= Solution
{c}
Write the <complex potential of a two-dimensional potential flow> as $w=\phi+i\psi$, and the complex velocity as $q=w'=u-iv$. The lower solid wall has $\psi=0$. The upper solid wall and its continuing free streamline have constant $\psi$ equal to the far-field volume flux per unit depth, namely $a$. Consequently the potential domain is the infinite strip
$$
\boxed{0<\operatorname{Im}w<a.}
$$
The points $A,E$ correspond to $\phi\to-\infty$, $B,C$ to $\phi\to+\infty$, and $D$ is $w=ia$ by the chosen potential normalization.
In the <hodograph plane>, the horizontal wall has $q$ real positive, the vertical wall has $q$ imaginary positive because the physical velocity is downward, and the <fluid free boundary> has $|q|=1$. The upstream reservoir velocity tends to zero, the outgoing velocity is one, and the velocity at $D$ is $i$. Thus the hodograph is \b[the first-quadrant unit quarter disk], with $A,E$ at $q=0$, $B,C$ at $q=1$, and $D$ at $q=i$.
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