# Thread: Laurent series and residue

1. ## Laurent series and residue

Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point.

(1) $\displaystyle \frac{z^2}{z^2-1}$; $\displaystyle z_0=1$

Thanks

2. This is what I got so far:

$\displaystyle \frac{z^2}{z^2-1}$ = $\displaystyle \frac{1}{z^2-1} + 1$

= $\displaystyle 1 + \frac{A}{z-1}+\frac{B}{z+1}$

$\displaystyle A=1/2, B=-1/2$

=$\displaystyle 1+\frac{1}{2}(\frac{1}{z-1})-\frac{1}{2}(\frac{1}{z+1})$

$\displaystyle 1 +\frac{1}{2}\frac{1}{z-1} = \frac{1}{2}\frac{1}{z+1}$

Any idea what I do from there? Thanks!

3. You have $\displaystyle 1+1/2\frac{1}{z-1}-1/2\frac{1}{z+1}$ so need to express the last one in terms of $\displaystyle z-1$. That's not hard, just subtract and add one in the denominator:

$\displaystyle \frac{1}{z+1}=\frac{1}{z+1-1+1}=\frac{1}{2+(z-1)}=\frac{1}{2-(-(z-1))}=1/2\frac{1}{1-\frac{-(z-1)}{2}}$

Now, just finish it by expressing the last expression as a sum.

(1) $\displaystyle \frac{z^2}{z^2-1}$; $\displaystyle z_0=1$
The function has form $\displaystyle \frac{f(z)}{g(z)}$ and $\displaystyle g'(z_0) \not = 0$.
Therefore, the residue is given by $\displaystyle \frac{f(z_0)}{g'(z_0)} = \frac{1}{2}$.