 Solution

Choose the branch of $\sqrt{(\zeta +1)(\zeta -1)}$ holomorphic in the upper half-plane whose boundary value at zero is $+i$. Thenand integration gives $\boxed{z=\arcsin \zeta ,\zeta =\sin z}$ on this branch. The normalization and intended upper-half-plane image fix the otherwise ambiguous square-root sign.

To see the image, put $z=x+iy$. For $-\pi /2<x<\pi /2$ and $y>0$,The bottom edge maps monotonically onto $[-1,1]$. The right wall maps by $\sin (\pi /2+iy)=\cosh y$ onto $[1,\infty )$, and the left by $\sin (-\pi /2+iy)=-\cosh y$ onto $(-\infty ,-1]$. These boundaries describe the conformal map of a vertical half-strip by sine.

For completeness, the map is one-to-one in the interior: equality of two sine values forces arguments differing by $2\pi m$ or summing to $(2m+1)\pi$, and neither alternative yields two distinct points in this half-strip. It is onto the upper half-plane as well. Setting $w=e^{iz}$ gives $w^2-2i\zeta w-1=0$; the root product is $-1$, and no root lies on $|w|=1$ because that would make $\zeta$ real. Thus exactly one root has $|w|<1$. The identity $\mathrm{Im}\zeta =(\mathrm{Re}w)(|w{|}^{-2}-1)/2$ then gives $\mathrm{Re}w>0$, so $z=-i\log w$ belongs to the half-strip. The closure gives the stated boundary-inclusive region; holomorphy is understood in its open interior.

In this region the transformed function is $G(z)=\log (2+\sin z)$. Its only finite intrinsic branch point in the closure occurs where $2+\sin z=0$:It lies on the left boundary. There are no singularities in the open half-strip. The points $\zeta =\pm 1$, where $f^{\prime }$ diverges, map to the bottom corners; $G$ is holomorphic through those corners since $2+\sin z\ne 0$ there.

For the principal logarithm, the boundary segment $z=-\pi /2+iy$, $y\ge \mathrm{arcosh}2$, is the preimage of its cut $\zeta \le -2$. Points on that cut above $z_{\ast }$ admit local analytic continuation after changing branch; $z_{\ast }$ itself is the genuine logarithmic branch point.