Proving $\cot(\pi z)$ is bounded on the disk $|z|=R$, $R=N+\frac{1}{2}, N\in\Bbb N^+$ I am working on a proof in complex analysis. One of the steps involves showing that $|\cot(\pi Re^{i\theta})|$ is bounded as $R\to\infty$, and $\theta$ is not fixed. In other words, $|\cot(\pi z)|$ is bounded for $z$ on the disk $|z|=R$, as $R$ tend to infinity. I expanded and reduced the expression to $$|\cot(\pi Re^{i\theta})|=\left|1+\frac{2}{e^{2i\pi Re^{i\theta}}-1}\right|.$$
Please advise on how to proceed.
 A: Let $z=x+iy$.  Then, we can express the magnitude of the cotangent function as
$$\begin{align}
|\cot(z)|&=\left|\frac{e^{i2z}+1}{e^{i2z}-1}\right|\\\\
&=\left|\frac{(1+\cos(2x)e^{-2y})+i\sin(2x)e^{-2y}}{(1-\cos(2x)e^{-2y})-i\sin(2x)e^{-2y}}\right|\\\\
&=\sqrt{\frac{1+2\cos(2x)e^{-2y}+\cos^2(2x)e^{-4y}+\sin^2(2x)e^{-4y}}{1-2\cos(2x)e^{-2y}+\cos^2(2x)e^{-4y}+\sin^2(2x)e^{-4y}}}\\\\
&=\sqrt{\frac{1+e^{-4y}+2\cos(2x)e^{-2y}}{1+e^{-4y}-2\cos(2x)e^{-2y}}}\\\\
&=\sqrt{\frac{\cosh(2y)+\cos(2x)}{\cosh(2y)-\cos(2x)}}\\\\
|\cot(\pi z)|&=\bbox[5px,border:2px solid #C0A000]{\sqrt{1+\frac{2\cos(2\pi x)}{\cosh(2\pi y)-\cos(2\pi x)}}} \tag 1
\end{align}$$
Next, suppose $|z|=N+1/2$.  We will analyze the following two cases; (i) $N+1/2\ge |x|\ge N+1/4$ and (ii) $|x|\le N+1/4$.

CASE $1$: $\displaystyle N+\frac12\ge |x|\ge N+\frac14$
For $N+\frac12\ge |x|\ge N+1/4$, we see that $\cos(2\pi x)\le 0$.  Therefore, since $\cosh(2\pi y)\ge 1$ and $\cos(2\pi x)\le 1$, we find that
$$-1\le\frac{2\cos(2\pi x)}{\cosh(2\pi y)-\cos(2\pi x)}\le 0$$
whence we have from $(1)$
$$\bbox[5px,border:2px solid #C0A000]{|\cot(\pi z)|\le 1} \tag 2$$
for $N+\frac12\ge |x|\ge N+1/4$.

CASE $2$:
For $|x|\le N+1/4$, we find the following lower bound for $\cosh(2\pi y)$ by using the asymptotic relationship $\cosh(x)=1+\frac12 x^2+O(x^4)$.  Therefore, we can write
$$\begin{align}
\cosh(2\pi y)&\ge 1+\frac12(2\pi y^2)\\\\
&=1+2\pi^2 ((N+1/2)^2-x^2)\\\\
&\ge 1+2\pi^2 ((N+1/2)^2-(N+1/4)^2)\\\\
&\ge 1+3\pi^2/8
\end{align}$$
whence we have from $(1)$
$$\bbox[5px,border:2px solid #C0A000]{|\cot(\pi z)|\le \sqrt{1+\frac{16}{3\pi^2}}< 1.24 } \tag 3$$
for $|x|\le N+1/4$.


Putting $(2)$ and $(3)$ together, we have the uniform bound for $|\cot(\pi z)|$ for $|z|=N+1/2$ is


$$\bbox[5px,border:2px solid #C0A000]{|\cot(\pi z)|\le 1.24}$$

for all $|z|=N+1/2$
And we are done!
A: There are three cases ($R\to\infty$):
$(1)$ $\enspace \displaystyle 0<|e^{2i\pi e^{i\theta}}|<1 \enspace $ => $\enspace \displaystyle \frac{1}{e^{2i\pi Re^{i\theta}}-1}\to -1\enspace $ => $\enspace |\cot(\pi Re^{i\theta})|\to |1+\frac{2}{-1}|=1$
$(2)$ $\enspace \displaystyle |e^{2i\pi e^{i\theta}}|>1\enspace $ => $\enspace \displaystyle \frac{1}{e^{2i\pi Re^{i\theta}}-1}\to 0\enspace $ => $\enspace |\cot(\pi Re^{i\theta})|\to |1+0|=1$
$(3)$ $\enspace \displaystyle |e^{2i\pi e^{i\theta}}|=1\enspace $ => $\enspace \sin\theta =0 \enspace ;\enspace \displaystyle R:=x+k \enspace $, $\enspace k\in\mathbb{N}\enspace $ with $\enspace k\to\infty$
$\enspace \enspace \displaystyle 0<x_0\leq x\leq \frac{1}{2}\enspace $ or $\enspace \displaystyle \frac{1}{2}\leq x\leq x_0<1\enspace $ ($0<x_0<1$ has to be choosen) 
$\enspace \enspace => \enspace \displaystyle |\frac{1}{e^{2i\pi Re^{i\theta}}-1}|=|\frac{1}{e^{i2\pi x}-1}|\leq |\frac{1}{e^{i2\pi x_0}-1}|$ 
$\enspace \enspace \displaystyle |\cot(\pi Re^{i\theta})|\leq 1+2|\frac{1}{e^{2i\pi Re^{i\theta}}-1}|\leq 1+2|\frac{1}{e^{i2\pi x_0}-1}|$
$\enspace \enspace$ We have seen here, that the limit depends on the definition of $R$.
$\enspace \enspace$ E.g. $\enspace \displaystyle R:=\frac{1}{2}+k$ , $k\in\mathbb{N} \enspace$ => $\enspace \displaystyle |\cot(\pi Re^{i\theta})|=|1+\frac{2}{-2}|=0$
