Determining the Laurent Series

Question

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.

Answer 1

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

$$\frac{1}{(z-1)(z+5)}=\frac{1}{6} \sum_{n=1}^{\infty} (-1)^{n+1}(z-2)^{-n}-\frac{1}{42} \sum_{n=0}^{\infty} (-1)^n\left(\frac{z-2}{7}\right)^n$$

Answer 2

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?