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

  • 1
    $\begingroup$ It's an even function. What does that say about its Laurent expansion around $0$? $\endgroup$ – Daniel Fischer Jan 6 '17 at 12:44
  • $\begingroup$ 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. $\endgroup$ – user8469759 Jan 6 '17 at 12:44

What about the usual Laurent (Taylor) series for the exponential?:

$$e^{-\frac3{z^2}}=\sum_{n=0}^\infty\frac{(-1)^n3^n}{z^{2n}n!}\implies\text{ the residue is zero...as expected, since all the powers}$$

of $\;z\;$ are even.

| cite | improve this answer | |
  • $\begingroup$ The assertion $\lim_{z\to0}e^{-3/z^2}=0$ is wrong. Try approaching $0$ from other directions in the complex plane. $\endgroup$ – GEdgar Jan 6 '17 at 12:55
  • $\begingroup$ @GEdgar That's what happens when you stand up in the middle of trying to answer a question: I completely forgot this was complex analysis. Deleting that part. Thanks $\endgroup$ – DonAntonio Jan 6 '17 at 13:10

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