= Solution
{c}
On the <fluid free boundary> write $q=e^{-i\theta}$, with $-\pi<\theta<0$: the lip has $q=-1$ and the downstream limit has $q=1$. Then
$$
\frac{q-1}{q+1}=-i\tan(\theta/2).
$$
Substitution into the relation from part (ii) gives
$$
e^{\pi w/a}=-\csc^2(\theta/2).
$$
The surface lies on the top of the <velocity potential> strip, so its <logarithm> has <imaginary part> $\pi$, giving
$$
w=\frac a\pi\log\csc^2(\theta/2)+ia,\qquad
\frac{dw}{d\theta}=-\frac a\pi\cot(\theta/2).
$$
Finally $dz/dw=1/q=e^{i\theta}$, and therefore
$$
\boxed{\frac{dz}{d\theta}
=-\frac a\pi e^{i\theta}\cot(\theta/2).}
$$
The source's typeset $\cot\theta/2$ is interpreted as $\cot(\theta/2)$, as the derivation requires.
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