= Solution
Use the <conformal map of a vertical half-strip by sine> followed by a <Cayley transform (hyperbolic geometry)>:
$$
\boxed{F(z)=\frac{\sin z-i}{\sin z+i}.}
$$
We verify that this maps the entire half-strip bijectively onto the <unit disc>. First,
$$
\operatorname{Im}\sin(x+iy)=\cos x\sinh y>0,
$$
so the <sine> image lies in the upper half-plane. Also $\cos z\ne0$ in the domain, making the map locally a <conformal map>. If $\sin z=\sin w$, the identity for a difference of sines gives either $z-w=2\pi n$ or $z+w=(2n+1)\pi$. The real-part bounds force $n=0$ in the first case, while positive imaginary parts exclude the second. Thus <sine> is injective.
For surjectivity, given any $w$ in the upper half-plane, solve
$$
t^2-2iwt-1=0.
$$
No root is on the unit circle, since $w=(t-t^{-1})/(2i)$ would then be real. The product of the two roots is $-1$, so exactly one root has $|t|<1$. For this root,
$$
\operatorname{Im}w=\frac12\operatorname{Re}t\,(|t|^{-2}-1)>0,
$$
which forces $\operatorname{Re}t>0$. Choose its logarithm with argument in $(-\pi/2,\pi/2)$ and put $z=-i\log t$. Then $\operatorname{Im}z=-\log|t|>0$, its real part lies in the required interval, and $\sin z=w$. This proves that <sine> maps onto the upper half-plane.
Finally, for $\operatorname{Im}w>0$, $|w-i|<|w+i|$, and the <Möbius transformation> $(w-i)/(w+i)$ is a bijection to the <unit disc>. Its inverse is $w=i(1+\zeta)/(1-\zeta)$. Both constituent derivatives are nonzero, completing the proof for the displayed <conformal map>.
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