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I need to calculate the residue of $\text{res}_{z=\pi}\dfrac{e^{2z}}{(e^{iz} +1)}$ The expected answer is $\dfrac{e^{2z}}{(1-i)}$. Im getting $\dfrac{e^{2z}}{(-i)}$ as the answer.

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A residue is a number, not a function... – vonbrand May 11 '13 at 11:13
For some basic information about writing math at this site see e.g. here, here, here and here. I tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. – Julian Kuelshammer May 11 '13 at 11:15
If you plug in $z=\pi$, you get $ie^{2\pi}$, which is the correct residue. – robjohn May 11 '13 at 11:27
you can upvote as well as accept an answer. – robjohn May 11 '13 at 11:55
up vote 5 down vote accepted

Since the denominator $g(z)=e^{iz}-1$ has a simple zero at $z_0=\pi$, and numerator $f(z)=e^{2z}$ never vanishes, you can apply the following formula $$\operatorname{Res}\Big(\frac{f}{g},z_0\Big)=\frac{f(z_0)}{g'(z_0)}$$ giving in your case the following result $$\operatorname{Res}\Big(\frac{e^{2z}}{e^{iz}+1},\pi\Big)=\frac{e^{2\pi}}{ie^{i\pi}}=\frac{e^{2\pi}}{-i}=e^{2\pi}i$$

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Thanks for the lecture notes has the wrong answer – MikeMan May 11 '13 at 11:31

The pole is simple, thus

$$\text{res}_{z=\pi }(f)=\lim_{z\to \pi }(z-\pi )\frac{e^{2z}}{e^{iz}+1}\stackrel{\text{l'Hospital, for ex.}}=\frac{e^{2\pi }}{ie^{i\pi }}=\frac{e^{2\pi}}{-i}=ie^{2\pi} $$

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residue at $\pi$, not $\pi i$ – Federica Maggioni May 11 '13 at 11:23
$$ \lim_{z\to\pi}\frac{e^{2z}(z-\pi)}{e^{iz}+1}=\frac{e^{2\pi}}{ie^{i\pi}}=ie^{2\pi‌​} $$ – robjohn May 11 '13 at 11:23
Thanks @FedericaMaggioni, completely forgot when calculating all that thing. – DonAntonio May 11 '13 at 11:29
Indeed @robjohn, took the wrong limit. Thanks. – DonAntonio May 11 '13 at 11:30

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