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


1 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:


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 $0<|z|<2.$

  • 1
    $\begingroup$ 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). $\endgroup$ Jan 22, 2019 at 6:00

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