# How to find this Laurent series?

1. May 7, 2007

### Noone1982

I understand perfectly well how to do Taylor series, but I am foggy on these Laurent series. Say, we have something like,

$$f\left( z \right)\; =\; \frac{1}{z^{2}\cdot \sin \left( z \right)}$$

I think I need to use the taylor series expressions for sin(z) but otherwise, I am not sure what to do about that z^2. If I use that in a taylor series with z=0, then I get a singularity.

Since its 1/sin(z), do I just inverse the taylor series for sin(z) Yes, you can see I am very unknowledgeable about this. I turn here because the explanations I found in my book are totally lacking.

2. May 7, 2007

### AiRAVATA

It's clear that z=0 is a pole for your function, thats why the Taylor series is not well defined there. The Laurent series will tell you how this singularity behaves (how it grows to infinity).

A Laurent series has the form

$$\sum_{n=1}^\infty \frac{b_n}{z^n}+\sum_{n=0}^\infty a_nz^n,$$

so you are right, in order to calculate the series, you just have write down the series for $\csc z$ and multiply it by $z^{-2}$.