# Math Help - Laurent Series Problem

1. ## Laurent Series Problem

Hi

Can someone tell me what are the steps in finding the Laurent series expansion of $f(z) = \frac{1}{z(z-1)^2}$ about z=0 and z=1

i found the binomial expansion when z=0 is $\frac{1}{z}-2-3z-4z^2$

P.S

2. ## Re: Laurent Series Problem

Originally Posted by Paymemoney
i found the binomial expansion when z=0 is $\frac{1}{z}-2-3z-4z^2$
That is not correct. Consider $\frac{1}{1-z}=\sum_{n=0}^{+\infty}z^n$ and take derivatives in both sides. Around $z=1$ , use the substitution $u=z-1$.

3. ## Re: Laurent Series Problem

after doing the problem again i got $\frac{1}{z}-2+3z-4z^2$, however answers are $\frac{-1}{z}-2-3z-4z^2$