= Conformal map of a symmetric circular lens
{title2=$h(z)=\frac{z^3-(1-z)^3}{z^3+(1-z)^3}$}
The intersection of the unit disks centered at $1/2\pm i\sqrt3/2$ has endpoints zero and one. The <Möbius transformation> $w=z/(1-z)$ maps it onto the sector $|\arg w|<\pi/6$: the two disk conditions become $\operatorname{Re}(e^{\pm i\pi/3}w)>0$. Cubing gives the right half-plane, and the <Cayley transform between the half-plane and disk> gives the displayed holomorphic bijection to the unit disk. The sector width ensures that cubing is injective.
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