= Quarter-sector mapping of a circular lens
{title2=$F(z)=((1+z)/(1-z))^4$}
For the lens $\operatorname{Im}z>0$ and $|z+i|<\sqrt2$, 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 $\pm1$ are boundary points, so the pole and zero introduce no interior singularity or branching.
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