= Solution
On the first quadrant choose the holomorphic logarithm with $0<\arg w<\pi/2$. It maps $w=\rho e^{i\theta}$ to $\log\rho+i\theta$, so its image is precisely the strip $0<\operatorname{Im}\zeta<\pi/2$. Its inverse there is the <exponential function>; both derivatives are nonzero. Composing with part (i) gives
$$
\boxed{F(z)=\operatorname{Log}\!\left((1+i)\frac{1+z}{1-z}\right).}
$$
This is a bijective <conformal map> from the lens onto the strip, with inverse
$$
z=\frac{e^\zeta-(1+i)}{e^\zeta+(1+i)}.
$$
The logarithm's branch is fixed by its argument range on the quadrant. Using a different branch would shift the strip vertically by an integer multiple of $2\pi$. This is the <lens-to-strip logarithmic conformal map>.
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