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#### Ruun

##### New member

- Feb 1, 2012

- 10

Hi all!

I have to perform this complex integration over three curves, the first one is \( C=\{ z \in \mathbb{C} : |z|=2 \} \) and the function to integrate is

$$ f(z)=\frac{z^2}{e^{2z}+1}$$

If I do the usual change of variables \(z=2e^{i\theta} \) and integrate from \( \theta = 0 \rightarrow \theta = 2\pi \) the denominator term is gives me trouble, because of the double exponential.

Another way I was thinking, it was Cauchy's theorem. As \(f(z) \) has no poles then (I'm not sure of this) the sum of all residues is $0$ so the integral is $0$ too. But there are $z \in \mathbb{C}$ such that $e^{2z}+1 = 0 $, given by $z_{n}=\frac{1}{2}i(\pi + 2\pi n), n \in \mathbb{Z}$ and $z_{0}$ is inside $C$ wich I think that is no pole, but the function goes to infinity, so it is not analytical there.

Thanks for your time

I have to perform this complex integration over three curves, the first one is \( C=\{ z \in \mathbb{C} : |z|=2 \} \) and the function to integrate is

$$ f(z)=\frac{z^2}{e^{2z}+1}$$

If I do the usual change of variables \(z=2e^{i\theta} \) and integrate from \( \theta = 0 \rightarrow \theta = 2\pi \) the denominator term is gives me trouble, because of the double exponential.

Another way I was thinking, it was Cauchy's theorem. As \(f(z) \) has no poles then (I'm not sure of this) the sum of all residues is $0$ so the integral is $0$ too. But there are $z \in \mathbb{C}$ such that $e^{2z}+1 = 0 $, given by $z_{n}=\frac{1}{2}i(\pi + 2\pi n), n \in \mathbb{Z}$ and $z_{0}$ is inside $C$ wich I think that is no pole, but the function goes to infinity, so it is not analytical there.

Thanks for your time

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