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

[latentcorpse]

This may be because I am tired but i can't find the residue of

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

help!

 

[HallsofIvy]

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

 

[latentcorpse]

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?