Lens-to-half-plane map through the intersection points ($F(z)=i{\left(\frac{z-i}{z+i}\right)}^2$)

A useful way to map a circular lens is to send its two corner points to zero and infinity with a Möbius transformation. The two bounding circles then become straight rays, and a suitable power opens their sector into a half-plane.

For the lens $|z+1|<\sqrt{2}$, $|z-1|<\sqrt{2}$, the corners are $\pm i$. Put $w=(z-i)/(z+i)$, with inverse $z=i(1+w)/(1-w)$. For $w=u+iv$, substitution givesThus the image is precisely $u<-|v|$, or $3\pi /4<\arg w<5\pi /4$. Rotation through $-3\pi /4$ followed by squaring gives the displayed conformal map onto the upper half-plane. The open sector contains no opposite pair, so squaring is injective; it has nonzero derivative away from its excluded vertex. These facts also prove bijectivity and conformality, rather than merely identifying boundary images.