In an exam for a course of Complex Analysis last year they asked to compute the following integral:
$$\int_{\partial B_1 (0)} \frac{dz}{(e^{2\pi z}+1)^2}.$$ To compute it, we only have to calculate the residues at $\pm \frac{i}{2}$, the zeros of $(e^{2\pi z}+1)^2$ inside $B_1(0)$.
To do this we'll calculate the relevant coefficient $d_{-1}$ of the Laurent series of $\frac{1}{(e^{2\pi z}+1)^2}$ at $\pm \frac{i}{2}$. It's easy enough. Let $f(x)=e^{2\pi z}+1$. Let's suppose we know (And it's true that we know) how to calculate its Maclaurin series:
$$f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n$$
where the $z_0$ in the right side can be replaced by $\pm \frac{i}{2}$. Let $g(z)=\frac{1}{f(z)}$ have a Laurent series expansion
$$g(z)=\sum_{n=-1}^{\infty} b_n (z-z_0)^n$$
in a neighborhood $U$ of $z_0$ (We can do this since $z_0$ is a simple zero of $f$).
Then $f(z)g(z)=1$ in $U$. With this fact we'll compute the terms $b_{-1}=\Res_{z_0} g$ and $b_0$. In the product of the series, the coefficient of $z_0$ is $1=c_{0}=a_0b_0+a_1b_{-1}$. Since $a_0=0$, then $1=a_1 b_{-1}$ so that $b_{-1}=\frac{1}{a_1}=\frac{1}{f'(z_0)}$(This fact is well known). In the same way we can see that
$$a_0 b_1+a_1 b_0+a_2 b_{-1}=0$$
which results in
$$a_1 b_0+a_2 b_{-1}=0$$
or
$$b_0=-\frac{a_2}{a_1^2}=-\frac{1}{2}\frac{f''(z_0)}{f'(z_0)^2}.$$
We have now calculated enough terms. Now, knowing that
$$g(z)=\sum_{n=-1}^\infty b_n z^n$$
we must only calculate the residues of $g(z)^2$ at $z_0$.
But, as before, the coefficient $d_{-1}$ of $z^{-1}$ will be
$$d_{-1}=b_0 b_{-1}+b_{-1}b_0=2b_0 b_{-1}=-\frac{2}{f'(z_0)}\frac{1}{2}\frac{f''(z_0)}{f'(z_0)^2}=-\frac{f''(z_0)}{f'(z_0)^3}.$$
Now we're ready. See that $f'(z)=2\pi e^{2\pi z},f''(z)=(2\pi)^2 e^{2\pi z}$, and since $e^{2\pi z_0}=-1$, $f'(z_0)=-2\pi,f''(z_0)=-4\pi^2$.
So $\Res_{z_0} \frac{1}{e^{2\pi z}+1}=d_{-1}=-\frac{-4\pi ^2}{-2\pi (4\pi ^2)}=-\frac{1}{2\pi}$.
And last, by the residue theorem,
$$\int_{\partial B_1(0)} \frac{dz}{(e^{2\pi z}+1)^2}=2\pi i \frac{-1}{2\pi}+2\pi i \frac{-1}{2\pi}=-2i.$$
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