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?

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 $z\not=2$.
