Calculate $\oint_{\gamma} \tanh(z) dz$ on the curve $|z - \frac {\pi}{4}i|=\frac 12$

Question

Calculate $\oint_{\gamma} \tanh(z) dz$ on the curve $|z - \frac {\pi}{4}i|=\frac 12$. I am not sure if I computed this correctly:

I tried to rewrite $\tanh(z)= \frac {\frac {e^z-e^{-z}}{2i}}{\frac {e^z+e^{-z}}2}=i\frac {1-e^{2z}}{1+e^{2z}}.$ Then tried to find the poles and found out that $e^{2z}+1= 0\implies z = i\frac {\pi}2$, but then it gave me that this pole is not inside the curve so then the integral must be $0$.

Answer 1

You are right, but your argument is not correct. In fact,$$e^{2z}+1=0\iff z=\dfrac\pi2i+n\pi i\text{ for some }n\in\mathbb Z.$$However, none of these numbers belong to the regioin of $\mathbb C$ bounded by the curve and so, indeed, the integral is $0$.