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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.

```

```
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:

(-1)^2
--------------
(-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
(z-1)^(-1).

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

z^2-2z
F'(z) = -------
(z+1)^2

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
http://mathforum.org/dr.math/
```

```
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).

```

```
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
integrals.

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```

```
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
school.)

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
reasoning:

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
2pi*i.

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
http://mathforum.org/dr.math/
```
Associated Topics:
College Analysis
College Imaginary/Complex Numbers

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