An Engineers Guide to Complex Integration

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An Engineers Guide to Complex Integration

David Sirajuddin

Itcanbeshown.com

April 15, 2008


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CONTENTS CONTENTS

Contents

1 Introduction 3

2 Complex Variables and Functions 3

2.1 Number Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.2 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2.1 Complex Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2.2 Analyticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.3 Branch Cuts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.4 Power Series Representation . . . . . . . . . . . . . . . . . . . . . . . 5

3 Introduction to Complex Integration 6

4 Cauchys Integral Theorem 7

5 Cauchys Integral Formula 7

6 Residue Theorem 8

6.1 Definition of a Residue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

6.2 Finding Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

6.2.1 Functional replacement . . . . . . . . . . . . . . . . . . . . . . . . . . 10

6.2.2 Quotient rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

6.2.3 Multiplication of Series . . . . . . . . . . . . . . . . . . . . . . . . . . 106.2.4 Division of Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

6.2.5 Computing Coefficients Directly . . . . . . . . . . . . . . . . . . . . . 12

7 General Approach to Solving Complex Integrals 12

8 General Approach to Solving for Real Integrals 14

8.1 Indented Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

8.2 Paths About and Involving Branch Cuts . . . . . . . . . . . . . . . . . . . . 18

8.3 Proving Complex Integrals Tend to Zero . . . . . . . . . . . . . . . . . . . . 18

8.3.1 ML Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

8.3.2 Jordans Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

8.4 Integrals Containing Trigonometric Functions . . . . . . . . . . . . . . . . . 21

9 References 21

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CONTENTS CONTENTS

10 Acknowledgements 22

11 Appendix 22

11.1 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

11.2 Useful Series Representations . . . . . . . . . . . . . . . . . . . . . . . . . . 22

11.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

11.3.1 Real Integration: Functional Replacement and ML Bound . . . . . . 24

11.3.2 Real Integration: Functional Replacement and Jordans Lemma . . . 26

11.3.3 Real Integration: Integrals Containing Trigonometric Functions . . . 29

11.3.4 Real Integration: Strange Contours (Fresnel Integral) . . . . . . . . . 31

11.3.5 Complex Integration: Quotient Rule . . . . . . . . . . . . . . . . . . 35

11.3.6 Finding Residues: Multiplication of Series . . . . . . . . . . . . . . . 36

11.3.7 Finding Residues: Division of Series . . . . . . . . . . . . . . . . . . . 37

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2 COMPLEX VARIABLES AND . . .

1 Introduction

Complex integration is an intuitive extension of real integration. Since a complex numberrepresents a point on a plane while a real number is a number on the real line, the analog of asingle real integral in the complex domain is always a path integral. For some special functions

and domains, the integration is path independent, but this should not be taken to be thecase in general. Given the sensitivity of the path taken for a given integral and its result,parametrization is often the most convenient way to evaluate such integrals. Of particularimportance in many physics and engineering problems is that of the closed path integral,where both the beginning and ending points of the path are identical. As with real calculus,such integrals are called contour integrals, and are denoted by a circle in the integration sign(

). This guide will discuss solution strategies for complex closed contour integrals, withparticular emphasis on their use in solving difficult real-valued integrals. The material will bepresented to the reader assuming minimal knowledge of complex mathematics. While formaldefinitions and theorems will be provided so as to make this document self-contained, proofswill not be included. The reader is recommended to see Brown and Churchills book (see

References ) for discussion and proofs of theorems used throughout this document. Generalstrategies of complex integration are discussed, while fully worked examples are provided inthe Appendix.

