x = ◊ 2x = 2 2 2 ∂x 2 x + y x + y2 1 ∂u 1 y = ◊ 2y = 2 2 2 ∂y 2 ( x + y ) x + y2 u=

∂2 u ∂x 2 ∂2 u ∂y 2 ∂2 u ∂x 2

+

∂2 u ∂y 2

1 ◊ ( x 2 + y2 ) - x ◊ 2 x

=

( x 2 + y 2 )2 1 ◊ (xx 2 + y 2 ) - y ◊ 2 y

=

( x 2 + y 2 )2 y2 - x 2

=

( x 2 + y 2 )2

+

x x +y 2

2

-i

Putting x = z, y = 0, f ′( z ) =

z

z2 1 = z

- i (0 )

Integrating w.r.t. z, 1 f ( z ) = Ú dz z = log z + c

=

y2 - x 2 ( x 2 + y 2 )2 x 2 - y2 ( x 2 + y 2 )2

x 2 - y2 ( x 2 + y 2 )2

=0 Hence, u is a harmonic function. ∂u ∂v Now, f ′( z ) = +i ∂x ∂x ∂u ∂u = -i ∂x ∂y =

=

[ Using C- R equations ] y x + y2 2

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.71

u + iv = log( x + iy) + c 1 y = log( x 2 + y 2 ) + i tan -1 + c 2 x Comparing imaginary parts, v = i tan -1

y +c x

example 9 If w = u + iv is an analytic function and v = x 2 - y 2 +

x x + y2 2

Solution v = x 2 - y2 +

x x 2 + y2

∂v (1)( x 2 + y 2 ) - x(2 x ) y2 - x 2 = 2x + = 2x + 2 2 2 2 ∂x (x + y ) ( x + y 2 )2 ∂v x 2 xy = -2 y - 2 ◊ 2 y = -2 y - 2 2 2 ∂y (x + y ) ( x + y 2 )2 ∂u ∂v +i ∂x ∂x ∂v ∂v = +i ∂y ∂x

f ′( z ) =

[ Using C- R equations ]

È 2 xy ˘ È y2 - x 2 ˘ + i Í2 x + 2 = - Í2 y + 2 ˙ 2 2˙ ( x + y ) ˙˚ ÍÎ ( x + y 2 )2 ˙˚ ÍÎ Putting x = z, y = 0, Ê 0 - z2 ˆ f ′( z ) = 0 + i Á 2 z + 4 ˜ z ¯ Ë = i(2 z - z -2 ) Integrating w.r.t. z,

(

)

f ( z ) = Ú i 2 z - z -2 dz Ê z -1 ˆ = i Á z2 +c -1 ˜¯ Ë 1ˆ Ê = i Á z2 + ˜ + c Ë z¯

, find u.

2.72

Chapter 2 Analytic Functions

È 1 ˘ u + iv = i Í( x + iy)2 + ˙+c x + iy ˚ Î È x - iy ˘ = i Í x 2 + i 2 y 2 + 2ixy + 2 ˙+c x + y 2 ˚˙ ÍÎ = ix 2 - iy 2 + 2i 2 xy +

ix x 2 + y2

-

i2 y x 2 + y2

+c

Ê x ˆ y - 2 xy + 2 +c = i Á x 2 - y2 + 2 2˜ x +y ¯ x + y2 Ë ˆ Ê Ê y x ˆ =Á 2 - 2 xy˜ + i Á x 2 - y 2 + 2 ˜ +c 2 x + y2 ¯ ¯ Ë Ëx +y Comparing real parts, u=

y x + y2 2

- 2 xy + c

example 10 Find the analytic function w = u + iv when v = e-2 y ( y cos 2 x + x sin 2 x ) and find u. Solution v = e -2 y ( y cos 2 x + x sin 2 x ) ∂v = e -2 y (-2 y sin 2 x + sin 2 x + 2 x cos 2 x ) ∂x ∂v = -2e -2 y ( y cos 2 x + x sin 2 x ) + e -2 y (cos 2 x ) ∂y ∂u ∂v f ′( z ) = +i ∂x ∂x ∂v ∂v = +i [ Using C-R equations ∂x ∂y

]

= -2e -2 y ( y cos 2 x + x sin 2 x ) + e -2 y cos 2 x + ie -2 y (-2 y sin 2 x + sin 2 x + 2 x cos 2 x ) Putting x = z, y = 0, f ′( z ) = -2e0 (0 + z sin 2 z ) + e0 cos 2 z + ie0 (-0 + sin 2 z + 2 z cos 2 z ) = -2 z sin 2 z + cos 2 z + i(sin 2 z + 2 z cos 2 z )

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.73

Integrating w.r.t z, f ( z ) = Ú (cos 2 z - 2 z sin 2 z )dz + i Ú (sin 2 z + 2 z cos 2 z )dz =

È Ê cos 2 z ˆ sin 2 z Ê sin 2 z ˆ ˘ - 2 Íz Á - (1) Á ˙ ˜ Ë 2 2 ¯ 4 ˜¯ ˚ Î Ë Ê cos 2 z ˆ ¸˘ È cos 2 z Ï sin 2 z + i Í+2 Ì z - (1) Á ˝˙ + c Ë 4 ˜¯ ˛˙˚ 2 2 Î Ó

sin 2 z sin 2 z È cos 2 z cos 2 z ˘ + z cos 2 z + i Í+ z si n 2 z + +c 2 2 2 2 ˙˚ Î = z cos 2 z + iz sin 2 z + c = z(cos 2 z + i sin 2 z ) + c

=

= zei 2 z + c u + iv = ( x + iy)ei 2( x + iy ) + c = ( x + iy)e2ix - 2 y + c = ( x + iy)e -2 y e2ix + c = e -2 y ( x + iy)(cos 2 x + i sin 2 x ) + c = e -2 y [( x cos 2 x - y sin 2 x ) + i( y cos 2 x + x sin 2 x )] + c Comparing real parts, u = e -2 y ( x cos 2 x - y sin 2 x )

example 11 Show that u(x, y)= x2 – y2 is harmonic. Find the corresponding analytic function f (z) = u + iv. [Summer 2013] Solution (i) u(x, y) = x2 – y2 ∂u = 2 x, ∂x ∂2 u ∂x 2 ∂2 u ∂x 2

+

∂2 u ∂y 2

= 2, = 2-2 =0

Hence, u is harmonic.

∂u = -2 y ∂y ∂2 u ∂y 2

= -2

2.74

(ii)

Chapter 2 Analytic Functions

f ( z ) = u + iv ∂u ∂v f ¢( z ) = +i ∂x ∂x ∂u ∂u = -i ∂x ∂y = 2 x - i(-2 y) Putting x = z, y = 0, f¢(z) = 2z Integrating w.r.t. z

[ Using C- R equations ]

Ú f ¢(z) dz = Ú 2 z dz f (z) = z2 + c

example 12 Show that u(x, y) = x2 – y2 + x is harmonic. Find the corresponding analytic function f (z) = u + iv. [Winter 2013] Solution (i) u(x, y) = x2 – y2 + x ∂u = 2 x + 1, ∂x ∂2 u ∂x 2 ∂2 u ∂x 2

+

∂2 u ∂y 2

= 2,

∂u = -2 y ∂y ∂2 u ∂y 2

= -2

= 2-2

=0 Hence, u is harmonic. (ii)

f ( z ) = u + iv ∂u ∂v f ¢( z ) = +i ∂x ∂x ∂u ∂u = -i [ Using C- R equations ] ∂x ∂y = (2 x + 1) - i (-2 y) Putting x = z, y = 0, f¢(z) = 2z + 1

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.75

Integrating w.r.t. z,

Ú f ¢(z) dz = Ú (2 z + 1) dz = z2 + z + c

example 13 Prove that u = ex (x cos y – y sin y) is harmonic and, hence, find the analytic function f (z) = u + iv. Solution u = e x ( x cos y - y sin y) ∂u = e x ( x cos y - y sin y) + e x (cos y) ∂x ∂2 u ∂x

2

= e x ( x cos y - y sin y + cos y) + e x (cos y) = e x ( x cos y - y sin y + 2 cos y)

∂u = e x (- x sin y - sin y - y cos y) ∂y ∂2 u ∂y

2

= e x (- x cos y - cos y - cos y + y sin y) = e x (- x cos y - 2 cos y + y sin y)

∂2 u ∂x 2

+

∂2 u ∂y 2

= e x ( x cos y - y sin y + 2 cos y) + e x (- x coss y - 2 cos y + y sin y) =0

Hence, u is a harmonic function. ∂u ∂v Now, f ′( z ) = +i ∂x ∂x ∂u ∂u = -i ∂x ∂y Putting

[ Using C- R equations ]

= e x ( x cos y - y sin y + cos y) - ie x (- x sin y - sin y - y cos y) x = z, y = 0, f ′( z ) = e z ( z cos 0 - 0 + cos 0) - ie z (0 - 0 - 0) = e z ( z + 1)

2.76

Chapter 2 Analytic Functions

Integrating w.r.t. z, f ( z ) = Ú ( z + 1) e z dz = ( z + 1) e z - (1) e z + c = ze z + c

example 14 Prove that x2 – y2 + e–2x cos 2 y is harmonic and find its harmonic conjugate. Solution u = x 2 - y 2 + e -2 x cos 2 y

Let

∂u = 2 x - 2e-2 x cos 2 y, ∂x ∂2 u ∂x ∂ u 2

∂x

2

2

∂ u 2

+

∂y

2

∂u = -2 y - 2e -2 x sin 2 y ∂y ∂2 u

= 2 + 4e -2 x cos 2 y,

∂y

2

= -2 - 4e -2 x cos 2 y

= (2 + 4e -2 x cos 2 y) + (-2 - 4e -2 x cos 2 y)

=0 ∂u ∂v f ′( z ) = +i ∂x ∂x ∂u ∂u = -i ∂x ∂y

[ Using C- R equations ]

= (2 x - 2e-2 x cos 2 y) - i(-2 y - 2e -2 x sin 2 y) Putting x = z, y = 0, f ′( z ) = 2( z - e -2 z cos 0) - i(0) = 2( z - e -2 z ) Integrating w.r.t. z, f ( z ) = Ú 2( z - e -2 z ) dz Ê z 2 e -2 z ˆ = 2Á +c -2 ˜¯ Ë 2 = z 2 + e -2 z + c u + iv = ( x + iy)2 + e -2( x + iy ) + c = x 2 + i 2 y 2 + 2ixy + e -2 x e -2iy + c

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.77

= x 2 - y 2 + 2ixy + e -2 x (cos 2 y - i sin 2 y) + c = x 2 - y 2 + e -2 x cos 2 y + i(2 xy - e -2 x sin 2 y) + c Comparing imaginary parts, v = 2 xy - e -2 x sin 2 y + c

example 15 Find the analytic function u + iv, if u – v = (x – y) (x2 + 4xy + y2). [Winter 2012] Solution f (z) = u + iv i f (z) = iu + i2v = iu – v Adding Eqs (1) and (2), f ( z ) + i f ( z ) = (u + iv) + (iu - v) (1 + i ) f ( z ) = (u - v) + i (u + v)

...(1) ...(2)

F ( z ) = U + iV where F(z) = (1 + i) f (z), U = u – v, V = u + v Since, f (z) is analytic, F(z) is analytic. U = u-v = ( x - y)( x 2 + 4 xy + y 2 ) = x 3 + 4 x 2 y + xy 2 - x 2 y - 4 xy 2 - y3 = x 3 + 3 x 2 y - 3 xy 2 - y3 ∂U = 3 x 2 + 6 xy - 3 y 2 ∂x ∂U = 3 x 2 - 6 xy - 3 y 2 ∂y ∂U ∂V +i ∂x ∂x ∂U ∂U = -i ∂x ∂y

F ¢( z ) =

[ Using C- R equations ]

= (3 x 2 + 6 xy - 3 y 2 ) - i(3 x 2 - 6 xy - 3 y 2 ) Putting x = z, y = 0, F ¢( z ) = 3z 2 - i3z 2 = 3(1 - i ) z 2

2.78

Chapter 2 Analytic Functions

Integrating w.r.t. z, F ( z ) = Ú 3(1 - i )z 2 dz = 3 (1 - i )

z3 +c 3

(1 + i ) f ( z ) = (1 - i )z 3 + c f (z) = =

[ Resubstitutting F ( z )]

(1 - i ) 3 c z + 1+ i 1+ i c (1 - i )2 z3 + (1 + i )(1 - i ) 1+ i

c 1 + i 2 - 2i 3 z + 2 1+ i c 3 = -i z + 1+ i =

= -i z 3 + c ¢ where

c¢ =

c 1+ i

example 16 Determine the analytic function f (z) = u + iv such that u – v = ex(cos y – sin y). [Summer 2015] Solution f ( z ) = u + iv

 (1)

i f ( z ) = iu + i 2 v = iu - v

 (2 )

Adding Eqs (1) and (2), f ( z ) + i f ( z ) = (u + iv) + (iu - v) (1 + i ) f ( z ) = (u - v) + i(u + v) F ( z ) = U + iV where F ( z ) = (1 + i ) f ( z ), U = u - v, V = u + v Since f (z) is analytic, F(z) is also analytic. U = u - v = e x (cos y - sin y) ∂U ∂U = e x (cos y - sin y), = e x (- sin y - cos y) ∂x ∂y ∂U ∂V +i ∂x ∂x ∂U ∂U = -i ∂x ∂y

F ′( z ) =

[ Using C- R equations ]

= e x (cos y - sin y) - ie x (- sin y - cos y)

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.79

Putting x = z, y = 0, F ′( z ) = e z (cos 0 - sin 0) + ie z (sin 0 + cos 0) = e z (1 + i ) Integrating w.r.t. z, F ( z ) = Ú (1 + i ) e z dz = (1 + i ) e z + c (1 + i ) f ( z ) = (1 + i )e z + c c f (z) = ez + 1+ i = e z + c′ c where c′ = 1+ i

example 17 If u + v =

2 sin 2 x e

2y

+ e -2 y - 2 cos 2 x

, find the analytic function f (z) = u + iv.

Solution f (z) = u + v i f (z) = iu – v Adding Eqs (1) and (2), (1+ i ) f ( z ) = (u - v) + i(u + v) F ( z ) = U + iV where

F ( z ) = (1 + i ) f ( z ), U = u - v, V = u + v V = u+v =

2 sin 2 x

+ e -2 y - 2 cos 2 x 2 sin 2 x = 2 cosh 2 y - 2 cos 2 x sin 2 x = cosh 2 y - cos 2 x ∂V (2 cos 2 x )(cosh 2 y - cos 2 x ) - sin 2 x(2 sin 2 x ) = ∂x (cosh 2 y - cos 2 x )2 =

e

2y

2 cos 2 x cosh 2 y - 2 (cosh h 2 y - cos 2 x )2

...(1) ...(2)

2.80

Chapter 2 Analytic Functions

∂V sin 2 x =(2 sinh 2 y) ∂y (cosh 2 y - cos 2 x )2 ∂U ∂V +i ∂x ∂x ∂V ∂V = +i ∂y ∂x

F ′( z ) =

=-

[ Using C- R equations ]

2 sin 2 x sinh 2 y (cosh 2 y - cos 2 x )

2

+i

2(cos 2 x cosh 2 y - 1) (cosh 2 y - cos 2 x )2

Putting x = z, y = 0, F ′( z ) = -0 + i =i

2(cos 2 z ◊ cosh 0 - 1)

(cosh 0 - cos 2 z )2 2(cos 2 z - 1)

(1 - cos 2 z )2 2 = -i 1 - cos 2 z 2 = -i 2 sin 2 z = -cosec 2 z Integrating w.r.t. z, F ( z ) = Ú -cosec 2 z dz = cot z + c (1 + i ) f ( z ) = cot z + c f (z) =

cot z c + 1+ i 1+ i

(1 - i ) cot z + c′ 2 c where c′ = 1+ i =

example 18 Determine the analytic function f (z) = u + iv if 3u + 2 v = y 2 - x 2 + 16 xy. Solution f (z) = u + iv

....(1)

i f (z) = iu – v

...(2)

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.81

Multiplying Eq. (1) by 2 and Eq. (2) by 3 and adding, 2 f ( z ) + 3i f ( z ) = 2(u + iv) + 3(iu - v) (2 + 3i ) f ( z ) = (2u - 3v) + i(3u + 2 v) F ( z ) = U + iV where F ( z ) = (2 + 3i ) f ( z ), U = 2u - 3v, V = 3u + 2 v Since f (z) is analytic, F(z) is also analytic. V = 3u + 2 v = y 2 - x 2 + 16 xy ∂V ∂V = -2 x + 16 y, = 2 y + 16 x ∂x ∂y ∂U ∂V F ′( z ) = +i ∂x ∂x ∂V ∂V = +i [ Using C- R equations ] ∂y ∂x = (2 y + 16 x ) + i(-2 x + 16 y) Putting x = z, y = 0, F ′( z ) = 16 z + i(-2 z ) = (16 - 2i ) z Integrating, w.r.t. z, F ( z ) = Ú (16 - 2i )z dz z2 +c 2 = (8 - i )z 2 + c = (16 - 2i )

(2 + 3i ) f ( z ) = (8 - i )z 2 + c (8 - i ) 2 c f (z) = z + 2 + 3i 2 + 3i c (8 - i )(2 - 3i ) 2 = z + c′, where c′ = 2 + 3i 4+9 13 - 26i 2 = z + c′ 13 = (1 - 2i )z 2 + c′

example 19 Find the analytic function f ( z ) = u + iv if u + v =

x x + y2 2

and f (1) = 1.

2.82

Chapter 2 Analytic Functions

Solution

f ( z ) = u + iv

 (1)

i f ( z ) = iu + i v = iu - v Adding Eqs (1) and (2), f ( z ) + i f ( z ) = (u + iv) + (iu - v) (1 + i ) f ( z ) = (u - v) + i(u + v) F ( z ) = U + iV where F ( z ) = (1 + i ) f ( z ), U = u - v, V = u + v 2

Since f (z) is analytic, F(z) is also analytic. V = u+v x = 2 x + y2 ∂V ( x 2 + y 2 )(1) - x(2 x ) = ∂x ( x 2 + y 2 )2 =

x 2 + y2 - 2 x 2 ( x 2 + y 2 )2 y2 - x 2

=

( x 2 + y 2 )2 x ∂V =- 2 ◊ 2y ∂y ( x + y 2 )2 =

-2 xy ( x + y 2 )2 2

∂U ∂V +i ∂x ∂x ∂V ∂V = +i ∂y ∂x

F ′( z ) =

= =

-2 xy ( x 2 + y 2 )2

+i

y2 - x 2 ( x 2 + y 2 )2

-2 xy + i( y 2 - x 2 ) ( x 2 + y 2 )2

Putting x = z, y = 0, F ′( z ) = =-

iz 2 z4 i z2

 (2 )

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.83

Integrating w.r.t. z, F (z) = Ú -

i z2

dz

i +c z i (1 + i ) f ( z ) = + c z =

Putting z = 1, (1 + i )(1) = i + c

[∵

f (1) = 1]

c =1 i \ (1 + i ) f ( z ) = + 1 z 1 i + f ( z) = z(1 + i ) 1 + i 1+ i 1- i = + 2z 2

example 20 Determine the analytic function f (z) = u + iv such that Êp ˆ 3-i e y - cos x + sin x . u-v = where f Á ˜ = Ë 2¯ 2 cosh y - cos x Solution f ( z ) = u + iv

 (1)

i f ( z ) = iu + i v 2

= iu - v Adding Eqs (1) and (2),

wherre

f ( z ) + i f ( z ) = (u + iv) + (iu - v) (1 + i ) f ( z ) = (u - v) + i(u + v) F ( z ) = U + iV F ( z ) = (1 + i ) f ( z ), U = u - v, V = u + v

Since f (z) is analytic, F(z) is also analytic. U = u-v =

e y - cos x + sin x cosh y - cos x

 (2 )

2.84

Chapter 2 Analytic Functions

∂U (sin x + cos x )(cosh y - cos x ) - (e y - cos x + sin x )(sin x ) = ∂x (cosh y - coss x )2 ∂U e y (cosh y - cos x ) - (e y - cos x + sin x )sinh y = ∂y (cosh y - cos x )2 ∂U ∂V +i ∂x ∂x ∂U ∂U = -i ∂x ∂y

F ′( z ) =

=

[ Using C- R equations ]

(sin x + cos x )(cosh y - cos x ) - (e y - cos x + sin x )sin x (cosh y - cos x )2 -i

e y (cosh y - cos x ) - (e y - cos x + sin x )sinh y (cosh y - cos x )2

Putting x = z, y = 0, (sin z + cos z )(1 - cos z ) - (1 - cos z + sin z )sin z F ′( z ) = (1 - cos z )2 (1 - cos z ) - (1 - cos z + sin z ) ⋅ 0 -i (1 - cos z )2 = = =

sin z - sin z cos z + cos z - cos2 z - sin z + sin z cos z - sin 2 z - i(1 - cos z ) (1 - cos z )2 cos z - 1 - i(1 - cos z ) (1 - cos z )2 (-1 - i )(1 - cos z )

(1 - cos z )2 -(i + 1) = 1 - cos z -(i + 1) = z 2 sin 2 2 Integrating w.r.t. z, -(i + 1) dz 2 z 2 sin 2 (i + 1) z =cosec 2 dz 2 Ú 2 (i + 1) Ê zˆ =-2 cot ˜ + c 2 ÁË 2¯

F ( z) = Ú

2.11

Construction of Analytic Functions: Milne—Thomson Method

2.85

z +c 2 z (1 + i ) f ( z ) = (i + 1) cot + c 2 p Putting z = , 2 3-i p (1 + i ) ÁÊË 2 ˆ˜¯ = (i + 1) cot 4 + c = (i + 1) cot

\

c = 2 + i - i -1 =1 z 1 f ( z ) = cot + 2 i +1 z 1- i = cot + 2 2

example 21 Show that 2x (1 – y) can be the imaginary part of an analytic function. Solution Since real and imaginary parts of an analytic function are harmonic functions, they satisfy the Laplace equation. Let v = 2x (1 – y). ∂v = 2(1 - y), ∂x ∂2 v ∂x 2 ∂2 v ∂x

2

+

∂2 v ∂y 2

= 0,

∂v = -2 x ∂y ∂2 v ∂y 2

=0

=0

Hence, v can be the imaginary part of an analytic function.

example 22 Find the orthogonal trajectory of the family of the curves 3x2y – y3 = a. Solution Let u = 3x2y – y3 and f (z) = u + iv be an analytic function. Then v = b will be the orthogonal trajectory to u = a.

2.86

Chapter 2 Analytic Functions

u = 3 x 2 y - y3 ∂u ∂u = 6 xy, = 3 x 2 - 3 y2 ∂x ∂y ∂u ∂v f ′( z ) = +i ∂x ∂x ∂u ∂u = -i [ Using C- R equations ] ∂x ∂y = 6 xy - i(3 x 2 - 3 y 2 ) Putting x = z, y = 0, f ′( z ) = -i(3z 2 ) Integrating w.r.t. z, = f ( z ) = -i Ú 3z 2 dz = -i z 3 + c u + iv = -i( x + iy)3 + c = -i( x 3 + i 3 y3 + 3 x 2 iy + 3 xi 2 y 2 ) + c = -ix 3 + i 2 y3 - 3 x 2 i 2 y + 3ixy 2 + c = (- y3 + 3 x 2 y) + i(- x 3 + 3 xy 2 ) + c Comparing imaginary parts, v = 3xy2 – x2 + c Hence, the orthogonal trajectory is 3 xy 2 - x 3 + c = b 3 xy 2 - x 3 = c′ where c¢ = b – c

exercIse 2.5 1. Prove that the function v = e - x (x cos y + y sin y ) is harmonic and determine the corresponding analytic function f (z) = u + iv . 2. Construct the analytic function whose real part is (i)

y x2 + y2

(ii) e 2 x (x cos 2y - y sin 2y )

(iii) cos x cosh y (iv) e x ÈÎ(x 2 - y 2 )ccos y - 2 xy sin y ˘˚ [ans.: (i) log z + c (ii) ze 2 z + c (iii) cos z + c (iv) x 2 e z + c]

2.11

2.87

Construction of Analytic Functions: Milne—Thomson Method

3. Construct the analytic function whose imaginary part is (i)

(x - y ) (x 2 + y 2 )

(ii) - sin x sinh y

(ii) e - x (x sin y - y cos y )

(iv)

2 sin x sin y cos 2 x + cosh 2y

2 cos x cosh y È ˘ 1+ i z Í ans. :(i) z + c (ii)cos z + c (iii) z e + c (iv) cos 2 x + cosh 2y + c ˙ Î ˚ 4. Show that the function u = sin x cosh y + 2 cos x sinh y + x 2 - y 2 + 4 xy is harmonic and find the analytic function f(z) = u + iv. ÈÎ ans. : (1 - 2i)(sin z + z 2 ) + c ˘˚ 5. Find the analytic function f(z) = u + iv given that 2u + 3v = e x (cos y - sin y ). È ˘ (-1 + 5i) e z + c˙ Í ans. : 13 Î ˚ 6. Find the analytic function f(z) = u + iv, given that 2u + v = e2x {(2 x + y )cos 2y + (x - 2y )sin 2y }. ÈÎ ans. : f (z) = ze 2 z + c ˘˚ cos x + sin x - e - y 7. Find the analytic function z = u + iv if u - v = when 2 cos x - e y - e - y p Ê ˆ f Á ˜ = 0. Ë 2¯ 1 z˘ È Í ans. : 2 - cot 2 ˙ Î ˚ 8. Show that the function u = e -2 xy sin(x 2 - y 2 ) is harmonic. Find the conjugate function v and express u + iv as an analytic function of z. È ans. : v = -e -2 xy cos(x 2 - y 2 ) + c, f (z) = -ieiz + c ˘ Î ˚ 2

9. Find the orthogonal trajectories of the family of curves (i) x 3 y - xy 3 = c

(ii) e x - cos y - xy = c ÈÎ ans. :(i) x 4 - 6 x 2 y 2 + y 4 = c (ii) x 2 - y 2 + 2e x - sin y = c ˘˚

2.88

Chapter 2 Analytic Functions

points to remember Analytic Functions A function f(z) is said to be an analytic function if it is defined and differentiable at each point of a region R. Entire Function A function f(z) is said to be an entire function if it is analytic everywhere in the finite plane (complex plane). 1. Necessary Conditions for f(z) to be Analytic (Cauchy—Riemann Equations) The necessary conditions for a function f (z) = u + iv to be an analytic function at all the points in a region R are (i)

∂u ∂v = ∂x ∂y

(ii)

∂u ∂v =∂y ∂x

2. Sufficient Conditions for f(z) to be Analytic For a function f (z) = u + iv, if the partial derivatives continuous in the region R and

∂u ∂u ∂v ∂v , , and are ∂x ∂y ∂x ∂y

∂u ∂v ∂u ∂v = , =then the function of f (z) is ∂x ∂y ∂y ∂x

analytic.

Cauchy–Riemann Equations in Polar Form If f (z) = u + iv is an analytic function where u and v are functions of r ,q and z = reiq then ∂u 1 ∂v ∂u ∂v = , = -r ∂r r ∂q ∂q ∂r

Harmonic Functions A real function f of two variables x and y is said to be a harmonic function in a region R if it has continuous second-order partial derivatives and satisfies the Laplace equation ∂2f ∂x 2

+

∂2f ∂y 2

=0

—2f = 0, where — 2 =

∂2 ∂x 2

+

∂2 ∂y 2

Properties of Analytic Functions (i) If f (z) = u + iv is an analytic function, u and v are harmonic functions. (ii) If f (z) = u + iv is an analytic function then the family of curves u(x, y) = c1 and the family of curves v(x, y) = c2 cut orthogonally.

3 Complex Integration CHAPTER

chapter outline 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

3.1

Introduction Some Basic Definitions Line Integral Simply connected and Multiply connected Regions Cauchy’s Integral Theorem Cauchy’s Integral Formula Liouville’s Theorem Maximum Modulus Theorem

IntroductIon

The concept of a real line integral is extended to that of a complex line integral. Complex integration plays an important role in the evaluation of complicated real integrals. It is a powerful tool in evaluating certain integrals. In case of real integration, b

Ú f ( x) dx,

the path of integration is always along the x-axis from x = a to x = b. In

a

z2

case of complex integration,

Ú f ( z ) dz ,

the path of integration can be along any curve

z1

from z = z1 to z = z2 but the value of the integral does not depend upon the path if f(z) is analytic.

3.2

some basIc defInItIons

1. continuous arc The set of points (x, y) defined by x = f(t), y = y(t), where t is a parameter in the interval (a, b), represents a continuous arc (or curve) provided f and y are continuous functions.

3.2

Chapter 3

Complex Integration

2. smooth arc If f and y are differentiable in the closed interval a £ t £ b and z¢(t) π 0, the arc is said to be smooth. 3. simple curve The curve which does not intersect itself is said to be a simple curve. 4. simple closed curve The curve which does not intersect itself and has coincident end points [z(b) = z(a)] is said to be a simple closed curve. 5. contour

A contour is a continuous chain of a finite number of smooth arcs.

6. closed contour A closed contour is a piecewise smooth closed curve in which only the initial and final values of z(t) are same.

3.3

LIne IntegraL

Let f (z) be a continuous function of the complex variable z = x + iy defined at every point of a curve C whose end points are A and B (Fig. 3.1). Divide the curve C into n parts at the points. A = P0 ( z0 ), P1 ( z1 ), P2 ( z2 ),  Pi ( zi ) Pn ( zn ) = B Let dzi = zi – zi – 1 and let xi be a point on the arc Pi – 1 Pi. The limit of the n

sum

 f (xi )d zi

as n Æ • in such

i =1

a way that the length of the chord dzi approaches zero, is called the line integral of f (z) along the path C and is denoted by Ú f ( z ) d z. If C is a closed C

fig. 3.1

curve, i.e., if Po and Pn coincide, the integral is called the contour integral and is denoted by

Ú f (z) dz . C

3.3.1

Properties of Line Integrals

If f(z) and g(z) are integrable along a curve C then the following properties hold. 1. Linearity

Ú ÈÎk1 f (z) + k2 g(z)˘˚ dz = k1 Ú f (z)dz + k2 Ú g(z) dz

C

C

C

3.3

Line Integral

3.3

2. sense reversal b

Ú a

a

f ( z ) dz = - Ú f ( z ) dz b

3. Partitioning of Path If the curve C consists of the curves C1 and C2 then

Ú f (z) dz = Ú

C

f ( z ) dz +

C1

Ú

f ( z ) dz

C2

4. Integral Inequality

Ú f (z) dz £ Ú

C

f ( z ) dz

C

5. mL Inequality If f(z) is continuous on the curve C of length L and f(z) £ M then

| |

Ú

f ( z ) dz £ ML

C

example 1 2 Find an upper bound for the absolute value of the integral Ú z dz, where C

C is the straight-line segment from 0 to 1 + i.

