= Solution
{c}
Integrate real and imaginary parts from the lip value $z(-\pi)=i$:
$$
x(\theta)=-\frac a\pi\int_{-\pi}^\theta
\cos s\,\cot(s/2)\,ds,\qquad
y(\theta)=1-\frac a\pi\int_{-\pi}^\theta
\sin s\,\cot(s/2)\,ds.
$$
The identities $\sin s\cot(s/2)=1+\cos s$ and
$$
\int\cos s\cot(s/2)\,ds=\log(1-\cos s)+\cos s
$$
give the explicit parametrization
$$
x(\theta)=-\frac a\pi
\bigl[\log(1-\cos\theta)+\cos\theta-\log2+1\bigr],
$$
$$
y(\theta)=1-\frac a\pi(\theta+\pi+\sin\theta),
\qquad-\pi<\theta<0.
$$
As $\theta\uparrow0$, $x\to+\infty$ and $y\to1-a$. This must be the downstream height $a$, so
$$
\boxed{a=\frac12.}
$$
The surface initially moves slightly left of the lip before turning right. Its leftmost point occurs at $\theta=-\pi/2$, with $x_{\min}=-(1-\log2)/(2\pi)$. This is the <half-contraction free streamline at a horizontal canal lip>.
\Image[/past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2008-canal-flow.png]
{title=The <fluid free boundary> contracts to half the lip height and traces the upper <hodograph plane>}
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