Question

expand f(z)=z/(z-1)(2-z) in a laurent series valid 1<|z|<2

Answer 1

$$\dfrac{z}{(z-1)(2-z)}=\dfrac{A}{z-1}+\dfrac{B}{2-z}=\dfrac{2A-Az+Bz-B}{(z-1)(2-z)}$$

$z^0: 0=2A-B$

$z^1: 1=-A+B$

$A=1, B=2$

$$f(z)=\dfrac{z}{(z-1)(2-z)}=\dfrac{1}{z-1}+\dfrac{2}{2-z}$$

Geometric series

$$\dfrac{1}{z-1}=\dfrac{1}{z}(\dfrac{1}{1-\dfrac{1}{z}})=\displaystyle\sum_{i=1}^n\dfrac{1}{z^n}, |z|>1$$

$$\dfrac{2}{2-z}=\dfrac{1}{1-\dfrac{z}{2}}=\displaystyle\sum_{i=0}^n\dfrac{z^n}{2^n}, |z|<2$$

$$\dfrac{z}{(z-1)(2-z)}=1+\displaystyle\sum_{i=1}^n\bigg(\dfrac{1}{z^n}+\dfrac{z^n}{2^n}\bigg),1<|z|<2$$