= Lens-to-half-plane map through the intersection points
{title2=$F(z)=i\left(\frac{z-i}{z+i}\right)^2$}
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 $|z+1|<\sqrt2$, $|z-1|<\sqrt2$, the corners are $\pm i$. Put $w=(z-i)/(z+i)$, with inverse $z=i(1+w)/(1-w)$. For $w=u+iv$, substitution gives
$$
|z+1|^2-2=\frac{4(u-v)}{|1-w|^2},\qquad |z-1|^2-2=\frac{4(u+v)}{|1-w|^2}.
$$
Thus 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.
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