complex variable textbook

Complex Variables

Kevin W. CasselMechanical, Materials and Aerospace Engineering Department

Illinois Institute of Technology10 West 32nd StreetChicago, IL [email protected]

c 2013 Kevin W. Cassel

Contents

1 Functions of a Complex Variable 31.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Complex Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Complex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5 Analytic Functions and the Cauchy-Riemann Equations . . . . . 131.6 Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.6.1 Heat Conduction . . . . . . . . . . . . . . . . . . . . . . . 201.6.2 Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.7 Branch Points and Branch Cuts . . . . . . . . . . . . . . . . . . . 23

2 Conformal Mapping and BVPs 272.1 Conformal Mapping . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.1 Elementary Mapping Functions . . . . . . . . . . . . . . . 272.1.2 One-to-One Mappings . . . . . . . . . . . . . . . . . . . . 302.1.3 What Makes a Map Conformal? . . . . . . . . . . . . . . 332.1.4 Application of Conformal Mapping . . . . . . . . . . . . . 35

2.2 Solutions to Canonical Dirichlet Problems . . . . . . . . . . . . . 392.2.1 Dirichlet Problem in a Circular Disk . . . . . . . . . . . . 392.2.2 Dirichlet Problem in the Upper Half Plane . . . . . . . . 44

2.3 Application to Fluid Flows . . . . . . . . . . . . . . . . . . . . . 482.3.1 Potential Flow Formulation . . . . . . . . . . . . . . . . . 482.3.2 Basic Flows . . . . . . . . . . . . . . . . . . . . . . . . . . 522.3.3 Superposition of Basic Flows . . . . . . . . . . . . . . . . 562.3.4 The Use of Conformal Mapping . . . . . . . . . . . . . . . 59

3 Complex Integration and Residue Theory 633.1 Line Integrals of Complex Functions . . . . . . . . . . . . . . . . 633.2 Cauchys Integral Formula . . . . . . . . . . . . . . . . . . . . . . 733.3 The Dirichlet Problem Poissons Integral Formula . . . . . . . . 753.4 Infinite Series Expansions of Complex Functions . . . . . . . . . 77

3.4.1 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . 773.4.2 Laurent Series . . . . . . . . . . . . . . . . . . . . . . . . 80

3.5 Singularities, Poles and Residues . . . . . . . . . . . . . . . . . . 83

1

CONTENTS 2

3.5.1 Types of Singularities . . . . . . . . . . . . . . . . . . . . 833.5.2 Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . 853.5.3 Cauchys Residue Theorem . . . . . . . . . . . . . . . . . 88

3.6 Evaluation of Real Definite Integrals . . . . . . . . . . . . . . . . 903.6.1 Trigonometric Functions . . . . . . . . . . . . . . . . . . . 903.6.2 Rational Functions . . . . . . . . . . . . . . . . . . . . . . 923.6.3 Combinations of Trigonometric and Rational Functions . 953.6.4 Indented Contours . . . . . . . . . . . . . . . . . . . . . . 983.6.5 Integrals Involving Branch Cuts . . . . . . . . . . . . . . 102

A Linear Fractional Transformations 107

Chapter 1

Functions of a ComplexVariable

1.1 Introduction

As with matrices, the typical undergraduate engineering student has had somelimited exposure to complex variables in connection with differential equations,physics, and some engineering courses. It is useful, therefore, to extend thislimited exposure and treat the topic in a standalone fashion in order to betterappreciate complex variables as a branch of mathematics that is a powerful toolin applied mathematics and engineering.

We begin by addressing some common misconceptions pertaining to theimaginary number and complex variables:

1. There is nothing imaginary about the imaginary number1 = i; it

is just as real as

4 = 2, for example.2. Similarly, you have probably been told, at least by your calculator, that

the log of a negative real number does not exist. As we will see, it doesexist, but it is complex.

With respect to complex variables, the application of mathematics to solvingphysical problems falls into three categories1:

I) Real Problem Real Intermediate Steps Real Solution.II) Real Problem Complex Intermediate Steps Real Solution.

e.g. The spring-mass example from Matrices, and evaluation of real definiteintegrals (see section 3.6).

1Note the dual meaning of our use of the term real. We are talking about real physicalsystems that can be expressed mathematically in terms of real parameters and variables.

3

CHAPTER 1. FUNCTIONS OF A COMPLEX VARIABLE 4

III) Real Problem Complex Intermediate Steps Complex Solution(where the real and imaginary parts, which are real, have physical mean-ing). e.g. Harmonic functions (see sections 1.4 and 2.2).

For the most part, you have gotten by with (I); however, allowing for (II)and (III) significantly expands the range of applications we can address. Thatis, we no longer live in a positive integer world, for example, all n of mysheep have returned to the pen.

1.2 Definitions

A complex number is of the form

= a+ ib,

where a and b are real numbers, and i is the imaginary number satisfying

i2 = 1.The real and imaginary parts of the complex number are denoted by

a = Re(), b = Im().

Similarly, a complex variable is of the form

z = x+ iy,

where x and y are real variables.It is instructive to express a complex number or variable as a point on the

complex (Argand) plane as a means of visualizing complex numbers geometri-cally.

Cartesian coordinate (x, y):

real axis

imaginaryaxis

y

y

x

x z=x+iy


Polar coordinates (r, ):

y

y

x

x

r

z=x+iy

The modulus of z, i.e. the length of z as a vector, is

r = |z| = |x+ iy| =x2 + y2.

The angle from the x-axis is called the argument of z and is denoted by arg(z).The Cartesian coordinates x and y are related to the polar coordinates r and by

x = r cos , y = r sin .

Therefore, a complex variable can be expressed in any of the following equivalentforms

z = x+ iy = r(cos + i sin ) = rei,

where r = |z| (length of z), and = arg(z) (] from x-axis). The fact thatz = rei will be shown later in this section. Some texts also use the notationcos + i sin = cis = .

Note: The argument of z is multi-valued according to

arg(z) = = 0 + 2kpi, pi < 0 pi, k = 0,1,2, . . . ,

where 0 is the principle argument of z. The range of the principle argumentalso may be defined by 0 < 2pi, or any other 2pi range.

The complex conjugate, or simply conjugate, of z = x+ iy is

z = x iy = rei.

Remarks:

1. The absolute value of a real number is a special case of the modulus.

2. Polar coordinates (r, ) are two-dimensional cylindrical coordinates (r, , z).

3. To obtain the complex conjugate of any expression replace i with i.4. Graphically, the conjugate of z is its reflection about the real axis.


5. Two complex numbers are equal if their respective real and imaginaryparts are equal.

6. The symbol i = 1 was introduced by Leonhard Euler (17071783).7. Karl Friedrich Gauss (17771855) introduced the term complex number.

8. See An Imaginary Tale: The Story of1 by Paul J. Nahin for an

historical account of the development of the mathematics of complex vari-ables.

Given these definitions, we must revisit algebra and trigonometry (see sec-tions 1.3 and 1.4) and calculus (see section 1.5) for complex functions as com-pared to real functions.

1.3 Complex Arithmetic

For z1 = x1 + iy1 and z2 = x2 + iy2, we define the following operations:

Addition:z1 + z2 = (x1 + x2) + i(y1 + y2).

Subtraction:z1 z2 = (x1 x2) + i(y1 y2).

Multiplication:z1z2 = (x1 + iy1)(x2 + iy2) = (x1x2 y1y2) + i(x1y2 + x2y1).

zz = (x+ iy)(x iy) = (x2 + y2) + i:0

(xy xy) = |z|2 = |z|2.Thus, zz is real.

Division (z1 6= 0):z2z1

=x2 + iy2x1 + iy1

=(x2 + iy2)(x1 iy1)

x21 + y21

(=z2z1z1z1

)z2z1

=x1x2 + y1y2x21 + y

21

+ ix1y2 x2y1x21 + y

21

.

Complex Conjugates:z1 + z2 = (x1 + x2) i(y1 + y2) = (x1 iy1) + (x2 iy2) = z1 + z2.

z1 z2 = . . . = z1 z2.z1z2 = . . . = z1z2.(z2z1

)= . . . =

z2z1.


Observe that all of these operations reduce to their familiar forms for realnumbers if Im(z1) = Im(z2) = 0.

1.4 Complex Functions

Consider the following functions f(z) of a complex variable. Note that they allreduce to the familiar relations for real variables, i.e. with Im(z) = 0.

Trigonometric Functions:

sin z =eiz eiz

2i= z z

3

3!+z5

5! =

n=0

(1)n z2n+1

(2n+ 1)!. (1.1)

cos z =eiz + eiz

2= 1 z

2

2!+z4

4! =

n=0

(1)n z2n

(2n)!. (1.2)

tan z =sin z

cos z.

Because these are the same relationships as for real variables, the trigonometricfunctions of a complex variable satisfy the same identities as for real variables.

Adding i(1.1) and (1.2) gives

eiz = cos z + i sin z.

Because this holds for a complex variable z, it is also true for the argument ofz; thus,

ei = cos + i sin .

This is known as Eulers formula. Then

z = x+ iy = r(cos + i sin ) = rei,

and because (ei)n = ein

(cos + i sin )n = cos(n) + i sin(n).

This is De Moivres theorem. It follows that

zn = rnein = rn[cos(n) + i sin(n)].

Noting then thatei = cos i sin ,

we can also obtain the following familiar relations

cos =ei + ei

2, sin =

ei ei2i

.


In the previous section we multiplied two complex variables expressed inCartesian coordinates. If z1 = r1e

i1 and z2 = r2ei2 , then

z1z2 = r1r2ei(1+2);

therefore, z1z2 has length r1r2 and argument 1 + 2. Thus, because |ei| = 1for real, i.e. the length is 1, zei rotates the vector representing z through anangle in the complex plane. Also

z2z1

=r2r1ei(21), (z1 6= 0),

which rotates through a negative (clockwise) angle .

Note: An alternative derivation of Eulers formula is as follows:

Consider the Taylor series expansion for ei, where is real

ei = 1 + i +(i)2

2!+

(i)3

3!+

(i)4

4!+

(i)5

5!+

=

(1

2

2!+4

4!+

)+ i

(

3

3!+5

5!+

)ei = cos + i sin .

