<s>[INST] 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?
 [/INST] Here is a related problem. You are asked to find the Laurent series at $x=2$ in the region $|z-2|>4$. You have
$$ f(z) = \frac{1}{(z-2)(z+2)} - \frac{1}{z-6} = \frac{1}{(z-2)((z-2)+4)} - \frac{1}{{(z-2)-4}} $$
$$ = \frac{1}{(z-2)^2(1+\frac{4}{(z-2)})} - \frac{1}{(z-2)(1-\frac{4}{z-2})} $$
$$ = \sum_{k=0}^{\infty} \frac{(-1)^k 4^k}{(z-2)^{k+2}}- \sum_{k=0}^{\infty} \frac{4^k}{(z-2)^{k+1}} $$
The above two series have the region of convergence $|z-2|>4\,.$ It is very important to pay attention to the region (in your case $|z-2|>4$) where you are asked to construct Laurent series and based on it you find the Laurent series.
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