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Complex Integrals and the Residue Theorem

Date: 12/18/2000 at 03:02:18
From: Lydia Monroe
Subject: Complex Analysis: Integrals

To Dr. Math:

I would be grateful if you could help me solve this integral, which 
involves complex numbers:

     the integral over C of (z^2/((z-1)^2*(z+1)))dz

where C is the circle C = {z| |z - 2i| = 2}.

I thought maybe I could use the Taylor Series to solve this, but I'm 
not sure. If I wanted to integrate around the problem points using the 
Cauchy Integral Problem, how would I set the problem up? I'm confused 
by the (z-1)^2 part. If it weren't squared, I think I could figure it 
out, but do I integrate over three curves, one for each z-1 and one 
for z+1? I'm confused... It's a closed curve (I think) with 
counter-clockwise orientation.

Thank you for any help you can give.

Date: 12/19/2000 at 13:12:04
From: Doctor Mitteldorf
Subject: Re: Complex Analysis: Integrals

Dear Lydia,

The rule is to evaluate the function at each of the poles contained 
within the contour of integration, then multiply by 2pi*i.

The situation at z = -1 is straightforward. The function behaves like:

     (-2)^2 * (z+1)

so the residue is (-1)^2/(-2)^2 = 1/4.

The situation at z = 1 is a little more complicated. (z-1)^(-2) by 
itself would not produce any residue at all; however, there's a part 
of the behavior at z = 1 that actually behaves like (z-1)^(-1). To 
find this, we must expand the rest of the function in a Taylor series 
about z = 1, and find the part the term that is linear in (z-1). This, 
when multiplied by (z-1)^(-2), gives a term that behaves like 

So we need to evaluabe the first derivative of F(z)= z^2/(z+1) at the 
point z = 1. I calculate that:

     F'(z) = -------

so that F'(1)= -1/4. Hence the residue is +1/4, exactly canceling the 
residue from the simple pole at z = -1.

My calculation is that the integral is zero.

- Doctor Mitteldorf, The Math Forum   

Date: 12/20/2000 at 17:03:32
From: Lydia Monroe
Subject: Re: Complex Analysis: Integrals

I don't know about poles and residues (though I'm going to look them 
up now) and though I got your answer, it's a little confusing for me. 
I tried to work it out with 'brute force' and I got a different 
answer. I guess I must have made an error. (I got (-pi*i - pi) as the 
answer.) I used the Cauchy Integral theorem and integrated over the 
problem point i (-1 is not inside or on C).

I got that the integral of (z^2/(z+1))/(z-1)^2, which is the same as 
the original integral, is equal to 2pi*i (-1/(i+1)), which comes to 
2pi*i (-1/2+(1/2)i) or ((-pi*i) - pi). Where did I go wrong?  
I also did this same integral over c: {z| |z+1| = (1/2)} and got pi, 
and also over c: {z| |z| = 3}, which includes both problem points, 
so I just added the results from my two earlier integrals and got 
(- pi*i).

Thank you for your help.

Date: 12/20/2000 at 17:32:05
From: Doctor Mitteldorf
Subject: Re: Complex Analysis: Integrals

Lydia -

The residue theorem is really the heart of what makes these complex 
integrals fun and interesting. It also simplifies computation 
enormously, and makes many integrals computable that would otherwise 
be opaque. (There are even some real integrals that can most easily be 
calculated by thinking in terms of integrals over a contour including 
half a plane.)

Thanks for affording me the opportunity to review the subject and 
remember things I haven't thought about in many years. I'd be 
interested to see how you set up the "brute force" version of these 

- Doctor Mitteldorf, The Math Forum   

Date: 01/11/2001 at 02:06:27
From: Lydia Monroe
Subject: Complex Analysis: Integrals

To Dr. Mitteldorf -

I recently wrote you about another complex analysis problem that I was 
struggling with and you suggested that I use poles and residues to 
solve the problem. Consequently I have been studying that area of 
complex analysis on my own and wondered if you could tell me whether 
the answer I have calculated for one particular problem is correct.

The problem goes: Find the residue of (sin z)/z^4 at z = 0. 
Now (sin z)/z^4 is the same as (1/z^4)*(sin z), which can be written 
as the multiple of a series: (1/z^4)(z-((z^3)/3!)+...) and this is the 
same as 1/z^3 + higher terms. So I thought perhaps then the residue 
would be 1/z^2, but I'm very uncertain about residues and such.

I took a class in complex analysis last semester but it suddenly sped 
up at the end of the semester and I only remember my professor 
mentioning residues once. There is no second semester course on 
complex analysis at my school. Is this unusual? (It is a top-ranked 

Thank you for your time. This forum is of tremendous help.

Date: 01/11/2001 at 06:32:54
From: Doctor Mitteldorf
Subject: Re: Complex Analysis: Integrals

Dear Lydia,

The logic of residue calculus is based on the following chain of 

1) If a function is analytic everywhere, then the integral over any 
   path is 0.

2) Therefore you can distort any path integral, as long as you're 
   moving it only within a region where the function is analytic.

3) Suppose you have an integral of a function with no singularities 
   except isolated poles: you can use the above principle to change 
   your path integral into a sum of infinitesimal circular loops 
   around the poles.

4) You can evaluate these infinitesimal loop integrals around poles by 
   analyzing the function at the pole in terms of its Taylor series. 
   Use the fact that the loop integral of z^n is identically zero for 
   every value of n, positive or negative, except n = -1.

5) For n = 1 the value of the integral is 2pi*i. It follows that you 
   can evaluate the contribution from any pole by representing the 
   function at the pole as a Taylor series, ignoring all Taylor terms 
   except n = -1, and then multiplying the n = -1 coefficient by 

These ideas are strange and defy our intuition. Each step of the above 
requires some analysis in order to convince yourself it's true; in 
addition, it helps to go through some examples so that you're 
comfortable with the ideas. I encourage you to go through this process 
on your own if there isn't a teacher available. Perhaps you can find a 
text that will help. The text I was taught from years ago was Ahlfors: 
_Complex Analysis_.

In the case of your function (1/z^4)*(sin z), you've done the analysis 
correctly: write it as (1/z^4)(z-((z^3)/3!)+...). Then the only term 
that matters is the second term, which multiplies out to be (1/6)z^-1. 
The residue is 1/6, so the integral is 1/3 pi*i.

- Doctor Mitteldorf, The Math Forum   
Associated Topics:
College Analysis
College Imaginary/Complex Numbers

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