[Summer 2014]

Solution The length of the line segment OA from 0 to 1 + i (Fig. 3.2) is L = (1 - 0)2 + (1 - 0)2 = 2 Let

f ( z ) = z 2 = ( x + iy)2 = x 2 - y 2 + 2ixy

Along the line OA, y = x f ( z ) = x 2 - x 2 + 2ix 2 = 2ix 2 f ( z ) = 2ix 2 = 2x £2 M=2

fig. 3.2

2

[∵ 0 £ x £ 1]

3.4

Chapter 3

Complex Integration

By ML inequality,

Ú f (z) dz

£ ML

C

Úz

2

dz £ 2 2

C

example 2 Find an upper bound for the absolute value of the integral

ez dz , Ú C z +1

[Summer 2013]

where C is |z| = 4. Solution

||

The length of the circular arc z = 4 (Fig. 3.3) is L = 2p r = 2p (4) = 8p

Let

f (z) =

ez z +1

f ( z) =

ez z +1

£

ez z +1

e4 £ 3 M=

e4 3

By ML inequality,

Ú f (z) dz

£ ML

C

ez

Ú z + 1 dz C

£

fig. 3.3

8p e 4 3

È∵ max ( z ) = 4, e z £ e 4 ˘ Í ˙ Í min ( z ) = -4, z + 1 ≥ 3, 1 £ 1 ˙ Í z + 1 3 ˙˚ Î

3.3

example 3 Find an upper bound for the integral

Line Integral

3.5

ez

Ú z 2 dz, where C is the line segment

C

from z = i to z = 2. Solution

Let z = i is the point A(0, 1) and z = 2 is the point B (2, 0). The length of the line segment AB (Fig. 3.4) is

y

L = (0 - 2)2 + (1 - 0)2 = 5 The equation of the line AB is 1- 0 y-0 = ( x - 2) 0-2 -2 y = x - 2

A(0,1)

x + 2y = 2 Let

f (z) = = = f (z) =

= = Along the line AB, y =

f (z) =

=

x

O (0, 0)

ez

B(2,0)

fig. 3.4

z2 e x + iy ( x + iy)2 e x eiy ( x + iy)2 e x eiy ( x + iy)2 e x eiy x + iy

2

ex

[∵ e x ≥ 0 for all values of x ]

x 2 + y2 2- x 2 ex Ê 2 - xˆ x2 + Á Ë 2 ˜¯

2

4e x 4 x2 + 4 + x2 - 4 x

3.6

Chapter 3

Complex Integration

=

4e x 5x2 - 4 x + 4

For maximum value, d f (z) = 0 dx 4e x (5 x 2 - 4 x + 4) - 4e x (10 x - 4) (5 x 2 - 4 x + 4)2

=0

5 x 2 - 14 x + 8 = 0 x = 2, But f(z) is maximum at x =

| |

4 5

4 . 5 f (z) =

4e x 5x2 - 4 x + 4 4

£

ez z2

£

4e 5 2

Ê 4ˆ Ê 4ˆ 5Á ˜ - 4 Á ˜ + 4 Ë 5¯ Ë 5¯ 4 e5 4

M = e5 By ML inequality,

Ú f (z) dz

£ ML

C

ez

Ú z2

4

dz £ 5 e 5

C

3.3.2

evaluation of Line Integrals

Let

f(z) = u + iv z = x + iy,

dz = dx + idy

Ú f (z) dz = Ú (u + iv) (dx + idy)

C

C

= Ú (udx - vdy) + i Ú (vdx + udy) C

C

Here, u and v are functions of x and y. To evaluate the line integral, it is converted to a single variable x or y using the equation of the curve C.

3.3

Line Integral

3.7

note If f(z) is analytic in a simply connected domain D then the line integral is independent of the path, i.e., z2

Ú f (z) dz = F (z2 ) - F (z1 )

z1

where

F¢(z) = f (z)

example 1 2 +i

z 2 dz along the line y =

Ú

Evaluate

0

x . 2

[Summer 2013]

Solution f (z) = z2 f ¢( z ) = 2 z Since f (z) is differentiable for all finite values of z, it is analytic in the z-plane. 2 +i

Hence,

Ú

z 2 dz will be independent of the path.

0

2 +i

Ú 0

z3 z dz = 3

2 +i

2

0

1 (2 + i )3 3 1 = (8 + i 3 + 6i 2 + 12i ) 3 1 = (8 - i - 6 + 12i ) 3 1 = (2 + 11 i ) 3 =

example 2 1+i

Evaluate

Ú

( x - y + ix 2 ) dz along the line from z = 0 to z = 1 + i.

0

Solution Let

z = x + iy dz = dx + idy

3.8

Chapter 3

Complex Integration

The equation of the line OA (Fig. 3.5) is 1- 0 ( x - 0) y-0 = 1- 0 y=x 1+ i

Ú

1+ i

Ú ( x - y + ix

( x - y + ix 2 ) dz =

0

2

) (dx + idy)

2

) (dx + idy)

0

Ú ( x - y + ix

=

0A

Along OA, y = x, dx = dy, x varies from 0 to 1. 1+ i

fig. 3.5

1

Ú ( x - y + ix

2

0

)dz = Ú ix 2 (dx + idx ) 0

1

= Ú ix 2 (1 + i ) dx 0

1

= (-1 + i ) Ú x 2 dx 0

= (-1 + i )

x3 3

1

0

-1 + i = 3 1 1 =- + i 3 3

example 3 Evaluate

Ú Re z dz,

C

where C is the shortest path from 1 + i to 3 + 2i. [Summer 2014] y

Solution Let

z = x + iy Re z = x dz = dx + idy The shortest distance from A(1 + i) to B(3 + 2i) is the line joining these points (Fig. 3.6). The equation of the line AB is y -1 =

2 -1 ( x - 1) 3 -1

B(3,2)

A(1,1) x

O (0,0)

fig. 3.6

3.3

Line Integral

3.9

2( y - 1) = x - 1 x - 2 y = -1

Ú Re z dz = Ú x (dx + i dy)

C

C

Along AB, y =

x +1 1 , dy = dx, x varies from 1 to 3. 2 2 3

1 ˆ Ê Ú Re z dz = Ú x ÁË dx + i 2 dx˜¯ 1

C

iˆ Ê = Á1 + ˜ Ë 2¯

3

Ú x dx 1

i ˆ x2 Ê = Á1 + ˜ Ë 2¯ 2

3

1

i ˆ Ê 9 1ˆ Ê = Á1 + ˜ Á - ˜ Ë 2¯ Ë 2 2¯ iˆ Ê = Á1 + ˜ 4 Ë 2¯ = 4 + 2i

example 4

Úz

Evaluate

C

2

dz, where C is the line joining the points (0, 0) and (4, 2). [Winter 2013]

Solution

y

Let

z = x + iy dz = dx + idy The equation of the line joining O(0, 0) and A(4, 2) (Fig. 3.7) is 2-0 y-0 = ( x - 0) 4-0 2y = x Along OA, x = 2y, dx = 2dy, y varies from 0 to 2.

0

x

O (0,0)

fig. 3.7

4 + 2i

Ú

A(4,2)

z dz = Ú ( x + iy) (dx + i dy) 2

2

C

= Ú ( x 2 - y 2 + 2ixy) (dx + idy) C

3.10

Chapter 3

Complex Integration

2

= Ú (4 y 2 - y 2 + 4iy 2 ) (2dy + idy) 0 2

= Ú (3 + 4i ) y 2 (2 + i ) dy 0

= (6 + 3i + 8i + 4i 2 )

y3 3

2

0

Ê 23 ˆ = (2 + 11 i ) Á ˜ Ë 3¯ = Aliter: Let

8 (2 + 11 i ) 3

f (z) = z2 f¢(z) = 2z

Since f (z) is differentiable for all finite values of z, it is analytic in the z-plane. 4 + 2i

Hence,

Ú

z 2 dz will be independent of the path.

0

4 + 2i

Ú 0

z3 z dz = 3

4 + 2i

2

=

0

1 (4 + 2i )3 3

23 (2 + i )3 3 8 = (8 + i 3 + 6i 2 + 12i ) 3 8 = (8 - i - 6 + 12i ) 3 8 = (2 + 11i ) 3 =

example 5 Evaluate

Ú z dz ,

where C is along the sides of the triangle having

C

vertices z = 0, 1, i.

[Summer 2015]

3.3

Line Integral

3.11

Solution z = x + iy, z = x - iy dz = dx + i dy

Let

Ú z dz = Ú ( x - iy)(dx + i dy)

C

C

The equation of the line joining A(1, 0) and B(0, 1) (Fig. 3.8) is 1- 0 y-0 = ( x - 1) 0 -1 y = -x +1 x+ y =1

fig. 3.8

Along OA, y = 0, dy = 0, x varies from 0 to 1. 1

Ú z dz = Ú x dx

C

0

x2 = 2 1 = 2

1

0

...(1)

Along AB, y = –x + 1, dy = – dx, x varies from 1 to 0. 0

Ú z dz = Ú [x - i(- x + 1)](dx - idx)

C

1

0

= Ú [(1 + i ) x - i ](1 - i ) dx 1

x2 = (1 - i ) (1 + i ) - ix 2 È (1 + i ) ˘ = (1 - i ) Í+ i˙ 2 Î ˚ È -1 - i + 2i ˘ = (1 - i ) Í ˙˚ 2 Î (1 - i )(i - 1) = 2 =i Along BO, x = 0, y varies from 1 to 0.

0

1

...(2)

3.12

Chapter 3

Complex Integration

0

Ú z dz = Ú (-iy) i dy 1

C

0

= Ú y dy 1

y2 2 1 =2

0

=

1

...(3)

Adding Eqs (1), (2), and (3), 1

1

Ú z dz = 2 + i - 2

C

=i

example 6 Find the value of the integral

Ú z dz ,

where C is the right-hand half

C

pˆ Ê p z = 2e Á - £ q £ ˜ of the circle z = 2, from z = –2i to z = 2i. Ë 2 2¯ [Winter 2014]

||

iq

Solution z = 2eiq dz = 2ieiq dq z = 2e - iq p 2

Ú z dz = Ú 2e

- iq

◊ 2ieiq dq

p 2 p 2

C

-

=

Ú

4i dq

p 2

= 4i q

p 2 -

fig. 3.9 p 2

3.3

Line Integral

3.13

Êp pˆ = 4i Á + ˜ Ë 2 2¯ = 4ip

example 7 Show that

dz = p i, where C is the right half of the circle z = 2. z C

||

Ú

Solution Let z = reiq For the circle z = 2, r = 2

||

z = 2eiq dz = 2ieiqdq

||

For the right half of the circle z = 2, q varies from p p - to (Fig. 3.10). 2 2 dz Úz = C

p 2

Ú

-

p 2

fig. 3.10

2ieiq dq 2eiq

p 2

=

Ú i dq

-

p 2

=i q

p 2 -

p 2

Êp pˆ = iÁ + ˜ Ë 2 2¯ = pi

example 8 Evaluate

Ú

z dz, where C is the left half of the unit circle |z| = 1 from

C

z = –i to z = i.

3.14

Chapter 3

Complex Integration

Solution z = reiq = eiq

Let

[∵ r = 1 for | z | = 1]

iq

dz = ie dq In the left half of the unit circle, q varies from 3p p to (Fig. 3.11). 2 2 p 2

Ú z dz = Ú 1◊ ie

iq

dq

3p 2

C

p 2

eiq =i i =e

i

fig. 3.11

3p 2

p 2

i

3p

-e 2 p p 3p 3p = cos + i sin - cos - i sin 2 2 2 2 = 0 + i - 0 - i(-1) = 2i

example 9

Ú (3z

Evaluate

2

+ 2 z + 1) dz, where C is the arc of the cycloid

C

x = a(q + sin q), y= a(1 – cos q), between q = 0 to = 2p. Solution Let f (z) = 3z2 + 2z + 1 Since f (z) is a polynomial, it is analytic everywhere. Hence, the integral is independent of the path. z = x + iy = a(q + sin q ) + ia(1 - cos q ) At q = 0, At q = 2p,

z=0 z = 2ap 2 ap

Ú

C

f ( z ) dz =

Ú 0

(3z 2 + 2 z + 1) dz

3.3

= z3 + z2 + z 2

3.15

Line Integral

2 ap 0 2

= 2 ap (4 a p + 2p a + 1)

example 10 Evaluate Ú ( x 2 - iy 2 ) dz along the parabola y = 2x2 from (1, 2) to (2, 8). C

y

Solution Along y = 2x2, dy = 4x dx, x varies from 1 to 2 (Fig. 3.12). dz = dx + i dy = dx + i ◊ 4 x dx

B(2,8) A(1,2)

= (1 + 4ix ) dx

O (0,0)

2 2 2 2 4 Ú ( x - iy ) dz = Ú ( x - i ◊ 4 x ) (1 + 4ix) dx

x

fig. 3.12

1

C

2

= Ú ( x 2 + 4ix 3 - 4ix 4 - 16i 2 x 5 ) dx 1 2

= Ú ( x 2 + 4ix 3 - 4ix 4 + 16 x 5 ) dx 1

2

x3 x4 x5 x6 + 4i - 4i + 16 3 4 5 6 1 1 4i 8 = (8 - 1) + i(16 - 1) - (32 - 1) + (64 - 1) 3 5 3 =

7 504 Ê 124 ˆ + + i Á 15 ˜ Ë 3 3 5 ¯ 511 49 = -i 3 5 =

example 11 1+i

Evaluate

Ú

2 ( x 2 + iy) dz along the path (i) y = x, (ii) y = x . Is the line

0

integral independent of the path?

3.16

Chapter 3

Complex Integration

Solution (i) Along the path y = x y = x, dy = dx, x varies from 0 to 1 (Fig. 3.13). 1+ i

Ú 0

1

( x 2 + iy) dz = Ú ( x 2 + iy) (dx + idy) 0

fig. 3.13

1

= Ú ( x + ix )(1 + i ) dx 2

0

= (1 + i )

x3 x2 +i 3 2

1

0

Ê1 iˆ = (1 + i ) Á + ˜ Ë 3 2¯ (2 + 3i ) 6 2 + 2i + 3i - 3 = 6 -1 + 5i = 6 = (1 + i )

(ii) Along the path y = x2, y = x2, dy = 2xdx, x varies form 0 to 1. dz = dx + i2x dx 1+ i

Ú 0

1

( x 2 + iy) dz = Ú ( x 2 + iy)(dx + 2ix dx ) 0

1

= Ú ( x 2 + ix 2 ) (1 + 2ix ) dx 0

1

= (1 + i )Ú ( x 2 + 2ix 3 ) dx 0

x3 x4 = (1 + i ) + 2i 3 4 Ê1 iˆ = (1 + i ) Á + ˜ Ë 3 2¯ -1 + 5i 6 Both the line integrals are equal. =

1

0

3.3

f ( z ) = x 2 + iy u + iv = x 2 + iy Comparing real and imaginary parts, u = x2 , ∂u = 2 x, ∂x ∂u = 0, ∂y

v=y ∂v =0 ∂x ∂v =1 ∂y

Since C–R equations are not satisfied, f (z) is not analytic. Hence, the line integral is not independent of the path.

example 12 4 +2i

Evaluate

Ú

z dz along the curve z = t2 + it.

0

Solution f (z) = z Along the curve z = t2 + it, f ( z ) = z = t 2 - it dz = 2t dt + idt = (2t + i ) dt When z = 0, t 2 + it = 0, t = 0 When z = 4 + 2i, t 2 + it = 4 + 2i, t = 2 \ t varies from 0 to 2. 4 + 2i

Ú 0

2

z dz = Ú (t 2 - it ) (2t + i ) dt 0 2

= Ú (2t 3 + it 2 - i 2t 2 - i 2 t ) dt 0

2t 4 t3 t2 = -i + 4 3 2 8 ˆ Ê = Á 8 - i + 2˜ Ë 3 ¯ = 10 - i

8 3

2

0

Line Integral

3.17

3.18

Chapter 3

Complex Integration

example 13 Integrate f(z) = x2 + ixy from (1, 1) to (2, 4) along the curve x = t, y = t2. Solution f (z) = x2 + ixy Along the curve x = t, y = t2 f (z) = t2 + i(t) (t2) = t2 + it3 dz = dx + idy = dt + i2t dt = (1 + 2it) dt When x = 1, t=1 When x = 2, t=2 \ t varies from 1 to 2. 2

Ú

C

f ( z ) dz = Ú (t 2 + it 3 ) (1 + 2it ) dt 1

2

= Ú (t 2 + 2it 3 + it 3 + 2i 2 t 4 ) dt 1 2

= Ú [t 2 - 2t 4 + 3it 3 ] dt 1

t3 t5 t4 = - 2 + 3i 3 5 4

2

1

16 ˆ Ê 1 2 3i ˆ Ê 8 64 =Á + 3i ˜ - Á - + ˜ Ë3 5 4 ¯ Ë3 5 4¯ =-

151 45 +i 15 4

exercIse 3.1 2 +i

1. Evaluate

Ú (2x + iy + 1)dz

along (i) the straight line joining (1 — i) to

1- i

(2 + i), and (ii) x = t + 1, y = 2t2 — 1. 25 ˘ È Í ans.: (i) 4(1 + 2i), (ii) 4 + 3 i ˙ Î ˚

3.3

2. Evaluate

Line Integral

3.19

2z + 3 dz, where C is (i) the upper half of the circle |z| = 2, z C

Ú

(ii) the lower half of the circle |z| = 2, and (iii) the whole circle in the anti-clockwise direction. [ans.: (i) 2(3pi — 4), (ii) 2(4 — 3pi), (iii) 12] 3. Show that

Ú log zdz = 2p i, where C is the unit circle in the Z-plane. C

4. Evaluate

Ú (z - z )dz, 2

where C is the upper half of the circle |z| = 1.

C

2˘ È Í ans.: 3 ˙ Î ˚

5. Evaluate Ú| z |2 dz, where C is the boundary of the square C with vertices C

(0, 0), (1, 0), (1, 1), (0, 1). [ans.: —1 + i]

Ú (2z

6. Evaluate

3

+ 8 z + 2) dz,

where C is the arc of the cycloid

C

x = a (q — sin q) y = a (1 — cos q) between the points (0, 0) and (2pa, 0). [ans.: 4pa(2p3a3 + 4pa + 1)] 7. Evaluate

Ú zdz

from z = 0 to z = 4 + 2i, where C is the curve given by

C

z = t2 + it.

-2 + i

8. Prove that

Ú

-2

9. Evaluate

Ú (z

2

8 ˘ È Í ans.: 10 - 3 i ˙ Î ˚ i (2 + z)2 = dz = - . 3 - 2 z + 1)dz, where C is the circle x2 + y2 = 2.

C

10. Evaluate

Ú z dz 2

[ans.: –8pi] from P(1, 1) to q (2, 4), where C is the curve x = t, y = t2.

C

86 È ˘ Í ans.: - 3 - 6i ˙ Î ˚

3.20

Chapter 3

3.4

Complex Integration

sImPLy connected and muLtIPLy connected regIons

3.4.1

simply connected region

A simply connected region R is the region enclosed by a simple curve, e.g., interior of a circle, rectangle, triangle, ellipse, etc., (Fig. 3.14).

3.4.2

R C

multiply connected region

fig. 3.14 A multiply connected region is the region enclosed by more than one simple curve, e.g., annulus region, regions with holes, etc., (Fig. 3.15a). A multiply connected region can be converted into a simply connected region by introducing cross-cuts (Fig. 3.15b).

fig. 3.15

3.4.3

Independence of Path

If f (z) is analytic in a simply connected region then the line integral is independent of the path.

3.5

cauchy’s IntegraL theorem

If f(z) is an analytic function and f ¢(z) is continuous at each point inside and on a closed curve C then

Ú f (z) dz = 0 C

Proof Let

f ( z ) = u + iv, z = x + iy dz = dx + idy

Ú f (z) dz = Ú (u + iv)(dx + i dy) C

C

= Ú (u dx - v dy) + i Ú (vdx + udy) C

C

3.5

3.21

Cauchy’s Integral Theorem

Ê ∂v ∂u ˆ Ê ∂u ∂v ˆ = ÚÚ Á - - ˜ dx dy + i ÚÚ Á - ˜ dx dy Ë ∂x ∂y ¯ Ë ∂x ∂y ¯ R R Ê ∂u ∂u ˆ Ê ∂u ∂u ˆ dx dy = ÚÚ Á - ˜ dx dy + i ÚÚ Á Ë ∂x ∂x ˜¯ y y ∂ ∂ Ë ¯ R R

[ Using Green’s theorem] [ Using C-R equations]

=0

3.5.1

cauchy–goursat theorem

If f (z) is analytic at all points inside and on a simple closed curve C contained in a simply connected domain D then

Ú f (z) dz = 0 C

3.5.2

extension of cauchy’s Integral theorem to a multiply connected region

If f (z) is analytic in the region R between two simple closed curves C1 and C2 then

Ú

f ( z ) dz =

C1

Ú

C1

f ( z ) dz

C2

C2

A

B

Proof The multiply connected region is made simply connected region by introducing a cross-cut AB (Fig. 3.16). The path of integration along AB and C2 is in the clockwise direction, and along BA and C1 is in the fig. 3.16 anticlockwise direction. By Cauchy’s integral theorem, in a simply connected region C1 ABC2 ,

Ú f (z) dz = 0 Ú f (z) dz + Ú f (z) dz + Ú

f ( z ) dz +

Ú

f ( z ) dz +

AB

C2

BA

Ú

f ( z ) dz = 0

Ú

f ( z ) dz = 0

C1

C2

C1

È ˘ Í∵ Ú f ( z ) dz = - Ú f ( z ) dz ˙ ÍÎ AB ˙˚ BA

Reversing the direction of the integral around C2, - Ú f ( z ) dz + C2

Ú

f ( z ) dz = 0

Ú

f ( z ) dz =

C1

C1

Ú

C2

f ( z ) dz

3.22

note

Chapter 3

Complex Integration

If C1 , C2 , C3 ,… , Cn be n closed

curves within C (Fig. 3.17) then

Ú f ( z ) dz = Ú

C

f ( z ) dz +

C1

Ú

f ( z ) dz

C2

+ +

Ú

f ( z ) dz

Cn

fig. 3.17

example 1 Evaluate

Úe

z

dz, where C is z = 1.

C

Solution f (z) = ez

(i) Let

(ii) Since f (z) is differentiable, it is analytic inside and on C. f¢(z) is continuous inside C. (iii) By Cauchy’s integral theorem,

Úe

z

dz = 0

C

example 2 Evaluate

Ú e

z2

dz, where C is any closed contour. Justify your answer.

C

[Winter 2013]

Solution f (z) = ez

(i) Let

2 2

f ¢( z ) = e z ◊ 2 z (ii) Since f (z) is differentiable for all finite values of z, it is analytic in the z-plane. f¢(z) is continuous in the z-plane. (iii) By Cauchy’s integral theorem,

Ú e C

z2

dz = 0

3.5

Cauchy’s Integral Theorem

3.23

example 3 If C is any simple closed contour, in either direction then show that

Ú exp(z

3

) dz = 0 .

[Winter 2014]

C

Solution (i) Let f ( z ) = exp ( z 3 ) = e z

3

3

f ¢( z ) = e z (3z 2 ) (ii) Since f (z) is differentiable for all finite values of z, it is analytic in the z-plane. f¢(z) is also continuous in the z-plane. (iii) By Cauchy’s integral theorem,

Ú exp (z

3

) dz = 0

C

example 4

Úe

Evaluate

sin z 2

dz, where C is z = 1.

C

Solution (i) Let f ( z ) = esin z

2

(ii) Since f (z) is differentiable, it is analytic inside and on C. f¢ (z) is continuous inside C. (iii) By Cauchy’s integral theorem,

Úe

sin z 2

dz = 0

C

example 5 Evaluate

Ú ( z

2

+ 3) dz, where C is any closed contour. Justify your

C

answer.

[Summer 2013]

3.24

Chapter 3

Complex Integration

Solution f (z) = z2 + 3 f ¢( z ) = 2 z (ii) Since f (z) is differentiable for all finite values of z, it is analytic in the z-plane. f¢(z) is also continuous in the z-plane. (iii) By Cauchy’s integral theorem, (i) Let

Ú (z

2

+ 3) dz = 0

fig. 3.18

C

example 6 2 Evaluate Ú ( z + 2 z ) dz , where C is |z| = 1. C

Solution 2 (i) Let f ( z ) = z + 2 z

(ii) Since f (z) is a polynomial, it is analytic inside and on C. f¢(z) is continuous inside C. (iii) By Cauchy’s integral theorem,

Ú (z

2

+ 2 z ) dz = 0

C

example 7 Evaluate

dz

Ú z + 4,

where C is the circle z = 2.

C

y

Solution (i) Let f ( z ) =

1 z+4

f (z) is not analytic at z = –4.

C z = −4

(0, 0)

(ii) C is the circle z = 2 with the centre at (0, 0) and a radius of 2 (Fig. 3.19).

fig. 3.19

z=2

x

3.5

Cauchy’s Integral Theorem

3.25

(iii) For z = -4, z = -4 = 4 > 2 Hence, z = –4 lies outside C. (iv) f ( z ) isanalytic inside and on C. f ′( z ) is continuous inside C. (v) By Cauchy’s integral theorem, dz

Ú z+4 = 0

C

example 8 Evaluate

1 3z 2 + 7 z + 1 Ú z + 1 dz, where C is z = 2 . C

Solution 3z 2 + 7 z + 1 z +1 f ( z ) is not analytic at z = -1.

(i) Let f ( z ) =

1 1 with the centre at (0, 0) and a radius of (Fig. 3.20). 2 2 1 y (iii) For z = -1, z = -1 = 1 > 2 Hence, z = -1 lies outside C C (iv) f ( z ) is analytic inside and on C. f ′( z ) is continuous inside C (0, 0) z = −1 z= 1 (ii) C is the circle z =

2

( v) By Cauchy’s integral theorem, 3z 2 + 7 z + 1 dz = 0 z +1 C

Ú

fig. 3.20

example 9 Evaluate

dz

Ú ( z - 3)2 ,

C

where C is the circle |z| = 1.

x

3.26

Chapter 3

Complex Integration

Solution (i) Let f ( z ) =

y

1

( z - 3)2 f ( z ) is not analytic at z = 3.

C

(ii) C is the circle z = 1 with the centre at (0, 0) and a radius of 1 (Fig. 3.21).

z=1

(0, 0)

z=3

x

(iii) For z = 3, z = 3 = 3 > 1. Hence, z = 3 lies outside C. (iv) f ( z ) is analytic inside and on C. f ′( z ) is continuous inside C. ( v) By Cauchy’s integral theorem, dz Ú z-3 2 = 0 ) C(

fig. 3.21

example 10 Evaluate

z dz

Ú ( z - 1)( z - 2) ,

C

1 where C is the circle z = . 2

Solution (i) Let f ( z ) =

z ( z - 1)( z - 2)

f ( z ) is not analytic at z = 1 and z = 2. 1 (ii) C is the circle z = with the centre at (0, 0) 2 1 and a radius of (Fig. 3.22). 2 1 (iii) For z = 1, z = 1 = 1 > . 2 Hence, z = 1 lies outside C. 1 For z = 2, z = 2 = 2 > 2 Hence, z = 2 lies outside C. (iv) f ( z ) is analytic inside and on C. f ′( z ) is continuous inside C.

y

C z=1 (0, 0)

z= 1 2

fig. 3.22

z=2

x

3.5

Cauchy’s Integral Theorem

3.27

(v) By Cauchy’s integral theorem, z dz

Ú (z - 1)(z - 2)

=0

C

example 11 Evaluate

ez Ú z + i dz, where C : |z – 1| = 1. C

[Summer 2015]

Solution (i) Let f ( z ) =

ez z+i

|

|

(ii) C is a circle z – 1 = 1 with the centre at (1, 0) and a radius of 1 (Fig. 3.23). (iii) For z = -i, z - 1 = -i - 1 = 2 > 1 Hence, z = –i lies outside C. (iv) f(z) is analytic inside and on C. f ¢(z) is continuous inside C.

fig. 3.23

(v) By Cauchy’s integral theorem, ez Ú z + i dz = 0 C

example 12 Evaluate

e2 z

Ú z 2 + 1 dz ,

C

1 where C is z = . 2

Solution (i) Let f ( z ) =

y

e2 z z2 + 1

=

e2 z ( z + i )( z - i )

f (z) is not analytic at z = ±1. 1 (ii) C is the circle z = with the centre at (0, 0) 2 1 and a radius of (Fig. 3.24). 2 1 2 Hence, z = -i lies outside C.

z=i C

z= 1 2

(0, 0)

z= −i

(iii) For z = -i, z = -i = 1 >

fig. 3.24

x

3.28

Chapter 3

Complex Integration

For z = i, z = i = 1 >

1 2

Hence, z = i lies outside C. (iv) f ( z ) is analytic inside and on C. f ′( z ) is continuous inside C. (v) By Cauchy’s integral theorem, e2 z

Ú z2 + 1 dz = 0

C

example 13 Evaluate

z +1

Ú ( z 2 + 2 z + 4)2 dz,

where C is z - 1 + i = 2.

C

Solution (i) Let f ( z ) = =

z +1 2

y 2

( z + 2 z + 4) z +1

È( z + 1 - 3i )( z + 1 + 3i )˘ Î ˚

z = −1+ √3i 2

C x

f ( z ) is not analytic at z = -1 + 3i and z = -1 - 3i.

