 Solution

The first region is the lens around zero bounded by the two circular arcs through $-1$ and $1$. Writing $z=x+iy$, its inequalities combine to $x^2+y^2+2|y|<1$. The second region is the open first quadrant, and the third is the open horizontal strip between imaginary parts zero and $\pi /2$.

Set $t=(1+z)/(1-z)=u+iv$. The inverse is $z=(t-1)/(t+1)$, andHence the lens condition is exactly $u>|v|$. Multiplying by $1+i$ gives real part $u-v$ and imaginary part $u+v$, both positive exactly under that condition. Thereforeis a Möbius transformation mapping the lens bijectively to the first quadrant. Its pole at one lies on the boundary, and its inverse is $z=(w-(1+i))/(w+(1+i))$, well-defined throughout the quadrant. The derivative $2(1+i)/(1-z{)}^2$ never vanishes in the lens. This is a lens-to-quadrant Möbius map.