Calculating residue

I really don't know too much about residue or how to calculate it, but just out of curiosity how would I find the residue of $e^{-1/z^2}$ at $z=0$? Would it make sense to do so? And would this value have any effect on some series representation around zero? I really don't know too much about any of this so forgive me if these questions are a little off target.

Thanks!

• For that function, just expanding the exponential series $$e^{-1/z^2} = \sum_{k=0}^\infty \frac{(-1)^k}{k!}z^{-2k}$$ is simple enough (unless you directly see that the residue is $0$ since it's an even function). – Daniel Fischer Sep 7 '13 at 20:55
• $e^{-1/z^2}=1-\frac{1}{z^2}+\frac{1}{2!z^4}-\ldots$. Residue is the coefficient of $\frac{1}{z}$, which is $0$. – njguliyev Sep 7 '13 at 20:55
• @njguliyev Post it as an answer? – Git Gud Sep 7 '13 at 21:04
• I see, so I find the series and look for the coefficient of $\frac{1}{z}$ which in the case of the function I mentioned is 0. Thank you both! – Twiltie Sep 7 '13 at 21:05

The Laurent expansion of the function $e^{-1/z^2}$ is $$e^{-1/z^2}=1-\frac{1}{z^2}+\frac{1}{2!z^4}-\ldots$$ $\operatorname{res}_{z=0} e^{-1/z^2}$ is the coefficient of $\frac1z$, which equals $0$.