= Solution
{c}
Separate the real and imaginary parts of the derivative and integrate from $D=(0,1)$:
$$
x(\theta)=-\frac{2a}{\pi}\int_{-\pi/2}^{\theta}\cos s\cot s\,ds,\qquad
y(\theta)=1-\frac{2a}{\pi}\int_{-\pi/2}^{\theta}\sin s\cot s\,ds.
$$
Because $\cos s\cot s=\csc s-\sin s$ and $\sin s\cot s=\cos s$, these integrals are
$$
x(\theta)=-\frac{2a}{\pi}\left[\log|\tan(\theta/2)|+\cos\theta\right],\qquad
y(\theta)=1-\frac{2a}{\pi}(1+\sin\theta).
$$
As $\theta\to0^-$, $x\to+\infty$ and $y\to1-2a/\pi$. This must equal the outgoing stream depth $a$, so
$$
\boxed{a=\frac{\pi}{\pi+2}.}
$$
\Image[/past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2009-draining-flow.png]
{title=Draining free surface and its potential-strip and hodograph representations}
The solution has a vertical tangent at $D$ and approaches the horizontal line $y=a$ downstream, agreeing with both solid-wall and free-streamline conditions.
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