2
$\begingroup$

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!

$\endgroup$
  • 3
    $\begingroup$ 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). $\endgroup$ – Daniel Fischer Sep 7 '13 at 20:55
  • 5
    $\begingroup$ $e^{-1/z^2}=1-\frac{1}{z^2}+\frac{1}{2!z^4}-\ldots$. Residue is the coefficient of $\frac{1}{z}$, which is $0$. $\endgroup$ – njguliyev Sep 7 '13 at 20:55
  • $\begingroup$ @njguliyev Post it as an answer? $\endgroup$ – Git Gud Sep 7 '13 at 21:04
  • $\begingroup$ 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! $\endgroup$ – Twiltie Sep 7 '13 at 21:05
2
$\begingroup$

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 $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.