Find Laurent Serie in all three ranges.
What I did:
$\frac{1}{z(z-1)(z-2)}=\frac{1}{2}\frac{1}{z}+\frac{1}{z-1}+\frac{1}{2}\frac{1}{z-2}$
$\frac{1}{z-1}= -\sum\limits_{n=0}^{\infty} z^n, |z|<1$
$\frac{1}{z-1}= \sum\limits_{n=1}^{\infty} z^{-n}, |z|>1$
$\frac{1}{z-2}= -\frac{1}{2}\sum\limits_{n=0}^{\infty} (\frac{z}{2})^n, |z|<2$
$\frac{1}{z-2}= \frac{1}{2}\sum\limits_{n=0}^{\infty} (\frac{2}{z})^n, |z|>2$
My questions:
What should I do with first part of function $\frac{1}{z}$?
How to combine these in three ranges?
I suppose that you're to find the laurent series around $z=0$, then it's as simple as:
$${1\over z} = z^{-1}$$
that is just one term for $1/z$ (for $|z|>0$).
As for combining them, you just add them together termwise. You will have the ranges $0<|z|<1$, $1<|z|<2$ and $|z|>2$ and select the variants that are applicable for respective range.
For example for the range $0<|z|<1$ you have $\sum c_nz^n$ where:
$$c_n = \begin{cases} - 1 - 2^{-n}/4 & \text{ if } n\ge 0 & \text{ because of }1/(z-1)\text{ and } 1/2(z-2) \\ 1/2 & \text{ if } n=-1 &\text{ because of }1/2z\\ 0 & \text{ otherwise } \end{cases}$$