Laurent Series Problem

**Paymemoney** · October 21st 2011, 09:31 PM

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

**FernandoRevilla** · October 21st 2011, 10:33 PM

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 $u=z-1$ .

**Paymemoney** · October 24th 2011, 03:03 AM

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

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