I am trying to find two distinct Laurent expansions of $f(z)=\frac{1}{z^2(1-z)},$ in powers of $z$.
My attempt:
I believe I have found one Laurent expansion of the region $0<|z|<1$. \begin{align} f(z)&=\frac{1}{z^2(1-z)} \\ &=\frac{1}{z}+\frac{1}{z^2}+\frac{1}{1-z} \\ &=\sum_{n=-2}^{\infty} z^n \end{align} But I cannot find a second expansion. I know the region for this expansion is $|z|>1$. The answer provided is $\sum_{n=-\infty}^{-3} (-1)^nz^n$, but I don't understand the method of getting to this answer.