 Conformal map of a vertical half-strip by sine

The sine maps $\{-\pi /2<\mathrm{Re}z<\pi /2,\mathrm{Im}z>0\}$ bijectively onto the upper half-plane: its imaginary part is $\cos (\mathrm{Re}z)\sinh (\mathrm{Im}z)>0$, and the equality of two sine values forces either equal arguments modulo $2\pi$ or arguments summing to an odd multiple of $\pi$. Only equality is possible in this domain. For surjectivity, solve $t^2-2iwt-1=0$ for $t=e^{iz}$. When $\mathrm{Im}w>0$, exactly one root has modulus below one and it has positive real part. The inverse $z=-i\log t$ therefore belongs to the half-strip. Composing with $(w-i)/(w+i)$ gives a conformal map onto the unit disc.