= Solution
{c}
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 consequently
$$
f'(Z)=C\frac{\sqrt{Z^2-1}}{Z}.
$$
Choose $R(Z)=\sqrt{Z^2-1}$ 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 gives
$$
\frac d{dZ}\left(R+\tan^{-1}\frac1R\right)
=\frac ZR-\frac1{ZR}=\frac RZ.
$$
Hence $f=A+C H$ 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 $A=0$, $C=2/\pi$:
$$
\boxed{z=\frac2\pi\left(\sqrt{Z^2-1}
+\tan^{-1}(Z^2-1)^{-1/2}\right).}
$$
Along the positive imaginary axis, $H$ is purely imaginary and tends to $-i\infty$ as $Z\to0$; 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>.
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