Complex_Analysis-Cain.pdf

Chapter Ten

Poles, Residues, and All That

10.1. Residues. A point z 0 is a singular point of a function f if f not analytic at z 0 , but is

analytic at some point of each neighborhood of z 0 . A singular point z 0 of f is said to be

isolated if there is a neighborhood of z 0 which contains no singular points of f save z 0 .In

other words, f is analytic on some region 0  | z  z 0 |  .

Examples

The function f given by

f  z  

z  z 2  4 

has isolated singular points at z  0, z  2 i , and z   2 i .

Every point on the negative real axis and the origin is a singular point of Log z , but there

are no isolated singular points.

Suppose now that z 0 is an isolated singular point of f . Then there is a Laurent series



f  z   j 

c j  z  z 0  j

valid for 0  | z  z 0 |  R , for some positive R . The coefficient c  1 of  z  z 0   1 is called the

residue of f at z 0 , and is frequently written

Res f .

Now, why do we care enough about c  1 to give it a special name? Well, observe that if C is

any positively oriented simple closed curve in 0  | z  z 0 |  R and which contains z 0

inside, then

c  1  1

2  i  C f  z  dz .

10.1

z  z 0

This provides the key to evaluating many complex integrals.

Example

We shall evaluate the integral

 C e 1/ z dz

where C is the circle | z |  1 with the usual positive orientation. Observe that the integrand

has an isolated singularity at z  0. We know then that the value of the integral is simply

2  i times the residue of e 1/ z at 0. Let’s find the Laurent series about 0. We already know

that



e z  j  0

j ! z j

for all z . Thus,



e 1/ z  j  0

j ! z  j  1  z  1

z 2 

The residue c  1  1, and so the value of the integral is simply 2  i .

Now suppose we have a function f which is analytic everywhere except for isolated

singularities, and let C be a simple closed curve (positively oriented) on which f is analytic.

Then there will be only a finite number of singularities of f inside C (why?). Call them z 1 ,

z 2 ,  , z n . For each k  1, 2,  , n , let C k be a positively oriented circle centered at z k and

with radius small enough to insure that it is inside C and has no other singular points inside

it.

10.2

Then,

 C f  z  dz   C 1

f  z  dz   C 2

f  z  dz   C n f  z  dz

 2  i

Res f  2  i

Res f  2  i

Res f

z  z n

 2  i k  1

Res f .

This is the celebrated Residue Theorem . It says that the integral of f is simply 2  i times

the sum of the residues at the singular points enclosed by the contour C .

Exercises

Evaluate the integrals. In each case, C is the positively oriented circle | z |  2.

1. 

e 1/ z 2 dz .

2. 

sin  z  dz .

3. 

cos  z  dz .

4. 

sin  z  dz .

5. 

z cos  z  dz .

10.3

z  z 1

z  z 2

z  z k

10.2. Poles and other singularities. In order for the Residue Theorem to be of much help

in evaluating integrals, there needs to be some better way of computing the

residue—finding the Laurent expansion about each isolated singular point is a chore. We

shall now see that in the case of a special but commonly occurring type of singularity the

residue is easy to find. Suppose z 0 is an isolated singularity of f and suppose that the

Laurent series of f at z 0 contains only a finite number of terms involving negative powers

of z  z 0 . Thus,

f  z  

 z  z 0  n 

 z  z 0  n  1  c  1

 z  z 0   c 0  c 1  z  z 0  .

Multiply this expression by  z  z 0  n

  z    z  z 0  n f  z   c  n  c  n  1  z  z 0  c  1  z  z 0  n  1  .

What we see is the Taylor series at z 0 for the function   z    z  z 0  n f  z  . The coefficient

of  z  z 0  n  1 is what we seek, and we know that this is

  n  1   z 0 

 n  1  !

The sought after residue c  1 is thus

c  1  z  z 0

Res f    n  1   z 0 

 n  1  !

where   z    z  z 0  n f  z  .

Example

We shall find all the residues of the function

f  z  

z 2  z 2  1  .

e z

First, observe that f has isolated singularities at 0, and  i . Let’s see about the residue at 0.

Here we have

10.4

c  n  1

c  n

  z   z 2 f  z  

 z 2  1  .

e z

The residue is simply    0  :

   z    z 2  1  e z  2 ze z

 z 2  1  2

Hence,

Res f     0   1.

z  0

Next, let’s see what we have at z  i :

  z    z  i  f  z  

z 2  z  i  ,

e z

and so

Res f  z     i    e i

2 i

z  i

In the same way, we see that

Res f  e  i

2 i

z  i

Let’s find the integral 

z 2  z 2  1  dz , where C is the contour pictured:

10.5

e z

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