2 Complex Variables and Functions

2.1 Number Sets

At this point, it is worth explicitly delineating the fundamental differences between realand complex numbers. A number x is said to be an element of the real number set, denoted

by xR, if it represents any number on the real number line. Alternatively, a number z issaid to be a complex number, written as zC, if it is of the form z = x + iy, where {x, y}Rand the imaginary number i =

1.This notation is perhaps more readable when it is informally written down in vector

form, or as an ordered pair: z = xx + iyy (x, y), where x and y are unit vectors inthe x, and y directions respectively. This representation is not conventional for complexnumbers, but it may be helpful at first to keep these notations in mind. Furthermore,these representations convey an important distinction between real and complex numbers:while real numbers denote scalar numbers on a number line, complex numbers representpoints on a complex plane. As in the case of ordered pairs, it is also possible to equivalentlyexpress complex numbers in polar form. Thus, a complex number z = x + iy = rei canbe fully represented in either cartesian or polar coordinates, where r is the radius measuredfrom the origin, and is the angle of z measured counterclockwise from the positive x-axis.Numbers on the complex plane that lie on the abscissa, are the conventional real numbers,while numbers on the ordinate axis are called imaginary numbers (owing to these numbersbeing multiples of the imaginary unit i). That is to say, real numbers are a subset of thecomplex number set (R C). The observation that complex numbers extend a scalar to apoint on a plane immediately admits the result that one extra dimension is inherited by the

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2 COMPLEX VARIABLES AND . . . 2.2 Functions

adoption of this number set. This extra dimension can sometimes change things considerablyin regards to complex mathematics and its real counterpart. In general, most results fromreal mathematics carry over into complex mathematics; however, the few aspects that aredifferent are grossly so. Thus, the reader is cautioned when assuming any property fromreal numbers is translatable into complex numbers.

2.2 Functions

A complex function f of an independent complex variable z = x + iy = rei, where{r 0, (r,,x,y)R}, is of the general form:

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

the latter representation of z involving radius r and angle of a complex number followsfrom the transformation from cartesian to polar coordinates, and the use of Eulers identityto replace the resulting trigonometric functions with a complex exponential. By convention,

the variable z is used to denote a complex variable. It is often convenient to group a functionin both real and imaginary parts. These parts are represented in the contained functionsu, and v, which are both real-valued, but themselves correspond to the real and imaginaryparts of the function f respectively, and both of which are functions of either x and y or rand .

2.2.1 Complex Differentiability

A function f(z) has a derivative f(z0) at a point z0, given by the following limit providedit exists:

f(z0) = limzz0 f(z) f(z0)z z0 (1)Since the complex number z0 = x0 + iy0 can be approached from any number z = x + iy byway of an infinite number of paths, the limit is said to exist if and only if the same, uniquelimit is obtained when approached from any path that originates from any complex numberz. The criteria that a function f(x, y) = u(x, y) + iv(x, y) = u(r, ) + iv(r, ) must meet inorder for this to be true are known as the Cauchy-Riemann Equations. These equations canbe expressed in either cartesian or polar form.

u

x=

v

yand

u

y= v

xCartesian Form (2)

u

r=

1

r

v

and

1

r

u

= v

rPolar Form (3)

Thus, while real differentiability only demands the continuity of a function at a point x0in order for it to be said that its derivative exists at that point, a complex-valued functionis complex differentiable at a point z0 if and only if it is both analytic and satisfies theCauchy-Riemann equations at that point.

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2 COMPLEX VARIABLES AND . . . 2.2 Functions

2.2.2 Analyticity

Complex differentiability is a much more robust property than real differentiability. Ifa function is complex differentiable at a point z0, it is also said to be analytic at z0. Afunction satisfying these conditions inherits another consequence in that the function isinfinitely differentiable. The terms regular and holomorphic are often used interchangeablywith analytic.

2.2.3 Branch Cuts

By virtue of the polar representation of complex numbers, certain functions are inherentlymultiple-valued. This is to say, that in the sense, that there exists an infinite number of anglesi that correspond to an angle 0 of a complex number z0 = r0e

i0. This is to say that, anangle i is equivalent to an angle 0 ifi = 0 + 2ni, where n is an integer. Thus, functionscan take on the same value for different coordinates. Functions that meet this qualificationlose their analytic properties, since they are no longer one-to-one. However, what is often

done so as to restore a functions analyticity is to define a curve of discontinuity, or branchcut. Branch cuts are often made in regards to the angle, and are usually taken to be eitherthe positive/negative real or imaginary axis. A common example is the complex logarithmfunction log z = ln r + i, where the value of the angle is restricted so that only values overan equivalent interval of 2 are permitted. Under the adoption of a branch cut, a functionwill hold distinct values for distinct radii and angles representing numbers in the complexplane and is hence one-to-one, and thereby analytic.