Hyperbolic Functions:

sinh z =ez ez

2= z +

z3

3!+z5

5!+ =

n=0

z2n+1

(2n+ 1)!. (1.3)

cosh z =ez + ez

2= 1 +

z2

2!+z4

4!+ =

n=0

z2n

(2n)!. (1.4)

tanh z =sinh z

cosh z.

Comparing (1.1), (1.2), (1.3) and (1.4), we see that

sinh(iz) = i sin z, cosh(iz) = cos z,

sin(iz) = i sinh z, cos(iz) = cosh z.

Note the similarity to the relationships sin(x) = sinx, cos(x) = cosx, etc.for sine odd and cosine even.

The real and imaginary parts of sin z and cos z are then:

sin z = sin(x+ iy)

= sinx cos(iy) + cosx sin(iy)

sin z = sinx cosh y + i cosx sinh y,


cos z = cos(x+ iy)

= cosx cos(iy) sinx sin(iy)cos z = cosx cosh y i sinx sinh y.

Exponentials:

Taylor series expansion:

ez = 1 + z +z2

2!+z3

3!+ =

n=0

zn

n!,

andez1ez2 = ez1+z2 ,

(ez)n = enz.

Real and imaginary parts of ez:

ez = ex+iy = exeiy = ex(cos y + i sin y).

Polynomials: n = 0, 1, 2, 3, . . .

f(z) = zn = (x+ iy)n

= (rei)n

= rnein

zn = rn[cos(n) + i sin(n)].

Then a polynomial of degree N can be defined as

f(z) =

Nn=0

Anzn = A0 +A1z +A2z

2 + +Anzn + +ANzN ,

where the coefficients An may be complex.

General Power Function: a = a1 + ia2

za = elog za

= ea log z.

General Exponential Function: a = a1 + ia2

az = elog az

= ez log a, a 6= 0, 1.

Logarithms:

Ifw = log z,


thenew = elog z,

orew = z,

where log represents the natural logarithm. With w = u + iv, the real (u) andimaginary (v) parts of log z can be determined from

z = ew

x+ iy = eu+iv

= eueiv

x+ iy = eu(cos v + i sin v).

Equating real and imaginary parts gives

x = eu cos v, y = eu sin v. (1.5)

Squaring both and adding, we have

e2u(cos2 v + sin2 v) = x2 + y2

e2u = x2 + y2 = |z|2 = r2.Then eu = r, and the real part of log z is

u = log r = log |z|,and from (1.5), the imaginary part of log z is

cos v =x

eu=x

r= cos

v = .Therefore, we may express log z as follows

log z = log |z|+ i = log r + i(0 + 2kpi), k = 0,1,2, . . . , (z 6= 0),where the principle argument is defined by pi < 0 pi, for example.

Thus, log z is multi-valued.

}}

log |z|y

x2pi

2pi


Remarks:

1) All values of log z are imaginary except when = 0, i.e. for positive realnumbers.

2) Other functions involving log are multi-valued, e.g. za = ea log z.

Example: Evaluate log z for z = i.

y

i

x

The modulus and argument of z are

|z| = 1, 0 = pi2,

respectively. Thus,

log i =*0log 1 + i

(pi2

+ 2kpi)

=4k + 1

2pii, k = 0,1,2, . . . ,

and all the values of log i are located on the imaginary axis.

Fractional Powers:

Using the log function, we may develop an expression for fractional powers ofa complex variable zn/m, where m and n are integers with no common factors,i.e. n/m is irreducible.

zn/m = elog zn/m

= e(n/m) log z

= e(n/m)[log r+i(0+2kpi)]

= elog rn/m

ei(n/m)(0+2kpi)

zn/m = rn/mei(n/m)(0+2kpi), k = 0, 1, 2, . . . ,m 1,The range for k results from the fact that other values of k produce pointsalready represented in the given range. Therefore, zn/m = (z1/m)n has mdistinct roots.

Example: Find the roots of z1/3 if z = 8 + i0.


Note that r = |z| = 8, 0 = arg z = 0 and m = 3 (n = 1) k = 0, 1, 2; thus,z1/3 = 81/3ei(1/3)(0+2kpi), 0 = 0

z1/3 = 2

[cos

(

3

)+ i sin

(

3

)], = 0, 2pi, 4pi

The three roots are then

= 0 : z1/3 = 2

= 2pi : z1/3 = 1 + i3 = 4pi : z1/3 = 1 i3

Note that the first root is the only one on the real axis and the second two arecomplex conjugates. Also note that this means that, just as 23 = 8, so also(1 i3)3 = 8. Try it!

Inverse Trigonometric Functions:

Ifw = sin1 z = arcsin z,

thenz = sinw,

and from equation (1.1)

z =eiw eiw

2i.

Multiplying by 2ieiw gives

(eiw)2 2iz(eiw) 1 = 0.This expression is quadratic in eiw; therefore, from the quadratic equation

ax2 + bx+ c = 0 x = b+b2 4ac

2a,

we have

eiw =2iz +

(2iz)2 4(1)(1)

2(1)

eiw = iz + (1 z2)1/2,where (1 z2)1/2 has two values. Solving for w (take the log of both sides)

w = sin1 z =1

ilog[iz + (1 z2)1/2],

which has an infinite number of values due to the log.


Similarly,

cos1 z =1

ilog[z + (z2 1)1/2].

tan1 z =1

2ilog

i zi+ z

.

sinh1 z = log[z + (1 + z2)1/2].

cosh1 z = log[z + (z2 1)1/2].

tanh1 z =1

2log

1 + z

1 z .

1.5 Analytic Functions and the Cauchy-RiemannEquations

Sections 1.3 and 1.4 address algebra and trigonometry of complex functions; wenow turn our attention to the calculus of such functions.

Derivatives of complex functions are defined just as for real functions. Thatis,

df(z)

dz= f (z) = lim

z0f(z + z) f(z)

z. (1.6)

Therefore, derivatives of complex functions are the same as for real functions,for example,

d

dx(sinx) = cosx d

dz(sin z) = cos z.

In addition, the product rule, quotient rule, chain rule, and LHopitals rule arethe same for complex functions as for real functions.

Recall that for real functions f(x), i.e. with Im(z) = 0, differentiability off(x) at a point x0 requires that the derivative, i.e. the limit in (1.6), exists andis finite. For the limit to exist, it must be the sam from the right and left alongthe real axis. Also, if a function is differentiable, then it is continuous (thereverse, however, is not necessarily true).

For complex functions f(z), we must extend the differentiability requirementto that of analyticity. A complex function f(z) is analytic at a point z = z0 iff(z) is differentiable at z0 and in some neighborhood of z0, i.e. within a circle offinite radius surrounding z0.

Differentiability requires that the derivative f (z0) exists, i.e. the limitexists in (1.6), in which case f(z) is continuous at z0.

Existence of the derivative f (z0) requires that its value be finite andunique, i.e. single valued from any direction.

Analyticity at a point is a stronger requirement than differentiability as itrequires differentiability in a neighborhood, not simply at a point.


A function f(z) is analytic in a region R if it is analytic at each point in R.f(z) is said to be an entire function if it is analytic in the entire finite complexplane.

Goursats Theorem: If f(z) is analytic at a point z0, then f(z) is continu-

ous at z0 (again, the reverse is not necessarily true).

Recall that for real functions f(x) to be differentiable, the limit defining f (x)in (1.6) must exist and be unique from the left and right along the real axis.For a complex function w = f(z) to be analytic at z, however, the limit

dw

dz= f (z) = lim

z0f(z + z) f(z)

z= lim

z0w

z(1.7)

must exist and be unique, i.e. single valued, as z 0 from any direction inthe complex plane.

Example: Is w = f(z) = x iy = z analytic?w

z=

x iyx+ iy

.

If we take z 0 along the x-axis y = 0 and

limz0

w

z=

x

x= 1.

If we take z 0 along the y-axis x = 0 and

limz0

w

z=yy

= 1.

Therefore, f(z) = z is not analytic.

Now suppose that the derivative (1.7) does exist uniquely for

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

How must the real part u(x, y) and imaginary part v(x, y) be related for w =f(z) to be analytic?

From (1.7)dw

dz= lim

z0w

z= lim

x0y0

u+ iv

x+ iy.

Path 1: If z 0 along the x-axis y = 0 anddw

dz= lim

x0u+ iv

x= lim

x0

(u

x+ i

v

x

)


dwdz

= f (z) =u

x+ i

v

x. (1.8)

Path 2: If z 0 along the y-axis x = 0 anddw

dz= lim

y0u+ iv

iy= lim

y0

(v

y iu

y

)

dwdz

= f (z) =v

y iu

y. (1.9)

For w = f(z) to be analytic, therefore, (1.8) and (1.9) must be equal

u

x+ i

v

x=v

y iu

y.

Equating real and imaginary parts requires that

u

x=v

y,

v

x= u

y. (1.10)

These are the Cauchy-Riemann (C-R) equations. They must be satisfiedat all points z = x + iy in the region R for w = f(z) to be analytic in R. Thepartial derivatives in (1.10) must be continuous in R, i.e. continuity of deriva-tives is a necessary, but not sufficient, condition for analyticity.

Remarks:

1) The Cauchy-Riemann equations provide a necessary and sufficient condi-tion for the existence of the derivative f (z) and for f(z) to be analytic.

2) The expressions (1.8) and (1.9) for f (z) are useful for obtaining the deriva-tive of a complex function and will be used in subsequent considerations.

3) For a function expressed in polar form f(z) = u(r, )+iv(r, ), the Cauchy-Riemann equations are (see PS#6, problem 6)

u

r=

1

r

v

,

v

r= 1

r

u

.

It can also be shown that the derivative of an analytic function expressedin polar coordinates (analogous to (1.8) and (1.9)) satisfies (see PS#6,problem 7)

f (z) =(u

r+ i

v

r

)(cos i sin ),

f (z) =(u

+ i

v

)(ir

)(cos i sin ).


4) Recall LHopitals Rule:

If f(z0) = 0 and g(z0) = 0, then

limzz0

f(z)

g(z)=f (z0)g(z0)

(g(z0) 6= 0),

or if the first n derivatives of f(z) and g(z) are zero at z0, then

limzz0

f(z)

g(z)=f (n+1)(z0)

g(n+1)(z0)(g(n+1)(z0) 6= 0).

