Finding the Laurent series (complex numbers) I have
$$
f(z)={\frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=\infty$.
I did
$$
{\frac{1}{z(1-z)}} = {\frac{A}{z}}+{\frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{\frac{1}{z}}+{\frac{1}{1-z}} = {\frac{1}{z}} + \sum z^n
$$
But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.

Hints :
For the $z=1$ case :
You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :
$$f(z) = \frac{1}{z(1-z)} = \frac{1}{z} + \frac{1}{1-z} = \frac{1}{1+(z-1)} + \frac{1}{1-z}  $$
$$=$$
$$\frac{1}{(z-1)\left(\frac{1}{z-1} + 1\right)} - \frac{1}{z-1} = \frac{1}{z-1}\left(\frac{1}{\frac{1}{z-1} + 1}\right)$$
Now, recall the geometric series $\frac{1}{1+w} = \sum_{n=1}^\infty (-1)^nw^n$. Let $w = \frac{1}{z-1}$. Thus :
$$f(z) = \frac{1}{z-1}\sum_{n=0}^\infty (-1)^n \left(\frac{1}{z-1}\right)^n =\sum_{n=1}^\infty (-1)^{n-1}\left(\frac1{z-1}\right)^{n+1}$$
For the $\infty$ case :
Recall the geometric series $\frac{1}{1-w} = \sum_{n=1}^\infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :
$$f(z) = \frac{1}{z(1-z)}= -\frac{1}{z^2(1-\frac{1}{z})}=-\sum_{n=0}^{\infty}z^{-n-2}$$
Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z \to \infty$.