(1, −1) z = −1− √3i

(ii) C is the circle z - 1 + i = 2 with the centre at (1, - 1) and a radius of 2 (Fig. 3.25). (iii) For z = -1 + 3i, z - 1 + i = -1 + 3i - 1 + i = 3.39 > 2 Hence, z = -1 + 3i lies outside C. For z = -1 - 3i, z - 1 + i = -1 - 3i - 1 + i = 2.13 > 2 Hence, z = - 1 - 3i lies outside C. (iv) f ( z ) isanalytic inside and on C. f ′( z ) is continuous inside C. (v) By Cauchy’s integral theorem, z +1 Ú (z2 + 2 z + 4)2 dz = 0 C

fig. 3.25

3.5

Cauchy’s Integral Theorem

3.29

example 14 Evaluate Ú sec z dz, where C is z = 1. C

y

Solution 1 cos z f ( z ) is not analytic at cos z = 0, p 3p i.e., z = ± , ± , 2 2 (ii) C is the circle z = 1 with the centre (i) Let f ( z ) = sec z =

C

z=− p 2

(0, 0)

z=1

z= p 2

x

at (0, 0) and a radius of 1 (Fig. 3.26). (iii) For z = ±

p 3p ,± ,…… 2 2

fig. 3.26

z >1 Hence, all these points lie outside C. (iv) f (z) is analytic inside and on C. f¢ (z) is continuous inside C. (v) By Cauchy’s integral theorem,

Ú sec z dz = 0

C

example 15 Evaluate

Ú tan z dz,

where C is z = 1.

C

Solution (i) Let f ( z ) = tan z =

sin z cos z

p 3p ,± , 2 2 (ii) C is the circle z = 1 with the centre at (0, 0) f ( z ) is not analytic at cos z = 0, i.e.., z = ±

and a radius of 1 (Fig. 3.27). p 3p 5p (iii) For z = ± , ± ,± ,… 2 2 2 z >1

y C z= p 2

z=− p 2

(0, 0)

Hence, all these points lie outside C. fig. 3.27

z=1

x

3.30

Chapter 3

Complex Integration

(iv) f (z) is analytic inside and on C. f¢ (z) is continuous inside C. (v) By Cauchy’s integral theorem,

Ú tan z dz = 0

C

exercIse 3.2 evaluate the following integrals using cauchy’s integral theorem: 1.

Úz

z+3 , where C is z - 1 = 1 - 2z + 5

2

C

[ans.: 0] 2.

z

Ú z - 2 dz, where C is

z =1

C

[ans.: 0] 3.

1

Ú 2z - 3 dz, where C is

z =1

C

[ans.: 0] 4.

e2z 1 dz, where C is z = 2 +1 2

Úz C

[ans.: 0] 5.

3z - 1 dz, where C is z = 2 3 -z

Úz C

[ans.: 0] 6.

Ú (x

2

- y 2 + 2ixy ) dz, where C is z = 2

C

[ans.: 0] 7.

e2z 1 ÚC z - 1 dz, where C is z = 2 [ans.: 0]

8.

Ú cot z dz, where C is C

z+

1 1 = 2 3 [ans.: 0]

3.6

3.6

3.31

Cauchy’s Integral Formula

cauchy’s IntegraL formuLa

If f (z) is analytic inside and on a closed curve C and if a is any point inside C then 1 f (z) Ú z - a dz 2p i  C

f (a) =

C C1

Proof Since f (z) is analytic inside and on C,

f (z) is also z-a

r a

analytic inside and on C except at z = a. Draw a small circle C1 with the centre at z = a and the radius r lying entirely inside C (Fig. 3.28). fig. 3.28

f (z) Now, is analytic in the region between C and C1. z-a By Cauchy’s integral theorem for a multiply connected region, f (z)

f ( z)

Ú z - a dz = Ú z - a dz C

C1

iq iq For the circle C1, let z - a = re ,dz = rie dq 2p

f (z)

Ú z - a dz = Ú

0

C

f (a + reiq ) reiq

ireiq dq

2p iq =i Ú f (a + re ) dq 0

In the limiting position, as r Æ 0, the circle C1 shrinks to the point a. Hence, Eq. (3.1) reduces to 2p

f (z) Ú z - a dz = i Ú f (a) dq 0 0 2p

= i f (a) q 0

= i f (a )(2p - 0) = 2p i f (a ) \

f (a) =

1 2p i

f (z)

Ú z - a dz

...(3.1)

3.32

Chapter 3

3.6.1

Complex Integration

cauchy’s Integral formula for the derivative of an analytic function

If a function f (z) is analytic in a region R then its derivative at any point z = a of R is also analytic in R and is given by 1 f (z) dz f ¢( a ) =  Ú 2p i 0 ( z - a )2 where C is any closed curve in R surrounding the point z = a. Proof By Cauchy’s integral formula, f (a) =

1 2p i

f (z)

Ú z - a dz C

Differentiating w.r.t. a using Differentiation Under the Integral Sign (DUIS), f ¢( a ) =

1 2p i

∂ È f (z) ˘

Ú ∂a ÍÎ z - a ˙˚ dz C

È

1

˘

1 = 2p i

Ú f (z) Í- (z - a)2 (-1)˙ dz ˚ Î

1 = 2p i

Ú (z - a)2

C

f (z)

dz

C

Differentiating again w.r.t. a, f ≤(a) =

2! 2p i

Ú (z - a)3 dz

f (z)

3! 2p i

Ú (z - a)4 dz

n! 2p i

Ú (z - a)n+1 dz

C

Similarly, f ′″ (a ) =

f (z)

C

In general, f ( n ) (a) =

3.7

f (z)

C

LIouvILLe theorem

If f (z) is analytic and bounded in the entire z-plane for all z then f (z) is constant.

3.8

maxImum moduLus theorem

If f (z) is nonconstant and analytic inside and on a simple closed contour C then the maximum value of f (z) occurs on C.

| |

3.6

Cauchy’s Integral Formula

3.33

example 1 Prove that

dz

Ú z - a = 2p i

and

C

Ú ( z - a)

n

dz = 0 (n is an integer and

C

n π –1) where C is the circle |z – a| = r.

[Summer 2014]

Solution (a) (i) Let I = Ú

C

dz z-a

|

|

(ii) C is the circle z – a = r with the centre at (a, 0) and a radius of r (Fig. 3.29).

|

| |

|

(iii) For z = a, z – a = a – a = 0 < r Hence, z = a lies inside C. (iv) Let f (z) = 1 f (z) is analytic inside and on C. (v) By Cauchy’s integral formula, f (z)

Ú z - a dz = 2p if (a)

C

Ú

C

dz = 2p i (1) z-a

fig. 3.29

= 2p i (b) (i) Let f(z) = (z – a)n (ii) Since f (z) is a polynomial, it is analytic inside and on C. f¢(z) is continuous inside C. (iii) By Cauchy’s integral theorem,

Ú ( z - a)

n

dz = 0

C

example 2 Evaluate

z

Ú z - 2 dz ,

C

3 where C is the circle z - 2 = . 2

Solution (i) Let I = Ú

C

z dz z-2

(ii) C is the circle z - 2 =

3 3 with the centre at (2, 0) and a radius of (Fig. 3.30). 2 2

3.34

Chapter 3

Complex Integration

(iii) For z = 2, z - 2 = 2 - 2 = 0
2 Hence, z = 2 lies outside C. cos p z 2 z-2 f ( z ) isanalytic inside and on C.

(iv) Let f ( z ) =

cos p z

2

Ú (z - 1))(z - 2) = Ú

C

C

cos p z 2 z - 2 dz z -1

( v) By Cauchy’s integral formula, f (z) Ú z - a = 2p i f (a) C

y

C

(0, 0)

z=2 z=1 z=— 3 2

fig. 3.34

x

3.6

Ú

C

cos p z 2 2 z - 2 dz = 2p i È cos p z ˘ Í ˙ z -1 ÍÎ z - 2 ˙˚

cos p z 2

Cauchy’s Integral Formula

3.37

z =1

È cos p ˘

Ú (z - 1)(z - 2) dz = 2p i ÍÎ 1 - 2 ˙˚

C

= 2p i (1) = 2p i

example 7 Evaluate

dz

Ú z 2 - 7z + 12 , where C is the circle

z = 3.5.

C

Solution (i) Let I = Ú

C

dz 2

z - 7 z + 12

y

dz =Ú ( z )( z - 3) 4 C C

(ii) C is the circle z = 3.5 with the centre at (0, 0) and a radius of 3.5 (Fig. 3.35). (iii) For z = 4, z = 4 = 4 > 3.5

z=4 (0, 0)

z = 3 z = 3.5

Hence, z = 4 lies outside C. For z = 3, z = 3 = 3 < 3.5 Hence, z = 3 lies inside C. 1 z-4 f ( z ) isanalytic inside and on C. 1 dz z-4 Ú z2 - 7z + 12 = Ú z - 3 dz C C

(iv) Let f ( z ) =

(v) By Cauchy’s integral formula, f (z)

Ú z - a dz = 2p i f (a)

C

fig. 3.35

x

3.38

Chapter 3

Ú

C

Complex Integration

1 z - 4 dz = 2p i È 1 ˘ Íz - 4˙ z-3 Î ˚ z =3 dz

Ú z2 - 7z + 12

C

È 1 ˘ = 2p i Í ˙ Î3 - 4 ˚ = -2pii

example 8 Evaluate

1

Ú z 2 - 1 dz ,

where C is the circle with the centre at (1, 0) and the

C

radius = 1. Solution (i) Let I = Ú

C

1 z2 - 1

dz y

1 dz ( )( 1 z + z - 1) C

=Ú

C

(ii) C is the circle with the centre at (1, 0) and a radius of 1 (Fig. 3.36). \ | z - 1| = 1

z=1 z = −1 (0, 0)

(1, 0)

(iii) For z = -1, z - 1 = -1 - 1 = 2 > 1 Hence, z = -1 lies outside C. For z = 1, z - 1 = 1 - 1 = 0 < 1. Hence, z = 1 lies inside C. 1 z +1 f ( z ) isanalytic inside and on C. 1 1 z +1 Ú z2 - 1 dz = Ú z - 1 dz C C

(iv) Let f ( z ) =

( v) By Cauchy’s integral formula, f (z) Ú z - a dz = 2p i f (a) C

fig. 3.36

x

3.6

Ú

C

Cauchy’s Integral Formula

3.39

1 z + 1 dz = 2p i È 1 ˘ Í z + 1˙ z -1 Î ˚ z =1 È 1 ˘

1

Ú z2 - 1 dz = 2p i ÍÎ1 + 1 ˙˚

C

= pi

example 9 z2 + 1

Ú z 2 - 1 dz ,

Evaluate

where C is z - 1 = 1.

C

Solution (i) Let I = Ú

C

z2 + 1 z2 - 1

dz y

2

z +1 dz 1)( z - 1) ( + z C

=Ú

C

(ii) C is the circle z - 1 = 1 with the centre z=1

at (1, 0) and a radius of 1 (Fig. 3.37). (iii) For z = -1, z - 1 = -1 - 1 = 2 > 1.

z = −1 (0, 0)

(1, 0)

Hence, z = -1 lies outside C. For z = 1, z - 1 = 1 - 1 = 0 < 1 Hence, z = 1 lies inside C. 2

z +1 z +1 f ( z ) isanalytic inside and on C.

(iv) Let f ( z ) =

2

z +1

Ú z2 - 1 dz = Ú

C

C

z2 + 1 z + 1 dz z -1

fig. 3.37

x

3.40

Chapter 3

Complex Integration

(v) By Cauchy’s integral formula, f (z) Ú z - a = 2p i f (a) C

Ú

C

z2 + 1 2 z + 1 dz = 2p i È z + 1 ˘ Í ˙ z -1 ÍÎ z + 1 ˙˚ 2

z +1

z =1

È1 + 1 ˘

Ú z2 - 1 dz = 2p i ÍÎ1 + 1 ˙˚

C

= 2p i

example 10 Evaluate

z+4

Ú z 2 + 2 z + 5 dz,

where C is the circle z + 1 + i = 2.

C

Solution (i) Let I = Ú

C

z+4 2

z + 2z + 5

y

dz

z+4 =Ú dz ( z + 1 + 2i )( z + 1 - 2i ) C

z = −1+ 2i

C

(ii) C is the circle z + 1 + i = 2 with the centre at (-1, -1) and a radius of 2 (Fig. 3.38). (iii) For z = -1 - 2i,

x (0, 0) (−1, −1) z = −1−2i

z + 1 + i = -1 - 2i + 1 + i = -i = 1 < 2 Hence, z = -1 - 2i lies inside C. For z = -1 + 2i, z + 1 + i = -1 + 2i + 1 + i = 3i = 3 > 2 Hence, z = -1 + 2i lies outside C.

fig. 3.38

3.6

Cauchy’s Integral Formula

3.41

z+4 z + 1 - 2i f ( z ) isanalytic inside and on C.

(iv) Let f ( z ) =

Ú

C

z+4 = Ú z + 1 - 2i dz z 2 + 2 z + 5 C ( z + 1 + 2i ) z+4

( v) By Cauchy’s integral formula, f (z)

Ú z - a dz = 2p i f (a)

C

Ú

C

z+4 z + 1 - 2i dz = 2p i È z + 4 ˘ Í z + 1 - 2i ˙ z + 1 + 2i Î ˚ z = -1- 2i z+4

Ê

-1 - 2i + 4 ˆ

Ú z2 + 2 z + 5 dz = 2p i ÁË -1 - 2i + 1 - 2i ˜¯

C

Ê 3 - 2i ˆ = 2p i Á Ë -4i ˜¯ =

p (2i - 3) 2

example 11 Evaluate

z

Ú z 2 + 1 dz,

where C is the circle z + i = 1.

C

Solution (i) Let I = Ú

C

z 2

z +1

dz

y

z dz ( z + i )( z - i) C

=Ú

(ii) C is the circle z + i = 1 with the centre at (0, -1) and a radius of 1 (Fig. 3.39). (iii) For z = -i, z + i = -i + i = 0 < 1

z=i (0, 0)

x C

(0, −1) z = −i

Hence, z = -i lies inside C. For z = i, z + i = i + i = 2i = 2 > 1 Hence, z = i lies outside C.

fig. 3.39

3.42

Chapter 3

Complex Integration

z z-i f ( z ) isanalytic inside and on C. z z z i Ú z2 + 1 dz = Ú z + i dz C C

(iv) Let f ( z ) =

(v) By Cauchy’s integral formula, f (z) Ú z - a dz = 2p i f (a)

C

Ú

C

z z - i dz = 2p i È z ˘ Íz -i˙ z+i Î ˚ z =- i È -i ˘

z

Ú z2 + 1 dz = 2p i ÍÎ -i - i ˙˚

C

Ê 1ˆ = 2pii Á ˜ Ë 2¯ = pi

example 12 Evaluate

4 - 3z

Ú z( z - 1)( z - 2) dz,

C

3 where C is the circle z = . 2

Solution (i) Let I = Ú

C

4 - 3z dz z( z - 1)( z - 2)

3 with the centrre 2 3 at (0, 0) and a radius of (Fig. 3.40). 2 3 (iii) For z = 0, z = 0 < 2 Hence, z = 0 lies inside C. 3 For z = 1, z = 1 = 1 < 2 Hence, z = 1 lies inside C.

y

(ii) C is the circle z =

C z=0 (0, 0)

z=2 z=1 z= 3 2

fig. 3.40

x

3.6

Cauchy’s Integral Formula

3.43

3 2 Hence, z = 2 lies outside C. 4 - 3z (iv) Let f ( z ) = z-2 f ( z ) isanalytic inside and on C. For z = 2, z = 2 = 2 >

Ê 4 - 3z ˆ ÁË z - 2 ˜¯ 4 - 3z = z( z - 1)( z - 2) z( z - 1) Ê 4 - 3z ˆ È z - ( z - 1) ˘ =Á Í ˙ Ë z - 2 ˜¯ ÍÎ z ( z - 1) ˙˚ =

Ú

C

4 - 3z È 1 1˘ - ˙ Í z - 2 Î z -1 z ˚

Ê 4 - 3z 4 - 3z ˆ Á ˜ 4 - 3z dz = Ú Á z - 2 - z - 2 ˜ dz z( z - 1)( z - 2) z -1 z ˜ CÁ Ë ¯ =Ú

C

4 - 3z 4 - 3z z - 2 dz - z - 2 dz Ú z z -1 C

(v) By Cauchy’s integral formula, f (z)

Ú z - a dz = 2p i f (a)

C

Ú

C

4 - 3z 4 - 3z z - 2 dz - z - 2 dz = 2p i f (1) - 2p i f (0) Ú 2 z -1 C 4 - 3z

Ê 4 - 3ˆ

Ê 4 - 0ˆ

Ú z(z - 1)(z - 2) dz = 2p i ÁË 1 - 2 ˜¯ - 2p i ÁË 0 - 2 ˜¯

C

= -2p i + 4p i = 2p i

example 13 If C is the circle |z| = 3 and if g( z0 ) =

2z2 - z - 2 Ú z - z dz then find g(2). 0 C

3.44

Chapter 3

Complex Integration

Solution 2z2 - z - 2 dz z - z0 C

(i) g( z0 ) = Ú

(ii) C is the circle z = 3 with the centre at (0, 0) and a radius of 3 (Fig. 3.41). (iii) z0 = 2 lies inside C . (iv) Let f ( z ) = 2 z 2 - z - 2 f ( z) isanalytic inside and on C. (v) By Cauchy’s integral formula,

y

C

f (z)

Ú z - a dz = 2p i f (a)

C

2z2 - z - 2 2 Ú z - 2 dz = 2p i ÈÎ2 z - z - 2˘˚ z =2 C

z=2

(0, 0)

z=3

x

= 2p i ÈÎ2(2)2 - 2 - 2 ˘˚ = 2p i(4)

fig. 3.41

= 8p i

example 14 Evaluate

z-2

Ú z( z - 1) dz, where C is the circle |z| = 3.

C

Solution (i) Let I = Ú

C

z-2 dz z( z - 1)

(ii) C is the circle z = 3 with the centre at (0, 0) and a radius of 3 (Fig. 3.42). (iii) For z = 0, z = 0 < 3

y

Hence, z = 0 lies inside C. For z = 1, z = 1 < 3 Hence, z = 1 lies inside C. (iv) Let f ( z ) = z - 2 f ( z ) isanalytic inside and on C. 1 z - ( z - 1) = z( z - 1) z( z - 1) 1 1 = z -1 z

C z=0 (0, 0)

z=1

fig. 3.42

z=3

x

3.6

z-2

Ê 1

Cauchy’s Integral Formula

3.45

1ˆ

Ú z(z - 1) dz = Ú ÁË z - 1 - z ˜¯ dz

C

C

=Ú

C

1 1 dz - Ú dz z -1 z C

(v) By Cauchy’s integral formula, f (z) Ú z - a dz = 2p i f (a) C z-2

Ú z(z - 1) dz = 2p i f (1) - 2p i f (0)

C

= 2p i(1 - 2) - 2p i (0 - 2) = -2p i + 4p i = 2p i

example 15 Evaluate

sin p z 2 + cos p z 2 Ú ( z - 1)( z - 2) dz, where C is z = 3. C

Solution sin p z 2 + cos p z 2 dz ( z - 1)( z - 2) C

(i) Let I = Ú

(ii) C is the circle z = 3 with the centre at (0, 0) and a radius of 3 (Fig. 3.43). (iii) For z = 1, z = 1 = 1 < 3

y

Hence, z = 1 lies inside C. For z = 2, z = 2 = 2 < 3 Hence, z = 2 lies inside C. (iv) Let f ( z ) = sin p z 2 + cos p z 2 f ( z ) isanalytic inside and on C.

C z=2 z=3 (0, 0) z = 1

1 ( z - 1) - ( z - 2) = ( z - 1)( z - 2) ( z - 1)( z - 2) 1 1 fig. 3.43 = z - 2 z -1 Ê sin p z 2 + cos p z 2 sin p z 2 + cos p z 2 ˆ sin p z 2 + cos p z 2 dz - Ú dz˜ Ú (z - 1)(z - 2) dz = Ú ÁË z 2 z 1 ¯ C C C

x

3.46

Chapter 3

Complex Integration

(v) By Cauchy’s integral formula, f (z)

Ú z - a dz = 2p i f (a)

C

sin p z 2 + cos p z 2 Ú (z - 1)(z - 2) dz = 2p i f (2) - 2p i f (1) C = 2p i(sin 4p + cos 4p ) - 2p i(sin p + cos p ) = 2p i(0 + 1) - 2p i(0 - 1) = 2p i + 2p i = 4pi

example 16 Evaluate

z

Ú ( z - 1)3 dz, where C is

z = 2.

C

Solution (i) Let I = Ú

C

z ( z - 1)3

y

dz

(ii) C is the circle z = 2 with the centre at (0, 0)

C

and a radius of 2 (Fig. 3.44). (0, 0)

(iii) For z = 1, z = 1 < 2

z=1 z=2

Hence, z = 1 lies inside C. (iv) Let f ( z ) = z f ( z ) isanalytic inside and on C. f ′( z ) = 1 f ″( z) = 0 (v) By Cauchy’s integral formula for derivative, f (z) 2p pi Ú (z - a)3 = 2! f ″ (a) C z

Ú (z - 1)3 dz =

C

2p i f ″ (1) 2!

= p i (0 ) =0

fig. 3.44

x

3.6

Cauchy’s Integral Formula

3.47

example 17 Evaluate

sin 6 z

Ú

pˆ CÊ ÁË z - 6 ˜¯

3

dz, whereC is z = 1.

Solution (i) Let I = Ú

C

sin 6 z pˆ Ê ÁË z - 6 ˜¯

3

dz

y

C

(ii) C is the circle z = 1 with the centre at (0, 0) and a radius of 1 (Fig. 3.45). p p (iii) For z = , z = = 0.52 < 1 6 6 Hence, z =

p lies inside C. 6

(iv) Let f ( z ) = sin 6 z f ( z ) isanalytic inside and on C. f ′ ( z ) = 6 sin 5 z cos z f ″ ( z ) = 6(5 sin 4 z cos2 z - sin 6 z ) (v) By Cauchy’s integral formula for derivative, f (z)

Ú (z - a)3 dz =

C

ÚÊ

C

(0, 0) z = p 6

sin 6 z

pˆ ÁË z - 6 ˜¯

3

dz =

2p i f ″ (a) 2! 2p i È 6(5 sin 4 z cos2 z - sin 6 z )˘˚ p z= 2! Î 6

p p pˆ Ê = 6p i Á 5 sin 4 cos2 - sin 6 ˜ Ë 6 6 6¯

fig. 3.45

z=1

x

3.48

Chapter 3

Complex Integration

example 18 Evaluate

e2 z

Ú ( z + 1)4 dz,

where C is the circle z = 2.

C

Solution (i) Let I = Ú

C

e2 z ( z + 1)4

dz

y

(ii) C is the circle z = 2 with the centre C

at (0, 0) and a radius of 2 (Fig. 3.46). (iii) For z = -1, z = -1 = 1 < 2 z = −1

Hence, z = -1 lies inside C.

(0, 0)

z=2

x

(iv) Let f ( z ) = e2 z f ( z ) isanalytic inside and on C. f ′( z ) = 2e2 z

fig. 3.46

f ′′( z ) = 4e2 z f ′″( z ) = 8e2 z (v) By Cauchy’s integral formula for derivative, f (z)

2p i f ″′(a ) 3!

e2 z

2p i f ″′(-1) 3!

Ú (z - a)4 dz =

C

Ú (z + 1)4 dz =

C

2p i È 2 z ˘ 8e ˚ z =-1 3! Î 2p i È -2 ˘ = 8e ˚ 6 Î =

=

8pi 3e2

example 19 Evaluate

È 3

6

˘

Ú Í z - i - ( z - i)2 ˙ dz, where C : |z| = 2.

CÎ

˚

[Summer 2015]

3.6

Cauchy’s Integral Formula

3.49

Solution 6 ˘ È 3 (i) Let I = Ú Í ˙ dz z i ( z - i )2 ˚ CÎ

|| | | ||

(ii) C is the circle z = 2 with the centre at (0, 0) and a radius of 2 (Fig. 3.47). (iii) For z = i, z = i = 1 < 2. Hence, z = i lies inside C. (iv) Let f1(z) = 3, f2(z) = 6 f1(z) and f2(z) are analytic inside and on C. (v) By Cauchy’s integral formula,

Ú

C

f (z) dz = 2p i f (a ) and z-a È 3

f (z)

Ú ( z - a )2

fig. 3.47

dz = 2p if ¢(a )

C

˘

6

Ú ÍÎ z - i - (z - i)2 ˙˚ dz = 2p i ÈÎ f1 (z)˘˚ z = i - 2p i ÈÎ f2¢ (z)˘˚ z = i

C

= 2p i (3) - 2p i (0) = 6pi

example 20 Evaluate

dz

Ú z 4ez ,

where C is z = 1.

C

Solution (i) Let I = Ú

C

=Ú

C

dz z 4 ez e- z z4

dz

(ii) C is the circle z = 1 with the centre

y

at (0, 0) and a radius of 1 (Fig. 3.48) . (iii) For z = 0, z = 0 < 1 Hence, z = 0 lies inside C. -z

(iv) Let f ( z ) = e f ( z ) isanalytic inside and on C.

C z=0 (0, 0)

f ′( z ) = - e - z f ″( z ) = e- z f ″′( z ) = -e - z

fig. 3.48

z=1

x

3.50

Chapter 3

Complex Integration

(v) By Cauchy’s integral formula for derivative, f (z)

Ú (z - a)4 dz =

C

dz

Ú z 4 e4

=

C

2p i f ″′(a ) 3! 2p i È - z ˘ -e ˚ z =0 3! Î

2p i (-1) 6 pi =3 =

example 21 Evaluate

dz

Ú ( z + 1)2 ( z - 2) , where C is the circle

C

3 z = . 2

Solution (i) Let I = Ú

C

dz ( z + 1)2 ( z - 2)

3 3 with the centre at (0, 0) and a radius of (Fig. 3.49). 2 2 3 (iii) For z = -1, z = -1 = 1 < 2 y Hence, z = -1 lies inside C. 3 For z = 2, z = 2 = 2 > C 2 Hence, z = 2 lies outside C. (ii) C is the circle z =

1 (iv) Let f ( z) = z-2 f ( z ) isanalytic inside and on C. 1 f ¢( z ) = ( z - 2 )2 1 dz 2 z Ú (z + 1)2 (z - 2) = Ú (z + 1)2 dz C C

z = −1

(0, 0)

fig. 3.49

z=2 z= 3 2

x

3.6

Cauchy’s Integral Formula

3.51

(v) By Cauchy’s integral formula for derivative, f (z) Ú (z - a)2 = 2p i f ′(a) C 1 È ˘ z - 2 dz = 2p i Í- 1 ˙ Ú 2 Í ( z - 2 )2 ˙ C ( z + 1) Î ˚ z =-1 È

˘ ˙ ˙ 1 2 ( ) Î ˚ Ê 1ˆ = 2p i Á - ˜ Ë 9¯

dz

Ú (z + 1)2 (z - 2) = 2p i ÍÍ-

C

=-

1

2

2p i 9

example 22 Evaluate

z

Ú ( z - 1)( z - 2)2 dz,

C

1 where C is z - 2 = . 2

Solution (i) Let I = Ú

C

z ( z - 1)( z - 2)2

dz

1 1 with the centre at (2, 0) and a radius of (Fig. 3.50). 2 2 1 (iii) For z = 1, z - 2 = 1 - 2 = 1 > y 2 Hence, z = 1 lies outside C. (ii) C is the circle z - 2 =

For z = 2, z - 2 = 2 - 2 = 0 < Hence, z = 2 lies inside C. z (iv) Let f ( z ) = z -1 f ( z ) isanalytic inside and on C.

C

1 2

(2, 0) z=1

z=2

fig. 3.50

x

3.52

Chapter 3

Complex Integration

f ′ (z) = =

( z - 1)(1) - z(1) ( z - 1)2 z -1- z

( z - 1)2 1 =( z - 1)2 z z z -1 Ú (z - 1)(z - 2)2 dz = Ú (z - 2)2 dz C C (v) By Cauchy’s integral formula, f (z)

Ú (z - a)2 dz = 2p i f ′ (a)

C

z È 1 ˘ z -1 Ú (z - 2)2 dz = 2p i - ÍÎ (z - 1)2 ˙˚ C z =2 È

z

1

˘

Ú (z - 1)(z - 2)2 dz = 2p i - ÍÎ (2 - 1)2 ˙˚

C

= -2p i

example 23 Evaluate

z +1

Ú z 4 - 4 z3 + 4 z 2 dz,

where C is the circle z - 2 - i = 2.