Example: Reconsider whether w = f(z) = z = x iy is analytic.

With w = u+ iv, we haveu = x, v = y.

Evaluating the Cauchy-Riemann equations

u

x= 1,

v

y= 1,

v

x= 0, u

y= 0.

The second equation is satisfied, but the first equation is not; therefore, z is notanalytic, in agreement with the previous example.

Example: Is the following function analytic

w = f(z) = (x y)2 + 2i(x+ y)?

Evaluating the Cauchy-Riemann equations

u

x= 2(x y), v

y= 2,

v

x= 2, u

y= 2(x y).

Thus, the Cauchy-Riemann equations are only satisfied along the line xy = 1,and there is no neighborhood around the points on the line in which f(z) isanalytic, i.e no R in which f(z) is analytic.

We can also use the Cauchy-Riemann equations to determine the imaginary(real) part of an analytic complex function if the real (imaginary) part is given.


Example: Given the imaginary part

v(x, y) = xy3 x3y,determine the real part u(x, y) such that w = f(z) = u(x, y)+iv(x, y) is analytic.

Cauchy-Riemann equations:

u

x=v

y= 3xy2 x3,

u

y= v

x= y3 + 3x2y.

Integrating the first equation with respect to x leads to

u(x, y) =

(3xy2 x3)dx+ C(y)

u(x, y) =3

2x2y2 1

4x4 + C(y).

Observe that the constant of integration must be allowed to be a function ofy due to the partial differentiation with respect to x. Substituting this into thesecond Cauchy-Riemann equation gives

3x2y + C (y) = y3 + 3x2y C (y) = y3.

Integrating

C(y) = 14y4 + C.

Thus, the real part is

u(x, y) =3

2x2y2 1

4x4 1

4y4 + C.

For an analytic function w = f(z) = u+ iv, the question arises as to how u(x, y)and v(x, y) are related graphically? To determine this, consider the slopes ofthe curves u(x, y) = U1 = constant and v(x, y) = V1 = constant at the point(x0, y0), where U1 = constant and V1 = constant intersect.

Taking the total differential of u(x, y) along the curve u(x, y) = U1 gives

du =u

xdx+

u

ydy = 0,

which is zero because u is a constant along the curve. The slope of u = U1 isthen (

dy

dx

)u=U1

= u/xu/y

. (1.11)


Similarly, the total differential of v(x, y) = V1 is

dv =v

xdx+

v

ydy = 0.

The slope of v = V1 is then (dy

dx

)v=V1

= v/xv/y

.

Alternatively, from the Cauchy-Riemann equations, we can write this expressionin terms of u as (

dy

dx

)v=V1

=u/y

u/x. (1.12)

Comparing (1.11) and (1.12), we see that at a point (x0, y0) where the curvesintersect (

dy

dx

)v=V1

= 1(dy

dx

)u=U1

.

That is, the curves u(x, y) = U1 and v(x, y) = V1 have negative reciprocalslopes; therefore, they are orthogonal at any point (x0, y0) where they intersect.

For example, consider lines of constant u(x, y) (solid) and v(x, y) (dashed) fromthe previous example with

f(z) = u(x, y) + iv(x, y) = 3x2y2/2 x4/4 y4/4 + i(xy3 x3y).

-2 -1 0 1 2-2

-1

0

1

2

The orthogonality of lines of constant u(x, y) and v(x, y) is very useful in appli-cations, i.e. when u(x, y) and v(x, y) have physical meaning.


1.6 Harmonic Functions

The mathematical result encapsulated in the Cauchy-Reimann equations leadsdirectly to our first major application of complex variables. From the Cauchy-Riemann equations, which show how the real and imaginary parts of an analyticfunction are related to one another, we can obtain separate equations that thereal and imaginary parts of an analytic function, i.e. u(x, y) and v(x, y), respec-tively, must satisfy.

Recall the Cauchy-Riemann equations

u

x=v

y, (1.13)

v

x= u

y. (1.14)

Taking /x of (1.13) and /y of (1.14) yields

2u

x2=

2v

xy,

2v

xy=

2u

y2.

Subtracting gives2u

x2+2u

y2= 0. (1.15)

Similarly, taking /y of (1.13) and /x of (1.14) gives

2u

xy=2v

y2,

2v

x2=

2u

xy.

Adding yields2v

x2+2v

y2= 0. (1.16)

Thus, the real and imaginary parts of an analytic function each must satisfyLaplaces Equation

2u = 0, 2v = 0. Solutions of Laplaces equation, in this case u(x, y) and v(x, y), are known

as harmonic functions.

If u(x, y) is harmonic, then v(x, y) is a harmonic conjugate of u(x, y), andvice versa.

This mathematical result is what makes complex variable theory of such practicalinterest as illustrated in the following sections and section 2.3 for potential fluidflow.


1.6.1 Heat Conduction

Consider heat conduction in a solid. The general governing equation is

(kT ) + q = cp Tt,

where is the gradient operator, andT = temperature,

k = thermal conductivity,

= density,

cp = specific heat,

q = heat generation.

If we have steady, 2-D heat conduction with no heat generation and k = constant(homogeneous medium), the governing equation becomes

2T = 2T

x2+2T

y2= 0,

i.e. the temperature satisfies Laplaces equation.

So let T (x, y) be the real part of an analytic function

(z) = T (x, y) + i(x, y),

where (z) is called the complex temperature. What is (x, y)?

We know that lines of constant (x, y) are orthogonal to lines of constantT (x, y), i.e. isotherms (or equipotential lines). Therefore, (x, y) = 1 = con-stant is a heat flux line that shows the direction that the heat conducts.

x

y

T3>T2

isotherms (equipotential lines)

heat flux lines

3

T2>T1

2

T1

1


In order to determine what (z) represents physically, recall equation (1.8)in section 1.3 for w = f(z) = u(x, y) + iv(x, y)

dw

dz= f (z) =

u

x+ i

v

x.

Thus, the derivative of the complex temperature is

d

dz=T

x+ i

x,

or from the Cauchy-Riemann equations (/x = T/y)d

dz= (z) =

T

x iT

y.

Taking the complex conjugate

d

dz= (z) =

T

x+ i

T

y= 1

k(qx + iqy),

where according to Fouriers law

qx = kTx

= heat flux in the x-direction,

qy = kTy

= heat flux in the y-direction.

Thus, we see that not only do the real and imaginary parts of (z) have physicalmeaning, the real and imaginary parts of (z) do as well.

1.6.2 Electrostatics

Electromagnetic fields in homogeneous and isotropic media are governed by thewave equations (obtained from Maxwells equations)

2E

t2= 2E,

2H

t2= 2H,

where E is the electric field vector, H is the magnetic field vector, is thedielectric (permittivity) tensor, and is the permeability tensor. If the electro-magnetic field is steady, i.e. static, the equations reduce to Laplace equationsfor the electric and magnetic fields

2E = 0, 2H = 0.In general, these are vector equations with up to three scalar equations corre-sponding to each component of the electric or magnetic field vectors.

Let us consider electrostatics in two dimensions, wherein there is only oneequation for the scalar electric field intensity E(x, y), i.e. 2E = 0. In this


case, we may regard E(x, y) as the real part of an analytic function, and theimaginary part as being related to the flux of the electrostatic field similar tothe heat conduction case. For example, one could define a complex function as

(z) = E(x, y) + i(x, y),

where E(x, y) is the electrostatic potential, and lines of constant (x, y) arecurrent flow lines.

Remarks:

1. Note the dual meaning of x and y here, where they represent the physicalcoordinates as well as the real and imaginary parts of the point z on thecomplex plane.

This highlights the primary limitation of complex variable theory,which is that it only applies in two-dimensional settings.

2. Laplaces equation for u(r, ) in polar coordinates is

2u = 2u

r2+

1

r

u

r+

1

r22u

2= 0.

3. The real and imaginary parts of the complex functions (z) and (z)take on physical meaning.

4. This approach may be applied in any physical situation governed byLaplaces equation in two dimensions and will be applied to incompressiblefluid flows in section 2.3 potential flow theory.

5. Recall that a scenario governed by Laplaces equation with fixed boundaryconditions is known as the Dirichlet problem.

Problem Set # 6


1.7 Branch Points and Branch Cuts

Armed with a requirement for analyticity, let us revisit multi-valued functionsrecalling that a function must be single-valued to be analytic.

Consider the multi-valued function

log z = log |z|+ i(0 + 2kpi), k = 0,1,2, . . . ,where pi < 0 pi.

Starting at z1 = 1 + i0 = eipi, i.e. k = 0, move counterclockwise along apath with r = |z| = 1, i.e. the unit circle, and consider values of log z.

z5

x

z4

z3

z2

z1, z6

y

k = 0:

z1 = eipi = 1 + i0 log z1 =*

0

log 1 + i(pi + 0) = ipi,z2 = e

ipi/2 = 0 i log z2 = ipi2 ,z3 = e

i0 = 1 i0 log z3 = 0,z4 = e

ipi/2 = 0 + i log z4 = ipi2 ,z5 = e

ipi = 1 + i0 log z5 = ipi.Thus, while z5 = z1, log z5 6= log z1.

We see that log z is not continuous across the negative real axis y = 0, x 0for k = 0. In order for the function to remain continuous, one must switch to adifferent branch of log z, i.e. k = 1, in going from z5 to z6 for 0 = pi.


k = 1:

z6 = ei(pi+2pi) = eipi = 1 + i0 (= z5 = z1) log z6 = log z5 = ipi.

Note that z5 and z6 are the same point in the complex plane but are on differentbranches of log z.

Branch: A single-valued and analytic portion of a multi-valued function.

e.g. log z with k = 0 is the principle branch (sometimes denoted by Log z),log z with k = 1 is another branch, etc. There are an infinity of branchesof log z.

Each branch of log z is analytic on the entire z-plane except for y = 0, x 0 (recall that f(z) must be continuous to be analytic).

The origin z = 0 is a branch point of log z, which is a point that must be insideany path that is necessary to change branches of a function.

The negative real axis (y = 0, x 0) is a branch cut of log z, which is a curvealong which the function is discontinuous if it is to remain single-valued, i.e. donot allow for changing branches.

