 Solution

The disk automorphism satisfies $f(w)=0$ and maps the unit circle bijectively to itself. Its derivative $(1-|w{|}^2)/(1-\overline{w}z{)}^2$ 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 $\mathbb{E}(t\wedge T)\le (1-|w{|}^2)/2$. Boundedness of $f^{\prime }$ 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 $\varphi (\Theta )$ modulo $2\pi$ is uniform, where $\varphi$ is strictly increasing and increases by exactly $2\pi$ in one circuit. The one-dimensional change-of-variables formula therefore givesIts 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.