C

Solution

y

(i) Let I = Ú

C

=Ú

C

=Ú

C

z +1 4

z - 4 z3 + 4 z2 z +1 2

2

z ( z - 4 z + 4) z +1 z 2 ( z - 2 )2

dz

C

dz

dz

(2, 1) z=0 (0, 0)

z=2

(ii) C is the circle z - 2 - i = 2 with the centre at (2, 1) and a radius of 2 (Fig. 3.51). fig. 3.51

x

3.6

(iii) For z = 0, z - 2 - i = 0 - 2 - i = 5 > 2 Hence, z = 0 lies outside C. For z = 2, z - 2 - i = 2 - 2 - i = 1 < 2 Hence, z = 2 lies inside C. (iv) Let f ( z ) =

z +1

z2 f ( z ) is analytic inside and on C. z 2 (1) - ( z + 1) 2 z

f ′ (z) =

z4

=

- z2 - 2z

=-

z4 z+2

z +1

Ú z 4 - 4 z3 + 4 z2 dz = Ú z2

C

C

z3 z +1

( z - 2)2

z +1 2 = Ú z 2 dz C (z - 2) (v) By Caucchy’s integral formula,

Ú

C

f (z)

( z - a )2

dz = 2p i f ′ ( a )

z +1 È z +2˘ z2 Ú z - 2 2 dz = 2p i ÍÎ- z3 ˙˚ ) z =2 C( z +1

È 2+2˘ ˙ 23 ˚

Ú z 4 - 4 z3 + 4 z2 dz = 2p i ÍÎ-

C

Ê 1ˆ = 2p i Á - ˜ Ë 2¯ = -p i

example 24 Evaluate

1 z2 + 5 dz, where C is z = 4. Ú 2pi C z - 3

Cauchy’s Integral Formula

3.53

3.54

Chapter 3

Complex Integration

Solution (i) By Cauchy’s integral formula, f (z)

Ú z - a dz = 2p i f (a)

C

f (z) 1 dz = f ( a ) Ú 2p i C z - a

 (1)

1 z2 + 5 dz 2p i CÚ z - 3

Given

(2)

Comparing integrand of Eqs (1) and (2), f (z) = z2 + 5

||

(ii) C is the circle z = 4 with the centre at (0, 0) and a radius of 4 (Fig. 3.52).

y

(iii) For z = 3, z = 3 = 3 < 4

C

Hence, z = 3 lies inside C. (iv) f (z) is analytic inside and on C. (0, 0) 2

(v)

z=3

z=4

x

1 z +5 dz = f (3) 2pi CÚ z - 3 = [ z 2 + 5] z = 3

fig. 3.52

= 32 + 5 = 14

example 25 3z 2 + 7 z + 1 Ú z - a dz, where C is the circle z = 2, find the values C of f (3), f ¢(1 + i ) and f ≤(1 + i ). If f (a ) =

Solution (i) By Cauchy’s integral formula, f (z)

Ú z - a dz = 2p i f (a)

C

f (a) =

f (z) 1 dz Ú 2p i z - a

 (1)

3.6

Cauchy’s Integral Formula

3z 2 + 7 z + 1 dz z-a C

 (2 )

f (a) = Ú

Given

3.55

Comparing the integrands of Eqs (1) and (2), f ( z) = 3z 2 + 7 z + 1 2p i f ( z ) = 2p i(3z 2 + 7 z + 1) (ii) C is the circle z = 2 with th he centre at (0, 0) and a radius of 2 (Fig. 3.53). (iii) For z = 3, z = 3 = 3 > 2 Hence, z = 3 lies outside C. f (z) (iv) isanalytic inside and on C. z-3

y

( v) By Cauchy’s integral theorem, f (z) Ú z - 3 dz = 0 C

z = 1+ i

2

3z + 7 z + 1 dz = 0 z-3 C

Ú

C

f (3) = 0

(0, 0)

[∵

z=2 z=3

x

a = 3]

( vi) f ′( z ) = 2p i(6 z + 7) f ″( z ) = 2p i (6 ) = 12p i

fig. 3.53

For z = 1 + i, z = 1 + i = 2 < 2 Hence, z = 1 + i lies inside C. f ¢(1 + i ) = 2p i [6(1 + i ) + 7] = 2p i(13 + 6i ) = 2p (-6 + 13i ) f ¢¢(1 + i) = 12 pi

exercIse 3.3 evaluate the following integrals using cauchy’s integral formula: 1.

z dz

Ú (z - 1)(z - 2) , where C is z - 2 C

=

1 2 [ans.: 4pi]

3.56

2.

Chapter 3

Complex Integration

sin p z 2 + cos p z 2 ÚC (z - 2)(z - 3) dz, where C is z = 4 [ans.: —4pi]

3.

Ú (z

2

C

dz , where C is z - i = 2 + 4)2 p ˘ È Í ans. : 16 ˙ Î ˚

4.

sin p z 2 + cos p z 2 ÚC (z - 1)2 (z - 2) dz, where C is z = 3 [ans.: 4p (p + 1) i ]

5.

ez ÚC (z + 2)(z + 1)2 dz, where C is z = 3 2pi ˘ È Í ans. : e 2 ˙ Î ˚

6.

Úz

2

C

7.

z +1 dz, where C is z + 1 - i = 2 + 2z + 4 [ans.: pi]

z3 - z ÚC (z - 2)3 dz, where C is z = 3 [ans.: 12 pi]

8.

Úz

2

C

z+4 dz, where C is z + 1 - i = 2 + 2z + 5 p È ˘ Í ans.: 2 (3 + 2i)˙ Î ˚

9.

cos p z 2 ÚC (z - 1)(z - 2) dz, where C is z = 3 [ans.: 4pi]

10.

dz

Ú z (z + 4), where C is z 3

=2

C

2pi ˘ È Í ans.: 27 ˙ Î ˚

Points to Remember

11.

3.57

ez ÚC (z - 1)(z - 4) dz, where C is z = 2 2pie ˘ È Í ans.: - 3 ˙ Î ˚

12.

z2 + 4 ÚC (z - 2)(z + 3i) dz, where C is z - 2 = 2 16pi ˘ È Í ans.: 2 + 3i ˙ Î ˚

13.

Úz C

z +1 dz, where C is z = 1 - 2z

3

3pi ˘ È Í ans.: - 2 ˙ Î ˚ 14.

15.

ze 2 z ÚC (z - 1)3 dz, where C is z + i = 2 z -1

Ú (z + 1) (z - 2) dz, where C is z - i 2

[ans.: 8pie2] =2

C

2pi ˘ È Í ans.: - 9 ˙ Î ˚

Points to remember Simply Connected Region A simply connected region R is the region enclosed by a simple curve.

Multiply Connected Region A multiply connected region is the region enclosed by more than one simple curve. A multiply connected region can be converted into a simply connected region by introducing cross-cuts.

Independence of Path If f (z) is analytic in a simply connected region then the line integral is independent of the path.

3.58

Chapter 3

Complex Integration

Cauchy’s Integral Theorem If f(z) is an analytic function and f ¢(z) is continuous at each point inside and on a closed curve C then

Ú f (z) dz = 0 C

1. Cauchy–Goursat Theorem If f (z) is analytic at all points inside and on a simple closed curve C contained in a simply connected domain D then Ú f (z) dz = 0 C

2. Extension of Cauchy’s Integral Theorem to a Multiply Connected Region If f (z) is analytic in the region R between two simple closed curves C1 and C2 then

Ú f (z) dz = Ú f (z) dz

C1

C2

Cauchy’s Integral Formula If f (z) is analytic inside and on a closed curve C and if a is any point inside C then f (a) =

1 f (z) dz  Ú 2p i C z - a

1. Cauchy’s Integral Formula for the Derivative of an Analytic Function If a function f (z) is analytic in a region R then its derivative at any point z = a of R is also analytic in R and is given by 1 f (z) f ¢( a ) = Ú ( z - a ) 2 dz 2p i  0 where C is any closed curve in R surrounding the point z = a.

Liouville Theorem If f (z) is analytic and bounded in the entire z-plane for all z then f (z) is constant.

Maximum Modulus Theorem If f (z) is non-constant and analytic inside and on a simple closed contour C then the maximum value of f (z) occurs on C.

| |

4 Power Series CHAPTER

chapter outline 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11

4.1

Introduction Sequences and Series Power Series Convergence of a Power Series Taylor’s Series Laurent’s Series Singular Points Residues Cauchy’s Residue Theorem Argument Theorem Rouche’s Theorem

IntroductIon

Series expansions are used to approximate a large class of arbitrary functions, which are often difficult to analyze, with simpler, well-understood functions such as Taylor series. Power series are natural generalizations of Taylor polynomials. A power series may represent a function, in the sense that wherever the series converges, it converges as well. Any polynomial can be easily expressed as a power series around any centre. In this chapter, we will study Taylor’s series, Laurent’s series and Cauchy’s residue theorem. The Laurent series of a complex function f(z) is a representation of that function as a power series which includes terms of negative degree. It may be used to express complex functions where a Taylor series expansion cannot be applied. Residues obtained in Laurent’s series are used to evaluate complex integrals by summing the residues at the isolated singularities of a function within the closed contour by Cauchy’s residue theorem.

4.2

sequences and serIes

A sequence of complex numbers z1, z2, z3, ...zn is a function which assigns a complex number zn to each positive integer n. A sequence is denoted by {zn}.

4.2

Chapter 4

Power Series

A sequence {zn} is said to be convergent if lim zn = l

n Æ•

where l is a finite quantity. An infinite series of complex numbers z1, z2, z3, ..., zn, ... is the sum of the terms of a sequence {zn} given as z1 + z2 + z3 +  + zn +  •

It is denoted as

 zn . n =1

The nth partial sum of a series is the sum of the first n terms of the series and is denoted as n

Sn = z1 + z2 + z3 +  + zn =

 zm

m =1

A series is said to be convergent if lim Sn = S

n Æ•

otherwise, the series is divergent. A necessary condition for the convergence of a complex series is lim zn = 0

n Æ•

where zn is the nth term of the series.

4.3

Power serIes

A power series is an infinite series of the form •

 an (z - z0 )n = a0 + a1 (z - z0 ) + a2 (z - z0 )2 + a3 (z - z0 )3 +  + an (z - z0 )n + 

n=0

where a0, a1, a2, ..., an,... are complex or real constants, z0 is a constant known as the centre of the series. This series is also known as the power series about the point z = z0. The power series about z = 0, i.e., the origin is given as •

 an z n = a0 + a1z + a2 z2 + a3 z3 +  + an z n + 

n=0

circle and radius of convergence of a Power series A power series may converge for all z or only at the point z = z0 or everywhere inside a circular disk z – z0 = R (i.e., in the region z –z0 < R) and diverges everywhere outside

|

|

|

|

4.4

|

4.3

Convergence of a Power Series

|

|

circular disk (i.e., in the region z –z0 > R). The circular disk z – z0 = R is known as the circle of convergence and R is known as the radius of convergence. The set of all points for which the series converges is known as the region of convergence. A power series is uniformly convergent in the region which lies entirely within its circle of convergence.

4.4

convergence of a Power serIes •

If the power series

 an (z - z0 )n

n=0

converges at a point z = a then it converges

|

| |

|

absolutely in the open circular disk z – z0 < a – z0 . A power series

•

 an (z - z0 )n

n=0

with a radius of convergence R > 0 converges uniformly inside and on every circular disc z – z0 = r, where r < R.

|

|

•

theorem 1 A power series

 an (z - z0 )n

and the k times differentiated series

n=0 •

 n(n - 1) (n - k + 1) an (z - z0 )n- k

n=k

have the same radius of convergence. •

theorem 2 Let

 an (z - z0 )n

be a power series with radius of convergence R,

n=0

then R = lim

n Æ•

an an +1

•

Absolute Convergence The series

 un (z)

is said to be absolutely convergent if

n =1

•

the series

Â

un ( z ) is also convergent.

n =1 •

Uniform Convergence

The series

 un (z)

is said to be uniformly convergent in

n=0

a region R if there exists a convergent series of positive real constants

•

 Mn

n=0

that un ( z ) £ M n for all z in R

such

4.4

Chapter 4

Power Series

example 1 Find the radii of convergence and region of convergence of the following: •

(i)

Â

zn

n =1

•

(ii)

2 +1 n

n =1

•

•

(-1)n (iii) Â ( z - 2i )n n n =1

an =

1 2 +1 n

Radius of convergence is R = lim

n Æ•

= lim

an an +1 2 n +1 + 1

2n + 1 1 2+ n 2 = lim 1 n Æ• 1+ n 2 2+0 = 1+ 0 =2 n Æ•

||

Region of convergence is z < 2. (ii)

an =

n!

nn Radius of convergence is R = lim

n Æ•

= lim

n Æ•

= lim

n Æ•

n! n

n

zn 2

Ê 6n + 1 ˆ n (iv)  Á ˜¯ ( z - 2i ) Ë 2 n + 5 n =1 [Summer 2015]

Solution (i)

Â

an an +1 n ! (n + 1)n +1 ◊ n n (n + 1)! (n + 1)n +1 n n (n + 1)

4.4

= lim

Convergence of a Power Series

(n + 1)n nn

n Æ•

Ê 1ˆ = lim Á 1 + ˜ n Æ• Ë n¯

n

=e Region of convergence is z < e.

||

(-1) n Radius of convergence is n

(iii)

an =

R = lim

n Æ•

= lim

n Æ•

an an +1 (-1)n (n + 1) ◊ n (-1)n +1

Ê 1ˆ = lim - Á 1 + ˜ Ë n¯ n Æ• = | -1 | =1 Region of convergence is z – 2i < 1.

|

(iv)

Ê 6n + 1 ˆ an = Á Ë 2 n + 5 ˜¯

|

2

Radius of convergence is R = lim

n Æ•

an an +1 2

Ê 6n + 1 ˆ Ê 2n + 2 + 5 ˆ = lim Á ˜ ◊ ÁË ˜ n Æ• Ë 2 n + 5 ¯ 6n + 6 + 1 ¯ 2

1ˆ Ê 7ˆ Ê 6+ 2+ Á ˜ Á n n˜ = lim Á ◊ 5˜ Á 7˜ n Æ• Á2+ ˜ Á6+ ˜ Ë n¯ Ë n¯ 2

Ê 6 + 0ˆ Ê 2 + 0ˆ =Á Ë 2 + 0 ˜¯ ÁË 6 + 0 ˜¯ =1

|

|

2

Region of convergence is z – 2i < 1.

2

2

4.5

4.6

Chapter 4

Power Series

example 2 •

Discuss the convergence of and circle of convergence.

Â

(2 n)!

n = 0 ( n !)

2

( z - 3i )n and also find the radius

Solution an =

(2 n)! (n !)2

Radius of convergence is an R = lim n Æ• an +1 (2 n)! [(n + 1)!] = lim ◊ n Æ• ( n !)2 [2( n + 1)]! 2

= lim

n Æ•

= lim

n Æ•

(2 n)! (n !)2

◊

(n + 1)2 (n !)2 (2 n + 2) (2 n + 1) (2 n)!

(n + 1)2 (2 n + 2)(2 n + 1) 2

Ê 1ˆ ÁË 1 + ˜¯ n = lim n Æ• Ê 2ˆ Ê 1ˆ ÁË 2 + ˜¯ ÁË 2 + ˜¯ n n 1+ 0 (2 + 0) (2 + 0) 1 = 4 =

1 The series is convergent in the region z - 3i < and is divergent in the region 4 1 z - 3i > . 4 1 Hence, the circle of convergence is z - 3i = . 4

example 3 •

n

1 ◊ 3 ◊ 5 (2 n - 1) Ê 1 - z ˆ Discuss the convergence of the  ÁË ˜ . n! z ¯ n=0

4.4

Convergence of a Power Series

Solution 1 ◊ 3 ◊ 5 (2 n - 1) n! Radius of convergence is an R = lim n Æ• an +1 an =

=

1 ◊ 3 ◊ 5 (2 n - 1) (n + 1)! ◊ 1 ◊ 3 ◊ 5 (2 n - 1)(2 n + 1) n!

n +1 2n + 1 1 1+ n = lim 1 n Æ• 2+ n = lim

n Æ•

1+ 0 2+0 1 = 2 The series is convergent in the region 1- z 1 < z 2 1- z 1 < z 2 =

2

Ê 1ˆ 2 2 1- z < Á ˜ z Ë 2¯ 1 2 2 1 - x - iy < x + iy 4 1 2 2 2 (1 - x ) + y < ( x + y 2 ) 4 1 1 + x 2 - 2 x + y2 < ( x 2 + y2 ) 4 4 + 4 x 2 - 8 x + 4 y2 < x 2 + y2 3 x 2 + 3 y2 - 8 x + 4 < 0 8 4 x 2 + y2 - x + < 0 3 3 2

4ˆ 4 Ê 2 16 ÁË x - ˜¯ + y - + < 0 3 9 3 2

4ˆ 4 Ê 2 ÁË x - ˜¯ + y < 3 9

4.7

4.8

Chapter 4

Power Series

Ê4 ˆ Hence, the series is convergent inside the circle with its centre at Á , 0˜ and a radius Ë3 ¯ of

2 4 2 4 2 , i.e., in the region z - < and is divergent in the region z - > . 3 3 3 3 3

4.5 taylor’s serIes If f (z) is analytic inside a circle C with centre at z = a then for each z inside C, f (z) can be expanded as a power series about z = a as f ( z ) = f ( a ) + ( z - a ) f ′( a ) +

( z - a )2 ( z - a )n ( n ) f ′′(a ) +  + f (a) +  2! n!

Proof Let z be any point inside the circle C. Draw a circle C1 inside C with the centre at z = a, enclosing the point z (Fig. 4.1). C Let w be any point on C1. C1

z-a < w-a z-a w-a

z w a

1 f (z) is analytic in the region z > 1 about z = 0 (Fig. 4.5).

fig. 4.4

||

    ∵

1 1,

-1

1 1 1 Ê 1ˆ + - Á1 - ˜ z z2 z Ë z¯ 1 1 1Ê 1 1 1 1 ˆ = + 2 - Á 1 + + 2 + 3 + 4 + ˜ ¯ z z zË z z z z 1 1 1 1 1 1 1 = + 2 - - 2 - 3 - 4 - 5 - z z z z z z z 1 1 1 = - 3 - 4 - 5 - z z z =

fig. 4.5

example 4 Expand f ( z ) =

1 in Laurent’s series in the following regions: ( z + 1)( z - 2)

(i) |z| < 1 (ii) 1 < |z| < 2 (iii) |z| > 2.

[Summer 2015]

4.24

Chapter 4

Power Series

Solution 1 ( z + 1)( z + 2) 1 ( z + 1) - ( z - 2) = 3 ( z + 1)( z - 2)

f (z) =

=

1È 1 1 ˘ Í 3 Î z - 2 z + 1 ˙˚

f (z) is not analytic at z = 2 and z = –1. (i) z < 1 f (z) is analytic in the region z < 1 about z = 0 (Fig. 4.6). z 1, f ( z) =

1 1, < 1, 2. 2 z( z + z - 2) Solution f (z) =

4z + 1

z( z 2 + z - 2) 4z + 1 = z( z - 1)( z + 2) A B C = + + z z -1 z + 2 4 z + 1 = A( z + 2)( z - 1) + Bz( z + 2) + Cz( z - 1) Putting z = 0, 1 = (2)(-1) A 1 A=2 Putting z = 1, 5 = 3B 5 B= 3

x

4.28

Chapter 4

Power Series

Putting z = -2,

y

-7 = (-2)(-3)C 7 C=6 11 5 1 7 1 \ f (z) = + 2 z 3 z -1 6 z + 2

z=2

x

||

f (z) is analytic in the region z > 2 about z = 0 (Fig. 4.12). ∵ f (z) = -

=-

z > 2 > 1, 1 + 2z

2 < 1, z

1 3 and 1 < z < 3.

Evaluate f ( z ) =

Solution 1 ( z + 1)( z + 3) A B = + z +1 z + 3 1 = A( z + 3) + B( z + 1)

f (z) =

4.6

Laurent’s Series

4.29

Putting z = –1, 1 = A(-1 + 3) 1 A= 2 Putting z = –3, 1 = B(-3 + 1) 1 B=2 1 1 1 1 \ f (z) = 2 z +1 2 z + 3

y

f (z) is not analytic at z = –1 and z = –3.

||

(i) z > 3 f (z) is analytic in the region z > 3 about z = 0 (Fig. 4.13). 3 1 ∵ z > 3 > 1, < 1, < 1 z z 1 1 1 1 f ( z) = 2 Ê 1ˆ 2 Ê 3ˆ z Á1 + ˜ z Á1 + ˜ Ë Ë z¯ z¯

||

=

1 Ê 1ˆ 1+ ˜ 2 z ÁË z¯

-1

-

1 Ê 3ˆ 1+ ˜ 2 z ÁË z¯

z=3

z = −3

x

fig. 4.13

-1

2 ˘ ˘ 1 È 3 Ê 3ˆ 2 1 È 1 Ê 1ˆ Í1 - + Á ˜ - ˙ - Í1 - + Á ˜ - ˙ 2z Í z Ë z ¯ ˙˚ ˙˚ 2 z ÍÎ z Ë z ¯ Î ˆ Ê 1 ˆ Ê 1 1 1 3 9 = Á - 2 + 3 - ˜ - Á - 2 + 3 - ˜ ¯ Ë 2z 2z ¯ Ë 2z 2z 2z 2z 1 4 = 2 - 3 - z z

=

|| ||

(ii) 1 < z < 3 f (z) is analytic in the annular region 1 < z < 3 about z = 0 (Fig. 4.14). 1< z ,

1 2, f (z) =

4 < 1, z

2 3. Solution f (z) =

Let

z2 - 1 z2 - 1 = 2 ( z + 2)( z + 3) z + 5z + 6

Since the degrees of the numerator and denominator are same, partial fraction cannot be applied. Dividing to reduce the degree of the numerator, -5z - 7 f (z) = 1 + ( z + 2)( z + 3) -5z - 7 A B = + ( z + 2)( z + 3) z + 2 z + 3 -5z - 7 = A( z + 3) + B( z + 2) Putting z = –2, -5(-2) - 7 = A(-2 + 3) A=3 Putting z = –3, -5(-3) - 7 = B(-3 + 2) B = -8 3 8 \ f (z) = 1 + z+2 z+3 f (z) is not analytic at z = –2 and z = –3. Let

y

(i) 2 < z < 3 f (z) is analytic in the annular region 2 < z < 3 about z = 0 (Fig. 4.18). 2< z,

2 3 about z = 0 (Fig. 4.19). ∵

z > 3 > 1, f ( z) = 1 +

3 < 1, z 3

Ê 2ˆ z Á1 + ˜ Ë z¯

3 Ê 2ˆ = 1 + Á1 + ˜ zË z¯

-1

1 3 z - 2z - 3 2

È ˆ 1 Ê 1 z z2 1 1 1 1 5 ˘ -  - 2 + 3 - ˜ (ii) + 2 + 3 + ˙ Í ans.: (i) Á - - 2 3 9 27 z z z z z Ë ¯ ˚ ÍÎ 9. f (z) =

1 ;0 < z < 2 z (z - 2) 2

È ˆ˘ 1 Ê z z2 + ˜ ˙ Í ans.: - 2 Á 1 + + 2z Ë 2 4 ¯ ˙˚ ÍÎ 10. f (z) =

1 + 2z ;0 < z 0 about z = 1. Let z – 1 = t, then z = t + 1. f (z) = =

ez ( z - 1)2 et +1

t2 e = 2 (e t ) t =

ˆ eÊ t2 t3 1 + t + + + ˜ 2 Á 2 ! 3! t Ë ¯

e e t + + + e + t 2 6 t e e e ( z - 1) = + + +e + 2 z -1 2 6 ( z - 1) =

e

2

Res [ f ( z ); z = 1] = Coefficient of

1 z -1

=e

example 18 Find the residues of f ( z ) =

z2

at its isolated singularities ( z - 1)2 ( z + 2)2 using Laurent’s series expansion. Also, state the valid region.

4.72

Chapter 4

Power Series

Solution f (z) =

z2

( z - 1)2 ( z + 2)2 A B C D = + + + 2 ( z - 1) ( z - 1) z + 2 ( z + 2 )2

z 2 = A( z - 1)( z + 2)2 + B( z + 2)2 + C ( z + 2)( z - 1)2 + D( z - 1)2 Putting z = 1,

Putting z = – 2,

1 = 9B 1 B= 9 4 = 9D 4 D= 9

Equating the coefficient of z3, 0=A+C A = –C Putting z = 0, 0 = - 4 A + 4 B + 2C + D 4 4 0 = - 4 A + + 2C + 9 9 8 4 A - 2C = 9 8 -4C - 2C = 9 8 -6C = 9 4 C=27 4 A= 27 4 1 1 1 4 1 4 1 \ f (z) = + + 2 27 z - 1 9 ( z - 1) 27 z + 2 9 ( z + 2)2 z = 1 and z = –2 are poles of order 2. Since poles are isolated singularities, z = 1 and z = –2 are isolated singularities.

4.8

Residues

(i) For Laurent’s series about z = 1, Let z – 1 = t then z = t + 1 4 1 1 1 4 1 4 1 + 2+ 27 t 9 t 27 (t + 1 + 2) 9 (t + 1 + 2)2 4 1 4 1 4 1 = + + 2 tˆ 9 2Ê 27t 9t 2 27 Ê tˆ 3 Á1 + ˜ 3 1 + ÁË 3 ˜¯ Ë 3¯

f (z) =

4 1 4Ê tˆ 1+ = + 27t 9t 2 81 ÁË 3 ˜¯ = =

4Ê tˆ + Á1 + ˜ 81 Ë 3 ¯

-2

ˆ ˆ 4 Ê 2t 3t 2 4 1 4Ê t t2 + 2 - Á 1 - + - ˜ + Á 1 - + - ˜ 27t 9t 81 Ë 3 9 3 9 ¯ ¯ 81 Ë ˘ 4 1 4 È z - 1 ( z - 1)2 + - Í1 + - ˙ 2 27( z - 1) 9( z - 1) 81 ÎÍ 3 9 ˙˚ 2 ˘ 4 È 2( z - 1) 3( z - 1) + - ˙ + Í1 3 9 81 ÍÎ ˙˚

Res [f (z); z = 1] = Coefficient of =

-1

1 z -1

4 27

(ii) For Laurent’s series about z = –2, Let z + 2 = t then z = t – 2. f (z) =

4 1 1 1 4 1 4 1 + + 2 27 (t - 2 - 1) 9 (t - 2 - 1) 27 t 9 t 2

È ˘ È ˘ Í ˙ Í ˙ 4 Í 1 1 1 Í ˙- 4 + 4 ˙+ = 2 Í ˙ 27t 9t 2 tˆ˙ 9 27 Í Ê Í (-3)2 ÊÁ 1 - t ˆ˜ ˙ Í -3 ÁË 1 - 3 ˜¯ ˙ Ë 3 ¯ ˙˚ ÍÎ Î ˚ 4Ê tˆ = - Á1 - ˜ Ë 81 3¯ =-

-1

1 Ê tˆ + Á1 - ˜ Ë 81 3¯

-2

-

4 4 + 2 27t 9t

ˆ 1 Ê 2t 3t 2 ˆ 4Ê t t2 4 4 1 + + + ˜ + Á 1 + + + ˜ + 2 Á 81 Ë 3 9 3 9 ¯ 81 Ë ¯ 27t 9t

4.73

4.74

Chapter 4

Power Series

=-

2 ˘ 1 È 2( z + 2) ( z + 2 )2 ˘ 4 È z + 2 ( z + 2) Í1 + + + ˙ + Í1 + + + ˙ 81 Í 3 9 3 3 ˙ 81 Í ˙ Î ˚ Î ˚ 4 4 + 27( z + 2) 9( z + 2)2

Res [f (z); z = –2] = Coefficient of =-

1 z+2

4 27

exercIse 4.4 find the residues of f (z) at the singular points: 1.

z (z - 1)2 1˘ È Í ans.: - 2i ˙ Î ˚

2.

z+2 (z + 1)2

3.

z -z (z + 1)2 (z 2 + 4)

[ans.: 1] 2

11 11 ± 2i ˘ È Í ans.: - 25 , 50 ˙ Î ˚ 4.

5.

6.

ez (z - 1)3

sin2 z z3

e˘ È Í ans.: 2 ˙ Î ˚

1˘ È Í ans.: 4 ˙ Î ˚

z+3 z(z - 1)(z + 2) 3 4 1˘ È Í ans. : - 2 , 3 , 6 ˙ Î ˚

4.9

7.

8.

Cauchy’s Residue Theorem

e2z z2 + p 2

4.75

1 1 ˘ È Í ans.: 2p i , - 2p i ˙ Î ˚

z3 (z + a)2 [ans.: 3a2]

9.

z2 + 1 z(z - 2)

10. z cos

1 5˘ È Í ans.: - 2 , 2 ˙ Î ˚

1 z [ans. : — 4]

4.9

cauchy’s resIdue theorem

If f (z) is analytic inside and on a simple closed curve C except at a finite number of singular points inside C then

Ú f (z) dz = 2p i

(sum of residues)

C1

C

where the integral is taken in the anticlockwise direction around C. Proof Let z1, z2, . . ., zn be finite numbers of singular points inside C (Fig. 4.36). Enclose each of the singular point in a small circle such that no other singular point lies inside this circle. The curve C along with these small circles C1, C2, C3, . . ., Cn form a multiply connected region and f (z) is analytic in this region. By Cauchy’s integral theorem for a multiply connected region,

C

z1

C2 z2

Cn C2 zn z3

fig. 4.36

Ú f (z) dz = Ú f (z) dz + Ú f (z) dz +  + Ú f (z) dz C

C1

C2

Cn

= 2p i ÈÎRes. at z1 ˘˚ + 2p i ÈÎRes. at z2 ˘˚ +  + 2p i ÈÎRes. at zn ˘˚ = 2p i (sum of residues)

4.76

Chapter 4

Power Series

example 1 z2 - 4z + 4 Ú z + i dz where C is |z| = 2. C

Evaluate

[Winter 2013]

Solution z2 - 4z + 4 z+i The poles are given by z = –i. (ii) C is the circle z = 2 with the centre at (0, 0) and a radius of 2 (Fig. 4.37). (iii) For z = –i, z = –i = 1 < 2 Hence, z = –i lies inside C. (iv) z = –i is a simple pole. Res [ f ( z ); z = -i ] = lim ( z + i ) f ( z ) (i) Let f ( z ) =

|| || | |

z Æ- i

= lim ( z 2 - 4 z + 4) z Æ- i

(v)

= i 2 + 4i + 4 = 3 + 4i By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues)

fig. 4.37

C

Ú

C

z2 - 4z + 4 = 2p i (3 + 4i ) z+i = 2p (3 i - 4)

example 2 Evaluate

1

Ú ( z - 1)2 ( z - 3) dz

C

where C is |z| = 2.

Solution (i)

f (z) =

1

( z - 1) ( z - 3) The poles are given by z = 1, z = 3. (ii) C is a circle z = 2 with the centre at (0, 0) and a radius of 2 (Fig. 4.38). (iii) For z = 1, z = 1 = 1 < 2 Hence, z = 1 lies inside C. For z = 3, z = 3 = 3 > 2 2

||

|| || || ||

fig. 4.38

4.9

Cauchy’s Residue Theorem

4.77

Hence, z = 3 lies outside C. (iv) z = 1 is a pole of order 2. 1 d È 2 Res [ f ( z ); z = 1] = lim Î( z - 1) f ( z )˘˚ (2 - 1)! z Æ1 dz È 1 ˘ Í z - 3˙ Î ˚ 1 ˘ È = lim Í2˙ z Æ1 Î ( z - 3) ˚ 1 =4 By Cauchy’s residue theorem, Ú f (z) dz = 2p i (sum of residues) = lim

z Æ1

(v)

d dz

C

Ê 1ˆ

1

Ú (z - 1)2 (z - 3) dz = 2p i ÁË - 4 ˜¯ C

=-

pi 2

example 3 Evaluate

sin p z 2 + cos p z 2 Ú ( z - 1)( z - 2) dz , where C is |z| = 3. C

Solution (i) Let f ( z ) =

sin p z 2 + cos p z 2 ( z - 1)( z - 2)

y

The poles are given by z = 1, z = 2. (ii) C is a circle z = 3 with the centre at (0, 0) and a radius of 3 (Fig. 4.39). (iii) For z = 1, z = 1 = 1 < 3 Hence, z = 1 lies inside C. For z = 2, z = 2 = 2 < 3 Hence, z = 2 lies inside C. (iv) z = 1 is a simple pole.

