Laurent series for $f(z) = \frac{1}{(z-4)(z+2)}$ Find the Laurent series for $f(z) = \frac{1}{(z-4)(z+2)}$, valid for $2<|z|<4$.
first i did partial fraction expansion: 
$\frac{1}{(z-4)(z+2)} = \frac{A}{(z-4)} + \frac{B}{(z+2)}$
The result i got : $\frac{1}{6}[\frac{1}{(z-4)} - \frac{1}{(z+2)}]$
For the annuls centered at $0$, I solved for the laurent series:
$\frac{1}{(z-4)}= \frac{1}{4}\frac{1}{-1+\frac{z}{4}} = \frac{1}{4} \sum_{n=0}^{\infty}\frac{(-1)^nz^n}{4^n}$ is the power series for $|z|<4$
What I am supposed to do about $\frac{1}{(z+2)}$ since it is not within $2<|z|<4$|?

You need to write the term such that the region of convergence changes:
$$
\frac{1}{z+2} = \frac{1}{z} \, \frac{1}{1 + 2z^{-1}}
$$
which can be series expanded (in powers of $1/z$) for $|2z^{-1}| < 1$, i.e. for $|z| > 2$.
