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?
 A: $$\frac{1}{z+2}=\frac{1}{(z-2)+4}=\frac{1}{4}\frac{1}{1+\left(\frac{z-2}{4}\right)}=\frac{1}{4}\left(1-\frac{z-2}{4}+\frac{(z-2)^2}{16}-\frac{(z-2)^3}{64}+\cdots\right)$$
$$\frac{1}{6-z}=-\frac{1}{(z-2)-4}=\cdots$$
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 $\,f(z)\,$
A: To expand as a Laurent series about $z_0=2$, you want to write the function as a power series in $z-2$, possibly with negative powers of $z-2$. Therefore, we can use $w=z-2$ and $z=w+2$:
$$
\begin{align}
\frac1{z^2-4}+\frac1{6-z}
&=\frac{1}{w^2+4w}+\frac1{4-w}\\
&=\frac1w\left(\frac14-\frac1{16}w+\frac1{64}w^2-\frac1{256}w^3+\frac1{1024}w^4-\dots\right)\\
&+\left(\frac14+\frac1{16}w+\frac1{64}w^2+\frac1{256}w^3+\dots\right)\\
&=\frac1{4w}+\frac3{16}+\frac5{64}w+\frac3{256}w^2+\frac5{1024}w^3+\dots\\
&=\frac1{4(z-2)}+\frac3{16}+\frac5{64}(z-2)+\frac3{256}(z-2)^2+\frac5{1024}(z-2)^3+\dots
\end{align}
$$
We have expanded the series in $w$ (aka $z-2$) about $w=0$ (aka $z=2$). Both of the power series in $w$ converge for $0<|w|<4$, which is $0<|z-2|<4$, which is $-2<z<2$ or $2<z<6$, which is $-2<z<6$ but $z\not=2$.
A: 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.
