I have tot determine the Laurent series of $f(z)=\frac{z}{(z-1)(2-z)}$ for the regions $|z-1|>1$ and $0<|z-2|<1$.
I already know what to do for the regions $|z|<1$, $1<|z|<2$ and $|z|>2$ by using the partial fraction decomposition $f(z)=\frac{1}{z-1}+\frac{2}{2-z}$, but I fail to determine it with the other given regions. I hope somebody can help me out with this.
Let $z-1 = u$, then $|u| > 1$ and $\frac{z}{(z-1)(2-z)} = \frac{u+1}{u(1-u)} = \frac{2}{1-u} + \frac{1}{u}$.
As $|u| > 1, \frac{2}{1-u} = \frac{-2}{u}.\frac{1}{1-\frac{1}{u}} = \frac{-2}{u} \sum\limits_{n \geq 0}\frac{1}{u^n} = \sum\limits_{n \geq 0} \frac{-2}{u^{n+1}}$.
Thus, $\frac{z}{(z-1)(2-z)} = \frac{2}{1-u} + \frac{1}{u} = \sum\limits_{n \geq 0} \frac{-2}{u^{n+1}} + \frac{1}{u} = \sum\limits_{n \geq 0} \frac{-2}{(z-1)^{n+1}} + \frac{1}{z-1}$.
The second region is similar.