Find the Laurent series for $f$ about $2$ which converges

Question

Let $f(z) = \frac{1}{(z-1)(z-5)}$ and $\gamma = C(2,2)$, a circle of radius $2$, centre $2$.

a) Find the largest open annulus about $2$ containing $\gamma$ in which $f$ is analytic.

Find the Laurent series for $f$ about $2$ which converges to $f(z)$ in this annulus

I'm struggling to figure out the Laurent series! I know the annulus is $1<|z-2|<3$ but can't manipulate it. Whats the best method to approach this?

Answer 1

Partial fractions give you $$f(z)=\frac{A}{z-1}+\frac{B}{z-5}.$$ I'm sure you have no problems finding $A$ and $B$.

Now, $$ \frac{1}{z-1}=\frac{1}{(z-2)+1}=\frac{1}{z-2}\Bigg[\frac{1}{1+\big(\frac{1}{z-2}\big)}\Bigg] $$ $$=\frac{1}{z-2}\Bigg[1-\frac{1}{z-2}+\frac{1}{(z-2)^2}-\frac{1}{(z-2)^3}+\cdots\Bigg], $$ which is valid for $|z-2|>1$.

Similarly, $$\frac{1}{z-5}=-\frac{1}{3}\Bigg(\frac{1}{1-\big(\frac{z-2}{3}\big)}\Bigg)=-\frac{1}{3}\Bigg[1+\frac{z-2}{3}+\bigg(\frac{z-2}{3}\bigg)^2+\bigg(\frac{z-2}{3}\bigg)^3+\cdots\Bigg],$$ which is valid for $|z-2|<3$.

Can you take it from here?

Answer 2

$f(z) = \frac{-1}{4}\sum _{n=0}^∞ \frac{(-1)^n-3^n}{(z-2)^{n+1}}$