Note: The branch cut of log z may be moved by changing the definition ofthe principle argument 0. For example, if 0 0 < 2pi, then the branch cut oflog z is the positive real axis.

y

x

In applications, the branch cut is often aligned with a natural discontinuityin the physical system if possible.

Example: Find the branch points and branch cuts for the fractional power ofz

f(z) = z23 ,


with the principle argument defined by 0 0 < 2pi.

Recall that

zn/m = rn/mei(n/m)(0+2kpi), k = 0, 1, 2, . . . ,m 1;

therefore, zn/m has m branches (cf. log z has an infinity of branches).

Starting at z1 = 1 + i0, k = 0 and following the unit circle, evaluate z2/3 at

z = 1 + i0 on its various branches.

1

2

3

4

5

6

y

x

For k = 0:

At point 1:r1 = 1, 0 = 0,

z2/31 = (1)2/3ei(2/3)(0+0) = 1.

At point 2:r2 = 1, 0 = 2pi

z2/32 = (1)

2/3ei(2/3)(2pi+0)

= ei4pi/3

= cos

(4pi

3

)+ i sin

(4pi

3

)z

2/32 =

1

2 i

3

2.

As with log z, we see that while z1 = z2, z2/31 6= z2/32 because they approach the

branch cut from opposite sides.


To remain continuous on y = 0, x 0, we must move to the next branchwith k = 1.

For k = 1:

At point 3 (r = 1, 0 = 0):

z2/33 = (1)e

i(2/3)(0+2pi) = ei4pi/3 = z2/32 .

At point 4 (r = 1, 0 = 2pi):

z2/34 = e

i(2/3)(2pi+2pi) = ei8pi/3 = 12

+ i

3

2.

For k = 2:

At point 5 (r = 1, 0 = 0):

z2/35 = e

i(2/3)(0+4pi) = ei8pi/3 = z2/34 .

At point 6 (r = 1, 0 = 2pi):

z2/36 = e

i(2/3)(2pi+4pi) = ei4pi = 1.

Therefore, the mth branch, i.e. k = 2, transitions back to the first branch, i.e.k = 0. The branch point is z = 0, and y = 0, x 0, i.e. the positive real axis,is the branch cut.

See the Mathematica notebook ComplexVariables.nb for illustrations ofhow to manipulate complex functions in Mathematica.

Chapter 2

Conformal Mapping andBoundary-Value Problems

2.1 Conformal Mapping

Based on Chapter 1, we can solve for harmonic functions in simple domains,for example, using the method of separation of variables (see Matrices). Inthis chapter, we seek to extend the solution of harmonic functions to morecomplicated domains using conformal mapping.

2.1.1 Elementary Mapping Functions

Previously, we considered the complex function

w = f(z) = u+ iv,

where z = x+ iy, and

u = Re[f(z)], v = Im[f(z)].

We may also view f(z) as a mapping or transformation function from the z-plane to the w-plane:

z-plane: w-plane:

y

(x, y) (u, v)

image of (x, y)f

x

v

u

27

CHAPTER 2. CONFORMAL MAPPING AND BVPS 28

The point (u, v) in the w-plane is called the image of the point (x, y) in thez-plane. If w = f(z) is single valued, then each point z = x+ iy is mapped to asingle point w = u+iv in the w-plane, and the mapping is said to be one-to-one.In many cases the inverse mapping function z = F (w) from the w-plane to thez-plane may also be obtained.

Consider, for example, the following elementary mapping functions:

1) Translation:w = z + a,

where a is a complex constant.

2) Stretching:w = bz,

where b is a real constant. Because z = rei, |w| = br and Arg(w) =Arg(z).

3) Rotation:w = eiz = eirei = rei(+),

where is real and is the angle through which z is rotated ( > 0 rotatesin the counterclockwise direction).

Note that the mapping w = cz, where c is a complex constant, accom-plishes both a stretching, b = |c|, and a rotation, = Arg(c).

4) Inversion:

w =1

z.

If z = rei and w = ei, then

ei =1

rei,

and

=1

r, = .

Therefore, all the points inside the unit circle in the z-plane are mappedto the region outside the unit circle in the w-plane ( = 1/r) and reflectedabout the real axis ( = ).

For example,

z =i

2 w = 1

z=

2

i= 2i,

or equivalently

=1

r=

1

1/2= 2, = = pi

2.


w-plane

1x

y

i

2 1u

v

f(z)

2i

z-plane

Note: The inversion mapping function transforms circles into circles orlines, i.e. circles with r =.

5) Linear Fractional (Bilinear or Mobius) Transformation:

w =az + b

cz + d,

where a, b, c and d are complex constants such that ad 6= bc.1 Note thatthe inverse map,

z =dw + bcw a ,

is also a linear fractional transformation.

In order to map three specified points z1, z2, z3 in the z-plane into threespecified image points w1, w2, w3 in the w-plane, the constants a, b, c, andd may be determined by solving

(w w1)(w3 w2)(w w2)(w3 w1) =

(z z1)(z3 z2)(z z2)(z3 z1) (2.1)

for w = f(z). If a point is specified to be at infinity, set any ratio offactors that includes that point equal to one. For example, if w1 = ,then (w w1)/(w3 w1) = 1.

Depending on how the points are chosen, i.e. three collinear points de-fine a line and three noncollinear points define a circle, we can map linesinto lines, circles into circles, lines into circles and circles into lines usinga linear fractional transformation.

Remarks:

1. The translation, stretching, rotation and inversion mapping functions areeach special cases of the fractional transformation.

2. See the Appendix for more details regarding linear fractional transforma-tions and a derivation of the implicit relationship (2.1).

1This condition ensures that the inverse map z = F (w) exists.


3. These elementary mapping functions may be combined together to obtainmore general mappings.

4. Many references contain tables of useful mappings. See, for example,McQuarrie 2003, Saff and Snider 2003, and the CRC Handbook.

5. Conformal mapping software is available. See, for example,http://www.lascauxsoftware.com/.

6. Although any complex function may be regarded as a mapping, it is typ-ically desired that mappings be one-to-one and conformal as described inthe following sections.

2.1.2 One-to-One Mappings

In order to obtain a criteria for a mapping to be one-to-one, let us consider theinverse mapping function z = F (w) from the w-plane to the z-plane.

In general, we will need expressions for /x and /y in terms of /u and/v, for example, to transform a partial differential equation from one planeto the other. Thus, in order to transform (x, y) (u, v), where u = u(x, y) andv = v(x, y), we have the following transformation laws from the chain rule

x=u

x

u+v

x

v, (2.2)

y=u

y

u+v

y

v. (2.3)

Now consider the inverse mapping (u, v) (x, y), where x = x(u, v) and y =y(u, v). In order to determine the inverse transformation, view (2.2) and (2.3)as two equations for the two unknowns /u and /v

u

x

v

x

u

y

v

y

u

v

=

x

y

.We may solve this system of equations using, for example, Cramers rule (Matrices,section 1.6)

u=

x

v

x

y

v

y

J

,

v=

u

x

x

u

y

y

J

,


where J is the Jacobian given by

J =

u

x

v

x

u

y

v

y

=u

x

u

y

v

x

v

y

=u

x

v

y uy

v

x.

For a unique solution to exist, the Jacobian may not be zero. If the mappingf(z) is analytic, then from the Cauchy-Riemann equations, the Jacobian can bewritten as

J =

(v

y

)2+

(u

y

)2.

Recall from equation (1.9) in section 1.3 that

f (z) =v

y iu

y;

therefore,

J = |f (z)|2 6= 0.Thus, if

1) f(z) is analytic at z0, and

2) f (z0) 6= 0,then a unique, i.e. single-valued, inverse mapping function z = F (w) exists atz0, and w0 = f(z0) is a one-to-one mapping at z0. Points where f

(z0) = 0 arecalled critical points at which the

1) angle is not preserved by mapping (see subsequent discussion of conformalmapping),

2) inverse mapping function z = F (w0) is not analytic at w0.

Example: Under the mapping

w = f(z) = z +1

z,

obtain the images of the curves

a) r = 2 (circle),

b) =pi

4(straight line).


To obtain the critical points of the above mapping, differentiate and set equalto zero as follows

f (z) = 1 1z2

= 0;

therefore, the critical points are located at z = 1, at which the mapping is notone-to-one.

First, we wish to find the real and imaginary parts of the mapping f(z). Rewrit-ing the mapping with z = rei

w = z +1

z

= rei +1

rei

w = r(cos + i sin ) +1

r(cos i sin ),

then the real and imaginary parts are

u = Re(w) =

(r +

1

r

)cos ,

v = Im(w) =

(r 1

r

)sin .

Now consider the images of the two curves, in which we want the equations ofthe curves (a) and (b) in terms of u and v only:

a) r = 2:

Substituting r = 2 into the above relationships for u and v gives

u =5

2cos , v =

3

2sin .

We want to express the image in the w-plane; therefore, to eliminate recall that cos2 + sin2 = 1. Thus,(

u

5/2

)2+

(v

3/2

)2= 1,

which is an ellipse in the w-plane.

b) =pi

4:

Substituting gives

u =

(r +

1

r

)12, v =

(r 1

r

)12.


Again, we want to eliminate r. Adding gives

2 (u+ v) = 2r, (2.4)

and subtracting gives 2 (u v) = 2

r. (2.5)

Substituting (2.4) for r into (2.5), we have

(u+ v) (u v) = 2,

oru2 v2 = 2,

which is a hyperbola in the w-plane.

2.1.3 What Makes a Map Conformal?

Let us take a further look at how a curve and its image are related under amapping.

Consider points z0 and z0 + dz on curve C in the z-plane that is mapped topoints w0 and w0 + dw on its image curve in the w-plane.

z0 + dz

y

f(z)

x u

v

w-planez-plane

C

w0 + dw

w0z0

Note that dz and dw are along C and , respectively, and and are theangles that tangents to the curves C and form with the horizontal at z0 andw0, respectively.

We want to determine how the behavior of the curves C and near z0 andw0, respectively, are related if f(z) is analytic. We have the mapping

w = f(z);


thus, differentiating yieldsdw = f (z)dz.