C

(0, 0) z = 1 z = 2

fig. 4.39

z=3

x

4.78

Chapter 4

Power Series

Res [ f ( z ); z = 1] = lim( z - 1) f ( z ) z Æ1

sin p z 2 + cos p z 2 z Æ1 z-2 sin p + cosp = -1 =1 z = 2 is a simple pole. Res [ f ( z ); z = 2 ] = lim( z - 2) f ( z ) = lim

zÆ2

sin p z 2 + cos p z 2 zÆ2 z -1 sin 4p + cos 4p = 1 =1 (v) By Cauchy’s residue theorem, Ú f (z) dz = 2p i (sum of residues) = lim

C

sin p z + cos p z 2 dz = 2p i(1 + 1) ( z 1 )( z 2 ) C

Ú

2

= 4p i

example 4 Evaluate

Ú

C

5z + 7 z + 2z - 3 2

dz, where C is |z – 2| = 2.

[Summer 2013]

Solution (i) Let

5z + 7

f (z) =

z + 2z - 3 5z + 7 = ( z + 3)( z - 1) 2

The poles are given by z = –3, z = 1. (ii) C is a circle z – 2 = 2 with the centre at (2, 0) and a radius of 2 (Fig. 4.40). (iii) For z = –3, z –2 = –3 – 2 = 5 > 2 Hence, z = –3 lies outside C. For z = 1, z – 2 = 1 – 2 = 1 < 2 Hence, z = 1 lies inside C. (iv) z = 1 is a simple pole.

|

|

|

|

| | | |

|

|

fig. 4.40

4.9

Cauchy’s Residue Theorem

Res [ f ( z ); z = 1] = lim ( z - 1) f ( z ) z Æ1

5z + 7 z+3 5+7 = 1+ 3 =3 By Cauchy’s residue theorem, Ú f (z) dz = 2p i (sum of residues) = lim

z Æ1

(v)

C

Ú C

5z + 7 z2 + 2z - 3

dz = 2p i (3) = 6p i

example 5 Evaluate

3z 2 + z - 1

Ú ( z 2 - 1)( z - 3) dz,

where C is the circle |z| = 2.

C

Solution (i) Let f ( z ) =

3z 2 + z - 1 ( z 2 - 1)( z - 3)

=

3z 2 + z - 1 ( z + 1)( z - 1) ( z - 3)

The poles are given by z = –1, z = 1, z = 3. (ii) C is a circle z = 2 with the centre at (0, 0) and a radius of 2 (Fig. 4.41).

||

(iii) For z = –1, z = -1 = 1 < 2 Hence, z = –1 lies inside C. For z = 1, z = 1 = 1 < 2 Hence, z = 1 lies inside C. For z = 3, z = 3 = 3 > 2 Hence, z = 3 lies outside C. (iv) z = –1 is a simple pole. Res [ f ( z ); z = -1] = lim ( z + 1) f ( z ) z Æ-1

3z 2 + z - 1 = lim z Æ-1 ( z - 1)( z - 3) 3(-1)2 + (-1) - 1 (-1 - 1) (-1 - 3) 1 = 8

=

fig. 4.41

4.79

4.80

Chapter 4

Power Series

z = 1 is a simple pole. Res [ f ( z ); z = 1] = lim( z - 1) f ( z ) z Æ1

3z 2 + z - 1 z Æ1 ( z + 1)( z - 3) 3 +1-1 = (1 + 1) (1 - 3) 3 =4 (v) By Cauchy’s residue theorem, = lim

Ú f (z) dz = 2p i (sum of residues)

C

3z 2 + z - 1

Ê1

3ˆ

Ú (z2 - 1)(z - 3) dz = 2p i ÁË 8 - 4 ˜¯

C

Ê 5ˆ = 2p i Á - ˜ Ë 8¯ =-

5p i 4

example 6 Using the residue theorem, evaluate

Ú

C

ez + z z -z 3

p . 2 [Summer 2015]

dz, where C : | z | =

Solution (i)

Let f ( z ) =

ez + z

=

ez + z z( z + 1)( z - 1)

z3 - z The poles are given by z = 0, z = –1, z = 1.

p = 1.57 with the centre (0, 2 p 0) and a radius of (Fig. 4.42). 2 p (iii) For z = 0, z = 0 < 2 Hence, z = 0 lies inside C. p For z = –1, z = -1 = 1 < 2 (ii) C is the circle z =

fig. 4.42

4.9

Hence, z = –1 lies inside C. p For z = 1, | z | = | 1 | = 1 < 2 Hence, z = 1 lies inside C. (iv) z = 0 is a simple pole. Res [ f ( z ); z = 0 ] = lim z f ( z ) zÆ0

= lim

zÆ0

ez + z ( z + 1)( z - 1)

e0 -1 = -1 z = –1 is a simple pole. Res [ f ( z ); z = -1] = lim ( z + 1) f ( z ) =

z Æ-1

= lim

z Æ-1

ez + z z( z - 1)

e -1 - 1 -1(-2) 1- e = 2e z = 1 is a simple pole. Res [ f ( z ); z = 1] = lim ( z - 1) f ( z ) =

z Æ1

ez + z z Æ1 z( z + 1) e +1 = 2 By Cauchy’s residue theorem, = lim

(v)

Ú f (z) dz = 2p i (sum of residues)

C

ez + z

Ê

Ú z3 - z dz = 2p i ÁË -1 +

C

1 - e e + 1ˆ + ˜ 2e 2 ¯

Ê -2e + 1 - e + e2 + e ˆ = 2p i Á ˜¯ Ë 2e pi 2 (e - 2e + 1) e pi = (e - 1)2 e =

Cauchy’s Residue Theorem

4.81

4.82

Chapter 4

Power Series

example 7 Use residues to evaluate the integrals of the function the circle |z| = 3 in the positive sense.

exp (-z )

around z2 [Winter 2012]

Solution (i)

Let f ( z ) =

exp (- z ) 2

=

e- z

z z2 The poles are given by z = 0. (ii) C is the circle z = 3 with the centre at (0, 0) and a radius of 3 (Fig. 4.43). (iii) For z = 0, z = 0 < 3 Hence, z = 0 lies inside C. (iv) z = 0 is a pole of order 2.

||

||

Res [ f ( z ); z = 0 ] =

1 d È 2 lim Î z f ( z )˘˚ (2 - 1)! z Æ0 dz

= lim e - z zÆ0

(v)

= e0 =1 By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues)

C

Ú

C

e- z z2

dz = 2p i (1) = 2p i

example 8 Evaluate

e2 z

Ú ( z - p i)3

dz, where C is |z – 2i| = 2.

C

Solution (i) Let f ( z ) =

e2 z ( z - p i )3

The poles are given by z = p i. (ii) C is a circle z – 2i = 2 with the centre at (0, 2) and a radius of 2 (Fig. 4.44).

|

|

fig. 4.43

4.9

Cauchy’s Residue Theorem

4.83

(iii) For z = 2p, z - 2i = p i - 2i = p - 2 = 1.14 < 2 Hence, z = pi lies inside C. (iv) z = pi is a pole of order 3. Res [ f ( z ); z = p i ] =

y 2

1 d lim 2 ÈÎ( z - p i )3 f ( z )˘˚ (3 - 1)! z Æp i dz

1 d2 lim 2 (e2 z ) 2 ! z Æp i dz d 1 = lim (2 e 2 z ) 2 z Æp i dz 1 = lim 4e2 z 2 z Æp i = 2e 2p i =

= 2 (cos 2p + i sin 2p ) =2 (v) By Cauchy’s residue theorem,

Ú

C

z = pi (0, 2)

x

(0, 1)

fig. 4.44

f ( z ) dz = 2p i (sum of residues)

C 2z

e

Ú (z - p i)3 dz = 2p i(2)

C

= 4p i

example 9 Evaluate

e z dz

Ú ( z 2 + p 2 )2 , where C is the circle |z| = 4.

C

y

Solution ez

(i) Let f ( z ) =

=

ez

( z 2 + p 2 )2 ( z + p i )2 ( z - p i )2 The poles are given by z = –p i, z = p i. (ii) C is a circle z = 4 with the centre at (0, 0) and a radius of 4 (Fig. 4.45).

||

(iii) For z = -p i, z = -p i = p < 4

C z = pi

z=4

(0, 0) z = −p i

Hence, z = –pi lies inside C. For z = p i, z = p i = p < 4 Hence, z = pi lies inside C.

fig. 4.45

x

4.84

Chapter 4

Power Series

(iv) z = –pi is a pole of order 2. 1 d È Res [ f ( z ); z = -p i ] = lim ( z + p i )2 f ( z )˘˚ (2 - 1)! z Æ-p i dz Î ˘ d È ez Í ˙ 2 z Æ- p i dz Í ( z - p i ) ˙ Î ˚

= lim

È ( z - p i )2 e z - e z ◊ 2( z - p i )(1) ˘ = lim Í ˙ z Æ- p i Í ( z - p i )4 ˙˚ Î È e z ( z - p i )( z - p i - 2) ˘ = lim Í ˙ z Æ- p i Í ( z - p i )4 ˙˚ Î È ez (z - p i - 2) ˘ ˙ = lim Í 3 z Æ- p i Í ˙ z p i ( ) Î ˚ = = = = =

e -p i ( -p i - p i - 2 )

(-p i - p i )3

- (cos p - i sin p ) (2p i + 2 ) -8p 3i 3 (p i + 1) 4p 3i -i(p i + 1) 4p 3 p -i 4p 3

Similarly, replacing i by – i in Res [ f ( z ); z = -p i ] , Res [ f ( z ); z = p i ] =

p +i

4p 3 (v) By Cauchy’s residue theorem, Ú f (z) dz = 2p i (sum of residues) C z

Êp -i

e

Ú (z2 + p 2 )2 dz = 2p i ÁË 4p 3

C

=

2p i(2p )

=

i p

4p 3

+

p + iˆ ˜ 4p 3 ¯

4.9

Cauchy’s Residue Theorem

4.85

example 10 Evaluate

Ú

2z + 6 z2 + 4

C

dz where C is |z – i| = 2.

[Summer 2013]

Solution (i)

Let f ( z ) =

2z + 6

z2 + 4 2z + 6 = ( z + 2i )( z - 2i ) The poles are given by z = –2i, z = 2i. (ii) C is a circle z – i = 2 with the centre at (0, 1) and a radius of 2 (Fig. 4.46). (iii) For z = 2i, z – i = 2i – i = i = 1 < 2 Hence, z = 2i lies inside C. For z = –2i, z – i = –2i – i = –3i = 3 > 2 Hence, z = –2i lies outside C. (iv) z = 2i is a simple pole. Res [ f ( z ); z = 2i ] = lim [( z - 2i ) f ( z )]

|

|

| | | | || | | | | | | z Æ2i

2z + 6 z Æ 2 i z + 2i 2(2i ) + 6 = 2i + 2i 2i + 3 = 2i By Cauchy’s residue theorem, Ú f (z) dz = 2p i (sum of residues) = lim

(v)

fig. 4.46

C

2z + 6

Ê 2i + 3 ˆ ˜ 2i ¯

Ú z2 + 4 dz = 2p i ÁË C

= p (2i + 3)

example 11 Find the value of the integral

3z 2 + 2

Ú ( z - 1)( z 2 + 9) dz taken counterclockwise

C

around the circle C : |z – 2| = 2. Solution (i)

Let f ( z ) =

3z 2 + 2 ( z - 1)( z 2 + 9)

=

3z 2 + 2 ( z - 1)( z + 3i )( z - 3i )

[Winter 2012]

4.86

Chapter 4

Power Series

The poles are given by z = 1, z = –3i, z = 3i. (ii) C is the circle z – 2 = 2 with the centre at (2, 0) and a radius of 2 (Fig. 4.47).

|

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|

| |

| | |

(iii) For z = 1, z – 2 = 1 – 2 = –1 = 1 < 2 Hence, z = 1 lies inside C. For z = 3i, z - 2 = 3i - 2 = 9 + 4 = 13 > 2 Hence, z = 3i lies outside C. For z = –3i, z - 2 = -3i - 2 = 9 + 4 = 13 > 2 Hence, z = –3i lies outside C. (iv) z = 1 is a simple pole. Res [ f ( z ); z = 1] = lim ( z - 1) f ( z )

fig. 4.47

z Æ1

= lim

3z 2 + 2

z Æ1

z2 + 9

3+2 1+ 9 1 = 2 By Cauchy’s residue theorem, =

(v)

Ú f (z) dz = 2p i (sum of residues)

C

3z 2 + 2

Ê 1ˆ

Ú (z - 1)(z2 + 9) dz = 2p i ÁË 2 ˜¯

C

= pi

example 12 Evaluate

z2

Ú ( z - 1)2 ( z + 2) dz,

where C is the circle |z| = 2.5.

C

Solution (i) Let f ( z ) =

z2 ( z - 1)2 ( z + 2)

The poles are given by z = 1, z = –2.

[Summer 2014]

4.9

Cauchy’s Residue Theorem

||

(ii) C is a circle z = 2.5 with the centre at (0, 0) and a radius of 2.5 (Fig. 4.48).

4.87

y

(iii) For z = 1, z = 1 = 1 < 2.5 C

Hence, z = 1 lies inside C. For z = -2, z = -2 = 2 < 2.5 Hence, z = –2 lies inside C. (iv) z = 1 is a pole of order 2. Res [ f ( z ); z = 1] =

z = −2 (0, 0)

z=1

1 d lim È( z - 1)2 f ( z )˘˚ (2 - 1)! z Æ1 dz Î d È z2 ˘ Í ˙ z Æ1 dz Í z + 2 ˙ Î ˚

= lim

È ( z + 2)(2 z ) - z 2 (1) ˘ = lim Í ˙ z Æ1 Í ( z + 2 )2 ˙˚ Î = lim

2z2 + 4z - z2

( z + 2 )2 2 + 4 -1 = 9 5 = 9 z = –2 is a simple pole. z Æ1

Res [f (z); z = – 2] = lim ( z + 2) f ( z ) z Æ-2

= lim

z Æ-2

=

z2 ( z - 1)2

(-2)2

(-2 - 1)2 4 = 9 (v) By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues)

C

z2

Ê4

5ˆ

Ú (z - 1)2 (z + 2) dz = 2p i ÁË 9 + 9 ˜¯

C

= 2p i(1) = 2p i

fig. 4.48

z = 2.5

x

4.88

Chapter 4

Power Series

example 13 Evaluate

5z + 7

Ú z 2 + 2 z - 3 dz,

where C is |z – 2| = 2.

C

Solution (i)

5z + 7

Let f ( z ) =

z + 2z - 3 2

=

5z + 7 ( z + 3)( z - 1)

The poles are given by z = 1, z = –3.

|

|

(ii) C is the circle z – 2 = 2 with the centre at (2, 0) and a radius of 2 (Fig. 4.47). (iii)

|

| | |

For z = 1, z – 2 = –1 = 1 < 2 Hence, z = 1 lies inside C. For z = –3, z – 2 = –3 – 2 = –5 = 5 > 2 Hence, z = –3 lies outside C.

|

| |

| | |

fig. 4.49

(iv) z = 1 is a simple pole Res [ f ( z ); z = 1] = lim ( z - 1) f ( z ) z Æ1

= lim

z Æ1

5z + 7 z+3

12 = 4 =3 (v)

By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues) C

= 2p i (3) = 6p i

example 14 Evaluate

z -1

Ú ( z + 1)2 ( z - 2) dz,

C

where C is |z – i | = 2.

4.9

Cauchy’s Residue Theorem

4.89

Solution (i) Let f ( z ) =

z -1

y

( z + 1)2 ( z - 2)

The poles are given by z = –1, z = 2.

|

|

(ii) C is a circle z – i = 2 with the centre at (0, 1) and a radius of 2 (Fig. 4.50). (iii) For z = -1, z - i = -1 - i = 2 < 2 Hence, z = –1 lies inside C. For z = 2, z - i = 2 - i = 5 > 2

C

(0, 1) z = −1

Hence, z = 2 lies outside C. (iv) z = –1 is a pole of order 2. Res [ f ( z ); z = -1] =

1 d È lim ( z + 1)2 f ( z )˘˚ (2 - 1)! z Æ-1 dz Î

= lim

fig. 4.50

d È z -1 ˘ Íz -2˙ Î ˚ ( z - 2)(1) - ( z - 1)(1)

z Æ-1 dz

= lim

( z - 2 )2

z =-1

= lim

-1

z =-1 ( z - 2)2

1

=-

(-1 - 2)2 1 =9 (v) By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues)

C

z -1

Ú (z + 1)2 (z - 2) dz

C

Ê 1ˆ = 2p i Á - ˜ Ë 9¯ =-

2p i 9

example 15 Evaluate

dz

Ú ( z 2 + 4)2 , where C is the circle |z – i| = 2.

C

z=2

x

4.90

Chapter 4

Power Series

Solution

y

(i) Let f ( z ) =

1 ( z 2 + 4 )2

=

1 ( z + 2i )2 ( z - 2i )2

The poles are given by z = –2i, z = 2i. (ii) C is a circle z – i = 2 with the centre at (0, 1) and a radius of 2 (Fig. 4.51). (iii) For z = -2i, z - i = -2i - i = -3i = 3 > 2

|

|

Hence, z = –2i lies outside C. For z = 2i, z - i = 2i - i = i = 1 < 2 Hence, z = 2i lies inside C. (iv) z = 2i is a pole of order 2. 1 d È Res [ f ( z ); z = 2i ] = lim ( z - 2i )2 f ( z )˘˚ (2 - 1)! z Æ2i dz Î

C z = 2i (0, 1) x

z = −2i

fig. 4.51

˘ d È 1 = lim Í ˙ 2 z Æ 2 i dz ( z + 2i ) Î ˚ È ˘ 2 = lim Í3˙ z Æ2i Î ( z + 2i ) ˚ 2 =(4i )3 2 =-64i 1 = 32i (v) By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues)

C

dz

Ú ( z 2 + 4 )2

C

Ê 1 ˆ = 2p i Á Ë 32i ˜¯ =

p 16

example 16 Evaluate

z4

Ú ( z + 1)( z - i)2 dz,

residue theorem.

where C: 9x2 + 4y2 = 36 by using the [Summer 2015]

4.9

Cauchy’s Residue Theorem

Solution (i)

z4

Let f ( z ) =

( z + 1)( z - i )2 The poles are given by z = –1, z = i. x 2 y2 + = 1 (Fig. 4.52) (ii) C is an ellipse 4 9 (iii) For z = –1, x = –1, y = 0 1 x 2 y 2 (-1)2 \ + = +0 = 0

ÈÎ∵

f ( z ) > g( z ) ˘˚

Thus, f (z) + g(z) and f (z) are nonzero on C. By the argument-principle corollary, Nf = N f +g = N f - N f +g = = = =

1 f ¢( z ) dz 2p i CÚ f ( z ) 1 f ¢( z ) + g ¢( z ) dz 2p i CÚ f ( z ) + g( z ) 1 È f ¢( z ) f ¢( z ) + g ¢( z ) ˘ dz 2p i CÚ ÍÎ f ( z ) f ( z ) + g( z ) ˙˚

f ¢( z ) [ f ( z ) + g( z )] - f ( z ) [ f ¢( z ) + g ¢( z )] 1 dz Ú f ( z ) [ f ( z ) + g( z )] 2p i C

1 [ f ¢( z ) g( z ) - f ( z ) g ¢( z )] dz f ( z ) [ f ( z ) + g( z )] 2p i CÚ

1 [ f ¢( z ) g( z ) - f ( z ) g ¢( z )] dz 2p i CÚ 2 È ( f ( z ) + g( z ) ˘ [ f ( z )] Í f ( z ) ˙ Î ˚

=

1 2pi CÚ

=-

1 F ¢( z ) dz 2pi CÚ F ( z )

g( z ) f (z)

where

F (z) = 1 +

Let

w = F (z) = 1 +

È f ¢( z ) g( z ) - f ( z ) g ¢( z ) ˘ Í ˙ { f (z)}2 ÍÎ ˙˚ dz È g( z ) ˘ 1 + Í f ( z ) ˙˚ Î

g( z ) f (z)

...(4.9)

4.108

Chapter 4

Power Series

w -1 =

g( z ) f (z)

w -1 =

g( z ) 17 ≥ g( z ) f ( z ) > g( z )

||

4.11

Rouche’s Theorem

4.109

Since f (z) and g(z) are analytic within and on C, also f (z) and f (z) + g(z) are nonzero on C, by Rouche’s theorem, Nf + g = Nf f (z) = 2z5 has a zero of order 5 inside C. Nf = 5 \ Nf + g = 5 Hence, F(z) has five zeros inside C.

example 2 Use Rouche’s theorem to determine the number of zeros of the polynomial z6 – 5z4 + z3 – 2z inside the circle |z| = 1. [Summer 2012] Solution Let F(z) = z6 – 5z + z3 – 2z Let f (z) = –5z4, g(z) = z6 + z3 – 2z, C: z = 1 On circle C: z = 1, f (z) = –5z4 = 5

|| | | |

||

|

g( z ) = z 6 + z 3 - 2 z

and

£ z6 + z3 + 2 z =4 \

g( z ) £ 4

         \ f (z) = 5 > 4 ≥ g(z)          \ f (z) > g(z) Since f (z) and g(z) are analytic within and on C, also f (z) and f (z) + g(z) are nonzero on C, by Rouche’s theorem, Nf + g = Nf f (z) = –5z4 has a zero of order 4 inside C. Nf = 4 \ Nf + g = 4 Hence, F(z) has four zeros inside C.

| | | | | | | |

example 3 Use Rouche’s theorem to determine the number of zeros of the polynomial z4 – 8z + 10 in the annulus region 1 < |z| < 3. Solution Let

F(z) = z4 – 8z + 10

4.110

Chapter 4

Power Series

(i) Let f (z) = 10, g(z) = z4 – 8z, On circle C1 : z = 1,

||

C1: z = 1

f ( z ) = 10 and

g( z ) = z 4 - 8 z £ z4 + 8 z = 1+ 8 =9 g( z ) £ 9 f ( z ) = 10 > 9 ≥ g( z ) \

f ( z ) > g( z )

Since f (z) and g(z) are analytic within and on C1, also f (z) and f (z) + g(z) are nonzero on C1, by Rouche’s theorem, Nf + g = Nf f (z) = 10 has no zero inside C1. Nf = 0 \ Nf + g = 0 Hence, F(z) has no zero inside C1. (ii) Let f (z) = z4, g(z) = –8z + 10, C2: z = 3 On circle C2: z = 3,

||

||

f (z) = z 4 = 34 = 81 g( z ) = -8 z + 10

and

£ 8 z + 10 = 24 + 10 = 34 \

g( z ) £ 34 f ( z ) = 81 > 34 ≥ g( z )

\

f ( z ) > g( z )

Since f (z) and g(z) are analytic within and on C2, also f (z) and f (z) + g(z) are nonzero on C2, by Rouche’s theorem, Nf + g = Nf f (z) = z4 has a zero of order 4 inside C2. Nf = 4

4.11

\

Rouche’s Theorem

4.111

Nf + g = 4 Hence, F(z) has four zeros inside C2. Total number of zeros of F(z) inside 1 < z < 2 = [ N f + g ]C2 - [ N f + g ]C1 = 4-0 =4

example 4 Prove that all the roots of z7 – 5z3 + 12 = 0 lie between the circles |z| = 1 and |z| = 2, using Rouche’s theorem. Solution Let F(z) = z7 – 5z3 + 12 (i) Let f (z) = 12, g(z) = z7 – 5z3, C1: z =1 On circle C1: z = 1,

||

||

f ( z ) = 12 = 12 and

g( z ) = z 7 - 5z 3 £ z7 + 5 z3 = 1+ 5 =6

\

g( z ) £ 6 f ( z ) = 12 > 6 ≥ g( z )

\

f ( z ) > g( z )

Since f (z) and g(z) are analytic within and on C1, and also f (z) and f (z) + g(z) are nonzero on C1, by Rouche’s theorem, N f +g = N f f ( z ) = 12 has no zero inside C1 Nf = 0 \ N f +g = 0 Hence, F(z) has no zero inside C1. (ii) Let f ( z ) = z 7 , g( z ) = 12 - 5z 3 , C2 :| z | = 2

||

On circle C2: z = 2

4.112

Chapter 4

Power Series

f (z) = z7 = 27 = 128 and

g( z ) = 12 - 5z 3 £ 12 + 5 z 3 = 12 + 5 23 = 52

\

g( z ) > 52 f ( z ) = 128 > 52 ≥ g( z)

\

f ( z ) > g( z )

Since f (z) and g(z) are analytic within and on C2, and also f (z) and f (z) + g(z) are non-zero on C2, by Rouches’ theorem, Nf + g = Nf f (z) = z7 has a zero of order 7 inside C2. Nf = 7 \ Nf + g = 7 Hence, F(z) has four zeros inside C2.

(

Total number of zeros of F(z) inside 1 < | z | < 2 = N f + g

) - (N ) C2

f +g C 1

= 7-0 =7 Hence, all the roots of F(z) = z7 – 5z3 + 12 = 0 lie between the circles z = 1 and z = 2.

||

||

example 5 Use Rouche’s theorem to determine the number of zeros of the polyno3 mial z5 + 15z + 1 inside the annulus region < | z | < 2. 2 Solution Let

F(z) = z5 + 15z + 1

(i) Let f ( z ) = 15 z, g( z ) = z 5 + 1, C1 : | z |= On circle C1 : z =

3 , 2

3 2

4.11

Rouche’s Theorem

4.113

f ( z ) = 15 z = 15 | z | Ê 3ˆ = 15 Á ˜ Ë 2¯ = and

45 2

g( z ) = z 5 + 1 £ | z5 | + | 1 | 5

Ê 3ˆ = Á ˜ +1 Ë 2¯ 275 32 275 g( z ) £ 32 45 275 f (z) = > ≥ g( z ) 2 32 f ( z ) > g( z ) =

\

\

Since f (z) and g(z) are analytic within and on C1, also f (z) and f (z) + g(z) are nonzero on C1, by Rouche’s theorem, Nf + g = Nf f (z) = 15z has a zero of order 1 inside C1. Nf = 1 \ Nf + g = 1 Hence, F(z) has one zero inside C1. (ii) Let f ( z ) = z 5 , g( z ) = 15z + 1, C2 : | z | = 2 on C2 : | z | = 2,

||

On circle C2: z = 2, f (z) = z5 = 25 = 32 and

g( z ) = 15 z + 1 £ 15 z + 1 = 15 2 + 1 = 30 + 1 = 31

\

g( z ) £ | 3 |

4.114

Chapter 4

Power Series

f ( z ) = 32 > 31 > g( z ) Since f (z) and g(z) are analytic within and on C2, also f (z) and f (z) + g(z) are nonzero on C2, by Rouche’s theorem, Nf + g = Nf f (z) = z5 has a zero of order 5 inside C2 Nf = 5 Nf + g = 5 Hence, F (z) has five zeros inside C2. 3 Total number of zeros of F (z) inside < z < 2 = N f + g - N f +g C2 C1 2

(

) (

)

= 5 -1 =4

exercIse 4.6 use rouche’s theorem to determine the number of zeros of F(z) within the given region: 1. F (z) = z 4 - 7 z - 1, z = 1 [ans.: 1] 2. F (z) = z 7 - 5z 4 + z 2 - 2, z = 1 [ans.: 4] 3. F (z) = 2 z 4 - 2 z 3 + 2 z 2 - 2 z + 9, z = 1 [ans.: 0] 4. F (z) = 2 z 5 - z 3 + 3z 2 - z + 8, z = 1 [ans.: 0] 5. F (z) = z 7 - 4 z 3 + z - 1, z = 1 [ans.: 3] 6. F (z) = z 4 - 5z + 1, 1 < z < 2 [ans.: 3] 7. F (z) = z 7 - z 3 - 12, 1 < z < 2 [ans.: 7] 8. F (z) = 2 z 5 - 6 z 2 + z + 1, 1 < z < 2 [ans.: 3]

Points to Remember

4.115

Points to remember Power Series A power series is an infinite series of the form •

 an (z - z0 )n = a0 + a1 (z - z0 ) + a2 (z - z0 )2 + a3 (z - z0 )3 +  + an (z - z0 )n + 

n=0

where a0, a1, a2, ..., an,... are complex or real constants, z0 is a constant known as the centre of the series. The power series about z = 0, i.e., the origin is given as •

 an z n = a0 + a1z + a2 z2 + a3 z3 +  + an z n + 

n=0

Taylor’s Series If f (z) is analytic inside a circle C with centre at z = a then for each z inside C, f (z) can be expanded as a power series about z = a as f ( z ) = f ( a ) + ( z - a ) f ′( a ) +

( z - a )2 ( z - a )n ( n ) f ′′(a ) +  + f (a) +  2! n!

Laurent’s Series If f(z) is analytic on two concentric circles C1 and C2 with the centre at z = a and radii r1, r2(r2 < r1) and in the annular region R between C1 and C2 then for all z in R, f (z) =

•

•

 an ( z - a ) n + Â

n =1 ( z - a )

n=0

where

an =

bn n

1 f ( w) dw Ú 2p i C (w - a )n +1 1

1 f ( w) dw bn = Ú 2p i C (w - a )- n +1 2

Singular Points A point at which the function f (z) is not analytic is known as a singular point or singularity of the function. Types of Singularities 1. Isolated Singularity: A singular point z = a is known as an isolated singularity if there is no other singularity within a small circle surrounding the point z = a; otherwise it is called a non-isolated singularity.

