arctan maps the unit disk onto a band around the imaginary axis Let $D\subseteq\mathbb{C}$ be the unit disk; that is, $D=\{z\in\mathbb{C}:\ |z|<1\}$. Let $B\subseteq \mathbb{C}$ be some band around the imaginary axis: $B=\{z\in\mathbb{C}:\ |\text{Re}(z)|<\pi/4\}$.
Why does it hold that the principal branch of $\arctan$ maps $D$ conformally onto $V$?

Here's an attempt:
let $U=\left\{z\in\mathbb{C}:\ \text{Re}(z)>0\right\}$. Define
$g:V\to U$ as follows:
\begin{equation*}
  g(z) = g(x+yi) = \exp(-2y + 2xi).
\end{equation*}
We have that $g$ maps $V$ conformally to $U$. Now define $h:U\to D$ as
follows:
\begin{equation*}
  h(z) = \frac{i(1-z)}{1+z}
\end{equation*}
We have that $h$ maps $U$ conformally to $D$ (it is a Möbius transformation). It follows that
\begin{equation*}
  \tan{z} = \frac{\sin z}{\cos z}
  = \frac{i\left(1-e^{2iz}\right)}{1+e^{2iz}}
  = h\circ g (z)
\end{equation*}
maps $V$ conformally to $D$, thus the principal branch of $\arctan{z}$ maps
$D$ conformally to $V$.