If f (z0) 6= 0, then near z0dw = f (z0)dz. (2.6)

Now let us write each of the following in polar form

dz = |dz|ei, (2.7)

dw = |dw|ei, (2.8)f (z0) = |f (z0)|eiArg[f (z0)]. (2.9)

Substituting (2.7)(2.9) into (2.6) gives

|dw|ei = |f (z0)|eiArg[f (z0)]|dz|ei,

or|dw|ei = |f (z0)||dz|eiArg[f (z0)+].

Therefore,|dw| = |f (z0)||dz|,

and = Arg[f (z0)] + . (2.10)

Therefore, the mapping f(z) stretches the infinitesimal curve segment dz by afactor |f (z0)| and rotates it by an angle Arg[f (z0)].

Now consider two intersecting curves C1 and C2 in the z-plane that are mappedto 1 and 2, respectively, in the w-plane.

w0

y

f(z)

x u

v

w-plane

C1

C2 2 1

z0

z-plane

1

2

12

Note that the angles are to the tangents to the curves at z0 and w0. We seefrom (2.10) that

1 = Arg[f(z0)] + 1, 2 = Arg[f (z0)] + 2.


Therefore, if f(z) is analytic and f (z0) 6= 0, then

2 1 = 2 1,

and the included angle between two intersecting curves at z0 in the z-plane ispreserved both in magnitude and sense by the mapping w = f(z). Thus, we saythat the mapping is conformal. Note that the requirement for a mapping to beconformal is the same as that for it to be one-to-one.

2.1.4 Application of Conformal Mapping

As shown in section 1.4, the real and imaginary parts of an analytic complexfunction each satisfy Laplaces equation. The fact that many physical phenom-ena are governed by Laplaces equation allows us to use the mathematical powerof complex variable theory, including conformal mapping, to treat several ap-plications of practical interest. Here, we address the use of conformal mappingin conjunction with Laplaces equation.

If (x, y) is harmonic, i.e. satisfies Laplaces equation, in domain D in the z-plane, what is the governing equation in the image domain in the w-planeunder a conformal mapping?

C

y

f(z)

x u

v

z-plane w-plane

D

2 = 02 = 0

We could apply the transformation laws (2.2) and (2.3) directly to Laplacesequation for (x, y), i.e.

2

x2+2

y2= 0,

to obtain the transformed equation in terms of u and v, but this is very tedious(see, for example, Jeffrey pp. 906907).

A more concise derivation can be accomplished by using complex variabletheory. Because (x, y) is harmonic, it along with its harmonic conjugate(x, y), may be regarded as the real and imaginary parts of a complex function

(z) = (x, y) + i(x, y), (2.11)


which is analytic in D. If the mapping w = f(z) and its inverse z = F (w) areanalytic, then the complex function (z) is

(F (w)) = (w),

which is an analytic function of an analytic function and as a result is itself ananalytic function. Because (w) is analytic in , its real and imaginary partsare harmonic functions and satisfy Laplaces equation

(w) = (u, v) + i(u, v). (2.12)

Equating real and imaginary parts of (2.11) and (2.12), we see that

(x, y) = (u(x, y), v(x, y)) , (x, y) = (u(x, y), v(x, y)) .

Therefore, under a conformal mapping, the solutions in both planes are har-monic functions and satisfy Laplaces equation. The Laplace equation is uniquein this regard as transforming differential equations typically results in an equa-tion of the same type, but far more complicated. This is the trade-off for trans-forming a complex geometry to a simpler one, i.e. the complexity moves fromthe geometry to the equation itself. This is not the case for Laplaces equationunder conformal mappings owing to the properties of analytic complex func-tions.

Example: (Adapted from Wunsch (1994), pp. 535537)

Determine the temperature distribution due to heat conduction between twoisothermal circles that pass through the origin as shown below.

z1z3 z2

T = 0

D

x

T = 1

11/2

y

Recall from section 1.4 that steady heat conduction is governed by Laplacesequation

2T

x2+2T

y2= 0.

We seek the solution to Laplaces equation in the above, somewhat complicated,domain. To simplify the problem, we will use a conformal mapping to transform


the domain to a simple one.

Given that the domain is defined by circles in the z-plane, let us use a lin-ear fractional transformation to map the domain D into an infinite strip in thew-plane.

1

v

u

w3 =

w2 w1

T = 0 T = 1

The three specified points are mapped as follows:

z1 =1

2 w1 = 1,

z2 = 1 w2 = 0,z3 = 0 w3 = .

(2.13)

Substituting the points (2.13) into (2.1) gives

w 1w 0 =

(z 12 )(0 1)(z 1)(0 12 )

,

which simplifies to the linear fractional transformation

w = f(z) =1 zz

. (2.14)

You may verify that under this transformation, the boundaries are mapped asfollows z 14

= 14 u = 1,z 12 = 12 u = 0.

To confirm the map, select three points on each bounding circle in the z-plane,and show that they are collinear along the boundaries in the w-plane. Observethat the portion of the domain where the circles converge at the origin is mappedto + and .

Now we solve Laplaces equation

2T

u2+2T

v2= 0


in the domain subject to the boundary conditions

T = 0 at u = 0,

T = 1 at u = 1.

Observe that the domain shape and the boundary conditions do not change inthe v-direction. Therefore, the solution in the w-plane does not depend on v.Being one-dimensional, Laplaces equation simplifies to

d2T

du2= 0.

Integrating twice gives the linear solution

T (u, v) = Au+B.

Applying the boundary conditions requires that A = 1 and B = 0, which givesthe solution

T (u, v) = u. (2.15)

As you can see, obtaining the solution in the w-plane is greatly simplified ascompared to that in the z-plane. Although T (u, v) is only a function of u inthis case, we have indicated its dependence on v also in the more general case.

In order to transform the solution back to the z-plane, we need the real (u)and imaginary (v) parts of the mapping (2.14) in terms of x and y. In Carte-sian coordinates

w =1 zz

=1 (x+ iy)x+ iy

=x iy (x2 + y2)

x2 + y2.

Then

u(x, y) = Re(w) =x (x2 + y2)x2 + y2

=x

x2 + y2 1,

v(x, y) = Im(w) = yx2 + y2

.

Substituting into the solution (2.15) gives

T (x, y) = T (u(x, y), v(x, y)) =x

x2 + y2 1.

This is the solution for the temperature in the original domain D.


Plotting the isotherms, i.e. lines of constant temperature:

0 0.2 0.4 0.6 0.8 10.6

0.4

0.2

0

0.2

0.4

0.6

T = 1

T = 0

Recall that the harmonic conjugate of T (x, y) gives the heat flux lines, whichare perpendicular to the isotherms.

Note that the mapping in the above example, and the corresponding domain, is not unique. That is, other mappings may also work.

2.2 Solutions to Canonical Dirichlet Problems

If one can map a Dirichlet problem (Laplaces equation with fixed boundaryconditions) to the upper half plane or the interior of a circle, then a solutionmay be obtained using Poissons integral formula. These solutions are discussedin the following sections.

2.2.1 Dirichlet Problem in a Circular Disk

We seek the solution to the Laplace equation in a circular disk of radius Rcentered at the origin:


A point in the interior is denoted by z = rei, and a point on the boundary(r = R) by Rei. Laplaces equation in cylindrical coordinates is

2U

r2+

1

r

U

r+

1

r22U

2= 0, (2.16)

with the boundary conditions

U(r, ) = U(R,) = f() on r = R. (2.17)

As in section 3.6.1 of Matrices, we seek a solution to this Dirichlet problemusing the method of separation of variables. Thus, we write the solution as

U(r, ) = P (r)Q(). (2.18)

Substituting into (2.16) and collecting terms leads to

r2

P

d2P

dr2+r

P

dP

dr= 1

Q

d2Q

d2= 2.

As the left-hand-side only depends on r, and the right-hand-side only on ,they must each equal the same constant, say 2 to ensure that it is positive inorder to obtain Fourier series (otherwise get hyperbolic sine and cosine). Thus,equation (2.16) is separable, and we have reduced the single partial differentialequation into two ordinary differential equations, i.e. eigenproblems

r2d2P

dr2+ r

dP

dr 2P = 0, (2.19)

d2Q

d2+ 2Q = 0. (2.20)


Equation (2.19) is of the CauchyEuler form, which has solutions of the formP (r) = rm. Substituting requires that

m(m 1) +m 2 = 0,m2 2 = 0;

therefore,m = , 6= 0,

and the solution to (2.19) is

P (r) = Cr + Dr, 6= 0. (2.21)Let us now consider the case when = 0 in equation (2.19). The equationreduces to

rd2P

dr2+dP

dr= 0,

which may be solved using reduction of order to obtain

P (r) = A+ B ln r, = 0. (2.22)

Equation (2.20) for Q() is linear with constant coefficients and has the solutions

Q() = E + F , = 0, (2.23)

Q() = G cos() + H sin(), 6= 0. (2.24)From superposition, the most general solution (2.18) to (2.16) is then

U(r, ) = (A+B ln r)(E+F )+(Cr+Dr)[G cos() + H sin()

]. (2.25)

The solution must be bounded as r 0, which requires that B = 0 and D = 0(for > 0). Our solution (2.25) then reduces to

U(r, ) = E + F + r [G cos() +H sin()] . (2.26)

Note that to be single-valued, the solution must be 2pi-periodic in , requiringthat

U(r, + 2pi) = U(r, ).

Observe that this is not the case for the F term; thus, F = 0. For the cosineterm to be 2pi-periodic,

cos[( + 2pi)] = cos(),

orcos() cos(2pi) sin() sin(2pi) = cos().

Matching coefficients of the cos() and sin() terms requires that

cos(2pi) = 1 = 1, 2, 3, . . . ,


and

sin(2pi) = 0 = 12, 1,

3

2, 2, . . . .

Because both conditions must be satisfied, 2pi-periodicity requires that the eigen-values be

= n = 1, 2, 3, . . . . (2.27)

Consideration of 2pi-periodicity of the sin() term in (2.26) produces the samerequirement.