4.116

Chapter 4

Power Series

2. Removable Singularity: An isolated singular point z = a is known as a removable singularity if lim f ( z ) exists or there is no negative power term of (z – a) in LauzÆa

rent’s series of f (z), i.e., bn = 0. 3. Essential Singularity: A singular point z = a is known as an essential singularity if the number of negative power terms of (z – a) in Laurent’s series is infinite. 4. Pole of Order m: An isolated singularity z = a is known as a pole of order m if in Laurent’s series, all the negative powers of (z – a) after the mth power are zero, i.e., 1 highest power of is m. z-a Meromorphic Function: A function f (z) which is analytic everywhere in the finite plane except at a finite number of poles is called a meromorphic function.

Residues 1 in the Laurent series expansion of f (z) is known as the z-a residue of f (z) at z = a. The coefficient of

Evaluation of Residues 1. Residue at a Simple Pole (i) If f (z) has a simple pole at z = a then Res [ f ( z ); z = a ] = lim ( z - a ) f ( z ) zÆa

g( z ) , where g( z ) and h( z ) are analytic at z = a. h( z ) If h(a ) = 0 but h ′(a ) π 0 then z = a is a simple pole. If h(a ) = 0 but g(a ) π 0 then g( z ) Res [ f ( z ); z = a ] = lim z Æ a h ′( z )

(ii) If f (z) is of the form f ( z ) =

2. Residue at a Pole of Order m If f (z) has a pole of order m at z = a then Res [ f ( z ); z = a ] =

È d m -1 ˘ 1 lim Í m -1 ( z - a )m f ( z )˙ (m - 1)! z Æ a ÍÎ dz ˙˚

Cauchy’s Residue Theorem If f (z) is analytic inside and on a simple closed curve C except at a finite number of singular points inside C then

Ú f (z) dz = 2p i

(sum of residues)

C

where the integral is taken in the anticlockwise direction around C.

Points to Remember

4.117

Argument Theorem Let f (z) be analytic on and within a simple closed curve C except at a pole z = a of order P inside C. Also, suppose that f (z) has only one zero z = b of order N inside C. Then 1 f ¢( z ) dz = N - P  Ú 2p i C f ( z )

Rouche’s Theorem Let f (z) and g(z) be analytic in a domain D and have finite number of zeros in D. Suppose C is a simple closed contour in D such that no zeros of f (z) and g(z) lie on C and g(z) < f (z) on C. Then the total number of zeros of f (z) + g(z) is equal to the total number of zeros of f (z) inside C. i.e., Nf + g = Nf

| | | |

5 Applications of CHAPTER

Contour Integration chapter outline 5.1 5.2 5.3 5.4 5.5

5.1

Introduction Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q Evaluation of Improper Real Integral of a Rational Function Evaluation of Improper Real Integral of a Rational Function Including Trigonometric Functions Evaluation of Improper Integral When Simple Poles Lie on the Real Axis

introduction

The Cauchy’s residue theorem is very important in the development and applications of the theory of functions of a complex variable. It is a powerful tool to evaluate line integrals of analytic functions over closed curves. The residue theorem can be applied to real definite integrals. These real integrals are evaluated by expressing them in terms of complex functions over a suitable contour. The process of evaluation of these integrals is called contour integration.

5.2

Evaluation of a rEal dEfinitE intEgral of a rational function of cos q and sin q 2p

Let

I=

Ú

f (sin q , cos q ) dq

0

where f (sin q, cos q ) is a rational function of sin q and cos q. To evaluate the integral, putting z = eiq, dz = i eiq dq = i z dq 1Ê 1 ˆ z2 + 1 cos q = Á z + ˜ = 2Ë z¯ 2z sin q =

1Ê 1 ˆ z2 - 1 z- ˜ = Á 2i Ë z¯ 2iz

5.2

Chapter 5 Applications of Contour Integration

z = eiq = 1 Substituting these values in the integral, I= Ú f ( z ) dz C

||

where C is a unit circle z = 1 By Cauchy’s residue theorem, I = 2p i (sum of residues at all poles lying inside the circle z = 1)

||

Example 1 2p

Ú

Evaluate

0

dq using contour integration. 5 - sin q

Solution y 2p

(i) Let I =

dq

Ú 5 - sin q

z = (5 + 2√6 )

0

Consider the contour C as the unit circle z = 1 (Fig. 5.1).

||

C

dz z2 - 1 Let z = e , dq = , sin q = iz 2iz iq

I=

(ii)

dz iz Ê z 2 - 1ˆ 5-Á ˜ Ë 2iz ¯

Ú C

2

=Ú

2

C - z + 10iz + 1

= -2 Ú

C

(iii) Let f ( z ) =

O

dz

1 2

z - 10iz - 1

dz

1 2

z - 10 iz - 1

The poles are given by z 2 - 10iz - 1 = 0 z=

z = (5 − 2√6 )i

10i ± (-10i )2 + 4 2

z=1

x

fig. 5.1

...(1)

5.2  Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q

10i ± (-96) 2 = 5i ± 2 6i =

= (5 ± 2 6 ) i \ f (z) =

1 È z - (5 + 2 6 )i ˘ È z - (5 - 2 6 )i ˘ Î ˚Î ˚

(iv) For z = (5 + 2 6 )i, z = (5 + 2 6 ) i = 9.89 > 1 Hence, z = (5 + 2 6 ) i lies outside C. For z = (5 - 2 6 )i, z = (5 - 2 6 ) i = 0.10 < 1 Hence, z = (5 - 2 6 ) i lies inside C. (v) Res [ f ( z ); z = (5 - 2 6 )i ] = =

lim

z Æ ( 5 - 2 6 )i

lim

1

z Æ ( 5 - 2 6 )i

=

[ z - (5 - 2 6 )i ] f ( z )

z - (5 + 2 6 )i 1

(5 - 2 6i ) - (5 + 2 6 )i 1 =4 6i (vi) By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues)

C

Ê

dz

Ú z2 - 10 iz - 1 = 2p i ÁË - 4

C

=-

1 ˆ 6i ˜¯

p 2 6

Substituting in Eq. (1), Ê p ˆ I = -2 Á Ë 2 6 ˜¯ =

p 6

5.3

5.4

Chapter 5 Applications of Contour Integration

Example 2 2p

Evaluate

Ú 0

dq using contour integration. 13 + 5 sin q

Solution

y 2p

(i) Let I =

dq

Ú 13 + 5 sin q 0

C

||

Consider the contour C as the unit circle z = 1 (Fig. 5.2).

O

dz z2 - 1 z = e , dq = , sinq = iz 2iz

z =− i 5

iq

dz iz (ii) I = Ú Ê z 2 - 1ˆ C 13 + 5 Á ˜ Ë 2iz ¯ = 2Ú

C

x

≈ z = −5i

fig. 5.2

dz

…(1)

5z 2 + 26iz - 5

(iii) Let f ( z ) =

z=1

1 2

5z + 26iz - 5 The poles are given by z=

- 26i ± (26i )2 - 4(5)(-5) 10

- 26i ± -576 10 - 26i ± 24i = 10 i = - , - 5i 5 1 \ f (z) = i Ê ˆ 5 Á z + ˜ ( z + 5i ) Ë 5¯ =

i i 1 (iv) For z = - , z = - = < 1 5 5 5

È∵ ax 2 + bx + c = a( x - a )( x - b )˘ Î ˚

5.2  Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q

i lies inside C. 5 For z = -5i, z = -5i = 5 > 1 Hence, z = -

Hence, z = –5i lies outside C. ÈÊ ˘ i˘ iˆ È (v) Res Í f ( z ); z = - ˙ = lim ÍÁ z + ˜ f ( z )˙ 5 ˚ z Æ - i ÎË 5¯ Î ˚ 5

= lim zÆ-

i 5

1 5( z + 5i )

1 Ê i ˆ 5 Á - + 5i˜ Ë 5 ¯ 1 = 24i (vi) By Cauchy’s residue theorem, =

Ú f (z) dz = 2p i (sum of residues)

C

Ê 1 ˆ

dz

Ú 5z2 + 26iz - 5 = 2p i ÁË 24i ˜¯

C

p 12 Substituting in Eq. (1), =

Êpˆ I = 2Á ˜ Ë 12 ¯ =

p 6

Example 3 p

Evaluate

1

Ú 17 - 8 cos q dq , 0

Solution 2p

(i)

Let I =

1

Ú 17 - 8 cos q dq 0

by integrating around a unit circle.

5.5

5.6

Chapter 5 Applications of Contour Integration

||

Consider the contour C as the unit circle z = 1 (Fig. 5.3). dz z2 + 1 y Let z = eiq, dq = , cos q = iz 2z 1 dz (ii) I = Ú 2 C Ê z + 1ˆ iz C 17 - 8 Á ˜ Ë 2z ¯ 1 1 dz Ú i C 17 z - 4 z 2 - 4

=-

(iii)

O z= 1 4

1 1 dz Ú 2 i C 4 z - 17 z + 4

Let f ( z ) =

z=1

≈

=

x z=4

...(1) fig. 5.3

1 4 z 2 - 17 z + 4

The poles are given by 4 z 2 - 17 z + 4 = 0 17 ± 289 - 64 8

z=

17 ± 225 8 17 ± 15 = 8 1 = 4, 4 =

1 1ˆ Ê 4( z - 4) Á z - ˜ Ë 4¯

\ f (z) =

(iv)

È∵ ax 2 + bx = c = a( x - a )( x - b )˘ Î ˚

|| ||

For z = 4, z = 4 = 4 > 1 Hence, z = 4 lies outside C. 1 1 1 , z = = 1 Hence, z = -2 - 3 lies outside C. (v) Res È f ( z ); z = -2 + 3 ˘ = lim Î ˚ z Æ-2 + = = =

lim

3

z Æ-2 + 3

È( z + 2 - 3 ) f ( z )˘ Î ˚ 1 z+2+ 3 1

-2 + 3 + 2 + 3 1 2 3

z=1

z = 3.73

x

5.2  Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q

5.9

(vi) By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues)

C

Ê 1 ˆ 3 ˜¯

dz

Ú z2 + 4 z + 1 = 2p i ÁË 2

C

pi

=

3

Substituting in Eq. (1), I= =

2 Ê pi ˆ i ÁË 3 ˜¯ 2p 3

Example 5 2p

Evaluate a real integral

Ú

0

1 (2 + cos q )2

dq using residue. [Summer 2013]

Solution 2p

(i) Let I =

1

Ú (2 + cos q )2 dq

y

0

Consider the contour C as the unit circle z = 1 (Fig. 5.5).

||

iq Let z = e , dq =

dz z2 + 1 , cos q = iz 2z z = 0.27

z=1

dz CÊ z + 1ˆ iz + 2 ÁË ˜ 2z ¯

(ii) I = Ú

(iii)

1 2

2

=

4 z dz i CÚ (4 z + z 2 + 1)2

=

4 z dz Ú 2 i C ( z + 4 z + 1)2

Let

f (z) =

fig. 5.5

...(1) z

2

( z + 4 z + 1)2

z = 3.73

x

5.10

Chapter 5 Applications of Contour Integration

The poles are given by ( z 2 + 4 z + 1)2 = 0 -4 ± 16 - 4 2 = -2 ± 3

z=

\ f (z) = =

z È z - ( -2 + 3 )˘ È z - ( -2 - 3 )˘ Î ˚ Î ˚ z 2

2

( z + 2 - 3 )2 ( z + 2 + 3 )2

(iv) For z = -2 + 3 , z = -2 + 3 = 0.27 < 1 Hence, z = -2 + 3 lies inside C. For z = -2 - 3, z = -2 - 3 = 3.73 > 1 Hence, z = -2 - 3 lies outside C. (v)

z = -2 + 3 is a pole of order 2. Res ÈÎ f ( z ); z = -2 + 3 ˘˚ = = =

1 (2 - 1)!

lim

z Æ( -2 +

lim

z Æ( -2 + 3 )

lim

z Æ( -2 +

2 d È ˘ Î( z + 2 - 3 ) f ( z )˚ 3 ) dz z

d dz ( z + 2 + 3 )2

2z 1 È ˘ + Í 3 2˙ 3) ( z + 2 + 3 ) ˙˚ ÍÎ ( z + 2 + 3 )

2 ( -2 + 3 )

=-

+

(-2 + 3 + 2 + 3 ) (-2 + ( -4 + 2 3 ) + 1 =(2 3 )3 (2 3 )2 (-2 + 3 ) + 1 =3

12 3

= = =

2- 3 + 3 12 3 2 12 3 1 6 3

12

1 3 + 2 + 3)

2

5.2  Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q

(vi)

5.11

By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues)

C

Ê 1 ˆ 3 ˜¯

z

Ú (z2 + 4 z + 1)2 dz = 2p i ÁË 6

C

=

pi 3 3

Substituting in Eq. (1), I= =

4 Ê pi ˆ i ÁË 3 3 ˜¯ 4p 3 3

Example 6 2p

Ú

Evaluate

0

cos 3q dq using contour integration. 5 - 4 cos q

Solution 2p

cos 3q

Ú 5 - 4 cos q dq

(i) Let I =

y

0

Consider the contour C as the unit circle z = 1 (Fig. 5.6).

C

||

z = eiq, dq =

dz z2 + 1 , cosq = iz 2z

z3 = e3iq = cos 3q + i cos 3q

O z= 1 2

z=1

z=2

x

cos 3q = RP z3 (ii) I = RP

z3 dz 2 Ê z + 1ˆ iz C 5 - 4Á ˜ Ë 2z ¯

Ú

= RP

fig. 5.6

z3 1 dz i CÚ 5z - 2 z 2 - 2

Ê 1ˆ = RP Á - ˜ Ë i¯

z3

Ú 2 z2 - 5z + 2 dz

C

…(1)

5.12

Chapter 5 Applications of Contour Integration

z3

(iii) Let f ( z ) =

2 z 2 - 5z + 2 The poles are given by 2 z 2 - 5z + 2 = 0 2z2 - 4z - z + 2 = 0 2 z ( z - 2) - 1( z - 2) = 0 ( z - 2)(2 z - 1) = 0 1 2

z = 2, \

f (z) =

z3 1ˆ Ê 2( z - 2 ) Á z - ˜ Ë 2¯

(iv) For z = 2, z = 2 = 2 > 1 Hence, z = 2 lies outside C. For z =

1 1 1 , z = = 1 Hence, z = –2 lies outside C. 1 1 1 For, z = - , z = - = < 1 2 2 2 1 Hence, z = - lies inside C. 2 (v)

Res

[ f (z); z = 0]

= lim zf ( z ) zÆ0

= lim

zÆ0

=

=

z2 + z + 1 1ˆ Ê 2( z + 2 ) Á z + ˜ Ë 2¯

0 + 0 +1 1ˆ Ê 2( 0 + 2 ) Á 0 + ˜ Ë 2¯ 1 2

1˘ 1ˆ È Ê Res Í f ( z ); z = - ˙ = lim1 Á z + ˜ f ( z ) Ë 2 2¯ Î ˚ z Æ2

= lim z Æ-

1 2

z2 + z + 1 2 z( z + 2)

1 1 - +1 4 2 = Ê 1ˆ Ê 1 ˆ 2 Á - ˜ Á - + 2˜ Ë 2¯ Ë 2 ¯ =-

1 2

(vi) By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues)

C

z2 + z + 1

Ê1

1ˆ

Ú z(2 z2 + 5z + 2) dz = 2p i ÁË 2 - 2 ˜¯

C

=0 Substituting in Eq. (1), 1 I = (0 ) = 0 i

5.2  Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q

5.15

Example 8 2p

Ú

Evaluate

0

dq using contour integration. 3 - 2 cos q + sin q

Solution 2p

(i) Let I =

dq

Ú 3 - 2 cos q + sin q 0

||

Consider the contour C as the unit circle z = 1 (Fig. 5.8). 2

z = eiq, dq =

(ii) I =

Ú C

=Ú

C

=Ú

C

= 2Ú

dz z +1 z2 - 1 , cosq = , sinq = , iz 2z 2iz

y

dz iz Ê z 2 + 1ˆ Ê z 2 - 1ˆ 3 - 2Á ˜ ˜ +Á Ë 2 z ¯ Ë 2iz ¯ dz iz 2z2 + 2 z2 - 1 3+ 2z 2iz 2dz

C

O

z=2−i

6iz - (2 z 2 + 2)i + ( z 2 - 1)

fig. 5.8

dz (1 - 2i )z 2 + 6iz - (2i + 1)

(iii) Let f ( z ) =

x

z=1 z = 2−i 5

1 2

(1 - 2i )z + 6iz - (2i + 1) The poles are given by (1 - 2i )z 2 + 6iz - (2i + 1) = 0 z=

-6i ± (6i )2 + 4(1 - 2i )(2i + 1) 2(1 - 2i )

=

-6i ± -36 + 4(2i + 1 + 4 - 2i ) 2(1 - 2i )

-6i ± -36 + 20 2(1 - 2i ) -6i ± 4i = 2(1 - 2i ) =

…(1)

5.16

Chapter 5 Applications of Contour Integration

i -5i , 1 - 2i 1 - 2i 2-i = , 2-i 5 =

\

f (z) =

1 È Ê 2 -iˆ˘ (1 - 2i ) [ z - (2 - i )] Í z - Á ˜˙ Î Ë 5 ¯˚

[∵ ax2 + bx + c = a(x − a) (x − b)]

(iv) For z = 2 - i, z = 2 - i = 5 > 1 Hence, z = 2 – i lies outside C. For z =

2-i 2-i 5 , z = = = 0.45 < 1 5 5 5

Hence, z =

2-i lies inside C. 5

ÈÊ ˘ 2-i˘ 2 -iˆ È (v) Res Í f ( z ); z = ÍÁË z ˜¯ f ( z )˙ ˙ = lim 2 i 5 5 Î ˚ zÆ Î ˚ 5

= lim

2 -i zÆ 5

=

=

1 (1 - 2i )( z - 2 + i )

1 Ê 2-i ˆ (1 - 2i ) Á - 2 + i˜ Ë 5 ¯ 1 + 2i Ê -8 + 4i ˆ 5Á Ë 5 ˜¯

1 + 2i 4i(2i + 1) 1 = 4i (vi) By Cauchy’s residue theorem, =

Ú f (z) dz = 2p i (sum of residues)

C

Ê 1ˆ

dz

Ú (1 - 2i)z2 + 6iz - (2i + 1) = 2p i ÁË 4i ˜¯

C

=

p 2

5.2  Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q

5.17

Substituting in Eq. (1), Êpˆ I = 2Á ˜ = p Ë 2¯

Example 9 2p

Evaluate

Ú 0

dq 1 - 2 a cos q + a 2

,(0 < a < 1) using contour integration.

Solution 2p

(i) Let I =

dq

Ú 1 - 2a cos q + a2 0

||

Consider the contour C as the unit circle z =1 (Fig. 5.9). 2

Let z = eiq, dq =

dz z +1 , , cosq = iz 2z

y

dz iz (ii) I = Ê z 2 + 1ˆ 2 C 1 - 2a Á 2z ˜ + a ¯ Ë

Ú

=

1 dz Ú 2 i C z - az - a + a 2 z

=

1 dz i CÚ - az 2 + (a 2 + 1)z - a

=-

1 dz Ú 2 i C az - (a 2 + 1)z + a

(iii) Let f ( z ) =

C

z = − a1

…(1)

O

z=a

fig. 5.9

1 2

2

az - (a + 1)z + a

The poles are given by Ê a 2 + 1ˆ z2 - Á ˜ z +1 = 0 Ë a ¯ z= =

a 2 + 1 ± (a 2 + 1)2 - 4 a 2 2a a 2 + 1 ± a 2 + 2a 2 + 1 - 4a 2 2a

z=1

x

5.18

Chapter 5 Applications of Contour Integration

=

a 2 + 1 ± (a 2 - 1)2 2a

a 2 + 1 ± (a 2 - 1) 2a 1 = a, a =

\ f (z) =

1 1ˆ Ê a( z - a ) Á z - ˜ Ë a¯

(iv) For z = a, z = a = a < 1 Hence, z = a lies inside C. For z =

1 1 1 , z = = >1 a a a 1 lies outside C. a

Hence, z =

(v) Res [ f ( z ); z = a ] = lim ÈÎ( z - a ) f ( z )˘˚ zÆa = lim

zÆa

=

=

1 1ˆ Ê aÁz - ˜ Ë a¯

1 1ˆ Ê aÁa - ˜ Ë a¯ 1 2

a -1 (vi) By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum of residues)

C

Ê

dz

1 ˆ

Ú az2 - (a2 + 1)z + a = 2p i ÁË a2 - 1˜¯

C

Substituting in Eq. (1), 1 Ê 1 ˆ I = - 2p i Á 2 ˜ Ë a - 1¯ i =-

2p 2

a -1 2p ,0 < a < 1 = 1 - a2

[∵ ax2 + bx + c = a(x − a) (x − b)]

5.2  Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q

5.19

Example 10 2p

Evaluate

Ú 0

sin 2 q dq , a > b > 0 using contour integration. a + b cos q

Solution 2p

(i) Let I =

Ú 0

sin 2 q dq a + b cos q

2p

=

1 - cos 2q

Ú 2(a + b cos q ) dq 0

||

Consider the contour C as the unit circle z = 1 (Fig. 5.10). y dz z2 + 1 z = eiq, dq = , cosq = iz 2z 2 2iq z = e = cos 2q + i sin 2q cos 2q = RP z2 1 − cos 2q = RP (1 − z2) dz z = z2 z = z1 O (1 - z )2 iz (ii) I = RP È Ê z 2 + 1ˆ ˘ C 2 Ía + b Á 2z ˜ ˙ ¯ ˙˚ Ë ÍÎ

Ú

= RP

(1 - z 2 )dz 1 i CÚ 2 az + b( z 2 + 1)

= RP

(1 - z 2 )dz 1 i CÚ bz 2 + 2 az + b

(iii) Let f ( z ) =

1 - z2

bz 2 - 2 az + b The poles are given by bz 2 + 2 az + b = 0 z=

-2 a ± 4 a 2 - 4b2 2b

=

-2 a ± 2 a 2 - b2 2b

=

- a ± a 2 - b2 b

C

z=1

x

fig. 5.10

…(1)

5.20

Chapter 5 Applications of Contour Integration

- a + a 2 - b2 - a - a 2 - b2 , z2 = b b 2 1- z \ f (z) = b( z - z1 )( z - z2 ) Let z1 =

(iv) For z = z1 =

- a + a 2 - b2 b - a + a 2 - b2 a a2 = - 2 -1 < 1 b b b

z = z1 =

a È ˘ Í∵ a > b, b > 1˙ Î ˚

Hence, z = z1 lies inside C. For z = z2 =

- a - a 2 - b2 b

z = z2 =

- a - a 2 - b2 a a2 = + 2 -1 > 1 b b b

Hence, z = z2 lies outside C. (v) Res ÈÎ f ( z ); z = z1 ˘˚ = lim ÈÎ( z - z1 ) f ( z )˘˚ zÆ z 1

= lim

z Æ z1

=

=

= = =

(1 - z 2 ) b( z - z2 )

1 - z12 b( z1 - z2 ) È Ê 2 2 Í1 - - a + a - b Á Í Á b ÍÎ Ë Ê 2 a 2 - b2 bÁ ÁË b

ˆ ˜ ˜¯

2˘

˙ ˙ ˙˚

ˆ ˜ ˜¯

b2 - Èa 2 + (a 2 - b2 ) - 2 a a 2 - b2 ˘ ÎÍ ˚˙ 2b 2 a 2 - b 2 2(b2 - a 2 ) - 2 a a 2 - b2 2b 2 a 2 - b 2 - a 2 - b2 - a b2

Ê a + a 2 - b2 = -Á ÁË b2

ˆ ˜ ˜¯

a È ˘ Í∵ a > b, b > 1˙ Î ˚

5.2  Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q

5.21

Ú f (z) dz = 2p i (sum of residues)

(vi)

C

Ê a + a 2 - b2 d z = 2 p i Á Ú 2 ÁË b2 C bz - 2 az + b (1 - z 2 )

ˆ ˜ ˜¯

Substituting in Eq. (1), Ê a + a 2 - b2 1È I = RP Í-2p i Á ÁË iÍ b2 Î =

(

-2p a + a 2 - b2 b

)

ˆ˘ ˜˙ ˜¯ ˙ ˚

2

ExErcisE 5.1 Evaluate the following integrals using contour integration: 2p

1.

dq

Ú 5 + 4 cos q

2p ˘ È Í ans.: 3 ˙ Î ˚

0

2p

2.

dq

Ú 5 + 4 sinq

2p ˘ È Í ans.: 3 ˙ Î ˚

0

2p

3.

sin q

Ú 5 + 4 cos q dq 0

2p

4.

ÎÈans.: 0 ˘˚ cos 2q

Ú 5 - 4 cos q 0

p˘ È Í ans.: 6 ˙ Î ˚ 2p

5.

Ú 0

sin2 q dq 5 - 3 cos q 2p ˘ È Í ans.: 3 ˙ Î ˚

2p

6.

dq

Ú a + b sinq ,

a>b>0

0

˘ È 2p ˙ Í ans.: a 2 - b2 ˚ Î

5.22

Chapter 5 Applications of Contour Integration

2p

7.

dq

Ú 2 + cos q

È 2p ˘ Í ans.: ˙ 3˚ Î

0

2p

8.

dq

Ú 5 - 3 cos q

p˘ È Í ans.: 2 ˙ Î ˚

0

2p

9.

dq

Ú a + b cos q , a > b > 0 0

2p

10.

dq

Ú 1 + a sinq , a

È ˘ 2p Í ans.: ˙ 2 2 a -b ˚ Î

b > 0 2

0

2p

12.

È 2p a ˘ Í ans.: ˙ 3 Í 2 2 2 ˙ (a - b ) ˙ ÍÎ ˚

dq

Ú (5 - 4 cos q)

2

10p ˘ È Í ans.: 27 ˙ Î ˚

0

5.3

Evaluation of impropEr rEal intEgral of a rational function •

Let I =

Ú

f ( x ) dx

-•

P( x ) and the degree of Q( x ) Q(x) is greater than the degree of P(x) by at least 2 and Q(x) has no real roots. To evaluate the integral, consider Ú f ( z )dz, where f ( x ) =

y

C1

C

where C is the contour consisting of the upper semicircle C1 of radius R with centre at the origin and the part of real axis from –R to R (Fig. 5.11).

−R

O

fig. 5.11

R

x

5.3  Evaluation of Improper Real Integral of a Rational Function        5.23

R

Ú

C

f ( z ) dz = Ú f ( z )dz + C1

Ú

f ( x ) dx

-R

By Cauchy’s residue theorem,

Ú f (z) dz = 2p i (sum

of residues at poles in the upper half of the z-plane)

C R

Ú

f ( z ) dz +

Ú

f ( x ) dx = 2p i (sum of resides)

-R

C1

R

Taking limit R Æ •, lim

R Æ•

Ú

f ( z ) dz + lim

Ú

f ( z )dz = 0

By Cauchy’s lemma, lim

R Æ•

R Æ•

C1

Ú

f ( x ) dx = 2p i (sum of residues)

-R

C1

•

Ú

f ( x ) dx = 2p i (sum of residues)

-•

Example 1 •

Evaluate

Ú

-• ( x

x 2 dx 2

+ 1)( x 2 + 4)

using contour integration.

Solution (i) Let

y

z 2 dz

Ú f (z) dz = Ú (z2 + 1)(z2 + 4)

C

C

Consider the contour C consisting of the upper semicircle C1 of radius R and part of the axis from –R to R (Fig. 5.12). (ii) Let f ( z ) =

z2

( z 2 + 1)( z 2 + 4) The poles are given by (z2 + 1) (z2 + 4) = 0 z = ± i, ± 2 i \ f ( z) =

z2 ( z + i )( z - i )( z + 2i )( z - 2i )

(iii) The poles z = i and z = 2i lie inside C. (iv) Res [ f ( z ); z = i ] = lim ÈÎ( z - i ) f ( z )˘˚ z Æi

C1 z = 2i z=i −R

O

fig. 5.12

R

x

5.24

Chapter 5 Applications of Contour Integration

= lim z Æi

z2 ( z + i )( z + 2i )( z - 2i )

i2 (2i )(3i )(-i ) 1 =6i Res [ f ( z ); z = 2i ] = lim ÈÎ( z - 2i ) f ( z )˘˚ =

z Æ2i

= lim

z Æ2i

z2 ( z + i )( z - i )( z + 2i )

4i 2 (3i )(2i )(4i ) -4 = -12i 1 = 3i =

•

Ú

(v)

f ( x ) dx = 2p i (sum of residues)

-• •

x 2 dx

Ê 1 1ˆ = 2p i Á - + ˜ Ë 6i 3i ¯ -• ( x + 1)( x + 4)

Ú

2

2

Ê 1ˆ = 2p Á ˜ Ë 6¯ =

p 3

Example 2 •

Ú

Evaluate

x2 - x + 2

4 2 -• x + 10 x + 9

dx using contour integration. y

Solution (i) Let

Ú

C

f ( z )dz = Ú

C

z2 - z + 2 z 4 + 10 z 2 + 9

dz

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.13).

z = 3i

C1

z=i −R

O

fig. 5.13

R

x

5.3  Evaluation of Improper Real Integral of a Rational Function        5.25

z2 - z + 2

(ii) Let f ( z ) =

z 4 + 10 z 2 + 9 The poles are given by z 4 + 10 z 2 + 9 = 0 ( z 2 + 1)( z 2 + 9) = 0 z = ± i, ± 3 i \

f ( z) =

z2 - z + 2 ( z + i )( z - i )( z + 3i )( z - 3i )

(iii) The poles z = i and z = 3i lie inside C. Res [ f ( z ); z = i ] = lim ( z - i ) f ( z )

(iv)

z Æi

= lim z Æi

z2 - z + 2 ( z + i )( z + 3i )( z - 3i )

i2 - i + 2 (2i )(4i )(-2i ) -1 - i + 2 = 16 6i 1- i = 16i Res [ f ( z ); z = 3i ] = lim ( z - 3i ) f ( z ) =

z Æ 3i

= lim

z Æ 3i

z2 - z + 2 ( z + i )( z - i )( z + 3i )

(3i )2 - 3i + 2 (4i )(2i )(6i ) -9 - 3ii + 2 = - 48 i =

=

7 + 3i 48i

•

Ú

(v)

f ( x ) dx = 2p i (sum of residues)

-• •

Ú

-•

x2 - x + 2

Ê 1 - i 7 + 3i ˆ dx = 2p i Á + Ë 16i 48i ˜¯ x + 10 x + 9 4

2

È 3(1 - i ) + 7 + 3i ˘ = 2p i Í ˙ 48i Î ˚

5.26

Chapter 5 Applications of Contour Integration

10p 24 5p = 12 =

Example 3 •

Evaluate

Ú

x2

-• ( x

2

+ 1)2

dx using contour integration. y

Solution (i) Let

Ú

C

f ( z ) dz = Ú

C

z2 ( z 2 + 1)2

dz C1

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.14). (ii) Let f ( z ) =

z=i −R

z2 ( z 2 + 1)2

The poles are given by ( z 2 + 1)2 = 0 z = ± i, ± i \

f ( z) =

z2 ( z + i )2 ( z - i )2

(iii) The poles z = i of order 2 lies inside C. (iv) Res [ f ( z ); z = i ] =

1 d lim [( z - i )2 f ( z )] (2 - 1)! z Æi dz

= lim z Æi

= lim

d È z2 ˘ Í ˙ dz ÍÎ ( z + i )2 ˙˚ ( z + i ) 2 ( 2 z ) - z 2 2( z + i ) ( z + i )4

z Æi

= lim

( z + i )[( z + i )(2 z ) - 2 z 2 ] ( z + i )4

z Æi

= lim z Æi

2iz ( z + i )3

O

fig. 5.14

R

x

5.3  Evaluation of Improper Real Integral of a Rational Function        5.27

2i ◊ i

=

(i + i )3 -2 = -8i 1 = 4i •

Ú

(v)

f ( x ) dx = 2p i(sum of residues)

-• •

x2

Ê 1ˆ dx = 2p i Á ˜ Ë 4i ¯ -• ( x + 1)

Ú

2

2

=

p 2

Example 4 •

1

Ú 1 + x 2 dx

Evaluate

using contour integration.

