Find Laurent Series of $\frac{z}{1-z^2}$ with $z_0 = -1$ and determine region of convergence

Question

Just having a bit of difficulty with the solution to this question.

From my understanding and trying out a few different methods, my Laurent series has come out to $\sum_{n=0}^\infty z^{2n+1}$.

But I don't understand how the $z_0=-1$ comes into it other then it is the centre of $z$ and how to determine the region of convergence. Any help would be greatly appreciated.

Answer 1

Hint:

The Laurent series around $z_0 = -1$ is an expansion with terms of the form $a_k(z - (-1))^k = a_k(z+1)^k$ for $k \in \mathbb{Z}$, and

$$\frac{z}{1- z^2 } = \frac{z}{z+1}\frac{1}{1-z} = \frac{z+1 - 1}{z+1}\frac{1}{2 - (z+1)} \\ = \frac{1}{2}\left(1 - \frac{1}{z+1} \right)\frac{1}{1 - \frac{z+1}{2}}$$

What can you do with the last factor on the RHS?