I would like to do this exercise :
Let $\displaystyle h(z) = \pi \mathrm{cotan}(\pi z) = \pi \frac{\cos(\pi z)}{\sin(\pi z)}$. And for $q \in \mathbb{N}^{*}$, let $C_{q}$ be the square in the complex plane defined by its vertices (given turning counterclockwise from the first quadrant) : $$\big(q+\frac{1}{2} \big)(1+i), \; \big(q+\frac{1}{2} \big)(-1+i), \; \big(q+\frac{1}{2} \big)(-1-i), \; \big(q+\frac{1}{2} \big)(1-i).$$ I need to prove that $h$ is bounded on $\bigcup \limits_{q \in \mathbb{N}^{\ast}} C_{q}$.
I proved that $h$ is meromorphic on $\mathbb{C}$, with every integer $k \in \mathbb{Z}$ being a pole of order $1$. I also proved that :
$$ \forall z \in C_{q}, \, \big\vert e^{2i\pi z}-1 \big\vert \geq \frac{1}{2} $$
As a consequence,
$$ \forall z \in C_{q}, \, \vert h(z) \vert \leq 2\pi \big(1+e^{2\pi (q+\frac{1}{2})} \big) \tag{$\star$}$$
because, for all $z \in \mathbb{C} \smallsetminus \mathbb{Z}$, $\displaystyle h(z) = i\pi \frac{e^{2i \pi z}+1}{e^{2i \pi z}-1} $. It only proves that $h$ is bounded on $C_{q}$ but I don't think this proves that $h$ is bounded on $\displaystyle \bigcup \limits_{q \in \mathbb{N}^{\ast}} C_{q}$ because I need to prove that :
$$ \exists C \in \mathbb{R}^{+}, \; \forall z \in \bigcup \limits_{q \in \mathbb{N}^{\ast}}, \; \vert h(z) \vert \leq C $$
I don't think I can find such a constant $C$ from $(\star)$. Do you have any idea ?
