Find the Laurent series for $f(z)=\frac{1}{z(1-z)}$

Question

I am having difficulties finding Laurent series of the above function, around these two domains

a) $1<|z|$

b) $1<|z-1|$

For a) I do $$ \sum_{n=0}^\infty \frac{1}{z^{n+1}} - \frac{1}{z} $$

Is this correct? And how can I solve b)?

Answer 1

Since$$f(z)=\frac1z\cdot\frac1{1-z}$$and since, when $|z|>1$, you have$$\frac1{1-z}=-\sum_{n=-\infty}^{-1}z^n,$$you have, in same same region\begin{align}f(z)&=\frac1z\left(-\sum_{n=-\infty}^{-1}z^n\right)\\&=-\sum_{n=-\infty}^{-1}z^{n-1}\\&=-\sum_{n=-\infty}^{-2}z^n.\end{align}In the other region, use the fact that$$f(z)=\frac1{z-1}\cdot\frac1{1+(z-1)}$$