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The conformal map of a vertical half-strip by sine sends $-\pi /2<\mathrm{Re}z<\pi /2$, $\mathrm{Im}z>0$ to the upper half-plane. The complex potential of a two-dimensional potential flow $W(z)=\log (a+\sin z)$ therefore pulls back the radial logarithmic flow originating at $\zeta =-a$. Its complex velocity is $W^{\prime }=\cos z/(a+\sin z)$, with a boundary singularity at $z=-\pi /2+i\mathrm{arcosh}a$. For $z=x+iy$, define $D=(a+\sin x\cosh y{)}^2+{\cos }^{2}x{\sinh }^{2}y$. ThenThe bottom boundary carries rightward flow, the right boundary upward flow, and the left boundary flow away from the singular point. The velocity tends to $(0,1)$ at large height. The finite bottom corners are stagnation points, not singularities of $W$.

Use the conformal map of a vertical half-strip by sine followed by a Cayley transform:We verify that this maps the entire half-strip bijectively onto the unit disc. First,so the sine image lies in the upper half-plane. Also $\cos z\ne 0$ 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, solveNo 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,which forces $\mathrm{Re}t>0$. Choose its logarithm with argument in $(-\pi /2,\pi /2)$ and put $z=-i\log t$. Then $\mathrm{Im}z=-\log |t|>0$, its real part lies in the required interval, and $\sin z=w$. This proves that sine maps onto the upper half-plane.

Finally, for $\mathrm{Im}w>0$, $|w-i|<|w+i|$, 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.

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.