Find the Laurent series of $\frac{1}{z^2(1-z)}$ when $0< |z-1| < 1$.
I don't quite know how to deal with this particular sum. Usually, I would set $w = z-1$, and that would give me
$$-\frac{1}{w(w-1)^2}$$
and I would need to find the expansion of this in the region $0 < |w| < 1$. But I can't figure out how to handle this series! Any help would be appreciated.
I'm assuming that you want the Laurent series centered at $z=1$. I'll perform the same variable change as you, $w=z-1$. Let's warm up with the following series: $$ \frac{1}{(w+1)^2}=\left(\sum_{n\geq 0}(-w)^n\right)^2=\sum_{n\geq 0}(n+1)(-w)^n,\quad |w|<1. $$ Replacing $w=z-1$ yields $$ \frac{1}{z^2}=\sum_{n\geq 0}(n+1)(1-z)^n. $$ Therefore the Laurent series is given by $$ \frac{1}{z^2(1-z)}=\sum_{n\geq 0}(n+1)(1-z)^{n-1}. $$ Note that this includes a negative power of $(1-z)$.