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Evaluate the following along the unit circle: $$I=\oint \frac{\cos(z)}{z(e^{z}-1)}dz$$

I tried doing it by $$f(z)=\frac{\cos(z)}{e^{z}-1}$$

Then the integral would be: $$I=2\pi if(0)$$

The problem is that $f(0)$ gives me $\frac{1}{0}$.

So how do I solve it?

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  • $\begingroup$ On what curve are you integrating? $\endgroup$
    – Mark
    Jul 22 '20 at 0:30
  • $\begingroup$ Over which curve are you integrating? $\endgroup$ Jul 22 '20 at 0:30
  • $\begingroup$ the curve is a circle of radius 1 $\endgroup$ Jul 22 '20 at 0:31
  • $\begingroup$ centered at origin $\endgroup$ Jul 22 '20 at 0:31
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    $\begingroup$ find residue at pole $z=0$ $\endgroup$
    – user805287
    Jul 22 '20 at 0:47
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This integral can be expressed with help of the residue theorem: $$I=\oint \frac{\cos(z)}{z(e^{z}-1)}dz=2\pi i \sum_{k}\; \mathrm{Res}(f,a_k)$$
There is only one removable singularity at $z=0$ and you should be able to find the redsidue of $f$ at $z=0$, just find the coeficient at $z^1$ of the following series expansion of $z^2 f(z)$: $$ z^2 \frac{\cos(z)}{z\left(e^{z}-1\right)}=\frac{z \cos(z)}{e^{z}-1}=1-\frac{z}{2}+O(z^2)$$ i.e. $$\mathrm{Res}\left(\frac{\cos(z)}{z\left(e^{z}-1\right)},0\right)=-\frac{1}{2}$$ and so $$I=\oint \frac{\cos(z)}{z(e^{z}-1)}dz=- i \pi$$

For a bit more involved calculation based on the Cauchy integral theorem see e.g. this answer or using simply the residue theorem see this one.

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  • $\begingroup$ It is not a removable singularity at 0, is it? $\endgroup$
    – user581023
    Jul 22 '20 at 7:21
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    $\begingroup$ @rain1 After multiplying by (z-z0)^2 (here z0 =0) f(z) becomes holomorphic at z0, i.e. without singularity, this is why we expand it in Taylor series. Is it clear? $\endgroup$
    – Artes
    Jul 22 '20 at 8:08
  • $\begingroup$ I understand now, thank you $\endgroup$
    – user581023
    Jul 22 '20 at 8:30
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    $\begingroup$ @Sebasiano Thanks for the edit, however there should be $z^2 \frac{\cos(z)}{z \left(e^z -1\right)}= \frac{z \cos(z)}{ e^z -1}=1-\frac{z}{2}+O(z^2)$, this edit made a confusion. $\endgroup$
    – Artes
    Jul 22 '20 at 19:42

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