Thus far our eigenfunction solution is

U(r, ) = E +

n=1

rn[Gn cos(n) +Hn sin(n)], n = 1, 2, 3, . . . , (2.28)

and it remains to apply the boundary condition at r = R. Applying (2.28) atr = R ( = ) with the boundary condition (2.17) gives

U(R,) = f() = E +

n=1

Rn[Gn cos(n) +Hn sin(n)]. (2.29)

Observe that because Rn is a constant, (2.29) is simply the Fourier series ex-pansion of the boundary condition (see section 3.1 of Matrices) with

E =1

2pi

2pi0

f()d,

Gn =1

piRn

2pi0

f() cos(n)d, (2.30)

Hn =1

piRn

2pi0

f() sin(n)d.

For a given boundary condition f(), the constants (2.30) may be determinedand substituted into (2.28) to obtain the solution throughout the interior. Nor-mally, when using the method of separation of variables, we are content with asolution of this form, i.e. expressed as a Fourier series. In this particular case,however, we can obtain a closed form solution by summing the series expansionas follows.

Substituting (2.30) into (2.28) and switching the order of the summationand integration gives

U(r, ) =1

pi

2pi0

f()

{1

2+

n=1

( rR

)n [cos(n) cos(n) + sin(n) sin(n)

]}d

=1

pi

2pi0

f()

{1

2+

n=1

( rR

)ncos[n( )]

}d. (2.31)


Consider the summation and rewrite in the form

n=1

( rR

)ncos[n( )] = Re

{ n=1

( rR

)nein()

}= Re

{ n=1

[ rRei()

]n}.

Now recall the binomial expansion

1

1 x =n=0

xn = 1 +

n=1

xn, |x| < 1;

therefore, we may writen=1

xn = 1 + 11 x.

Applying this result to the above summation gives

n=1

[ rRei()

]n= 1 + 1

1 rRei()

= 1 + R2 Rr[cos( ) + i sin( )]R2 + r2 2Rr cos( ) .

Taking the real part and substituting into (2.31) leads to the Poisson integral formulafor the solution to the Dirichlet problem in a circular disk of radius R

U(r, ) =R2 r2

2pi

2pi0

U(R,)1

R2 + r2 2Rr cos( )d, (2.32)

where U(R,) = f() are the values of U(r, ) on the circular boundary atr = R.

Remarks:

1. A more concise derivation of the Poisson integral formula (2.32) may befound in section 3.3 based on the Cauchy integral formula. It may also beobtained using Greens functions (Asmar, section 6.5).

2. The Poisson integral formula allows us to obtain U(r, ) at all points inthe interior of the circle from the specified values U(R,), which must bepiece-wise continuous, on the surface.

3. Except for simple forms of U(R,) = f(), equation (2.32) must be eval-uated numerically. Alternatively, the solution may be found in terms of aFourier series given by equation (2.28) with (2.30).

4. The following can be shown to hold for any Dirichlet problem:

(a) Gausss mean value theorem: If C is a circle that is completely withina region R in which U(z) is analytic, i.e. satisfies Laplaces equation,then U(z) at the center of the circle is the average of the values ofU(z) along C.


(b) Maximum and minimum modulus theorems: Let U(z) be continuousand nonconstant throughout a closed bounded regionR and analytic,i.e. satisfies Laplaces equation, within the interior of R. Then themaximum value of |U(z)| occurs on the boundary of R. If U(z) isnowhere zero in R, then the minimum value of |U(z)| must also occuron the boundary.

For example, the temperature in the interior of a heat conduction problemis bounded by the minimum and maximum temperatures on the boundary.

2.2.2 Dirichlet Problem in the Upper Half Plane

Here, we obtain the solution to the Dirichlet problem in the upper half plane bytransforming the solution for the circular disk of radiusR using a conformal map.

Let us rewrite the Poisson integral formula (2.32) for the circular disk in theform

U(z) =1

2pi

2pi0

U(Rei)R2 r2

|Rei rei|2 d,

or for the unit disk with R = 1 (and z = rei)

U(z) =1

2pi

2pi0

U(ei)1 |z|2|ei z|2 d. (2.33)

We seek to map the unit circle in the z-plane to the upper-half-plane in thew-plane.

The (inverse) fractional transformation that maps w1, w2, and w3 to z1, z2, andz3 is

z = F (w) =w iw + i

. (2.34)

We take to be the image on the w-planes real axis of a point on the unit circlein the z-plane, which are the boundaries of the respective domains; therefore,

ei = F () = i + i

. (2.35)


Note that the limits of integration imply that 0 2pi in (2.33); however,this range is arbitrary, and we take it to be pi pi to correspond to in the w-plane according to the mapping (2.34).

To transform (2.33) to the w-plane, consider that:

i) The unit circle ei in the z-plane maps to the real axis in the w-plane;therefore,

U(ei) = U [F ()] = U(, 0). (2.36)

ii)

1 |z|2 = 1w iw + i

2= 1

(u+ iv) i(u+ iv) + i (u iv) i)(u iv) i2

=4v

u2 + (1 + v)2, (2.37)

where we have used the fact that

| |2 = [Re()]2 + [Im()]2.iii)

|ei z|2 = |F () F (w)|2 = i + i w iw + i

2= 4

(u )2 + v2(1 + 2)[u2 + (1 + v)2]

. (2.38)

iv) Differentiating (2.35)

ieid =

[1

+ i i

( + i)2

]d,

i i + i

d =2i

( + i)2d,

d = 22 + 1

d. (2.39)

Substituting (2.36)(2.39) into (2.33) gives

U(u, v) =1

2pi

U(, 0)4v

u2 + (1 + v)21

4

(1 + 2)[u2 + (1 + v)2]

( u)2 + v22

1 + 2d,

U(u, v) =v

pi

U(, 0)1

( u)2 + v2 d, v > 0, (2.40)which is the Poisson integral formula for the Dirichlet problem in the upper halfplane with U(, 0) being the values on the boundary v = 0.

Remarks:


1) Equation (2.40) may be obtained directly using Fourier transforms (As-mar, section 11.5) or Greens functions (Asmar, section 6.5).

2) To write (2.40) in the z-plane, exchange (u, v) (x, y).

Example: (Adapted from McQuarrie 2003)

Consider the electrostatic potential E(x, y) in the semi-infinite strip shown be-low

with the constant boundary conditions

E = E0 at x = 1,E = E1 at y = 0,

E = E2 at x = 1.

(2.41)

The governing equation for the electrostatic potential in a homogeneous andisotropic medium is the Laplace equation

2E

x2+2E

y2= 0. (2.42)

The semi-infinite vertical strip 1 x 1, 0 y


The governing equation in the w-plane is

2E

u2+2E

v2= 0, (2.44)

with the boundary conditions

E = E0 for u < 1, v = 0 ( < 1),E = E1 for 1 < u < 1, v = 0 (1 1),E = E2 for u > 1, v = 0 ( > 1).

(2.45)

From the Poisson integral formula (2.40), the solution in the upper half planeis

E(u, v) =v

pi

E(, 0)1

( u)2 + v2 d, v > 0, (2.46)

where E(, 0) are the values on the boundary v = 0 given by (2.45). BecauseE(, 0) is constant along each portion of the boundary, and b

a

d

( u)2 + v2 =[

1

vtan1

( uv

)]ba

,

equation (2.46) leads to the solution in the w-plane

E(u, v) =1

pi

{E0

[tan1

( uv

)]1

+ E1

[tan1

( uv

)]11

+E2

[tan1

( uv

)]1

}=

1

pi

{E0

[tan1

(1 uv

) tan1

( uv

)]+E1

[tan1

(1 uv

) tan1

(1 uv

)]+ E2

[tan1

( uv

) tan1

(1 uv

)]},

E(u, v) =E1 E0

pitan1

(u+ 1

v

)+E2 E1

pitan1

(u 1v

)+E0 + E2

2.

(2.47)In carrying out the last step, note that tan1(x) = tan1 x, tan1() =pi2 , and tan1() = pi2 . In order to transform the solution back to the physical


z-plane, recall the mapping (2.43)

w = sin(pi

2z)

= sin[pi

2(x+ iy)

]= sin

(pi2x)

cos(pi

2iy)

+ cos(pi

2x)

sin(pi

2iy)

w = sin(pi

2x)

cosh(pi

2y)

+ i cos(pi

2x)

sinh(pi

2y).

Thus,

u = Re[w] = sin(pi

2x)

cosh(pi

2y),

v = Im[w] = cos(pi

2x)

sinh(pi

2y),

(2.48)

which when substituted into (2.47) gives the solution for the electrostatic po-tential E(x, y).

Problem Set # 7

2.3 Application to Fluid Flows

As illustrated in the previous two examples, along with section 1.4, potentialtheory may be applied to steady heat conduction and electrostatics. In thissection we provide a more complete treatment of potential theory as appliedto ideal fluid flow with the understanding that these same techniques may beapplied to any context in which Laplaces equation is the governing equation.

2.3.1 Potential Flow Formulation

For two-dimensional incompressible flow, the divergence of the velocity field iszero, i.e.

V = vxx

+vyy

= 0. (2.49)

This is called the continuity equation and enforces conservation of mass. Here,the velocity vector is defined by V = vx(x, y)i + vy(x, y)j, where vx(x, y) andvy(x, y) are the streamwise and normal velocity components in the x and y di-rections, respectively.

Defining the streamfunction (x, y) by

vx =

y, vy =

x, (2.50)


the continuity equation (2.49) is exactly satisfied (substitute (2.50) into (2.49)).That is, defining the streamfunction in this manner effectively replaces the con-tinuity equation. Streamlines are contours of constant streamfunction. Considerthe total differential of the streamfunction (x, y) along a streamline

d =

xdx+

ydy = 0;

therefore, the slope of the streamline is

dy

dx=/x/y

=vyvx,

which is the direction of the velocity vector. Thus, the streamlines of a flowfield are everywhere tangent to the velocity vector, thereby indicating local flowdirection.

y

v

u

v

=

=

v

1

2

For any two-dimensional flow, the vorticity (x, y) is defined by the curl of thevelocity field

= V = vyx vx

y, (2.51)

which is a measure of the rotation rate of the fluid particles. For irrotationalflow

= V = 0,in which case there is a scalar function (x, y) for which

() = 0.

