Lens-to-quadrant Möbius map ($M(z)=(1+i)\frac{1+z}{1-z}$)

Let $A=\{z:|z-i|<\sqrt{2},|z+i|<\sqrt{2}\}$. Writing $t=(1+z)/(1-z)=u+iv$, its inverse gives $1-|z{|}^2=4u/|t+1{|}^2$ and $\mathrm{Im}z=2v/|t+1{|}^2$. 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.