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.

$$f(z) = \frac{z+2}{z^2-z-2} = \frac{z+2}{(z-2)(z+1)} = \frac{4}{3(z-2)} - \frac{1}{3(z+1)} = -\frac{4}6 \frac{1}{z(1-\frac{z}{2})} - \frac{1}3 \frac{1}{z(1-\frac{-1}{z})} = -\frac{2}{3z} \sum_{k=0}^{\infty}\frac{z^k}{2^k} - \frac{1}{3z} \sum_{k=0}^{\infty}\frac{(-1)^k}{z^k}$$
This should give you convergence inside the annulus $1 < |z| < 2$.