2.2.4 Power Series Representation

An analytic function can equivalently be expressed as a power series. It is discussed here

to provide a straightforward definition of a residue in subsequent sections, and because itcan sometimes be quicker to find residues by computing coefficients of a power series ratherthan through other methods (See Section 6.2).

The most general case is if a function f is analytic throughout an annular domain R1 R0;(iii) for all points z on CR there is a positive constant MR such that |f(z)| MR, where

limR

MR = 0

Then, for every positive constant a,

limR

CR

f(z)eiazdz = 0 (19)

This to say, if an appropriate function g(z) can be factored into a form of Eqn. (19), wherethe constituent function f(z) meets the above criterion, then the integral is equal to zero.

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11 APPENDIX

10 Acknowledgements

I would like to thank Benjamin Tong Yee, who at the time of this writing is attendingthe University of Michigan for his graduate degree in Nuclear Engineering & RadiologicalSciences, for his suggestion on how to typeset the Residue notation in LATEX. Furthermore, I

found it silly to scour the internet and/or paraphrase theorems. Thus, many of the theorems,and definitions cited in this guide are taken verbatim from Chuchills Complex Variables andApplications, and are the sole property of the authors. This was the book from which I stud-ied complex mathematics of a single variable, and I wish to thank the authors for a splendidtext. Finally, a special thanks to my Math 555 professor, Dr. David E. Barrett, at theUniversity of Michigan for making the course so accessible to an engineering, and not math-ematics, student like myself as well as for his excellent instruction. I have attempted to dofor all of you what he did for me. This guide, under no circumstances, is to be used for profit,especially since many of the theorems are attributed entirely to the authors. Please do nothesitate to send any enquiries, corrections, and/or concerns to me at [email protected].

11 Appendix

11.1 Glossary

1. Analytic: a function is analytic at a point z0 if it is both continuous, and satisfies theCauchy-Riemann equations at z0.

2. Branch cut: Typically this is a restriction that is applied on the angular value ofa multiple-valued function so as to enforce that it is single-valued (e.g. the complexlogarithm function), and inherits analytic properties. The cut can be either an infinite

ray (constant angle), or a curve. The only requirement is that the function along thecut is discontinuous such that it preserves its single-valuedness.

3. Complex differentiable: a function is complex differentiable if it is analytic. A complexdifferentiable function is also infinitely differentiable.

4. Holomorphic: See analytic, strictly speaking these two terms are not entirely synony-mous, although the small difference between them is of no consequence for applicationsof the type described in this guide.

5. Meromorphic: A function that is analytic inside an annular domain.

11.2 Useful Series Representations

The following series representations are useful in invoking the methods described in Sec-tions 6.2.3 and 6.2.4.

1. 11z

=

n=0 zn (|z| < 1)

2. ez =

n=0zn

n!(|z| < )

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11 APPENDIX 11.2 Useful Series Representations

3. sin z =

n=0(1)n z2n+1

(2n+1)!(|z| < )

4. cos z =

n=0(1)n z2n

(2n)! (|z| < )

5. sinh z = n=0

z2n+1

(2n+1)!(|z| < )

6. cosh z =

n=0z2n

(2n)! (|z| < )

These series can be obtained from each other. For instance, the cos z series is yielded fromdifferentiation of the sin z series, and that the cosh z expansion can be found via the relationcosh z = cos(iz). One can find series for tan z by polynomial division of the series expansionsfor sin z/ cos z, etc. Of the above, one of the more useful expressions is series (1), which canbe used to find expressions for any form of the reciprocal of any difference or sum involvinga constant and any power of z. This is accomplished by substitution of z for an appropriateform to find the desired expansion.

