Expanding Laurent Series with $(z-1)$ term on numerator

Question

$\frac{z-1}{z^3\cdot(z-2)}$. I want to expand series in regions $|z|<2$. I tried to just pull out $\frac{(z-1)}{z^3}$ and expand the $\frac{1}{(z-2)}$,the answer I got is similar to solution but the solution doesn't have the $(z-1)$ part. What do I do with that term? I've tried making it $\frac{1}{(z-1)^{-1}}$ and take partial fractions, do it from there but still wrong.

Answer 1

First step: find $A,B,C$ and $D$ such that

$$\frac{z-1}{z^3\cdot(z-2)}= \frac{A}{z}+\frac{B}{z^2}+\frac{C}{z^3}+\frac{D}{z-2}.$$

Second step:

$$\frac{D}{z-2}=\frac{D}{2}\frac{1}{\frac{z}{2}-1}=-\frac{D}{2}\sum_{n=0}^{\infty}\frac{z^n}{2^n}.$$

Then the Laurent expansion is given by

$$\frac{z-1}{z^3\cdot(z-2)}= \frac{A}{z}+\frac{B}{z^2}+\frac{C}{z^3}-\frac{D}{2}\sum_{n=0}^{\infty}\frac{z^n}{2^n}$$

for $0<|z|<2.$