Determining the Laurent Series I need to determine the Laurent series of this function:
$$\frac{1}{(z-1)(z+5)}$$ Inside the annulus: $$\left\{z|1<|z-2|<6\right\}$$ Any help appreciated.
 A: Using partial fraction expansion, we can write
$$\frac{1}{(z-1)(z+5)}=\frac{1/6}{z-1}-\frac{1/6}{z+5}$$
We seek a series about the center of the annulus $z=2$.  To that end,
$$\begin{align}
-\frac{1/6}{z+5}&=\frac16 \frac{1}{(z-2)+7}\\\\
&=-\frac{1}{42}\frac{1}{1+\frac{z-1}{7}}\\\\
&=-\frac{1}{42} \sum_{n=0}^{\infty} (-1)^n\left(\frac{z-2}{7}\right)^n
\end{align}$$
which converges within the entire annulus.  We also can write
$$\begin{align}
\frac{1/6}{z-1}&=\frac16 \frac{1}{(z-2)+1}\\\\
&=\frac{1}{6}\frac{1/(z-2)}{1+(z-2)^{-1}}\\\\
&=\frac{1}{6} \sum_{n=1}^{\infty} (-1)^{n+1}(z-2)^{-n}
\end{align}$$
which converges for $|z-2|>1$, which completely covers the annulus.  Thus, the appropriate Laurent series is
$$\frac{1}{(z-1)(z+5)}=\frac{1}{6} \sum_{n=1}^{\infty} (-1)^{n+1}(z-2)^{-n}-\frac{1}{42} \sum_{n=0}^{\infty} (-1)^n\left(\frac{z-2}{7}\right)^n$$
A: Here is just a set up: $\frac{1}{1-x}=1+x+x^2+x^3+x^4+.....$,then
$\frac{1}{1+x}=1-x+x^2-x^3+x^4-.....$ then $\frac{1}{1+x/5}=\frac{5}{5+x}=1-(x/5)+(x/5)^2-(x/5)^3+(x/5)^4-.....$ Now divide by 5 to get $\frac{1}{5+x}=1/5-(x/5)/5+(x/5)^2/5-(x/5)^3/5+...$ And then integrate. Can you now work out the details?
