Find the Laurent expansion in powers of $\,z\,$ and $\,1/z\,$ $f(z)=\large\frac{z+2}{z^2-z-2}$ in $1<|z|<2$ and then in $2<|z|<\infty$
I am unsure how the two different regions of $z$ affect the series expansion. 
Any help would be appreciated.
 A: Hints:
$$1<|z|<2\implies \begin{cases}\frac{1}{2}<\frac{1}{|z|}<1\\{}\\\frac{|z|}{2}<1\end{cases}\;\;\;,\;\;\;\;\text{so}$$
$$\frac{z+2}{(z-2)(z+1)}=\frac{1}{3}\left(\frac{4}{z-2}-\frac{1}{z+1}\right)=$$
$$=-\frac{1}{3}\left(\frac{2}{1-\frac{z}{2}}+\frac{1}{z}\frac{1}{1+\frac{1}{z}}\right)=-\frac{1}{3}\left(4\left(1+\frac{z}{2}+\frac{z^2}{2^2}+\ldots\right)+\frac{1}{z}\left(1-\frac{1}{z}+\frac{1}{z^2}-\frac{1}{z^3}+\ldots\right)\right)=$$
$$=-\frac{1}{3}\left(4+\frac{4z}{2}+\frac{4z^2}{2^2}+\ldots+\frac{4z^n}{2^n}+\ldots+\frac{1}{z}-\frac{1}{z^2}+\frac{1}{z^3}-\ldots\right)=$$
$$=\ldots-\frac{1}{3z^3}+\frac{1}{3z^2}-\frac{1}{3z}+\frac{5}{3}-\frac{4}{3\cdot 2}z-\frac{4}{3\cdot2^2}z^2-\ldots$$
A: $$f(z) = \frac{z+2}{z^2-z-2} = \frac{z+2}{(z-2)(z+1)} = \frac{4}{3(z-2)} - \frac{1}{3(z+1)} = -\frac{4}6 \frac{1}{z(1-\frac{z}{2})} - \frac{1}3 \frac{1}{z(1-\frac{-1}{z})} = -\frac{2}{3z} \sum_{k=0}^{\infty}\frac{z^k}{2^k} - \frac{1}{3z} \sum_{k=0}^{\infty}\frac{(-1)^k}{z^k}$$
This should give you convergence inside the annulus $1 < |z| < 2$.
