$\cot \pi iy$ tends to $-i$ as $y \to \infty$ For a fixed real number $x$, why does $\cot \pi z$ tends to $-i$ as $y \to \infty$? (Here, $z=x+iy$.) Can this be proved simply?
 A: Note the following:
$\cot(z) = \frac{\cos(z)}{\sin(z)}$
$\cos(x+iy) = \cos(x) \cosh(y) - i \sin(x) \sinh(y)$
$\sin(x+iy) = \sin(x) \cosh(y) + i \cos(x) \sinh(y)$
so
\begin{equation}
\cot(z) = \frac{\cos(x) \cosh(y) - i \sin(x) \sinh(y)}{\sin(x) \cosh(y) + i \cos(x) \sinh(y)}
\end{equation}
and multiply the numerator and denominator by the conjugate of the denominator to arrive at
\begin{equation}
\frac{(\cos(x) \cosh(y) - i \sin(x) \sinh(y))(\sin(x) \cosh(y) - i \cos(x) \sinh(y))}{\sin^2(x) \cosh^2(y) + \cos^2(x) \sinh^2(y)}
\end{equation}
and distribute and factor a little to arrive at 
\begin{equation}
\frac{\cos(x)\sin(x)(\cosh^2(x) - \sinh^2(x)) - i \cosh(y)\sinh(y)(\cos^2(x)+\sin^2(x))}{\sin^2(x) \cosh^2(y) + \cos^2(x) \sinh^2(y)}
\end{equation}
this simplifies by the Pythagorean identities: $\cos^2(x) + \sin^2(x) = 1$ and $\cosh^2(x) - \sinh^2(x) = 1$: 
\begin{equation}
\frac{\cos(x)\sin(x) - i \cosh(y)\sinh(y)}{\sin^2(x) \cosh^2(y) + \cos^2(x) \sinh^2(y)}
\end{equation}
We now focus on the denominator, expressing the $\sinh^2(y) = \cosh^2(y) - 1$ and $\sin^2(x) = 1 - \cos^2(x)$ we have 
\begin{equation}
\frac{\cos(x)\sin(x) - i \cosh(y)\sinh(y)}{(1-\cos^2(x)) \cosh^2(y) + \cos^2(x) (\cosh^2(y)-1)}
\end{equation}
and distributing we notice two cross terms cancel giving:
\begin{equation}
\frac{\cos(x)\sin(x) - i \cosh(y)\sinh(y)}{\cosh^2(y) - \cos^2(x)}
\end{equation}
Finally we ask about the limit, in the denominator $\cosh^2(y)$ grows without bound as $\pi y \rightarrow \infty$ and dominates the bounded $\cos^2(x)$. It also dominates the real part of the numerator. So we must only consider the ratio:
\begin{equation}
\lim_{y \rightarrow \infty} -i\frac{\cosh(y)\sinh(y)}{\cosh^2(y)} =  \lim_{y \rightarrow \infty} -i\tanh(y) 
\end{equation} 
Finally $\tanh(y) = \frac{e^y - e^{-y}}{e^y + e^{-y}}$ when $y$ is large and positive only the positive exponents matter. It should be clear that $\tanh(y)$ approaches one in this limit. So in fact the solution should be: 
\begin{equation}
\lim_{y \rightarrow \infty} \cot(x+iy) = \lim_{y \rightarrow \infty} -i\tanh(y) = -i
\end{equation} 
not $i$ as you suggested. 
A: Note 
$$\cot(\pi (x+iy)) = \frac{1-\tan(\pi x)\tan(i\pi y)}{\tan(\pi x)+\tan(i\pi y)}
=\frac{1-i\tan(\pi x)\tanh(\pi y)}{\tan(\pi x)+i\tanh(\pi y)}$$
Then, use $\lim_{y\rightarrow\infty} \tanh(\pi y) = 1$,
$$\lim_{y\rightarrow\infty} \cot(\pi z)
=\frac{1-i\tan(\pi x)}{\tan(\pi x)+i}=\frac{-i(i+\tan(\pi x))}{\tan(\pi x)+i}=-i$$
A: Hint 1: use that $\sin(z+w)=\sin(z)\cos(w)+\sin(w)\cos(z)$ and $\cos(z+w)=\cos(z)\cos(w)-\sin(z)\sin(w)$.
Hint 2: $\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$ for every $z$. Something similar holds for $\sin$.
A: I'm pretty sure the limit should be -i not i. By replacing $z$ by $\frac{z}{\pi}$, it suffices to show that $\cot(x+iy) \to -i$ as $y \to \infty$. 
Notice that $$\sin(x+iy) = \sin(x)\cos(iy)+\sin(iy)\cos(x) = \sin(x)\cosh(y)+i\cos(x)\sinh(y)$$
and similarily,
$$\cos(x+iy) = \cos(x)\cosh(y)-i\sin(x)\sinh(y)$$
so $$ \cot(x+iy) = \frac{\cos(x)\cosh(y)-i\sin(x)\sinh(y)}{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}$$ 
Through further algebraic manipulation and trigonometric identities, one can show that:
$$\Re[\cot(x+iy)] = -\frac{\sin(2x)}{\cos(2x)-\cosh(2y)}$$
and
$$\Im[\cot(x+iy)] = \frac{\sinh(2y)}{\cos(2x)-\cosh(2y)}$$
As $y \to +\infty$, $\cosh(2y) \to +\infty$, and $\sin(2x),\cos(2x)$ are bounded so $\Re[\cot(x+iy)] \to 0$.
Likewise, the imaginary part  is dominated by the behavior of $-\frac{\sinh(2y)}{\cosh(2y)}$ as $y \to +\infty$ so $\Im[\cot(x+iy)] \to -1$. It follows that $\cot(x+iy) \to -i$ as $y \to \infty$
