Proof that the function $\cot(\pi z)$ is uniformly bounded on the sides of the square with vertices $\pm(N+1/2)\pm i(N+1/2),n∈ℕ$.
My idea was that since those squares are compact and this function is continuous on those squares the image must be compact and therefore bounded.
But I'm not sure what to do with the "uniform" notion.
Uniformly bounded just means that the bound doesn't depend on $N$.
First note that
$$ |\cot \pi z|^{2} = \Big|\frac{\cos \pi z}{\sin \pi z} \Big|^{2}$$
$$ = \Big| \frac{\cos \pi z \cosh \pi y -i \sin \pi x \sinh \pi y}{\sin \pi x \cosh \pi y + i \cos \pi x \sinh \pi y }\Big|^{2}$$
$$ =\frac{\cos^{2} \pi x \cosh^{2} \pi y +\sin^{2} \pi x \sinh^{2} \pi y}{\sin^{2} \pi x \cosh^{2} \pi y + \cos^{2} \pi x \sinh^{2} \pi y }$$
$$ = \frac{\cos^{2} \pi(1+ \sinh^{2} \pi y) + \sin^{2} \pi x \sinh^{2} \pi y}{\sin^{2} \pi x(1+\sinh^{2} \pi y) + \cos^{2} \pi x \sinh^{2} \pi y} $$
$$ = \frac{\cos^{2} \pi x + \sinh^{2} \pi y}{\sin^{2} \pi x +\sinh^{2} \pi y}$$
Next note that $\cos^{2}\Big( \pi (N+\frac{1}{2}) \Big) = 0$ and $\sin^{2} \Big( \pi (N+ \frac{1}{2}) \Big) =1$.
So on the vertical sides of the square (where $x = N + \frac{1}{2}$ or $x = - N -\frac{1}{2}$),
$$ |\cot \pi z|^{2} =\frac{\sinh^{2} \pi y}{1+ \sinh^{2} \pi y} = \frac{\sinh^{2} \pi y}{\cosh^{2} \pi y} = \tanh^{2} \pi y $$
$$ \implies |\cot \pi z| \le |\tanh \pi y| \le 1 $$
And on the horizontal sides of the square (where $y = N + \frac{1}{2}$ or $y = -N - \frac{1}{2}$),
$$ |\cot \pi z|^{2} \le \frac{1 + \sinh^{2} \pi y}{\sin^{2} \pi x + \sinh^{2}\pi y} $$
$$ = \frac{\cosh^{2} \pi y}{\sin^{2} \pi x + \sinh^{2} \pi y} \le \frac{\cosh^{2}\pi y}{\sinh^{2} \pi y}= \coth^{2} \pi y$$
$$ \implies |\cot \pi z| \le |\coth \pi y| \le \coth \frac{\pi}{2} \approx 1.09$$
Thus $|\cot \pi z|$ is bounded by $\coth \frac{\pi}{2}$ on the sides the square, which implies that $\cot \pi z$ is uniformly bounded on the sides of the square.