0

Solution (i) Let

y

1

Ú f (z) dz = Ú z2 + 1 dz

C

C

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.15). (ii) Let f ( z ) =

1

z=i −R

z2 + 1 The poles are given by z2 + 1 = 0 z = ±i \

C1

f (z) =

fig. 5.15

1 ( z + i )( z - i )

(iii) The pole z = i lies inside C. (iv) Res[ f ( z ); z = i ] = lim( z - i ) f ( z ) z Æi

= lim z Æi

=

1 2i

O

1 z+i

R

x

5.28

Chapter 5 Applications of Contour Integration

•

Ú

(v)

f ( x )dx = 2p i(sum of residues)

-• •

Ê 1ˆ

1

Ú 1 + x 2 dx = 2p i ÁË 2i ˜¯

-•

=p •

2Ú 0

1 1 + x2

•

a È a ˘ Í∵ Ú f ( x )dx = 2 Ú f ( x )dx if f (- x ) = f ( x )˙ ÍÎ - a ˙˚ 0

dx = p p

1

Ú 1 + x 2 dx = 2 0

Example 5 •

Evaluate

dx

Ú (1 + x 2 )2

using contour integration.

0

Solution y

(i) Let

dz

Ú f (z)dz = Ú (1 + z2 )2

C

C

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.16). (ii) Let f ( z ) =

1

C1 z=i −R

fig. 5.16

(1 + z 2 )2 The poles are given by (1 + z 2 )2 = 0 1 + z2 = 0 z 2 = -1 z = ±i \ f (z) =

1 2

( z + i ) ( z - i )2

(iii) The pole z = i of order 2 lies inside C. (iv) Res[ f ( z ); z = i ] =

O

1 d lim [( z - i )2 f ( z )] (2 - 1)! z Æi dz

R

x

5.3  Evaluation of Improper Real Integral of a Rational Function        5.29

= lim z Æi

d È 1 ˘ Í ˙ dz Î ( z + i )2 ˚

È 2 ˘ = lim Í3˙ z Æi Î (z + i) ˚ 2 =(i + i )3 2 = 8i 1 = 4i •

Ú

(v)

f ( x ) dx = 2p i (sum of residues)

-• •

Ê 1ˆ dx = 2p i Á ˜ Ë 4i ¯ - • (1 + x ) 1

Ú

2 2

•

2Ú 0

1 (1 + x 2 )2

•

=

p 2

dx =

p 2

a È a ˘ Í∵ Ú f ( x ) dx = 2 Ú f ( x ) dx if f (- x ) = f ( x )˙ ÍÎ - a 0 ˚˙

p

1

Ú (1 + x 2 )2 dx = 4 0

Example 6 •

x 2 dx

Ú ( x 2 + a 2 )( x 2 + b2 ) , a > 0, b > 0

Evaluate

0

Solution (i) Let

Ú

C

f ( z ) dz = Ú

C

z 2 dz

y

( z 2 + a 2 )( z 2 + b2 )

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.17). (ii) Let

f (z) =

z 2

2

bi

C1

ai

2 2

2

( z + a )( z + b )

−R

O

fig. 5.17

R

x

5.30

Chapter 5 Applications of Contour Integration

The poles are given by ( z 2 + a 2 )( z 2 + b2 ) = 0 z = ± ai, ± bi \ f ( z) =

z2 ( z + ai )( z - ai )( z + bi )( z - bi )

(iii) The poles z = ai and z = bi lie inside C as a > 0, b > 0. Res [ f ( z ); z = ai ] = lim ( z - ai ) f ( z )

(iv)

z Æ ai

z2 z Æ ai ( z + ai )( z + bi )( z - bi )

= lim

a2i2 (2 ai )(ai + bi )(ai - bi ) a =2 2i ( b - a 2 ) Res [ f ( z ); z = bi ] = lim ( z - bi ) f ( z ) =

z Æ bi

z2 z Æ bi ( z + ai )( z - ai )( z + bi )

= lim

b2 i 2 (bi + ai )(bi - ai )(2bi ) b =2i(a 2 - b2 ) b = 2 2i(b - a 2 ) =

•

Ú

(v)

f ( x )dx = 2p i(sum of residues)

-• •

È ˘ a b + dx = 2p i Í2 2 2 2 ˙ Î 2i(b - a ) 2i(b - a ) ˚ -• ( x + a )( x + b ) È b-a ˘ =pÍ 2 ˙ Î b - a2 ˚ p = b+a

Ú

•

2Ú 0

•

x2

2

2

2

2

x2 ( x 2 + a 2 )( x 2 + b2 ) x2

dx =

a a ˘ p È Í∵ Ú f ( x ) dx = 2 Ú f ( x ) dx if f (- x ) = f ( x )˙ b + a ÍÎ - a ˙˚ 0

p

Ú ( x 2 + a2 )( x 2 + b2 ) dx = 2(b + a) 0

5.3  Evaluation of Improper Real Integral of a Rational Function        5.31

Example 7 •

Show that

dx

Ú 1 + x4

=

0

p 2 2

.

Solution (i) Let

dz

Ú f (z) dz = Ú 1 + z 4

C

C

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.18). 1 y (ii) Let f ( z ) = 4 1+ z The poles are given by 1 + z4 = 0 C1

z 4 = -1 z= =

1 (-1) 4 ip e4

,e

z=e -

ip 4

i 3p ,e 4

ip

(iii) The poles z = e 4 and z = e i

i 3p 4

,e

-

i 3p 4

−R

i

p

(iv) Res [ f ( z ); z = e 4 ] = lim ( z - e 4 ) f ( z ) i

p

zÆe 4

z-e

= lim

1+ z

p i zÆe 4

p 4

4

È0 ˘ Í 0 form ˙ Î ˚

1

= lim

4 z3

p zÆe 4 i

=

i

1 4e

3i

p 4 3p

1 -i 4 e 4 1Ê 3p 3p ˆ = Á cos - i sin ˜ 4Ë 4 4¯ =

=

z=e 4 O

fig. 5.18

lie inside C.

p

ip

i 3p 4

1 ˆ 1Ê 1 -iÁ ˜ 4Ë 2 2¯

R

x

5.32

Chapter 5 Applications of Contour Integration

=Res [ f ( z ); z =

3p i e 4

(1 + i ) 4 2

] = lim ( z - e

i

3p 4

3p zÆe 4

) f (z)

i

z-e

= lim

i

3p 4

1+ z

3p i zÆe 4

1

= lim

3p i zÆe 4

È0 ˘ Í 0 form ˙ Î ˚

4

[Applying L’Hospital rule ]

4 z3

1

=

4e

i

9p 4 9p

1 -i 4 e 4 1Ê 9p 9p ˆ = Á cos - i sin ˜ 4Ë 4 4 ¯ =

= =

1Ê 1 1 ˆ -i Á ˜ 4Ë 2 2¯ 1- i 4 2

•

Ú

(v)

-• •

f ( x ) dx = 2p i (sum of residues) È 1+ i

1

Ú 1 + x 4 dx = 2p i ÍÎ- 4

-•

2

+

1- i ˘ ˙ 4 2˚

Ê 2i ˆ = 2p i Á Ë 4 2 ˜¯

•

2Ú 0

•

1 1 + x4 1

=

p

dx =

p

2

Ú 1 + x 4 dx = 2 0

a È a ˘ Í∵ Ú f ( x ) dx = 2 Ú f ( x ) dx if f (- x ) = f ( x )˙ ÍÎ - a ˙˚ 0

2 p 2

5.3  Evaluation of Improper Real Integral of a Rational Function        5.33

Example 8 •

Using the theory of residues, evaluate

-•

Solution (i) Let

Ú

C

Ú

f ( z ) dz = Ú

C

eiz z2 + 1

cos x x2 + 1

dx.

[Summer 2015]

dz

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.19). (ii) Let f ( z ) =

eiz z2 + 1

y

The poles are given by C1

z2 + 1 = 0 z = ±i

i

eiz ( z + i )( z - i )

\ f (z) =

(iii) The pole z = i lies inside C. Res [ f ( z ); z = i ] = lim ( z - i ) f ( z )

(iv)

z Æi

= lim z Æi

eiz (z + i)

2

ei = 2i

e -1 2i 1 = 2ie =

•

Ú

(v)

f ( x ) dx = 2p i (sum of residues)

-• •

Ú

eix 2

-• x + 1 •

Ú

-•

cos x + i sin x 2

x +1

Ê 1 ˆ = 2p i Á Ë 2ie ˜¯

dx =

p e

–R

O

fig. 5.19

R

x

5.34

Chapter 5 Applications of Contour Integration

Comparing real parts, •

Ú

-•

cos x x2 + 1

dx =

p e

ExErcisE 5.2 Evaluate the following integrals using contour integration: •

1.

x 2 dx Ú0 (x 2 + 9)(x 2 + 4)

•

2.

Ú (x

dx ,a > b > 0 + a )(x 2 + b 2 )

2

2

0

•

3.

Úx

dx + 10 x 2 + 9

4

0

•

4.

Ú (x

dx + 1)3

2

-•

•

5.

Ú (x

2

0

•

6.

Úx

-•

•

7.

Ú (x 0

dx ,a > 0 + a 2 )2

x2 + x + 3 dx 4 + 5x 2 + 4

x2 dx + 1)3

2

p ˘ È Í ans.: 10 ˙ Î ˚

È ˘ p Í ans.: 2ab(a + b) ˙ Î ˚

p ˘ È Í ans.: 24 ˙ Î ˚

3p ˘ È Í ans.: 8 ˙ Î ˚

p ˘ È Í ans.: 4 a 3 ˙ Î ˚

5p ˘ È Í ans.: 6 ˙ Î ˚

p˘ È Í ans.: 6 ˙ Î ˚

5.4  Evaluation of Improper Real Integral of a Rational Function Including...        5.35

•

8.

x2 Ú 2 2 2 dx , a > 0 -• ( x + a )

3p ˘ È Í ans.: 2a ˙ Î ˚

•

9.

x2 dx Ú 2 2 -• ( x + 1)( x + 2 x + 2)

3p ˘ È Í ans.: 5 ˙ Î ˚

•

10.

5.4

x 2 dx Ú 2 2 -• ( x + 9)( x + 25)

p˘ È Í ans.: 8 ˙ Î ˚

Evaluation of impropEr rEal intEgral of a rational function including trigonomEtric functions •

•

cos mx dx or Let I = Ú ( ) Q x -•

sin mx dx ( ) Q x -•

Ú

y

where Q(x) is a polynomial in x. C1

To evaluate the integral, consider

Ú f (z) dz = Ú e

imz

g ( z ) dz

C

x 1 −R O R and C is the contour Q( z ) fig. 5.20 consisting of the upper semicircle C1 of radius R with centre at the origin and the part of the real axis from −R to R (Fig. 5.20).

where g( z ) =

R

Ú

f ( z ) dz =

C

Ú

f ( x ) dx +

-R

Ú

f ( z ) dz

C1

By Cauchy’s residue theorem,

Ú f (z) dz = 2p i

(sum of residues at poles in the upper half of z-plane)

C

R

Ú

-R

f ( x ) dx +

Ú

C1

f ( z ) dz = 2p i (sum of residues)

5.36

Chapter 5 Applications of Contour Integration

Taking R Æ •, R

lim

R Æ•

Ú

R Æ•

•

Ú

Ú

f ( x ) dx + lim

-R

f ( x ) dx + lim

R Æ•

-•

By Jordan’s lemma, lim

R Æ•

Úe

f ( z ) dz = 2p i (sum of residues)

C1 imz

g( z ) dz = 2p i(sum of residues)

C1

Úe

imz

g( z )dz = 0

C1

•

Ú

f ( x ) dx = 2p i (sum of residues)

-•

Example 1 •

Evaluate

cos mx

Ú x 2 + a 2 dx, a > 0, m > 0

using contour integration.

0

Solution (i) Let

Ú

C

f ( z ) dz = Ú

C

eimz z2 + a2

dz

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.21). (ii) Let f ( z ) =

y

eimz

z2 + a2 The poles are given by z2 + a2 = 0 z = ± ai eimz \ f (z) = ( z + ai )( z - ai )

(iii) The pole z = ai lies inside C. (iv) Res [ f ( z ); z = ai ] = lim ( z - ai ) f ( z ) z Æ ai

eimz z Æ ai ( z + ai )

= lim =

eim ( ai ) (ai + ai )

=

e - ma 2 ai

C1 z = ai −R

O

fig. 5.21

R

x

5.4  Evaluation of Improper Real Integral of a Rational Function Including...        5.37

•

Ú

(v)

f ( x ) dx = 2p i (sum of residues)

-• •

Ê e - ma ˆ x i d = 2 p Á 2 ai ˜ x2 + a2 Ë ¯ eimx

Ú

-•

= •

Ú

p e - ma a

(cos mx + i sin mx ) x2 + a2

-•

dx =

p e - ma a

Comparing real parts, •

Ú

-• •

cos mx x2 + a2

dx =

p e - ma a

p e - ma 2Ú 2 x d = 2 a 0 x +a •

cos mx

cos mx

Ú x 2 + a2 dx = 0

a È a ˘ Í∵ Ú f ( x )dx = 2 Ú f ( x )dx if f (- x ) = f ( x )˙ ÍÎ - a ˙˚ 0

p e - ma 2a

Example 2 •

Evaluate a real integral

Ú

0

x sin x x2 + 9

dx using residue.

[Winter 2013]

Solution (i) Let

Ú

C

f ( z ) dz = Ú

C

zeiz z2 + 9

dz

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.22). (ii) Let f ( z ) =

zeiz

y

z2 + 9 The poles are given by z2 + 9 = 0 z = ±3i \ f (z) =

zeiz ( z + 3i )( z - 3i )

C1 z = 3i

–R

O

fig. 5.22

R

x

5.38

Chapter 5 Applications of Contour Integration

(iii) The pole z = 3i lies inside C. (iv) Res [ f ( z ); z = 3i ] = lim ( z - 3i ) f ( z ) z Æ 3i

= lim

z Æ 3i

zeiz z + 3i 2

3ie3i = 3i + 3i = = =

3ie -3 6i e -3 2 1 2 e3

•

Ú

(v)

f ( x ) dx = 2p i (sum of residues)

-• •

Ú

-• •

Ú

xeix

Ê 1 ˆ dx = 2p i Á 3 ˜ Ë 2e ¯ x +9 2

x(cos x + i sin x ) 2

dx =

pi

x +9 e3 Comparing imaginary parts,

-•

•

x sin x

Ú

-• •

2Ú 0

•

x2 + 9 x sin x x2 + 9

dx = dx =

p e3 È Í∵ ÍÎ

p e3

a

Ú

-a

a ˘ f ( x ) dx = 2 Ú f ( x ) dx, if f (- x ) = f ( x )˙ ˙˚ 0

p

x sin x

Ú x 2 + 9 dx = 2 e 3 0

Example 3 •

Evaluate

Ú

0

x sin mx x2 + a2

dx, a > 0, m > 0.

Solution (i) Let

Ú

C

f ( z )dz =

zeimz z2 + a2

dz

5.4  Evaluation of Improper Real Integral of a Rational Function Including...        5.39

Consider the contour C consisting of upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.23). (ii) Let f ( z ) =

zeimz

y

2

2

z +a The poles are given by z2 + a2 = 0 z = ± ai \ f (z) =

C1

zeimz ( z + ai )( z - ai )

(iii) The pole z = ai lies inside C.

z = ai −R

O

R

fig. 5.23

(iv) Res [ f ( z ); z = ai ] = lim ( z - ai ) f ( z ) z Æ ai

zeimz z Æ ai z + ai

= lim

2

aiei ma = ai + ai =

aie - ma 2 ai

=

e - ma 2

•

Ú

(v)

f ( x ) dx = 2p i (sum of residues)

-• •

Ú

-•

Ê e - ma ˆ d x = p i 2 Á ˜ x2 + a2 Ë 2 ¯ xeimx

= pie - ma •

x(cos mx + i sin mx )

dx = p ie - ma 2 2 x + a -• Comparing imaginary parts,

Ú

•

Ú

-• •

2Ú 0

•

x sin mx x2 + a2

a È a ˘ - ma 2 f ( x ) if f ( x ) = f ( x ) x e f ( x ) x d p ∵ d = = Í ˙ Ú Ú x2 + a2 ÍÎ - a ˙˚ 0

x sin mx

x sin mx

Ú x2 + a2 0

dx = p e - ma

dx =

p e- ma 2

x

5.40

Chapter 5 Applications of Contour Integration

Example 4 •

Evaluate

cos mx

Ú (1 + x 2 )2 dx, m > 0

using contour integration.

0

Solution eimz

(i) Let f ( z )dz = Ú

dz (1 + z 2 )2 Consider the contour C consisting of the upper semicircle of radius R and the part of the real axis from –R to R (Fig. 5.24). C

(ii) Let f ( z ) =

eimz

y

(1 + z 2 )2

The poles are given by (1 + z 2 )2 = 0

C1

1 + z2 = 0

z= i

z = ±i \ f (z) =

eimz

−R

fig. 5.24

( z + i )2 ( z - i )2

(iii) The pole z = i of order 2 lies inside C. (iv) Res [ f ( z ); z = i ] =

1 d lim [( z - i )2 f ( z )] (2 - 1)! z Æi dz

= lim z Æi

= lim

d È eimz ˘ Í ˙ dz ÍÎ ( z + i )2 ˙˚ ( z + i )2 eimz ◊ im - eimz 2( z + i ) ( z + i )4

z Æi

= lim z Æi

=

[im( z + i ) - 2]eimz ( z + i )3

[(i + i )im - 2]ei (i + i )3

=

(-2m - 2)e - m -8i

=

(m + 1)e - m 4i

O

2

m

R

x

5.4  Evaluation of Improper Real Integral of a Rational Function Including...        5.41

•

Ú

(v)

f ( x ) dx = 2p i(sum of residues)

-• •

È (m + 1)e - m ˘ d x = 2 p i Í ˙ Ú 2 2 4i ÍÎ ˙˚ -• (1 + x ) eimx

p (m + 1)e - m 2

= •

Ú

cos mx + i sin mx (1 + x 2 )2

-•

dx =

p (m + 1)e - m 2

Comparing real parts, •

Ú

-•

cos mx (1 + x 2 )2

•

dx =

p (m + 1)e - m 2 a È a ˘ d = Í∵ Ú f ( x ) x 2 Ú f ( x )dx if f (- x ) = f ( x )˙ ÍÎ - a ˙˚ 0

p (m + 1)e- m x d 2Ú = 2 2 2 0 (1 + x ) cos mx

•

cos mx

Ú (1 + x 2 )2 dx = 0

p (m + 1)e - m 4

Example 5 •

Evaluate

Ú

-• ( x

cos xdx 2

+ a 2 )( x 2 + b2 )

, a > 0, b > 0 using contour integration.

Solution (i) Let

y

eiz dz

Ú f (z)dz = Ú (z2 + a2 )(z2 + b2 )

C

C

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from −R to R (Fig. 5.25). 1 (ii) Let f ( z ) = 2 2 ( z + a )( z 2 + b2 ) The poles are given by ( z 2 + a 2 )( z 2 + b2 ) = 0 z = ± ai, ±bi

z = bi

C1

z = ai −R

O

fig. 5.25

R

x

5.42

Chapter 5 Applications of Contour Integration

1 ( z + ai )( z - ai )( z + bi )( z - bi )

\ f (z) =

(iii) The poles z = ai and z = bi lie inside C. (iv) Res [ f ( z ); z = ai ] = lim ( z - ai ) f ( z ) z Æ ai

eiz z Æ ai ( z + ai )( z + bi )( z - bi)

= lim

ei ( ai ) (2ai )(ai + bi )(ai - bi )

=

e- a

=

-2ai(a 2 - b2 )

Res[ f ( z ); z = bi ] = lim ( z - bi ) f ( z ) z Æ bi

eiz z Æ bi ( z + ai )( z - ai )( z + bi )

= lim = =

ei (bi ) (bi + ai )(bi - ai )(2bi ) e- b 2bi(a 2 - b2 )

•

Ú

(v)

f ( x ) dx = 2p i (sum of residues)

-• •

È ˘ e- a e- b d x = 2 p i + Í Ú ( x 2 + a2 )( x 2 + b2 ) 2 2 2 2 ˙ ÍÎ -2 ai(a - b ) 2bi(a - b ) ˙˚ -• eix

•

Ú

-•

cos x + i sin x 2

2

2

2

( x + a )( x + b )

=

Ê e- b e- a ˆ Á a ˜¯ a 2 - b2 Ë b

dx =

Ê e- b e- a ˆ Á a ˜¯ a -b Ë b

p

p

2

2

Comparing real parts, •

Ú

-•

cos x dx 2

2

2

2

( x + a )( x + b )

=

Ê e- b e- a ˆ Á a ˜¯ a -b Ë b p

2

2

5.4  Evaluation of Improper Real Integral of a Rational Function Including...        5.43

Example 6 •

Evaluate

x 2 cos mx

Ú ( x 2 + a 2 )( x 2 + b2 ) dx, m > 0, a > b > 0 .

0

Solution (i) Let

y

z 2 eimz

Ú f (z)dz = Ú (z2 + a2 )(z2 + b2 ) dz

C

C

Consider the contour C consisting of the upper semicircle C1 of radius R and the part of the real axis from –R to R (Fig. 5.26). z e

The poles are given by ( z 2 + a 2 )( z 2 + b2 ) = 0 z = ± ai, ± bi z 2 eimz ( z + ai )( z - ai )( z + bi )( z - bi)

(iii) The poles z = ai and z = bi lie inside C. (iv) Res [ f ( z ); z = ai ] = lim ( z - ai ) f ( z ) z Æ ai

z 2 eimz z Æ ai ( z + ai )( z + bi )( z - bi )

= lim

2

= = =

(ai )2 ei ma (2 ai )(ai + bi )(ai - bi ) - a 2 e - ma 2 ai(b2 - a 2 ) - ae - ma

2i ( b 2 - a 2 ) Res [ f ( z ); z = bi ] = lim ( z - bi ) f ( z ) z Æ bi

z 2 eimz z Æ bi ( z + ai )( z - ai )( z + bi )

= lim

O

fig. 5.26

( z 2 + a 2 )( z 2 + b2 )

\ f ( z) =

C1

z = bi −R

2 imz

(ii) Let f ( z ) =

z = ai

R

x

5.44

Chapter 5 Applications of Contour Integration

2

= = =

(bi )2 ei mb (bi + ai )(bi - ai )(2bi ) -b2 e - mb (a 2 - b2 )(2bi ) be - mb 2i ( b 2 - a 2 )

•

Ú

(v)

f ( x )dx = 2p i (sum of residues)

-• •

È ae - ma be - mb ˘ 2 p i + d x = Í Ú 2 2 2 2 2 2 2 2 ˙ ÍÎ 2i(b - a ) 2i(b - a ) ˙˚ -• ( x + a )( x + b ) È be - mb - ae - ma ˘ =pÍ ˙ b2 - a 2 ÍÎ ˙˚ •

Ú

x 2 eimx

x 2 (cos mx + i sin mx ) ( x 2 + a 2 )( x 2 + b2 )

-•

dx =

p (be- mb - ae- ma ) b2 - a 2

Comparing real parts, •

Ú

-•

x 2 cos mx ( x 2 + a 2 )( x 2 + b2 )

•

2Ú 0

dx =

2

x cos mx ( x 2 + a 2 )( x 2 + b2 )

•

x 2 cos mx

Ú ( x 2 + a2 )( x 2 + b2 )

dx =

dx =

0

p (be- mb - ae - ma ) (b 2 - a 2 ) p (be

- mb

- ae

- ma

)

(b 2 - a 2 )

a È a Í∵ Ú f ( x )dx = 2 Ú f ( x )dx Í -a 0 Í Îif f (- x ) = f ( x )

˘ ˙ ˙ ˙ ˚

p (be - mb - ae - ma ) 2( b 2 - a 2 )

ExErcisE 5.3 Evaluate the following integrals using contour integration: •

1.

-•

•

2.

cos mx dx, a > 0, m > 0 2 + a 2 )2

Ú (x

cos x

Ú 1+ x

-•

2

p È - ma ˘ Í ans.: 4 a 3 (1 + ma) e ˙ Î ˚

dx p˘ È Í ans.: e ˙ Î ˚

5.5

•

3.

Úx

2

-•

•

4.

Úx

4

-•

•

5.

Úx

2

-•

•

6.

Úx

-•

5.5

4

Evaluation of Improper Integral When Simple Poles Lie on the Real Axis

x sin x dx + 2x + 2

5.45

p È ˘ Í ans.: e (sin1 + 1)˙ Î ˚

cos x dx + 5x 2 + 4

È 1 ˆ˘ Ê 1 - 2 ˜˙ Í ans.: p ÁË 3e 6e ¯ ˚ Î

sin x dx + 4x + 5

p È ˘ Í ans.: - e sin2˙ Î ˚

cos ax dx , a > 0 + 10 x 2 + 9

È p Ê - a e -3a ˆ ˘ Í ans.: ÁË e ˜˙ 8 3 ¯ ˙˚ ÍÎ

Evaluation of impropEr intEgral WhEn simplE polEs liE on thE rEal axis •

Let I =

Ú

f ( x ) dx

-•

To evaluate the integral, consider

Ú f (z) dz.

C

If f (z) has a simple pole on the real axis then this pole is deleted by indenting the contour. The contour is indented by drawing a small semi-circle C1 of radius r having the pole as its centre (Fig. 5.27). If f (z) has a simple pole at z = b on the real axis then lim

r Æ0

Ú

fig. 5.27

f ( z ) dz = p i Res [ f ( z ); z = b ]

C1

If f (z) has a simple pole z = b on the real axis and another simple pole z = z1 in the upper half of the z-plane then •

Ú

-•

f ( x ) dx = p i Res [ f ( z ); z = b ] + 2p i Res ÈÎ f ( z ); z = z1 ˘˚

5.46

Chapter 5 Applications of Contour Integration

Example 1 •

Find the principal value of

dx

Ú

-• ( x

2

- 3 x + 2)( x 2 + 1)

.

Solution (i) Let

dz

Ú f (z) dz = Ú (z2 - 3z + 2)(z2 + 1)

C

C

(ii) Let f ( z ) =

1 2

( z - 3z + 2)( z 2 + 1)

The poles are given by ( z 2 - 3z + 2)( z 2 + 1) = 0 ( z - 1)( z - 2)( z + i )( z - i ) = 0 z = 1, 2, ± i \ f (z) =

1 ( z - 1)( z - 2)( z + i )( z - i )

(iii) The poles z = 1, z = 2 lie on the real axis and the pole z = i lies in the upper half of the z-plane (Fig. 5.28). (iv) Res [ f ( z ); z = 1] = lim ( z - 1) f ( z ) z Æ1

= lim

z Æ1

1 ( z - 2)( z 2 + 1)

1 2 Res [ f ( z ); z = 2 ] = lim ( z - 2) f ( z ) =-

zÆ2

= lim

zÆ2

1

fig. 5.28 2

( z - 1)( z + 1)

1 5 Res [ f ( z ); z = i ] = lim ( z - i ) f ( z ) =

z Æi

= lim

1 2

( z - 3z + 2)( z + i ) 1 = (-1 - 3i + 2) 2i 1 = (1 - 3i ) 2i z Æi

5.5

Evaluation of Improper Integral When Simple Poles Lie on the Real Axis

5.47

1 (-i + 3) ¥ 2(i + 3) (-i + 3) -i + 3 = 2(1 + 9) 3-i = 20 =

•

Ú

(v) PV of

-•

dx 2

( x - 3 x + 2)( x 2 + 1)

{

}

= p i Res [ f ( z ); z = 1] + [ f ( z ); z = 2 ]

+ 2p i Res [ f ( z ); z = i ] Ê 1 1ˆ Ê 3-iˆ = p i Á - + ˜ + 2p i Á Ë 2 5¯ Ë 20 ˜¯ Ê 3-iˆ Ê 3ˆ = pi Á - ˜ + pi Á Ë 10 ˜¯ Ë 10 ¯ ==

3p i 3p i p i 2 + 10 10 10

p 10

Example 2 •

Show that the principal value of

Ú

-• (x

sin x dx 2

+ 4)( x - 1)

=

1ˆ pÊ ÁË cos1 - 2 ˜¯ . 5 e

Solution (i) Let

Ú

C

f ( z ) dz = Ú

C

(ii) Let f ( z ) =

eiz ( z 2 + 4)( z - 1)

dz

eiz ( z 2 + 4)( z - 1)

The poles are given by ( z 2 + 4) ( z - 1) = 0 z = ±2i, 1

fig. 5.29

eiz \ f (z) = ( z + 2i )( z - 2i )( z - 1) (iii) The pole z = 1 lies on the real axis and the pole z = 2i lies in the upper half of the z-plane (Fig. 5.29).

