 Solution

Let $a=e^{i\pi /3}$ and $b=e^{-i\pi /3}$. The Möbius map $T(z)=(z-a)/(z-b)$ sends the two circular arcs through $a,b$ to rays bounding a wedge of angle $2\pi /3$. After choosing the branch and a unimodular constant so the interior has positive argument, $f(z)=e^{i\varphi }T(z{)}^{3/2}$ maps that wedge bijectively onto the upper half-plane.