= Solution
{c}
For the <velocity potential> strip, $e^{\pi w/a}$ maps to the <upper half-plane>. The real <Möbius transformation>
$$
\zeta=-\frac1{1+e^{\pi w/a}}
$$
preserves that <half-plane> and sends its marked endpoints to $-1,\infty,0$ in the order $A/E,D,B/C$.
For the <hodograph plane>, first put $r=(q-1)/(q+1)$. Its image is the second quadrant, bounded by the negative real ray and positive imaginary ray. Squaring takes that quadrant to the lower <half-plane>, and negating restores the <upper half-plane>. Thus
$$
\zeta=-\left(\frac{q-1}{q+1}\right)^2
$$
has exactly the same three marked values. Uniqueness of a <conformal map> to a <half-plane> with three assigned boundary values makes these maps equal. Cancel the minus signs to obtain
$$
\boxed{\frac1{e^{\pi w/a}+1}
=\left(\frac{w'-1}{w'+1}\right)^2.}
$$
The sign in the intermediate map is important for identifying the upper, rather than lower, <half-plane>.
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