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The principal complex inverse sine onis the branchwhose derivative equals one at zero.

It is the complex sine member of the inverse trigonometric functions.

The inverse trigonometric functions invert trigonometric functions on branches where they are one-to-one. Standard real ranges are $[-\pi /2,\pi /2]$ for $\arcsin$, $[0,\pi ]$ for $\arccos$, and $(-\pi /2,\pi /2)$ for $\arctan$. Thus $\arcsin (1/2)=\pi /6$ but $\arcsin (\sin (5\pi /6))=\pi /6$: the inverse returns the selected branch. The notation ${\sin }^{-1}x$ usually means $\arcsin x$, not $1/\sin x$. Implicit differentiation gives $(\arcsin x{)}^{\prime }=(1-x^2{)}^{-1/2}$ for $|x|<1$.

The finite corners have interior angle $3\pi /2$, hence exterior-angle parameters $-1/2$. The channel end at $Z=0$ is logarithmic. Their Schwarz-Christoffel mapping is consequentlyChoose $R(Z)=\sqrt{Z^2-1}$ analytic in the upper half-plane with $R(Z)\sim Z$ at upper infinity. In particular $R(iy)=i\sqrt{1+y^2}$ for $y>0$. A direct differentiation givesHence $f=A+CH$ with $H=R+{\tan }^{-1}(1/R)$.

The inverse trigonometric functions here is the branch obtained by analytic continuation from $Z>1$ through the upper half-plane, not an independent Cauchy principal value evaluation at each point. Its endpoint values give $H(1)=\pi /2$, $H(-1)=-\pi /2$. Imposing $f(1)=1$ and $f(-1)=-1$ therefore gives $A=0$, $C=2/\pi$:Along the positive imaginary axis, $H$ is purely imaginary and tends to $-i\infty$ as $Z\to 0$; at upper infinity $H\sim Z$. The real intervals $(0,1)$ and $(-1,0)$ trace the channel walls $x=1$ and $x=-1$ downward, while the outer intervals trace the horizontal boundary. This verifies the stated end assignments as well as the constants. The map is the Schwarz-Christoffel map of a half-plane joined to a strip.