# Finding Laurent Series

1. Nov 8, 2013

### Hertz

1. The problem statement, all variables and given/known data

Specifically, I'm trying to find the laurent series for $f(z)=\frac{z^2}{z+1}$ around the point $z=-1$. My real problem is my procedure in general though. I'm not sure what I'm doing wrong on a lot of these Laurent Series but for some reason I'm struggling with them.

(Even more specifically, I'm trying to find the type of singularity and the residue at z=-1.)

2. Relevant equations

$\sum^{\infty}_{n=-\infty}{c_n (z-z_0)^n}$ where
$c_n=\frac{1}{2\pi i}\int_{C}{\frac{f(z)dz}{(z-z_0)^{n+1}}}$

3. The attempt at a solution

What I did is take the $z^2$ out and set $z_0=-1$ because that's the point I want to expand around. Then I set C such that $z=-1+e^{i\theta}$ where $-\pi < \theta < \pi$ and integrated.

I found the integral, which was $\frac{sin(n+1)\pi}{(n+1)\pi}$ which equals zero for all values of n besides $n=-1$.

I could probably find out more about the coefficient at -1 if I evaluated the integral for n=-1, but at this point I realized I still had the $x^2$ in there that would throw off the degree of my Laurent Series anyways...

So I basically feel like I have spent way more time on this problem than I should have and have almost no results to show for it. Clearly, my procedure is not spot on :\. Can anybody help me out? How should I start out on a problem like this?

Normally, when I try to find a Laurent Series, this is what I do:
1. Decide where it should be centered.
2. Think about other series representations that I've memorized to maybe do a quick easy substitution or break up the function into multiple parts.
3. If I have no success this far, I'll usually just result to the laurent series formula that I have posted above, but clearly, I'm running into problems with it.

2. Nov 8, 2013

### brmath

You have a simple pole at z = -1. That means you will have one negative exponent in your Laurent expansion. Start with the Taylor's series for $z^2$ at the point -1 (i.e. you are expanding in powers of z+1. You will get

$z^2 = \sum [stuff]$. This series converges throughout the complex plane. Now divide it by z+1 to get

$z^2 = \sum [stuff]/(z+1)$. This gives you the entire Laurent expansion.

In general finding Laurent expansions via the integral is a fraught process. Sometimes it's the only way, but if you can think of something simpler, use it. The Laurent series is unique, so if you have a series that converges to your function in a domain around a singularity, it has to be the Laurent series. (If there is no singularity it has to be the Taylor's series).