Supremum of $\cot(\pi z)$ where $z$ is on circle with radius $n+1/2$

Question

I try to estimate the supremum of $|\cot(\pi z)|$ and where $z=(n+1/2) e^{i t}$, $n\in\mathbb N$ and $t\in[0,2\pi)$. I should be a constant.

So far I did by wiriting it in exponential form and expanding it

$|\cot(\pi z)|=|1+\frac{2}{e^{2\pi i z}-1}|\le 1+\frac{2}{|e^{2\pi i z}-1|}$

I focus on $|e^{2\pi i z}-1|$ and put $z$ in.

$|e^{(2n+1)\pi i \cos t}e^{-(2n+1)\pi\sin t}-1|$

From plotting I know its $>1$ but how can I estimate that matheamtical?

Answer 1

Note that $$|\cot(\pi z)|^2={\cosh(2\pi y)+\cos(2\pi x)\over \cosh(2\pi y)-\cos(2\pi x)}\ .$$ It follows that $$|\cot(\pi z(t))|^2={1+q(t)\over 1-q(t)}$$ with $$q(t)={\cos\bigl((2n+1)\pi\cos t\bigr)\over \cosh\bigl((2n+1)\pi\sin t\bigr)}\ .$$ We have to determine the maximum of the function $q$. This $q$ starts off in the negative and then has a first peak where it is ever so small positive. You have to get hold of this peak.