Expand $f(z)=\frac{1}{z(z-3)}$ as laurent series in domain $1 < |z-4| < 4$

Question

Expand $f(z)=\frac{1}{z(z-3)}$ as laurent series in domain $1 < |z-4| < 4$

Any suggestion

i have

$\frac{1}{z-3}= \sum_{n=0}^{\infty} (-1)^n \frac{1}{(z-4)^{n+1}}$

and

$\frac{4}{z}= \sum_{n=0} ^{\infty}(-1)^n \ (\frac{z-4}{4})^n$

in the given domain. is this right ??. if yes how to multiply these series

Thanks in advanced

Answer 1

Start with

$ f(z)=\frac{1}{z(z-3)}=-\frac{1}{3}\left(\frac{1}{z}-\frac{1}{z-3}\right) =-\frac{1}{3}\left[\frac{1}{(z-4)+4}-\frac{1}{z-4+1}\right]$

$$\hspace{2cm}=-\frac{1}{3}\left[\frac{1}{4(1+\frac{z-4}{4})}-\frac{1}{(z-4)[1+\frac{1}{z-4}]}\right]$$

$$\hspace{4.8cm}=-\frac{1}{12}\left[\left(1+\frac{z-4}{4}\right)^{-1}\right]+\frac{1}{3(z-4)}\left[\left(1+\frac{1}{z-4} \right)^{-1}\right]\\\vdots$$