= Lens-to-quadrant Möbius map
{title2=$M(z)=(1+i)\frac{1+z}{1-z}$}
Let $A=\{z:|z-i|<\sqrt2,\ |z+i|<\sqrt2\}$. Writing $t=(1+z)/(1-z)=u+iv$, its inverse gives $1-|z|^2=4u/|t+1|^2$ and $\operatorname{Im}z=2v/|t+1|^2$. The lens condition is therefore exactly $u>|v|$. Multiplication by $1+i$ converts that to positive real and imaginary parts. Thus the displayed <Möbius transformation> maps the lens bijectively to the first quadrant. Its two boundary arcs meet at $-1,1$, which map to zero and infinity. The centre zero maps to $1+i$. The derivative never vanishes inside the lens.
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