Set $U:=\{ z\in \mathbb{C}\mid \arg(z)\in(-\alpha,\alpha)\}$ with $\alpha\in(0,\pi)$.Give explicitly the conformal mapping $f:U\rightarrow \mathbb{D}$ such that $f(U)=\mathbb{D}$ with $f(1)=0$ and $f'(1)>0$
This is one of many exercises for Riemann mapping. My idea is find a conformal function $g$ such that $g(U)=\mathbb{H}$ the superior half plane and take conformal mapping $h:\mathbb{H}\rightarrow \mathbb{D}$ (with the assumption indicated, $i.e.$ $h(z)=\frac{1}{2}\frac{z+i}{z-i}$). Since $\alpha\in(0,\pi)$, $g(z)=e^{i(\pi-\alpha)}z$ is biholomorphic between $U$ and $\mathbb{H}$. Take $f=g\circ h$ is alright?