5.48

Chapter 5 Applications of Contour Integration

(iv) Res [ f ( z ); z = 1] = lim ( z - 1) f ( z ) z Æ1

= lim

z Æ1

=

eiz ( z 2 + 4)

ei 5

Res [ f ( z ); z = 2i ] = lim ( z - 2i ) f ( z ) z Æ2i

= lim

z Æ2i

=

ei ( 2 i ) 4i(2i - 1)

=

e -2 -8 - 4i

=

e -2 (2 - i ) ¥ -4(2 + i ) (2 - i )

=

e -2 (2 - i ) -4(5)

= = •

(v) PV of

Ú

-• •

PV of

Ú

-•

eiz ( z + 2i )( z - 1)

e -2 (i - 2) 20 i-2 20 e2

eix 2

( x + 4)( x - 1)

dz = p i Res [ f ( z ); z = 1] + 2p i Res [ f ( z ); z = 2i ]

Ê ei ˆ Ê i-2 ˆ p x i d = Á ˜¯ + 2p i Á 2 Ë 5 ( x + 4)( x - 1) Ë 20 e2 ˜¯ cos x + i sin x

=

pi p (cos1 + i sin1) + (-1 - 2i ) 5 10 e2

=

p p (i cos1 - sin1) (1 + 2i ) 5 10 e2

=-

pÊ 1 ˆ Êp 2p ˆ sin1 + 2 ˜ + i Á cos1 Á 5Ë 2e ¯ Ë 5 10 e2 ˜¯

=-

pÊ 1 ˆ pÊ 1ˆ sin1 + 2 ˜ + i Á cos1 - 2 ˜ 5 ÁË 5Ë 2e ¯ e ¯

5.5

Evaluation of Improper Integral When Simple Poles Lie on the Real Axis

Comparing imaginary parts, •

PV of

Ú

sin x dx

-•

2

( x + 4)( x - 1)

=

pÊ 1ˆ cos1 - 2 ˜ Á Ë 5 e ¯

Example 3 •

Using indentation along a branch cut, show that

sin x p dx = . 2 0 x

Ú

Solution (i) Let

Ú

C

eiz dz z C

f ( z ) dz = Ú

(ii) Let f ( z ) =

eiz z

The pole is given by z=0 (iii) The pole z = 0 lies on the real axis (Fig. 5.30). (iv) Res [ f ( z ); z = 0 ] = lim z f ( z ) zÆ0

= lim eiz

fig. 5.30

zÆ0

=e

0

=1 •

eix Ú x dx = p i Res [f (z); z = 0] -•

(v) •

cos x + i sin x dx = p i (1) x -•

Ú

Comparing imaginary parts, •

sin x dx = p x -•

Ú

•

sin x dx = p x 0

2Ú

•

sin x p dx = x 2 0

Ú

È Í∵ ÎÍ

a

Ú

-a

a ˘ f ( x ) dx = 2 Ú f ( x ) dx, if f (- x ) = f ( x )˙ 0 ˚˙

5.49

5.50

Chapter 5 Applications of Contour Integration

ExErcisE 5.4 Evaluate the following integrals using contour integration. •

1.

1 - cos x dx x2 0

Ú

•

2.

x

Ú 8-x

3

p˘ È Í ans.: 2 ˙ Î ˚

dx

-•

È p 3˘ Í ans.: ˙ 6 ˙˚ ÍÎ

•

3.

sin mx dx , m > 0 x 0

Ú

p˘ È Í ans.: 2 ˙ Î ˚ •

4.

cosmx dx x -•

Ú

ÎÈans.: 0 ˘˚ •

5.

dx 2 x - ix -•

Ú

ÈÎans.: p ˘˚ •

6.

x2 dx -• x - 1

Ú

4

p˘ È Í ans.: 2 ˙ Î ˚ •

7.

dx 2 - • x + 3x - 4

Ú

4

p ˘ È Í ans.: - 10 ˙ Î ˚

Points to Remember

5.51

•

3x + 2 dx 2 - • x( x - 4)( x + 9)

Ú

8.

14 ˘ È Í ans.: - 75 p ˙ Î ˚

points to remember Evaluation of a Real Definite Integral of a Rational Function of cos q and sin q 2p

Ú 0

f (sin q , cos q ) dq = 2p i (sum of residues at all poles lying inside the circle z = 1)

||

Evaluation of Improper Real Integral of a Rational Function •

Ú

f ( x ) dx = 2p i (sum of residues at poles in the upper half of the z-plane)

-•

P( x ) and the degree of Q(x) is greater than the degree of P(x) by at Q( x ) least 2 and Q(x) has no real roots. where f ( x ) =

Evaluation of Improper Real Integral of a Rational Function Including Trigonometric Functions •

cos mx dx = 2p i (sum of residues at poles in the upper half of the z-plane) Q( x ) -•

Ú

•

sin mx dx = 2p i (sum of residues at poles in the upper half of the z-plane) Q( x ) -•

Ú

where Q(x) is a polynomial in x.

Evaluation of Improper Integral When Simple Poles Lie on the Real Axis If f (z) has a simple pole z = b on the real axis and another simple pole z = z1 in the upper half of the z-plane then •

Ú

-•

f ( x ) dx = p i Res [ f ( z ); z = b ] + 2p i Res ÈÎ f ( z ); z = z1 ˘˚

6 Conformal Mapping CHAPTER

and Its Applications chapter outline 6.1 6.2 6.3 6.4 6.5 6.6

6.1

Introduction Conformal Mapping Some Standard Transformations Some Special Transformations Schwarz—Christoffel Transformation Bilinear Transformation

IntroductIon

A conformal mapping, also called a conformal map or conformal transformation, is a transformation that preserves local angles. An analytic function is conformal at any point where it has a nonzero derivative. Conformal mappings are useful for solving problems in engineering and physics that can be expressed in terms of functions of a complex variable but exhibit inconvenient geometrics. By choosing an appropriate mapping, the inconvenient geometry can be transformed into a much more convenient geometry.

6.2

conformal mappIng

Conformal mapping transforms curves from one complex plane to another with respect to size and orientation. If the point z (x, y) describes some curve C in the z-plane, the point w (u, v) describes a corresponding curve C¢ in the w-plane. Hence, for each point (x, y) in the z-plane, there corresponds a point (u, v) in the w-plane. Thus, a curve C in the z-plane is mapped into the corresponding curve C¢ in the w-plane by the mapping w = f(z).

6.2

Chapter 6

Conformal Mapping and Its Applications

A mapping or transformation w = f(z) is said to be conformal at a point z0 if it preserves the angle between any two curves passing through z0 in magnitude and sense. Isogonal transformation A transformation which preserves the angle between any two curves passing through a point only in magnitude and not in sense is known as the isogonal transformation at that point. conformal property of analytic function If w = f(z) is an analytic function and f ¢(z) π 0 in a region R of the z-plane then the mapping w = f(z) is conformal at all points of the region R. critical points A point of the transformation w = f(z) at which f ¢(z) = 0 is known as the critical point. At this point, mapping is not conformal.

Example 1 At what points is the mapping w = z 2 +

1 z2

not conformal?

Solution The mapping is not conformal at critical points, i.e., where f¢(z) = 0. 1 w = f (z) = z2 + 2 z 2 f ¢( z ) = 2 z - 3 z f ¢( z ) = 0 2 2z - 3 = 0 z 2z4 - 2 = 0 z4 = 1 z 2 = ±1 z 2 = 1, z 2 = -1 = i 2 z = ±1, z = ±i Hence, the mapping is not conformal at z = ±1, z = ±i.

6.3

SomE Standard tranSformatIonS

6.3.1 translation The transformation w = z + c , where c is a complex constant, represents translation. Let z = x + iy, w = u + iv, c = a + ib

6.3

Some Standard Transformations

6.3

w = z+c u + iv = ( x + iy) + (a + ib) = ( x + a ) + i( y + b) Comparing real and imaginary parts, u = x + a,

v = y+b

Therefore, the image of the point (x, y) in the z-plane is the point ( x + a, y + b) in the w-plane. Hence, w = z + c is simply a translation of the axes and preserves shape and size of the region of the z-plane.

Example 1 Find the image of 2 x + y - 3 = 0 under the transformation w = z + 2i. Solution w = z + 2i u + iv = x + iy + 2i = x + i ( y + 2) Comparing real and imaginary parts, u = x, x = u,

v = y+2 y = v-2

2x + y - 3 = 0 2u + (v - 2) - 3 = 0

Given

2u + v - 5 = 0 Hence, the line 2 x + y - 3 = 0 in the z-plane is mapped onto a line 2u + v - 5 = 0 in the w-plane (Fig. 6.1). y

v (0, 5)

(0, 3)

2u + v − 5 = 0

2x + y − 3 = 0 3,0 2 O

5, 0 2 x

O

z-plane

w-plane

fig. 6.1

u

6.4

Chapter 6

Conformal Mapping and Its Applications

Example 2 Determine and sketch the image of |z| = 1 under the transformation w = z + i. Solution w = z+i u + iv = x + iy + i = x + i ( y + 1) Comparing real and imaginary parts, u = x, x = u, Given

v = y +1 y = v -1

|z| = 1 x 2 + y2 = 1 x 2 + y2 = 1 u2 + (v - 1)2 = 1

Hence, the image of the circle z = 1 in the z-plane is the circle u2 + (v – 1)2 = 1 with the centre at (0, 1) and a radius of 1 in the w-plane (Fig. 6.2).

||

fig. 6.2

Example 3 Find the image of |z| = 2 under the mapping w = z + 3 + 2i. Solution w = z + 3 + 2i u + iv = x + iy + 3 + 2i = ( x + 3) + i( y + 2) Comparing real and imaginary parts, u = x + 3, x = u - 3,

v = y+2 y = v-2

6.3

Some Standard Transformations

6.5

|z| = 2

Given

x 2 + y2 = 2 x 2 + y2 = 4 (u - 3)2 + (v - 2)2 = 4

||

Hence, the circle z = 2 in the z-plane is mapped onto a circle with the centre at (3, 2) and a radius of 2 in the w-plane (Fig. 6.3). v

y

(3, 2)

x

O

u

O z-plane

w-plane

fig. 6.3

Example 4 Find the image of the triangular region whose vertices are i, 1 + i, 1 - i under the transformation w = z + 4 - 2i . Show the region graphically. Solution w = z + 4 - 2i u + iv = x + iy + 4 - 2i = ( x + 4) + i ( y - 2) Comparing real and imaginary parts, u = x + 4, v = y - 2 (i) For vertex i,

z=i x + iy = i Comparing real and imaginary parts, x = 0, y = 1 u = 4, v = -1 Hence, the point A(0,1) is mapped onto the point A ¢(4, -1).

(ii) For vertex1 + i, z = 1+ i x + iy = 1 + i

6.6

Chapter 6

Conformal Mapping and Its Applications

Comparing real and imaginary parts, x = 1, y = 1 u = 5, v = -1 Hence, the point B(1, 1) is mapped onto the point B ¢(5, -1). (iii) For vertex 1 – i, z=1–i x + iy = 1 – i Comparing real and imaginary parts, x = 1, y = -1 u = 5, v = -3 Hence, the point C(1, –1) is mapped onto the point C¢ (5, –3). Hence, the image of the triangular region ABC in z-plane is the triangular region A¢B¢C¢ with vertices (4 – i), (5 – i) and (5 – 3i), in the w-plane (Fig. 6.4). v

y

A (0, 1)

B(1, 1)

O

A′ (4, −1)

u

B′(5, −1)

x

O C(1, −1) z-plane

w-plane

C′(5, −3)

fig. 6.4

Example 5 Find the image of the circle z - a = r by the mapping w = z + c, where c is a real constant. Solution w = z+c u + iv = x + iy + c = ( x + c) + iy Comparing real and imaginary parts, u = x + c, v = y x = u - c, y = v Given

z -a = r x + iy - a = r ( x - a ) + iy = r

6.3

Some Standard Transformations

6.7

( x - a )2 + y 2 = r ( x - a )2 + y 2 = r 2 (u - c - a )2 + v 2 = r 2

[u - (c + a )]2 + v2 = r 2 Hence, the circle z - a = r in the z-plane is mapped onto a circle [u - (c + a )] + v 2 = r 2 with the centre at [(c + a), 0] and a radius of r in the w-plane. 2

6.3.2

Magnification and Rotation

The transformation w = cz, where c is a complex constant represents magnification and rotation. Let w = Reif , z = reiq , c = reia w = cz if

Re = ( reia )(reiq ) i a +q = rre ( ) Comparing both the sides, R = rr , f = q + a Therefore, this transformation maps the point (r, q) in the z-plane onto the point (rr, q + a)n in the w-plane. The radius vector r is magnified by r and is rotated through an angle a. Hence, geometrically it maps any figure in the z-plane into a similar figure in the wplane. In particular, this transformation maps circles into circles.

Example 1 Find the map of the circle z = 3 under the transformation w = 2z. Solution w = 2z u + iv = 2( x + iy) Comparing real and imaginary parts, u = 2 x, u x= , 2 Given

z =3 x 2 + y2 = 3

v = 2y v y= 2

6.8

Chapter 6

Conformal Mapping and Its Applications

x 2 + y2 = 9 2

2

Ê uˆ Ê vˆ ÁË 2 ˜¯ + ÁË 2 ˜¯ = 9 u2 + v 2 = 36 Hence, the map of the circle z = 3 in the z-plane is a circle u2 + v2 = 36 with the centre at (0,0) and a radius of 6 in the w-plane (Fig. 6.5).

||

v y

u2 + v2 = 36 2+

x

y2 =

9

x

O

O

z-plane

u

w-plane

fig. 6.5

Example 2 Find the image of x2 + y2 = 4 under the transformation w = 3z. Solution w = 3z u + iv = 3( x + iy) Comparing real and imaginary parts, u = 3 x, u x= , 3

v = 3y v y= 3

x 2 + y2 = 4

Given 2

2

Ê uˆ Ê vˆ ÁË 3 ˜¯ + ÁË 3 ˜¯ = 4 u2 + v 2 = 36 Hence, the image of x2 + y2 = 4 in the z-plane is a circle u2 + v2 = 36 with the centre at (0,0) and a radius of 6 in the w-plane (Fig. 6.6).

6.3

Some Standard Transformations

6.9

v y u2 + v2 = 36

x2 + y2 = 4 O

x

u

O

z-plane w-plane

fig. 6.6

Example 3 Find and sketch the image of the region |z| > 1 under the transformation w = 4z. [Summer 2014] Solution w = 4z u + iv = 4( x + iy) Comparing real and imaginary parts,

u = 4 x, u x= , 4 z >1

v = 4y v y= 4

||

Given 2

x + y2 > 1 x 2 + y2 > 1 u2 v 2 + >1 16 16 u2 + v 2 > 16 Hence, the image of z > 1 (Exterior of the circle z = 1) in the z-plane is the exterior of the circle u2 + v2 = 16 in the w-plane (Fig. 6.7).

||

||

fig. 6.7

6.10

Chapter 6

Conformal Mapping and Its Applications

Example 4 Determine the region in the w-plane into which the triangle bounded by the lines x = 0, y = 0 and x + y = 1 in the z-plane is mapped under the transformation w = 4z. Solution w = 4z u + iv = 4( x + iy) Comparing real and imaginary parts, u = 4 x, v = 4y u v x= , y= 4 4 (i) When x = 0, u = 0 Hence, the line x = 0 is mapped onto the line u = 0. (ii) When y = 0, v = 0 Hence, the line y = 0 is mapped onto the line v = 0. (iii) When x + y = 1, u v + =1 4 4 u+v = 4 Hence, the line x + y = 1 is mapped onto the line u + v = 4. Hence, the triangle OAB in the z-plane is mapped onto the triangular region OA¢B¢ in the w-plane (Fig. 6.8). v

y

B′

x=0

B

x

+

u=0

u

y

=

v

=

4

1 A

O

+

y=0

x

A′ O

z-plane

v=0

u

w-plane

fig. 6.8

Example 5 Draw the image of the square whose vertices are at (0, 0), (1, 0), (1, 1), (0, 1) in the z-plane, under the transformation w = (1 + i)z. What has this transformation done to the original square?

6.3

6.11

Some Standard Transformations

Solution w = (1 + i )z u + iv = (1 + i )( x + iy) = x + iy + ix + i 2 y = ( x - y ) + i( x + y) Comparing real and imaginary parts, u = x - y, v= x+y Let ABCD be the square with vertices A (0, 0), B (1, 0), C (1, 1), D (0, 1) in the z-plane. (i) At (0, 0), x = 0, y = 0 u = 0, v = 0 Hence, the point A (0, 0 ) is mapped onto the point A′(0, 0). (ii) At (1, 0), x = 1, y = 0 u = 1, v = 1 Hence, the point B(1, 0) is mapped onto the point B′(1, 1). (iii) At (1, 1), x = 1, y = 1 u = 0, v = 2 Hence, the point C (1, 1) is mapped onto the point C ′(0, 2). (iv) At (0, 1), x = 0, y = 1 u = -1, v = 1 Hence, the point D(0, 1) is mapped onto the point D′(-1, 1). Hence, the image of the square ABCD in the z-plane is the square A¢B¢C¢D¢ in the p w-plane (Fig. 6.9). This transformation has rotated the square by an angle but has 4 preserved the size of the square. v y C′ (0, 2) D (0, 1)

C (1, 1) B′ (1, 1)

D′ (−1, 1)

A (0, 0)

B (1, 0)

x

A¢ (0, 0)

z-plane

w-plane

fig. 6.9

u

6.12

Chapter 6

Conformal Mapping and Its Applications

Example 6 Determine the region in the w-plane into which the region bounded by x = 0, y = 0, x = 1, y = 2 in the z-plane is mapped under the transformation w = (1 + i )z + (2 - i ). Solution w = (1 + i )z + (2 - i ) u + iv = (1 + i )( x + iy) + (2 - i ) = x + iy + ix - y + 2 - i = ( x - y + 2) + i( x + y - 1) Comparing real and imaginary parts, u = x - y + 2, x- y = u-2

v = x + y -1

x + y = v +1

...(1) ...(2)

Adding Eqs (1) and (2), 2x = u + v -1 1 x = (u + v - 1) 2 Subtracting Eq. (1) from Eq. (2), 2 y = (v + 1) - (u - 2) 1 y = (v - u + 3) 2 Let ABCD be the given region in the z-plane. (i) When x = 0, u + v = 1 Hence, the line x = 0 is mapped onto the line u + v = 1. (ii) When y = 0, u - v = 3 Hence, the line y = 0 is mapped onto the line u - v = 3. (iii) When x = 1, u + v = 3 Hence, the line x = 1 is mapped onto the line u + v = 3. (iv) When y = 2, u - v = -1 Hence, the line y = 2 is mapped onto the line u - v = -1. Hence, the region ABCD in the z-plane is mapped onto the region A¢B¢C¢D¢ in the w-plane (Fig. 6.10).

6.3

v

C′ u

x=1

u

x

v=

3

B′ +

v=

u

1

u

y=0 B

+

D′

3

x=0 A

u − v = −1

C

v=

y=2

−

y

D

6.13

Some Standard Transformations

A′ z-plane

w-plane

fig. 6.10

Example 7 Find the image of the infinite strip 0 £ x £ 1 under the transformation w = iz + 1. Sketch the region in the w-plane. [Summer 2015] Solution w = iz + 1 u + iv = i( x + iy) + 1 = ix - y + 1 = (1 - y) + ix Comparing real and imaginary parts, u = 1 - y, y = 1 - u, Given

v=x x=v

0 £ x £1 0 £ v £1

Hence, the image of the infinite strip 0 £ x £ 1 in z-plane is the infinite strip 0 £ v £ 1 in the w-plane (Fig. 6.11).

6.14

Chapter 6

Conformal Mapping and Its Applications

v y v=1 x=1 x

O

O

z-plane

u

w-plane

fig. 6.11

Example 8 Find the image of the semi-infinite strip x > 0, 0 < y < 2 when w = iz + 1. Sketch the strip and its image. Solution w = iz + 1 u + iv = i( x + iy) + 1 = ix - y + 1 = (1 - y) + ix Comparing real and imaginary parts, u = 1 - y, y = 1 - u, Given and

or

v=x x=v

x>0 v>0 0 -1 -1 < u < 1

Hence, the image of the semi-infinite strip x > 0, 0 < y < 2 in the z-plane is the semiinfinite strip v > 0, –1 < u < 1 in the w-plane (Fig. 6.12).

6.3

Some Standard Transformations

6.15

v y (0,2)

x

O

(1,0) O

z-plane

(–1,0)

u

w-plane

fig. 6.12

6.3.3

Inversion

1 represents inversion and reflection. z This transformation is conformal for all z π 0. The transformation w = Let w = Reif , z = reiq R= w,

where

w= Reif =

f = arg(w),

r= z,

q = arg( z )

1 z 1

reiq 1 = e - iq r Comparing both the sides, 1 R= , r 1 w = , z

f = -q arg(w) = - arg( z )

Ê1 ˆ Hence, this transformation maps the point (r, q) in the z-plane onto the point Á , -q ˜ Ër ¯ in the w-plane. Thus, this transformation represents inversion with respect to the unit circle z = 1 followed by the reflection in the real axis.

||

If

z < 1,

w >1

and

z > 1,

w c, when c > 0 under the transformation 1 w = . Show the regions graphically. z Solution 1 z 1 z= w

w=

= x + iy =

1 u + iv u - iv u2 + v 2

6.3

Some Standard Transformations

6.21

Comparing real and imaginary parts, x=

u u2 + v 2

,

Given

v

y=-

u2 + v 2 x > c, when c > 0

u

>c u + v2 u > u2 + v 2 c u u2 + v 2 < c u u2 + v 2 - < 0 c 2

Ê 1 ˆ 1 which represents a circle with the centre at Á , 0˜ and a radius of . Ë 2c ¯ 2c Hence, the image of x > c in the z-plane is the interior of the circle in the w-plane (Fig. 6.16). y

v u2 + v2 − cu < 0

x>c x=c O

(c, 0)

x

O

1 ,0 2c

u

w-plane

z-plane

fig. 6.16

Example 6 1 Find the image of the infinite strip 0 < y < under the transformation 2 1 w= . z Solution 1 z 1 z= w

w=

6.22

Chapter 6

Conformal Mapping and Its Applications

= x + iy =

1 u + iv u - iv u2 + v 2

Comparing real and imaginary parts, x=

u 2

u +v

2

1 2 v

0< y
0 which represents the exterior of the circle u2 + v 2 + 2 v = 0 with the centre at (0, –1) and a radius of 1. 1 Hence, the image of the strip 0 < y < in the z-plane is the lower half (below u-axis) 2 of the w-plane outside the circle u2 + v 2 + 2 v = 0 in the w-plane (Fig. 6.17). v y y= 1 2 O O

y= 0

u

x (0, −1)

z-plane w-plane

fig. 6.17

u2 + v2 + 2v = 0

6.3

Some Standard Transformations

6.23

Example 7 1 1 1 £ y £ under the transformation w = . z 4 2 Also, show the regions graphically. Find the image of the strip

Solution 1 z 1 z= w

w=

=

1 u + iv

x + iy =

u - iv u2 + v 2

Comparing real and imaginary parts, x=

u 2

2

,

u +v 1 1 £y£ 4 2 v 1 1 £- 2 £ 2 4 2 u +v

Given

u2 + v 2 £ - 4 v £ 2(u2 + v 2 )

y=-

v 2

u + v2

ÈMultiplying by 4(u2 + v 2 )˘ Î ˚

( i) u 2 + v 2 < - 4 v u2 + v 2 + 4v < 0 which represents the interior of a circle with the centre at (0, –2) and a radius of 2. (ii) - 4 v < 2(u2 + v 2 ) 0 < u2 + v 2 + 2v u2 + v 2 + 2v > 0 which represents the exterior of a circle with the centre at (0, –1) and a radius of 1. 1 1 £ y £ in the z-plane is the region between the circles 4 2 2 2 2 2 u + v + 2 v = 0 and u + v + 4 v = 0 in the w-plane (Fig. 6.18).

Hence, the image of the strip

6.24

Chapter 6

Conformal Mapping and Its Applications

v

O

y y= 1 2

u (0, −1)

(0, −2)

y= 1 4

u2 + v2 + 2v = 0

u2 + v2 + 4v = 0

x

O

w-plane

z-plane

fig. 6.18

Example 8 1 Find the image of the strip 1 < x < 2 under the transformation w = . z Solution 1 z 1 z= w

w=

= x + iy =

1 u + iv u - iv u2 + v 2

Comparing real and imaginary parts, x= Given

u 2

u +v

2

,

y=-

v 2

u + v2

1 0). (ii) Equation of the line BC is x = 2. Substituting in u and v, u = 4 - y2 ,

v = 4y v y= 4

Substituting y in u, u = 4-

v2 16

16u = 64 - v 2 v 2 = -16(u - 4) which represents a parabola with vertex (4, 0) and opening to the left of the vertex. The image of the line x = 2 is a parabola v 2 = -16 (u - 4). (iii) Equation of the line CD is y = 2. Substituting in u and v, u = x 2 - 4,

v = 4x v x= 4 Substituting x in u, u=

v2 -4 16

16u = v 2 - 64 v 2 = 16(u + 4) which represents a parabola with vertex (–4, 0) and opening to the right of the vertex. The image of the line y = 2 is the parabola v 2 = 16(u + 4).

6.4

Some Special Transformations

6.33

(iv) Equation of the line DA is x = 0. Substituting in u and v, v=0 u = - y2 , which represents the left part of the u-axis, where u < 0. The image of the line x = 0 is the left part of the u-axis (u < 0). v y C′ D(0, 2)

C (2, 2)

B(2, 0)

A(0, 0)

(0, 8)

v2 = 16 (u + 4)

x

D′ (−4, 0)

v2 = −16 (u − 4)

B′ A′(0, 0) (4, 0)

u

z-plane

w-plane

fig. 6.23

∵ x > 0, y > 0 v = 2 xy > 0 Hence, the image of the square ABCD in the z-plane is the curvilinear triangle B¢C¢D¢ in the upper half of the w-plane (Fig. 6.23).

Example 4 Find the image of z - 1 = 1 under the transformation w = z2. Solution w = z2

[Considering polar form ]

Reif = (reiq )2 2 2 iq

=r e \ Given

R = r 2 , f = 2q z -1 = 1 reiq - 1 = 1

6.34

Chapter 6

Conformal Mapping and Its Applications

r (cos q + i sin q ) - 1 = 1 (r cos q - 1) + ir sin q 2

2

2

=1

2

(r cos q - 1) + r sin q = 1 2

2

r cos q – 2r cos q + 1 + r2sin2q = 1 r2 = 2r cos q r = 2 cos q r2 = 4 cos2q = 2(1 + cos 2q) Substituting r2 and 2q r 2 and 2q, ,

[∵ cos 2q = 2cos2q – 1]

R = 2(1 + cos f ) which represents a cardioid. Hence, the image of the circle z - 1 = 1 in the z-plane is the cardioid R = 2(1 + cos f ) in the w-plane (Fig. 6.24). v

y r = 2 cos q

R = 2(1 + cos f) x

O

u

O

z-plane

w-plane

fig. 6.24

Example 5 p p £ q £ in the 6 3 2 z-plane under the transformation w = z . Show the regions graphically. Find the image of the region bounded by 1 £ r £ 2 and Solution w = z2 Reif = (reiq )2 = r 2 e2iq Comparing both the sides, R = r2, f = 2q Given 1 £ r £ 2

6.4

6.35

Some Special Transformations

1 £ r2 £ 4 1£R£4 p p £q £ 6 3 p f p £ £ 6 2 3 p 2p £f £ 3 3

and

p p Hence, the image of the region bounded by 1 £ r £ 2 and £ q £ in the z-plane is 6 3 p 2p the region 1 £ R £ 4 and £ f £ in the w-plane (Fig. 6.25). 3 3 v

y p q= – 3

O

p q= – 6 r=1 r=2

p f= – 3

2p f= – 3 x

O

z-plane

u R=1 R=4

w-plane

fig. 6.25

6.4.2

w = ez u + iv = e x + iy = e x eiy = e x (cos y + i sin y)

Comparing real and imaginary parts, u = e x cos y,

v = e x sin y

(i) Let x = constant = a, say Substituting x = a in u and v, u = e a cos y

 (6.1)

v = e a sin y Squaring and adding Eqs (6.1) and (6.2),

 (6.2)

u 2 + v 2 = e2 a which represents a circle with the centre at (0, 0) and a radius of ea. For a > 0, ea > 1, the radius of the circle is greater than 1. For a < 0, ea < 1, the radius of the circle is less than 1. For a = 0, ea = 1, the radius of the circle is 1.

6.36

Chapter 6

Conformal Mapping and Its Applications

Hence, the lines parallel to the y-axis in the z-plane are mapped onto the circles in the w-plane with the centre at (0, 0) and radius, r > 1 or r < 1 or r = 1 according to whether a > 0 or a < 0 or a = 0 respectively. (ii) Let y = constant = b, say Substituting y = b in u and v, u = e x cos b

 (6.3)

x

 (6.4)

v = e sin b Dividing Eq. (6.3) by Eq. (6.4), v e x sin b = x u e cos b = tan b v = u tan b

which represents a radial line. Hence, the lines parallel to the x-axis in the z-plane are mapped onto radial lines in the w-plane.

Example 1 Prove that the transformation w = ez transforms the region between the real axis and a line parallel to the real axis y = p into the upper half of the w-plane. Solution w = ez u + iv = e x + iy = e x eiy = e x (cos y + i sin y) Comparing real and imaginary parts, u = e x cos y,

v = e x sin y

Given region lies between real axis ( i.e., y = 0 ) and the line y = p . In the region 0 < y < p, (i) sin y > 0 È∵ e x is always positive ˘ Î ˚

e x sin y > 0 v>0 (ii) cos y > 0, 0 < y < and cos y < 0,

p 2

p < y