= Solution
{c}
The disk exit time is finite almost surely; for example, either coordinate of the planar <Brownian motion> eventually leaves $(-1,1)$. By path continuity, the exit lies on the unit circle. Starting at the origin, rotational invariance of <planar Brownian motion> makes the exit law invariant under every circle rotation. The unique rotation-invariant probability measure is normalized <arc length>. Consequently \b[$Z_{T_D}$ is uniform on the unit circle], or equivalently its angle has density $1/(2\pi)$ on $[0,2\pi)$.
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