= Solution
{c}
Use the branch of $\sqrt{1-Z^2}$ that is positive for $Z=iY$, $Y>0$. The <right-half-plane slit map>
$$
\boxed{f(Z)=(1-Z^2)^{-1/2}}
$$
is holomorphic in the <upper half-plane>, with <derivative> $Z(1-Z^2)^{-3/2}$, the required <Schwarz-Christoffel formula> form. Its <derivative> has no interior zero.
To verify the image and global injectivity, factor the construction. Squaring maps the <upper half-plane> bijectively onto $\mathbb C\setminus[0,\infty)$. Thus $1-Z^2$ ranges over $\mathbb C\setminus(-\infty,1]$. The positive-real-part square root maps this onto the <right half-plane> with the interval $(0,1]$ removed; reciprocation maps it onto the <right half-plane> slit along $[1,\infty)$. This also identifies the slit tip as a boundary point; the strict inequality in the printed slit description must be understood with that boundary convention.
Approaching the real axis from above gives
$$
\begin{array}{c|cccc}
Z&(-\infty,-1)&(-1,0)&(0,1)&(1,\infty)\\\hline
f(Z)&\text{negative imaginary axis}&\text{lower slit bank}&\text{upper slit bank}&\text{positive imaginary axis}.
\end{array}
$$
In particular $Z=-1$ maps to the lower infinite prime end $A$, $Z=0$ to the tip $B=1$, and $Z=1$ to the upper infinite prime end $C$. The real-axis point at infinity maps to zero on the vertical wall.
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