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 $V$.