Expand $f(z) = \frac{1}{z(z-1)(z-2)}$ in the region $0 < |z| < 1$:
Using partial fraction decomposition, $f(z) = \frac{1}{2} \cdot \frac{1}{z}-\frac{1}{z-1} + \frac{1}{2}\cdot\frac{1}{z-2}$.
So, using some algebra I have
$f(z) = \frac{1}{2} \cdot \frac{1}{z}+\frac{1}{1-z} - \frac{1}{4}\cdot\frac{1}{1-\frac{z}{2}} = {\frac{1}{2} \cdot \frac{1}{z} + \sum_{n=0}^\infty z^n -\frac{1}{4}\sum_{n=0}^\infty (\frac{z}{2})^n}$.
Also, I need to expand in the region $1 < |z| < 2$. How does this change the problem? The only thing I think I have to do different is rewrite the $-\frac{1}{z-1}$ term as $-\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}$, but I am not entirely sure.
Any help appreciated.
Actually, it's better to use the partial fraction decomposition$$\frac1{(z-1)(z-2)}=-\frac1{2-z}+\frac1{1-z}$$and to divide everything by $z$ at the end.
So, if $0<\lvert z\rvert<1$, then$$-\frac1{2-z}+\frac1{1-z}=-\frac12\cdot\frac1{1-\frac z2}+\frac1{1-z}=-\frac12\sum_{n=0}^\infty\frac{z^n}{2^n}+\sum_{n=0}^\infty z^n.\tag1$$Therefore, the Laurent series of $f$ is $(1)$ divided by $z$.
If $1<\lvert z\rvert<2$, then$$-\frac1{2-z}+\frac1{1-z}=-\frac12\cdot\frac1{1-\frac z2}+\frac1{1-z}=-\frac12\sum_{n=0}^\infty\frac{z^n}{2^n}-\sum_{n=-\infty}^{-1}z^n.\tag2$$Again, the Laurent series of $f$ is $(2)$ divided by $z$.