# Finding the residue of e^(-1/z^2) at z=0

1. Apr 1, 2009

### latentcorpse

This may be because im tired but i cant find the residue of

$e^{-\frac{1}{z^2}}$ at $z=0$

help!!!

2. Apr 2, 2009

### HallsofIvy

Staff Emeritus
Re: Residue

You know the Taylor's series for ez about z= 0 don't you? Replace z in that by -1/z2 to get a Laurent series for this function.

3. Apr 2, 2009

### latentcorpse

Re: Residue

hmmmm

$e^{-\frac{1}{z^2}}=1-\frac{1}{z^2}+\frac{1}{2z^4}+...$

so the residue is 0?

i was going to do this but i got confused - why is this an acceptable form of the laurent series - we derived it from the taylor series which doesn't admit negative n values. so surely there could be some other terms that we're not taking into account?