I'm trying to compute the following
$$ \int_C \frac{dz}{(e^{2 \pi z}+1)^2} \quad C=\{z\in \mathbb{C}: |z|=1 \textrm{ for } \Im(z)\ge0, \quad |\Re(z)|\leq 1 \textrm{ for }\Im(z)=0\}. $$
Solution: I know that $C$ is the upper-half circle and looking for the singularities $$f(z)=\frac{1}{e^{2 \pi z}+1} \Longrightarrow e^{2\pi z}=-1\iff z=\frac{1}{2}i.$$ Then it belongs to the interior of $C$. My issue is that I can't apply the Cauchy Integral Formula directly. On the other hand, I was thinking to use series expansion, but it doesn't look nice. For sure I should use the residue theorem, but for that I need to express $f(z)$ as a Laurent series.
thanks
We want the residue for $\frac{1}{(e^{2\pi z}+1)^2}$ about $i/2,$ so let $z=y+i/2$ and do the expansion about $y=0$. We have $$ e^{2\pi(y+i/2)}+1 = 1-e^{2\pi y}$$ so we just need to expand $(e^{2\pi y}-1)^{-2}.$ We get $$\frac{1}{(e^{2\pi y}-1)^2} \approx \frac{1}{(2\pi y+\frac{1}{2}(2\pi y)^2)^2} = \frac{1}{(2\pi y)^2}\frac{1}{(1+\pi y)^2} \approx \frac{1}{(2\pi y)^2}(1-2\pi y) \\= \frac{1}{(2\pi y)^2} - \frac{1}{2\pi y} $$ so we get a residue of $-\frac{1}{2\pi}.$ So assuming the closed half-circle is traversed counterclockwize, we get a result of $-i.$