Finding Laurent series expansion of $f(z) = \frac{z}{(z−1)(2−z)}$ where $|z-1| > 1$

Question

Write the Laurent series expansion of $f(z) = \frac{z}{(z−1)(2−z)}$ where $|z− 1| > 1$.

I am learning complex analysis by myself in this pandemic, but I am not able to solve this problem. Please help me to solve it.

Answer 1

$$f(z)=\frac{z}{(z-1)(2-z)}=\frac{z}{z-1}+\frac{z}{2-z}=\frac{z}{z-1}+\frac{z}{1+(1-z)}$$ $$=\frac{z}{z-1}+\frac{z}{(1-z)[1-(-\frac{1}{1-z})]}$$ $$=\frac{z}{z-1}+\frac{z}{1-z}\sum_{n=0}^\infty (-1)^n \frac{1}{(1-z)^n}$$ $$=\frac{z}{z-1}+\frac{z}{1-z} +\sum_{n=1}^\infty (-1)^n \frac{z}{(1-z)^{n+1}}$$ $$=-\sum_{n=2}^\infty \frac{z}{(z-1)^n}$$