Quarter-sector mapping of a circular lens ($F(z)=((1+z)/(1-z){)}^4$)

For the lens $\mathrm{Im}z>0$ and $|z+i|<\sqrt{2}$, the Möbius transformation $w=(1+z)/(1-z)$ maps its real boundary segment to the positive real axis and its circular boundary arc to the ray with argument $\pi /4$. Its interior maps bijectively to $0<\arg w<\pi /4$. The fourth power therefore maps this lens conformally onto the upper half-plane. Composing with $(F-i)/(F+i)$ maps it onto the unit disc. The endpoints $\pm 1$ are boundary points, so the pole and zero introduce no interior singularity or branching.