= 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 $z=(t-1)/(t+1)$, and
$$
1-|z|^2=\frac{4u}{|t+1|^2},\qquad\operatorname{Im}z=\frac{2v}{|t+1|^2}.
$$
Hence 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. Therefore
$$
\boxed{M(z)=(1+i)\frac{1+z}{1-z}}
$$
is 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>.
\Image[/past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2017-lens-quadrant-strip.png]
{title=A <Möbius transformation> followed by a logarithm maps the lens to a horizontal strip}
{description=The marked interior point travels from zero to $1+i$ to $(\log2)/2+i\pi/4$. Boundaries are excluded. Unbounded directions are indicated by arrows.}
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