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Wolfram Alpha tells me that the residue of

$$ \frac {1}{e^z-1} $$ at the point z= $ 2i\pi $ is 1.

Now i understand how the formula for residue works for simple poles , i just don't understand how $ (z-2i\pi) $ can cancel anything at the bottom. Basically i dont know how to prove this.

Could someone please help?

Thank you very much for your help and time.

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Note that\begin{align}\lim_{z\to2\pi i}\frac{e^z-1}{z-2\pi i}&=\lim_{z\to2\pi i}\frac{e^z-e^{2\pi i}}{z-2\pi i}\\&=\exp'(2\pi i)\\&=e^{2\pi i}\\&=1.\end{align}Therefore,$$\lim_{z\to2\pi i}\frac{z-2\pi i}{e^z-1}=1$$and this is enough to prove that $\operatorname{res}_{z=2\pi i}\frac1{e^z-1}=1$.

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You have to use L'Hopital's Rule. Since both the numerator and denominator of your limit approach zero as $z\to 2\pi i$,

$$\lim_{z\to {2\pi i}} \frac{z-2\pi i}{e^z-1}=\lim_{z\to 2\pi i} \frac{1}{e^z}=1$$

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$$\frac {1}{e^z-1} =\frac {1}{z+z^2/2 + z^3/6+... } = 1/z -1/2+z/12+...$$

The residue is the coefficient of $1/z$ thus it is $1$

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