Laurent Expansion $\frac{1}{z(1-z)^2}$

Question

Hi I was wondering if anyone could help me with this Laurent expansion

$ f(z)=\frac{1}{z(1-z)^2} $ about $z=1$

I don't think I have done it correctly but this is what I did: $f(z)=\frac{1}{z} \frac{1}{1-z} \frac{1}{1-z} =\frac{1}{z} \sum_{n=0}^{\infty}z^n \sum_{n=0}^{\infty}z^n =\frac{1}{z} \sum_{n=0}^{\infty}z^{2n} =\sum_{n=0}^{\infty}z^{2n-1}$

Thank you in advance:)

Answer 1

First, start with $\frac1z$:\begin{align}\frac1z&=\frac1{1+(z-1)}\\&=\sum_{n=0}^\infty(-1)^n(z-1)^n\end{align}if $\lvert z-1\rvert<1$. So\begin{align}\frac1{z(z-1)^2}&=\sum_{n=0}^\infty(-1)^n(z-1)^{n-2}\\&=\sum_{n=-2}^\infty(-1)^n(z-1)^n.\end{align}