 Conformal map of a symmetric circular lens

The intersection of the unit disks centered at $1/2\pm i\sqrt{3}/2$ 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 $\mathrm{Re}(e^{\pm i\pi /3}w)>0$. 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.