Write the Laurent Series of $$ \frac{1}{z(1-z)^2}$$
Now the singularities are $z=0$ and $z=1$.
My attempt:
For $z=0$ case $$= \frac{1}{z}.(1-z)^{-2}$$ $$=\frac{1}{z}.[1+2z+3z^2+...]$$
For $z=1$ case $$=\frac{1}{(1-z)^2}.\frac{1}{(1+z-1)}$$ $$=\frac{1}{(1-z)^2}.(1+(z-1))^{-1}$$ $$=\frac{1}{(1-z)^2}.[1+(z-1)+(z-1)^2+(z-1)^3+...]$$
I used the $(1+x)^{-n}$ expansion in the last line.
Am I doing it right? Is there a better approach?
Here's an approach that uses partial fraction decomposition: \begin{align} \frac{1}{z(1-z)^2} &= \frac{1}{z} + \frac{1}{1-z} + \frac{1}{(1-z)^2} \\ &= \frac{1}{z} + \sum_{n \ge 0} z^n + \sum_{n \ge 0} \binom{n+1}{1} z^n \\ &= \frac{1}{z} + \sum_{n \ge 0} (n+2) z^n \end{align}