Lens-to-strip logarithmic conformal map ($F(z)=\mathrm{Log}\left((1+i)\frac{1+z}{1-z}\right)$)

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 $0<\mathrm{Im}\zeta <\pi /2$. 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.