For example, in order to obtain the series representation for 1/(4 + z2), begin with series

(1)

1

1 w =n=0

wn (|w| < 1)

Enforcing the substitution w = z2/4, the above equation becomes

1

1 + (z2/4)=

n=0

z2n

4n(|z| < 2)

Multiplying and dividing both sides of the equation by a form of unity (i.e. 4 /4)

41

4 + z2=

n=0

4

4

z2n

4n

1

4 + z2=

n=0

z2n

4n+1(|z| < 1)

And, the desired expression has been obtained.

*N.B. Something that was not mentioned in this guide was the idea of the radius of conver-gence for power series expansions. This is to say that a function f(z) has a valid convergentpower series representation only within a certain radius |z|. The radius of convergence foreach of the above series has been noted parenthetically, and must be payed with close at-tention. It is important when finding series expansions to mind the radius of convergence,elsewise the series expression arrived at will not be correct. Luckily, this is of little setback,as all that is needed to do is to express the function in a fashion such that for all points

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11 APPENDIX 11.3 Examples

z of a considered domain, the expanded term holds values that are within the radius ofconvergence. This is easily accomplished by factorization.

11.3 Examples

11.3.1 Real Integration: Functional Replacement and ML Bound

The real integral 0

dx

2x2 + 8=

1

2

0

dx

x2 + 4

can be evaluated by first transposing the function f(x) contained in the integrand into acomplex function f(z),

f(z) =1

z2 + 4=

1

(z + 2i)(z 2i)The singularities at z = 2i are evident from the factorization on the RHS. A contour Cis chosen to be a semicircle in the upper-half plane centered at the origin with a radiusR chosen to be large enough to contain the singularity z = 2i. The semicircular arc thatexcepts the real line is denoted by CR. The factor of (1/2) in the problem statement will beneglected for now, and incorporated at the end. Residue theorem implies

C

dz

z2 + 4=

RR

dx

x2 + 4+

CR

dz

z2 + 4= 2i Res

z=2if(z)

Thus, the real integral can be expressed as

RR

dx

x2 + 4 = 2iResz=2i f(z) CRdz

z2 + 4

Letting R tend to infinity

dx

x2 + 4= 2iRes

z=2if(z) lim

R

CR

dz

z2 + 4(20)

An expression similar to the original integral is arrived at. Next, prove the the complexintegral on the RHS is equal to zero by way of an ML bound. That is to say, find theupperbound of the integral by way of the following inequality

CR

dz

z2 + 4 MLwhere M is the maximum value of the function on the contour, and L is the length of thecontour. Beginning with the representation of R before the limits are applied, by inspection,the length of the contour L = R. Furthermore, the maximum value of the function isobtained in the following manner:

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1z2 + 4 = 1|z2 + 4|

=1

|z|2

+ 4=

1

R2 + 4= M

Then the product ML, CR

dz

z2 + 4

ML = RR2 + 4As R , the ML bound becomes

ML = limR

R

R2 + 4

= limR

/R

1 + 4/R2

= 0

Since the upper bound of the integral is equal to zero, then the integral itself must be zero,CR

dz

z2 + 4= 0

The only term left to compute is residue in Eqn. (20). This is done by functional replacementof the complex function f(z)

f(z) =1

(z + 2i)(z 2i) =(z)

z 2iwhere

(z) =1

z + 2i

The residue at z = 2i is then equivalent to (2i)

Resz=2i

f(z) = (2i)

=1

(2i) + 2i

=1

4i

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f(z) =zeiz

z2 + 9

It is more convenient to transpose the sine function into a complex exponential rather thanto a complex-valued sine function, as the sine term can be easily extracted by taking the

imaginary part of this integral, while a complex sine function would provide no such conve-nience. The singular points are located at the zeros of the polynomial in the denominator,that is to say, when z2 + 9 = 0. Solutions to this equation are found easily by factorization

zeiz

z2 + 9=

zeiz

(z + 3i)(z 3i)Singularities are located at z = 3i. A semicircular contour C is chosen, centered at z = 0,with a sufficiently large radius R chosen such that the point z = 3i is contained within thesemicircular domain. The semicircular arc is denoted as CR. Note, that when choosing thiscontour, the point z = 3i will not contribute to the value of the integral and is therefore

neglected in the subsequent analysis.Applying residue theorem, and breaking up the integral into complex and real-valuedparts admits the following

