= Solution
Use the <Möbius transformation>
$$
w=T(z)=\frac{1+z}{1-z},\qquad z=\frac{w-1}{w+1}.
$$
The lens's boundary intersections are $z=\pm1$. Write $w=u+iv$. Substitution in the inverse gives $\operatorname{Im}z=2v/|w+1|^2$, and
$$
|z+i|^2<2\quad\Longleftrightarrow\quad |(1+i)w+i-1|^2<2|w+1|^2\quad\Longleftrightarrow\quad v<u.
$$
Consequently the two domain inequalities are exactly $0<v<u$, or $0<\arg w<\pi/4$. This algebra proves that $T$ maps the lens bijectively onto the quarter-sector, not merely that its boundary curves go to rays.
The map $w\mapsto w^4$ maps this sector bijectively onto the upper half-plane: the argument increases from $(0,\pi/4)$ to $(0,\pi)$, and the positive modulus map is bijective. Its inverse is the fourth-root branch with argument in $(0,\pi/4)$. Therefore the <quarter-sector mapping of a circular lens> is
$$
\boxed{F(z)=\left(\frac{1+z}{1-z}\right)^4:D\cap H\longrightarrow H}.
$$
Its derivative is nonzero in the lens, since $T'(z)=2/(1-z)^2$ and $T(z)\ne0$ there. Finally the Cayley transform $\zeta\mapsto(\zeta-i)/(\zeta+i)$ maps $H$ bijectively to the unit disc, because $|\zeta-i|<|\zeta+i|$ exactly when $\operatorname{Im}\zeta>0$. A required conformal bijection is thus
$$
\boxed{G(z)=\frac{\left(\dfrac{1+z}{1-z}\right)^4-i}{\left(\dfrac{1+z}{1-z}\right)^4+i}}.
$$
All poles and branching endpoints of the construction lie outside the open domain.
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