= Solution
{c}
The lower wall has <streamfunction> zero. The <streamline> through the lip and along the upper <fluid free boundary> has <streamfunction> equal to the downstream <volumetric flow rate>, namely $a$ because the downstream speed is one and its depth is $a$. The <complex potential of a two-dimensional potential flow> therefore fills
$$
\boxed{0<\operatorname{Im}w<a.}
$$
The chosen <velocity potential> origin gives $w(D)=ia$. The remote reservoir endpoints $A,E$ have <velocity potential> tending to $-\infty$; the downstream endpoints $B,C$ have <velocity potential> tending to $+\infty$.
Put $q=dw/dz=u-iv$. On the floor its values run from zero to one along the positive <real axis>. Above the plate the flow goes left toward the lip, so the plate <velocities> run along the negative <real axis> from zero at $E$ to $-1$ at $D$. On the <fluid free boundary> <Bernoulli's equation> and its constant <pressure> give $|q|=1$. Its downward <velocity> has $v<0$, hence $\operatorname{Im}q>0$. Consequently the <hodograph plane> domain is
$$
\boxed{|q|<1,\quad\operatorname{Im}q>0,}
$$
the upper unit <semicircle>. Its marked boundary values are $q(A)=q(E)=0$, $q(D)=-1$, and $q(B)=q(C)=1$.
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