C

zeizdz

z2 + 9= 2i Res

z=3if(z)

RR

xeixdx

x2 + 9+

CR

zeizdz

z2 + 9= 2i Res

z=3if(z)

RR

xeixdx

x2 + 9= 2i Res

z=3if(z)

CR

zeizdz

z2 + 9

Adjustments are then made to the limits of integration so that they fit those given in theoriginal problem, i.e. let R .

limR

RR

xeixdx

x2 + 9= 2i Res

z=3if(z) lim

R

CR

zeizdz

z2 + 9

xeixdx

x2 + 9= 2i Res

z=3if(z) lim

R

CR

zeizdz

z2 + 9(21)

Where the bounds have been informally represented as explicit limits of integration for thereal integral, and left in front of the complex integral on the RHS. The limit is not writtenas being applied to the residue term, as the residue contains no further dependence on theradius of the contour, provided that it was all ready contained in the domain interior toit before the limit was taken. In order to solve the RHS of the above equation, begin byproving the complex integral is equal to zero.

Since the complex integral contains a complex exponential, and given the limit of theradius R of the contour tending to infinity, it is evident that the integral cannot be easily

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shown to be zero by way of an ML bound. Jordans lemma is invoked to prove this, butbefore it can be used the function contained in the integrand must first be factored in thesame form as the statement of the theorem. That is to say, the function f(z) must be showncapable of being expressed as f(z) = g(z)eiz, for some complex function g. This is easilyaccomplished by examining the function in the integrand.

zeiz

z2 + 9= g(z)eiz

where

g(z) =z

z2 + 9

The function g immediately meets all the requirements of the theorem, there exists a radiusR0 < R for a positive semicircle in the upper half plane where g is analytic to all pointsexterior. Then, by way of Jordans lemma, it can be said that

limR

CR

zeiz

dzz2 + 9

= limR

CR

g(z)eizdz = 0

The residue at z = 3i is determined by the so-called functional replacement. Rewriting f(z)as

f(z) =zeiz

z2 + 9

=zeiz

(z + 3i)(z 3i)= (z)

z 3iwhere

(z) =zeiz

z + 3i

The resulting expression shows a simple pole, the residue of the function f is equivalent tothe value of (z) at z = 3i.

Resz=3i f(z) = (3i)

=(3i)e3i

2

(3i) + 3i

= i3e3

6i

= e3

2

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Inputting this result into Eqn. (21)

xeixdx

x2 + 9= 2i

e3

2

= ie3

Taking the imaginary part of the above equation,

Im

xeixdx

x2 + 9

= Im

ie3

x sin xdx

x2 + 9=

e3

The desired result is obtained.

11.3.3 Real Integration: Integrals Containing Trigonometric Functions

Prove the integration formula

d

5 + 4 sin =

3(22)

The function is difficult to integrate due to the sine functions location in the denominator.The standard method of replacing the real sine function with a complex exponential providesno easy method to isolate the sine function in order to retrieve the real-valued result. Thus,since the limits of integration encompass all 2 radians of a circle, the methods of Section

8.4 can be used. Letting z = ei, the following substitutions are identified:

sin =z z1

2i

and

d =dz

iz

inserting these expressions into the original integral allows the integral to be taken into thecomplex domain.

|z|=1

dz5 + (4/2i)(z z1) 1iz

The function contained in the integrand is manipulated until the residues can be recognized.

