Proof of inequality about $\cot \pi z$. I am trying to show that for $z=x+iy$ where $|y|\geq 1$,
$$|\cot\pi z|\leq\frac{1+\exp(-2\pi|y|)}{1-\exp(-2\pi|y|)}.$$
Here is my work so far:
\begin{align*}
|\cot\pi z|&=\left\vert\frac{\exp(i\pi z)-\exp(-i\pi z)}{\exp(i\pi z)-\exp(-i\pi z)}\right\vert\\
&=\left\vert\frac{\exp(i\pi x)\exp(-\pi y)+\exp(-i\pi x)\exp(\pi y)}{\exp(i\pi x)\exp(-\pi y)-\exp(-i\pi x)\exp(\pi y)}\right\vert\\
&=\left\vert\frac{\exp(-2\pi y)+\exp(-2i\pi x)}{\exp(-2\pi y)-\exp(-2i\pi x)}\right\vert\\
&\leq\frac{1+\exp(-2\pi y)}{1-\exp(-2\pi y)},
\end{align*}
where the last step is from the triangle inequality in the numerator and the reverse triangle inequality in the denominator. It appears that I have lost the absolute value sign of $y$; did I do anything wrong?
 A: The last step in your calculation is wrong for negative $y$. You have
$$
 |\exp(-2\pi y) + \exp(-2i\pi x)| \le 1 + \exp(-2\pi y) \\
 |\exp(-2\pi y) - \exp(-2i\pi x)| \ge 1 - \exp(-2\pi y) 
$$
but the right-hand side of the second inequality is negative if $y < 0$, so that you cannot divide the inequalities.
But your calculation is correct for $y > 0$. For negative $y$ you can write
\begin{align*}
|\cot\pi z|&=\left\vert\frac{\exp(i\pi z)-\exp(-i\pi z)}{\exp(i\pi z)-\exp(-i\pi z)}\right\vert\\
&=\left\vert\frac{\exp(i\pi x)\exp(-\pi y)+\exp(-i\pi x)\exp(\pi y)}{\exp(i\pi x)\exp(-\pi y)-\exp(-i\pi x)\exp(\pi y)}\right\vert\\
&=\left\vert\frac{\exp(2i\pi x)+\exp(2\pi y)}{\exp(2i\pi x)-\exp(2\pi y)}\right\vert\\
&\leq\frac{1+\exp(2\pi y)}{1-\exp(2\pi y)} = \frac{1+\exp(-2\pi |y|)}{1-\exp(-2\pi |y|)} 
\end{align*}
behause then $1-\exp(2\pi y) > 0$.
Alternatively, use that $|\cot |$ is an even function, so that for $y < 0$
$$
|\cot \pi (x+iy)| = |\cot \pi (-x-iy)| \leq\frac{1+\exp(-2\pi (-y))}{1-\exp(-2\pi (-y))}
= \frac{1+\exp(-2\pi |y|)}{1-\exp(-2\pi |y|)} \, .
$$
Note that this proves
$$
|\cot\pi z|\leq\frac{1+\exp(-2\pi|y|)}{1-\exp(-2\pi|y|)}.
$$
for all $z=x+iy$ with $y \ne 0$, not only for $|y| \ge 1$.
