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

$$\frac{z-1}{z^3\cdot(z-2)}$$. 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.

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 $$0<|z|<2.$$

• Thanks. I forgot about how to do partial fraction on $\frac{1}{z^3}$ repeated root. Also it is definitely easier to use geometric series expansion than binomial expansion(which i've been doing). Jan 22, 2019 at 6:00