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1

5 + (4/2i)(z z1)1

iz=

1

5 + (2/i)(z z1)1

iz

=1

5 2i(z z1

)

1

iz=

1

5 2iz + 2iz11

iz

= i5z 2iz2 + 2i

=i

2iz2 5z 2i=

1/2

z2 + (5i/2)z 1the quadratic formula is used to find the roots, z1 and z2 of the polynomial in the denomi-

nator.

z1 =1

4i and z2 = 11

4i

Since the contour is designated as |z| = 1, it is evident that the only root, and hencesingularity, that is interior to it is the root z1. Given these roots, the integral can beexpressed as

|z|=1

(1/2)dz

(z z1)(z z2)

which can be written in the form of functional replacement,(z)

z z1 where (z) =(1/2)

z z2Thus, the integral in the problem statement is equal to the product of 2i and the residueat z = z1. The residue is equivalent to (z1):

Resz=z1

(1/2)

(z z1)(z z2) = (z1)

=(1/2)

z1 z2= 1

6i

Thus, |z|=1

(1/2)dz

(z z1)(z z2) = 2i Resz=z1 f(z) = 2i(1

6i) =

3

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Since this integral is equivalent to the original integral, it can then be said that

d

5 + 4 sin =

3

11.3.4 Real Integration: Strange Contours (Fresnel Integral)The following integral

0

cos x2dx (23)

can prove difficult only due to the odd choice of contour one must use in order to obtaina solution. This integral is known as a Fresnel integral and arises in the field of opticsin the description of near field Fresnel diffraction. While the function is transcendentalwhen evaluating the integral over definite limits, a solution can be found when the boundsare treated as semi-infinite. In fact, the convergence of the real integral over semi-infinite

bounds is suggested when looking at a trace of the function (Figure 4).

Figure 4 - A plot is shown of the Fresnel cosine function. The frequency increases with x.

As evident from above, the frequency of the function increases with x, until the wavelengthof the function tends to zero as x

. The function oscillates above and below the x-axis,

suggesting that it is possible in the limit for large x, that sufficient contributions from thearea sweeped out by the function will be cancelled out by its negative and positive portions,leading to a finite result. This is precisely the case, and this finite value of the integral isfound via complex integration.

The real-valued function f(x) = cos x2 is transposed into the complex domain as acomplex exponential f(z) = exp(iz2). The complex function f is identified to hold nosingularities; however, this only suggests that the complex closed contour integral of thisfunction about any domain is zero, not that all path integrals making up the closed contour

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where C is used to denote the closed contour shown in Figure 5(c). Notice how since thereare no residues contained within the contour, the contour integral of the function is equalto zero. That is to say, residue theorem is reduced to Cauchys integral theorem in thisinstance. The integral can further be broken up into a summation of path integrals, wherethe sum of the paths is equivalent to the contour C.

C1

eiz2

dz I

+

C2

eiz2

dz II

C3

eiz2

dz III

= 0 (25)

The integrals have been labelled as I, II, and III for convenient referencing, and each pathC1, C2, and C3 are as shown in Figure 5(c). Each integral is evaluated below.

Integral I

The only work that needs to be done on this integral is to parameterize it along the real

line, and take limits. Thus, C1

eiz2

dz =

R0

eix2

dx

since f(z) = f(x, y), and y = 0 for all points on the real line. Letting R , Integral Ibecomes

0

eix2

dx. (26)

where it is noted that the real part of this integral can be taken to make this equation of thesame form as the original integral in the problem statement (Eqn. (22)).

Integral II

Since integral II has a complex-valued, nonconstant path, it would be convenient to provethis integral tends to zero. This integral does indeed turn out to be identical to zero, byway of Jordans lemma. This is shown by initially factoring the function in the integrand inthe following way

C2

eiz2

dz =

C2

ei(z2z)eizdz =

C2

g(z)eizdz.

The above equation is now in the form of the statement in Jordans lemma, where g(z) =eiz(z1). Since the function g is entire, that is, it is analytic for all points in the complexplane, it meets the requirements for Jordans lemma. Thus, it can be said that

C2

eiz2

dz = 0 (27)

Integral III

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Thus, demonstrating the importance of choosing an appropriate contour.

