= Solution
{c}
The <disk automorphism> satisfies $f(w)=0$ and maps the unit circle bijectively to itself. Its derivative $(1-|w|^2)/(1-\overline wz)^2$ is bounded above and away from zero on the closed disk. By <conformal invariance of planar Brownian motion>, the image path up to the exit time is a time-changed planar Brownian motion started at zero. Its exit point is $f(Z_T)$; the clock changes when it exits, but not where.
The original exit time is finite almost surely. For example, $|Z_{t\wedge T}|^2-2(t\wedge T)$ is a martingale, so $\mathbb E(t\wedge T)\leq(1-|w|^2)/2$. Boundedness of $f'$ makes the image exit clock finite too. By <rotational invariance of planar Brownian motion>, the image exit angle is uniform on $[0,2\pi)$.
Let $\Theta=\arg Z_T$. Parts (a) and (b) show that $\phi(\Theta)$ modulo $2\pi$ is uniform, where $\phi$ is strictly increasing and increases by exactly $2\pi$ in one circuit. The one-dimensional change-of-variables formula therefore gives
$$
\boxed{h(\theta)=\frac{\phi'(\theta)}{2\pi}
=\frac1{2\pi}\frac{1-|w|^2}{|w-e^{i\theta}|^2}.}
$$
Its integral is one because the total angle change is $2\pi$. This <harmonic measure> density is the <Poisson kernel for the unit disk> divided by $2\pi$. It is uniform for $w=0$ and concentrates near the boundary point closest to $w$ as $|w|$ increases.
\Image[/past-exam-of-the-brownian-motion-in-complex-analysis-course-of-the-university-of-oxford/2011-disk-exit-density.png]
{title=The disk exit-angle density for Brownian motion started at w=0.6+0.25i. Darker boundary arcs have greater density; the dashed horizontal line is the uniform density for a start at the centre.}
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