# Laurent Expansion of $f(z)=\frac{1}{z^2(1-z)}$

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.

$$f(z)=-\frac 1 {z^{3}} \frac 1 {1-\frac 1 z}=-\frac 1 {z^{3}} (1+\frac1 z +\frac 1 {z^{2}}+...)$$ for $$|z| >1$$.
• For $|z|>1$, do we only consider $\frac{1}{1-z}$? – user557493 Sep 24 '18 at 11:51
• for instance, if we have $$f(z)=\frac{3}{(z-2)(z+1)},$$ for the region $|z-1|<1$, we only consider $\frac{1}{z-2}$. Why is that? – user557493 Sep 24 '18 at 11:54