11.3.5 Complex Integration: Quotient Rule

To integrate

C

coth zdz =C

cosh z

sinh zdz

with the simple closed contour C taken to be the circle |z| = 1 is straightforward in therespect that the contour is defined, as well as the function to be integrated. Singular pointsof the function can be located by expressing the complex hyperbolic sine function in termsof x and y.

sinh z = sinh x cos y + i cosh x sin y

It is clear that the zeroes of this functions are periodic since it involves trigonometric func-

tions, with zeroes located at

sinh(in) = 0

where n is an integer. From this statement, all zeroes of the function lie on the y-axiswhereas only one singularity (corresponding to n = 0) is contained within the contour|z| = 1. Application of residue theorem implies

C

coth zdz = 2i Resz=0

coth z

where the residue at z = 0 can be found by invoking the quotient rule of Section 6.2.2 withp(z) = cosh z and q(z) = sinh z.

coth z =cosh z

sinh z

=p(z)

q(z)

Since p(0) = 0, q(0) = 0, and q(0) = p(0) = 0, using the theorem is valid and the residue isgiven by

Resz=0 coth z =

cosh(0)

cosh(0) = 1

Thus, |z|=1

coth zdz = 2i.

35


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11.3.6 Finding Residues: Multiplication of Series

Suppose the residues of the function

f(z) =ez

z(1 + z2)

are to be found. It is evident that these residues are located at z = 0 and z = i. While all ofthese residues can be found via the method of functional replacement, replete examples haveall ready been shown illustrating this strategy. Thus, rather than focusing on this method,this example will instead show how direct extraction of the residue from the functions seriesexpansion, corresponding to z = 0, is accomplished. The need for only this residue couldarise when considering a complex circular closed contour integral, centered at the origin,with a contour of radius R < 1.

The methodology is straightforward: find a series representation centered about z = 0,and locate the coefficient that corresponds to the series z1 dependence, this coefficient isthe residue at that point. In order to find the series expansion of the whole function, the

function is first written as the following for clarity:

f(z) =1

z ez 1

1 + z2

A series representation centered about z = 0 for the entire function above can be obtained bymultiplication of the known Maclaurin series among the constituent terms of the function.That is to say, known expansions for each term in the above equation can be multipliedtogether in order to obtain a combined series expansion for the whole function. From thelist given in Section 11.2, the following series are noted,

ez

=

n=0

zn

n!

and

1

1 + z2=n=0

(1)nz2n

where the latter series follows from the replacement of z with z2 in series (1). The term z1is all ready in the form of a series expansion centered about z = 0. Beginning by multiplyingthe first few terms of following two known Maclaurin series

ez 11 + z2

=

1 + z + 12

z2 + 16

z3 + . . .

1 z2 + z4 + . . .

= 1 + z +1

2z2 +

1

6z3 + . . .

z2 z3 + . . .

ez1

1 + z2= 1 + z 1

2z2 5

6z3 + . . .

36


38/38


Finally, the z1 term is multiplied through yielding the full expansion of the function f(z)

ez

z(1 + z2)=

1

z+ 1 1

2z 5

6z2 + . . .

Identifying the coefficient in front of the z1 term as the residue provides the desired result.

Resz=0

ez

z(1 + z2)= 1

Thus, while in this particular case it was not the quickest way to find the functions residue (asfunctional replacement would have required much less work), direct extraction of the residuefrom the power series expansion was shown to still be a possible method. Furthermore, forsome functions (e.g. exp(cos z)), this method can actually prove to be less tasking, so it isimportant to not forget where residues come from.

11.3.7 Finding Residues: Division of Series

The Laurent series representation, centered about z = 0, of

f(z) =1

ez 1can be found by division of series. Recall from the list in Section 11.2, that

ez =n=0

zn

n!= 1 + z +

1

2z2 +

1

6z3 + . . .

then the denominator of f becomes

ez 1 = z + 12

z2 + 16

z3 + . . .

the method proceeds by dividing this series into unity.

1z 12 + . . .

z + 12

z2 + 16

z3 + . . .

1

1 + 12

z + 16

z2 + . . .

12z 16z2 + . . .

12

z 14

z2 + . . .1

12

z2 + 1

24

z3 + . . .

The first term of the Laurent series that appears in the quotient of this division has thenecessary z1 dependence, its coefficient then must be the residue.

Resz=0

1

ez 1 = 1As discussed earlier, usually only a small number of terms of the series need be computed in

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An Engineers Guide to Complex Integration

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