Laurent expansion of $f(z)= \frac{z}{(z-1)^2(z+2)}$

Question

Compute the Laurent series of $f(z)= \frac{z}{(z-1)^2(z+2)}$ in a drilled disc of center 1. Specify the maximum radius of this disc.

I did the partial fraction decomposition of f: $f(z) = \frac{2}{9} \frac{1}{z-1} + \frac{1}{3} \frac{1}{(z-1)^2} - \frac{2}{9} \frac{1}{z+2}$. What shall I do next?

Answer 1

Note that\begin{align}\frac1{z+2}&=\frac1{3+(z-1)}\\&=\frac13\cdot\frac1{1+(z-1)/3}\\&=\frac13\sum_{n=0}^\infty\frac{(-1)^n}{3^n}(z-1)^n\text{ (when $|z-1|<3$)}\\&=\sum_{n=0}^\infty\frac{(-1)^n}{3^{n+1}}(z-1)^n.\end{align}Therefore\begin{align}\frac z{(z-1)^2(z+2)}&=\frac2{9(z-1)}-\frac1{3(z-1)^2}-\frac2{9(z+2)}\\&=-\frac1{3(z-1)^2}+\frac2{9(z-1)}-\sum_{n=0}^\infty\frac{2(-1)^n}{3^{n+3}}(z-1)^n.\end{align}