This is a vector identity, i.e. the curl of the gradient of any scalar function iszero. Therefore, from the last two equations, we can write the velocity vectoras the gradient of a scalar function (x, y) according to

V = ,

vxi+ vyj =

xi+

yj,


and

vx =

x, vy =

y. (2.52)

The scalar function (x, y) is called the velocity potential. Substituting into thecontinuity equation (2.49) gives

2

x2+2

y2= 0,

and the velocity potential satisfies the Laplace equation. Substituting the defi-nitions of the streamfunction (2.50) into that of the vorticity (2.51) gives

2

x2+2

y2= .

This is known as a Poisson equation. If the flow is irrotational, then (x, y)satisfies Laplaces equation (irrotational flow is typically, but not always, invis-cid, i.e. neglect friction). Therefore, both the velocity potential (x, y) and thestreamfunction (x, y) are harmonic functions.

We call two-dimensional, incompressible, inviscid and irrotational flow idealflow or potential flow. Potential flow is in common use today for flows in whichviscous, i.e. friction, effects are negligible or in regions of flow fields away fromareas where viscous effects are important, such as near solid surfaces.

Because the streamfunction and velocity potential are harmonic functionsand because they turn out to be harmonic conjugates, we may define the complexpotential function by

(z) = (x, y) + i(x, y),

where (z) is an analytic function of z = x + iy. Note that from (2.50) and(2.52), the Cauchy-Riemann equations are (u , v )

vx =

x=

y, vy =

y=

x.

Taking the derivative of (z), we have from section 1.3

d

dz=

y i

y=

x+ i

x.

Therefore,d

dz= vx(x, y) ivy(x, y) = W (z),

which is the complex conjugate of the complex velocity W (z) = vx(x, y) +ivy(x, y).


Remarks:

1) Recall from section 1.3 that lines of (x, y) = constant and (x, y) =constant are orthogonal at every point where they intersect. Therefore,the velocity potential drives the fluid along the streamlines from higher tolower potential.

1

1

2

2

3

3

equipotential lines

streamlines

2) Any streamline, i.e. line of constant (x, y), may be considered a solidsurface in potential flow (in viscous flow, all solid surfaces are streamlines,but not all streamlines may be considered a solid surface).

3) Stagnation points in potential flow occur where the velocity V = 0, i.e.W = ddz = 0.

4) In polar coordinates, the velocity vector is V = vr(r, )er + v(r, )e,where vr(r, ) and v(r, ) are the velocity components in the r and di-rections, respectively. The velocity components are related to the velocitypotential and streamfunction as follows:

vr =

r=

1

r

,

v =1

r

=

r,

which are the Cauchy-Reimann equations in polar coordinates. Laplacesequation in polar coordinates is

2 = 1r

r

(r

r

)+

1

r22

2= 0.


The complex conjugate of the complex velocity is

W (z) =d

dz= [vr(r, ) iv(r, )] ei,

where we note the ei factor.

5) Applications involving Laplaces equation and the role of (x, y) and(x, y) are summarized in the following table:

Application: (x, y) = constant (x, y) = constant

Heat conduction Isotherms Heat flux lines

Electrostatics Electrostatic potential Current flow lines

Potential flow Velocity potential Streamlines

Elasticity Strain function Stress lines

Gravitational fields Gravitational potential Force lines

Magnetism Magnetic potential Force lines

2.3.2 Basic Flows

We can define several basic flows by specifying the complex potential. In thenext section, we will show how to treat more complex flows by combining thefollowing basic flows:

1) Uniform Flow:

The complex potential for a uniform flow of speed U forming an angle with the positive x-axis is given by

(z) = Ueiz.

Given the complex potential, there are three ways to obtain the velocitycomponents: 1) from the velocity potential , 2) from the streamfunction, or 3) from the complex velocity W . For example, the complex conjugateof the complex velocity for uniform flow is

W =d

dz= Uei = U(cos i sin) = vx ivy,

and the velocity components are

vx = U cos, vy = U sin.


U

1

2

3y

x

2) Sources and Sinks:

The complex potential is

(z) =Q

2pilog z,

where Q is the strength (the volumetric flow rate) of the source or sink.

Q > 0 Source: Q < 0 Sink:y

x

y

x

In polar coordinates, the complex potential is

(z) =Q

2pilog(rei) =

Q

2pi(log r + i).

Evaluating the real and imaginary parts gives the velocity potential andthe streamfunction

= Re[] =Q

2pilog r, = Im[] =

Q

2pi,

respectively. The velocity components are then

vr =

r=

1

r

=

Q

2pir,

v =1

r

=

r= 0,


from the definitions in polar coordinates.

3) Point Vortex:

The complex potential is

(z) = i 2pi

log z,

where is the strength (the circulation) of the vortex, and > 0 corre-sponds to a vortex with counterclockwise rotation and < 0 correspondsto one with clockwise rotation.

y

x

Note that the presence of i changes the role of the real and imaginary partsas compared to the source/sink flow, thereby switching the equipotentialand streamlines.

4) Doublet:

A doublet is obtained by placing a source and a sink on the real axisa distance apart and taking 0 in a particular manner. The resultingvelocity potential is

=S

z,

where S is the strength of the doublet.


y

x

5) Corner and Wedge Flows:

The flow through and over corners and past wedges may be obtainedusing the complex potential

(z) =A

nzn, n 1

2,

where the geometry is determined by the choice of n. To see how n ischosen, let us obtain the velocity potential and streamfunction from thecomplex potential.

(z) =A

nzn

=A

n

(rei

)n=

A

nrnein

(z) =A

nrn [cos(n) + i sin(n)] .

Therefore, the velocity potential and streamfunction are

(r, ) =A

nrn cos(n),

(r, ) =A

nrn sin(n).

If we define surfaces as being locations where the streamfunction is zero,this occurs when sin(n0) = 0. Therefore, 0 = kpi/n, k = 0, 1, 2, . . .,which are rays from the origin separated by an angle pi/n. For example,n = 2 corresponds to the flow through a right-angle corner.


n > 1 => concave corner or wedge

n < 1 => convex corner

Let us also obtain the velocity components using the complex conjugateof the complex velocity, which is

W =d

dz

= Azn1

= A(rei

)n1= Arn1einei

W = Arn1 [cos(n) + i sin(n)] ei.

Thus, the velocity components are

vr(r, ) = Arn1 cos(n),

v(r, ) = Arn1 sin(n).Note that v = 0 along the surfaces where = 0, i.e. there is no flownormal to the streamlines. Also observe that if n < 1, i.e. the corner isconvex, then the velocity is singular at the corner r = 0.

Note: All of the above basic flows have been defined as being centered at theorigin. In order to center each at z0, substitute (z z0) for z in the expressionsfor the complex potential.

2.3.3 Superposition of Basic Flows

Because the governing equations for potential flow are linear, we can superim-pose the basic flows to obtain more complex flows

(z) = 1(z) + 2(z) + .

Example: Consider the flow past a circular cylinder. The complex potential is


the superposition of a uniform flow (with = 0) and a doublet according to

(z) = Uz +S

z

= Urei +S

rei

= Ur(cos + i sin ) +S

r(cos i sin )

(z) =

(Ur +

S

r

)cos + i

(Ur S

r

)sin .

Therefore, the velocity potential and streamfunction are

(r, ) = Re[] =

(Ur +

S

r

)cos ,

(r, ) = Im[] =

(Ur S

r

)sin .

Let us define a cylinder of radius R =S/U , and check if it is a streamline:

(R, ) =

(UR S

R

)sin

=

(U

S

U S

U

S

)sin

(R, ) = 0.

The streamfunction is constant on the cylinder of radius R; therefore, it is astreamline and may be considered a surface. Note that the fact that = 0along the cylinder is not important, i.e. any streamline may be regarded as asolid surface in potential flow.


A schematic plot of the streamlines appears as follows:

y

x

The actual streamlines are:

-2 -1 1 2

-1.5

-1

-0.5

0.5

1

1.5

cylinderstrmlines.nb 1

Note that there is a singularity at the origin, but this is not in the flow domain.

Remarks:

1) The flow around a rotating circular cylinder may be obtained by super-imposing a uniform flow, doublet and point vortex.

2) To model flows with a feature (e.g. vortex) above a plane surface, super-impose an image an equal distance below the surface. For example, for apoint vortex above a wall:


h

himagevortex

The symmetry line, which is a streamline, is considered a wall.

3) Any airfoil shape can be modeled by a series of sources, sinks and vortices.

2.3.4 The Use of Conformal Mapping

Using conformal mapping, we can treat more complicated geometries by trans-forming them into flows for which we can obtain the solution.

Physical Plane: f(z) Transformed Plane:

z = x+ iy F (w) w = u+ iv

(z) = (x, y) + i(x, y) (w) = (u, v) + i(u, v)

2

x2+2

y2= 0

2

u2+2

v2= 0

2

x2+2

y2= 0

2

u2+2

v2= 0

Example: Consider an alternative method of obtaining the solution for theflow past a circular cylinder using conformal mapping.

The mapping

w = f(z) = z +r20z


transforms the flow past a cylinder of radius r0 into the flow over a flat platealigned in the flow direction:

z-plane: w-plane:

y

r0

x

v

u

f(z)

F(w)

w - planez- plane

U

The solution for the potential flow past a flat plate aligned with the flow issimply uniform flow, for which the complex potential is

(w) = Uw.

Mapping this solution back to the z-plane, we have

(z) = U

(z +

r20z

),

which is the same complex potential as before, i.e. uniform flow plus a doublet(with S = Ur20). Then we proceed as before to obtain (r, ), (r, ), vr(r, )and v(r, ).

Consider the complex conjugate of the complex velocities in the z-plane andthe w-plane. In the z-plane

W (z) =d

dz,

and in the w-plane

W (w) =d

dw.

But from the chain rule

d

dz=d

dw

dw

dz=d

dwf (z);

therefore,

W (z) = W (w)f (z).

That is, the magnitudes of the velocities differ by a factor of f (z) in the z-and w-planes, while the values of and are the same at image points, i.e.


(z) = (w). From the above result, we see that the critical points of themapping, i.e. where f (z) = 0, are stagnation points in the physical domain (thez-plane), i.e. W (z) = 0. Note that streamlines may branch at stagnation points.

Example: Find the stagnation points for the flow about a circular cylinder.

The critical points of the transformation used in the previous example are ob-tained from

f (z) = 1 r20

z2=z2 r20z2

=(z + r0)(z r0)

z2= 0.

