= Lens-to-strip logarithmic conformal map
{title2=$F(z)=\operatorname{Log}\!\left((1+i)\frac{1+z}{1-z}\right)$}
Compose the <lens-to-quadrant Möbius map> with the logarithm having argument in $(0,\pi/2)$ on the first quadrant. Polar coordinates show that the latter maps bijectively to the horizontal strip $0<\operatorname{Im}\zeta<\pi/2$. Its inverse is the <exponential function>, so the combined inverse is $(e^\zeta-(1+i))/(e^\zeta+(1+i))$. Both derivatives are nonzero, proving <conformality>. The logarithm's branch fixes the strip's vertical position; adding $2\pi i$ shifts that position without changing the exponential inverse.
\Image[/past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2017-lens-quadrant-strip.png]
{title=The <lens-to-quadrant Möbius map> followed by a <logarithm branch>}
{description=The marked point shows the same interior point at each stage. The logarithm converts radial size into horizontal position and the quadrant's angle into the strip's height.}
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