Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ We have $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$. I want to expand this as a Laurent series in $z_0=2$ on $\{4<|z-2|<\infty\}$. 
The partial decomposition is:
$$f(z)=\frac{1}{4}\frac{1}{z-2}-\frac{1}{4}\frac{1}{z+2}+\frac{1}{6-z}$$
In my reference, they expand $\frac{1}{z+2}$ and $\frac{1}{6-z}$ with the help of the geometric series, but they leave $\frac{1}{z-2}$ as it is. Why?

$$\frac{1}{z+2}=\frac{1}{(z-2)+4}=\frac{1}{4}\frac{1}{1+\left(\frac{z-2}{4}\right)}=\frac{1}{4}\left(1-\frac{z-2}{4}+\frac{(z-2)^2}{16}-\frac{(z-2)^3}{64}+\cdots\right)$$
$$\frac{1}{6-z}=-\frac{1}{(z-2)-4}=\cdots$$
and now just mimic the first development above of the given fraction in powers of $\,z-2\,$
Pay attention to the fact that the first (and also the second one, btw) development is valid for
$$\left|\frac{z-2}{4}\right|<1\Longleftrightarrow|z-2|<4$$
Once you've done all the above just add the different fractions which form part of $\,f(z)\,$