Therefore, the critical points of the mapping, i.e. the stagnation points of thepotential flow, are at z = r0, which are at the front and back of the cylinderalong the symmetry line. In addition, observe that the mapping has a singular-ity at z = 0 as discussed previously.

Finally, we introduce two useful general purpose mappings:

1) Schwarz-Christoffel Transformation:

The Schwarz-Christoffel transformation maps the interior of a polygonwith n vertices in the z-plane to the upper half of the w-plane.

w-plane: z-plane:

u1 u2 uu

v

n

z = F(w) y

2

pi

11

3z

2z

1z

x

The inverse transformation z = F (w) is

z = A

(w u1)k1(w u2)k2 (w un)kndw +B,

where ki =ipi 1.

2) Joukowski Transformation:

In the Joukowski transformation, the roles of z and w are switched fromthat used to obtain the flow around a cylinder. The transformation fromthe w-plane to the z-plane is

z = F (w) = w +c2

w,


or from the z-plane to the w-plane

w = f(z) =1

2z

[(z2

)2 c2

]1/2,

where c is a real constant. By choosing appropriate values of c, thistransformation may be used to map:

z-plane: w-plane:Flat Plate

Ellipse f(z) Circular CylinderJoukowski Airfoil

Chapter 3

Integration in the ComplexPlane and Residue Theory

3.1 Line Integrals of Complex Functions

Recall from calculus the three types of integrals of real functions:

1) Indefinite integrals:

F (x) =

f(x)dx dF (x)

dx= f(x).

2) Definite integrals:

I =

ba

f(x)dx.

x

a b

f(x)

63

CHAPTER 3. COMPLEX INTEGRATION AND RESIDUE THEORY 64

3) Line integrals: C

F dr.

x

y

(x0,y0)

(x1,y1)

C

Recall that for the real line integral, if

F = P (x, y)i+Q(x, y)j,

andr = xi+ yj,

then C

F dr =C

(Pdx+Qdy).

This is an exact differential, i.e. it is independent of the path C, if

Pdx+Qdy = d =

xdx+

ydy,

where (x, y) is a scalar function. Thus,

P =

x, Q =

y.

Taking /y of the first equation and /x of the second, we have

P

y=

2

xy,

Q

x=

2

xy.

Equating givesP

y=Q

x. (3.1)

Because Pdx+Qdy may be written as the differential of a scalar function,C

F dr = (x1,y1)

(x0,y0)

d = (x1, y1) (x0, y0),

and the integral only depends upon the end points.


Now consider the integral of a complex function f(z) = u(x, y) + iv(x, y) alonga piecewise smooth contour in the complex plane. Being piecewise smooth re-quires that the contour is continuous over each portion (or piece).

x

y

Z0

Z1

C

Determine the real and imaginary partsC

f(z)dz =

C

(u+ iv)(dx+ idy) =

C

(udx vdy) + iC

(vdx+ udy). (3.2)

Therefore, the real and imaginary parts are real line integrals.

In order for the real part to be an exact differential (P = u,Q = v in (3.1))u

y= v

x. (3.3)

Similarly, for the imaginary part (P = v,Q = u)

v

y=u

x. (3.4)

But (3.3) and (3.4) are the Cauchy-Riemann equations.

Thus, forCf(z)dz to be independent of the path, the countour C must be

in a region R where f(z) is analytic, i.e. satisfies the Cauchy-Riemann equa-tions.

x

y

R

c1

c2

c3

z0

z1


C

f(z)dz =

z1z0

f(z)dz = F (z1) F (z0), (3.5)

where f(z)dz = dF (z). Therefore, f(z)dz is the exact differential of a functionF (z), and

dF (z)

dz= f(z).

Remarks:

1) We may specify any C in R from z0 to z1 for convenience in performingthe integration. This is called contour integration.

2) If z1 = z0, i.e. C is a closed contour, and f(z) is analytic inside and on C,then

C

f(z)dz = 0. (3.6)

which is Cauchys integral theorem, often referred to as the Cauchy-Goursattheorem. Note that the positive direction is counterclockwise.

x

y

C

Example: Evaluate the integral 1+2i0+i

z2dz

along the two contours:

Contour (a):

y = x2 + 1 (x =y 1).

x

y

z0=0+i

z1=1+2i

1

2


Contour (b):

x

y

z1

z01

2

II

I

Note that f(z) = z2 is not analytic; therefore, we must perform the integralsalong the paths given.

First, consider contour (a). Find the real and imaginary parts of f(z) = z2.

z2 = (x iy)(x iy) = x2 y2 2ixy

u = x2 y2, v = 2xy.From (3.2)

C

f(z)dz =

C

(udx vdy) + iC

(vdx+ udy)

=

(1,2)(0,1)

(x2 y2)dx+ (1,2)

(0,1)

2xydy

i (1,2)

(0,1)

2xydx+ i

(1,2)(0,1)

(x2 y2)dy.


Substituting the contours y = x2 + 1 or x =y 1 as appropriate (1,2)

(0,1)

z2dz =

10

[x2 (x2 + 1)2]dx+ 2 2

1

yy 1dy

2i 1

0

x(x2 + 1)dx+ i

21

(y 1 y2)dy

=

10

(x4 x2 1)dx+ 2 2

1

yy 1dy

2i 1

0

(x3 + x)dx+ i

21

(y2 + y 1)dy

=

[1

5x5 1

3x3 x

]10

+ 2

[2(3y + 2)

15

(y 1)3]

]21

2i[

1

4x4 +

1

2x2]1

0

+ i

[1

3y3 +

1

2y2 y

]21

= 2315

+32

15+ i

(3

2 11

6

) (1,2)

(0,1)

z2dz =3

5 i10

3.

Now consider contour (b).

Path I: y = 1 uI = x2 1, vI = 2x, dy = 0.Path II: x = 1 uII = 1 y2, vII = 2y, dx = 0.

Then from (3.2) (1,2)(0,1)

z2dz =

I

uIdx+ i

I

vIdxII

vIIdy + i

II

uIIdy

=

10

(x2 1)dx i 1

0

2xdx+

21

2ydy + i

21

(1 y2)dy

=

[1

3x3 x

]10

i [x2]10

+[y2]21

+ i

[y 1

3y3]2

1

= 23

+ 3 + i

(1 4

3

) (1,2)

(0,1)

z2dz =7

3 i7

3.

As expected, because f(z) = z2 is not analytic, integrating along different pathsresults in different values of the contour integral.


Example: Evaluate the integral C

dz

z,

where C is the straight line between z0 = 1 and z1 = i.

x

y

z1=i

z0=10

C

Note that f(z) = 1/z is analytic in any region that does not contain the origin,i.e.

f (z) = 1z2

exists everywhere except z = 0.

Therefore, integrating along any path between z0 and z1 that is within a simply-connected region (a closed curve that does not intersect itself) containing C butnot z = 0 will produce the same result.

Because the integral is independent of the path, it only depends on the endpoints; thus,

C

dz

z= [log z]i1

= log i log 1C

dz

z= log |1|+ i

(pi2

+ 2kpi), k = 0,1,2, . . . .

Taking the principle branch (k = 0), we have z1z0

dz

z=pi

2i.

Let us imagine that we did not know this, and instead evaluate the integralusing a particular contour.

A convenient contour is the first quadrant of the unit circle.


x

y

z1=i

z0=10

C1

Note that along C1

z = ei, 0 pi2,

anddz = ieid.

Thus, C1

dz

z=

pi2

0

ei(ieid)

=

pi2

0

id

= i

pi20

C1

dz

z=

pi

2i,

which is the same as above and would be equal to the integral along the straightline C.

Now consider C

dz

z,

where C is a closed contour. If C does not surround the origin, where f(z) = 1/zis not analytic, then from Cauchys integral theorem

C

dz

z= 0.

What if C does surround the origin? Let us consider the entire unit circle.C

dz

z=

2pi0

ei(ieid) = i2pi0

= 2pii. (3.7)

To show that this result is true for any simply connected closed contour sur-rounding the origin, consider another closed contour C1 that surrounds the


origin but does not intersect C:

0 1

C

C1

R

y

x

C

C1 may alsobe inside C

Note that C1 may also be inside C. In order to use Cauchys theorem, make acut between C and C1 to form a simply connected region R. Then f(z) = 1/z isanalytic in R, and from Cauchys theorem the integral along the closed contoursurrounding R equals zero.

As the gap C 0, the integrations from C1 C and C C1 cancel asthey are in opposite directions, and

C1

dz

zC

dz

z= 0.

The negative sign in the second term is because the integration along C is inthe clockwise, or negative, direction. Therefore, from (3.7)

C1

dz

z=

C

dz

z= 2pii.

Remarks:

1) This can be shown to be true even if C and C1 intersect.

C1 is any curve enclosing the origin, i.e. the point where f(z) = 1/zis not analytic.

2) In general, for f(z) = zn, where n is an integer (n 6= 1), integrating


around the unit circle gives|z|=1

zndz =

2pi0

ein(ieid)

= i

2pi0

ei(n+1)d

= i

2pi0

[cos(n+ 1) + i sin(n+ 1)]d

=i

n+ 1[sin(n+ 1) i cos(n+ 1)]2pi0

C

zndz = 0 (n 6= 1),

where C may be any closed contour surrounding the origin (by the previ-ous argument). This is true even though f(z) is not analytic at z = 0 forn negative.

Summarizing: C

zndz =

{0, n 6= 12pii n = 1

. (3.8)

3) Replacing z by z z0 in (3.8), we haveC

(z z0)ndz ={

0, n 6= 12pii n = 1

, (3.9)

where C encloses z = z0.

x

y

0

Z=Z0

This important result will be used extensively in subsequent sections.

4) Principle of Deformation of Contours:C1

f(z)dz =

C2

f(z)dz,

if f(z) is analytic on C1 and C2 and at all points crossed in deforming C1into C2.


x

y

0

C2

C1

f(z) analytical

3.2 Cauchys Integral Formula

We seek to determine the value of f(z) at a point z = z0 from f(z) along aclosed contour C surrounding z = z0:

z0

y

x

C

C

Note that:

f(z) is analytic on C and inside C, C is a small circle of radius centered at z = z0.

Thenf(z)

z z0

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