# Finding the residue of $e^{\frac {-3} {z^2}}$

I'm going through past exam papers and came across the following question

find the residue of $f(z)=e^{\frac {-3} {z^2}}$ at $z=0$

I know how to find the residue and the residue theorem but I'm unsure how to find it for this question. Any help would be greatly appreciated. Thanks

• It's an even function. What does that say about its Laurent expansion around $0$? – Daniel Fischer Jan 6 '17 at 12:44
• Sum of residual is equal to 0. What ever is going to be the residual at 0 this is equivalent to get the residual to infinity and changing the sign. – user8469759 Jan 6 '17 at 12:44

$$e^{-\frac3{z^2}}=\sum_{n=0}^\infty\frac{(-1)^n3^n}{z^{2n}n!}\implies\text{ the residue is zero...as expected, since all the powers}$$
of $\;z\;$ are even.
• The assertion $\lim_{z\to0}e^{-3/z^2}=0$ is wrong. Try approaching $0$ from other directions in the complex plane. – GEdgar Jan 6 '17 at 12:55