1. Laurent series

Hi

Complex Analysis....

Problem:
Find the Laurent series for $\displaystyle f(z)=\frac{z^{2}}{z^{2}-1}$ about the point $\displaystyle z_{0}=1$

I got that $\displaystyle f(z)=1+\frac{1}{2}\cdot \frac{1}{z-1}-\frac{1}{2}\cdot \frac{1}{z+1}$ , which I wrote as $\displaystyle 1+\frac{1}{2}\cdot \frac{1}{z-1}+\frac{1}{4}\cdot \sum_{k=0}^{\infty}\left(\frac{z-1}{4}\right)^{k}$

Need some input.

I guess all the terms need to be expressed as powers of $\displaystyle (z-1)$ .

Thx

2. $\displaystyle \frac{1}{z+1}=\frac{1}{2}\left(\frac{1}{1-(-\frac{z-1}{2})}\right)=\frac{1}{2}\sum_{k=0}^{\infty}\frac {(-1)^k (z-1)^k}{2^k}$

so:

$\displaystyle \frac{z^2}{z^2-1}=1+\frac{1}{2}(z-1)^{-1}-\frac{1}{4}\sum_{k=0}^{\infty}\frac{(-1)^k (z-1)^k}{2^k}$

Try checking your work in Mathematica. I'm not saying let the software do it for you but rather let the software check your work. Enter:

Code:
Series[z^2/(z^2-1),{z,1,5}]
and then compare that to what you obtain manually.

3. thanks, IŽll do that