= Solution
{c}
Write $Z_0=r_0e^{i\alpha_0}$ with $0<r_0<1$ and choose its initial argument once. The exponential map in the source is a covering of the punctured <disc>, rather than a bijective map onto the full <disc>. Its pathwise inverse is the continuous lift
$$
W_t=-i\bigl(\log r_0+i\alpha_0+L_t\bigr)
=\alpha_0+\theta_t-i\log|Z_t|.
$$
For $t<T_{\mathbb D}$ this lies in the <upper half-plane>, and at the <disc> exit it lies on the real line. Local <conformal invariance of planar Brownian motion>, or the <quadratic variation> of its analytic <derivative> $-i/Z$, shows that $W_t$ is <planar Brownian motion> run at the clock
$$
A_t=\int_0^t|Z_s|^{-2}ds.
$$
The <disc> exit time is finite almost surely. The path avoids zero on the <compact> interval through its exit, so its minimum modulus there is positive and $A_{T_{\mathbb D}}<\infty$. Before this terminal clock time the lifted imaginary coordinate is positive, and at it the coordinate is zero. Hence it is exactly the upper-half-plane exit time of the time-changed <Brownian motion>, not an additional arbitrary <stopping time>.
The starting point of the lift is $\alpha_0+ib$ with $b=-\log r_0>0$. Its real exit coordinate has <Cauchy distribution> $b/[\pi((s-\alpha_0)^2+b^2)]$. Since that coordinate is $\alpha_0+\theta_{T_{\mathbb D}}$, we obtain the stronger exact law
$$
\boxed{\theta_{T_{\mathbb D}}\ \sim\ \operatorname{Cauchy}(0,-\log|Z_0|).}
$$
Its first absolute moment diverges:
$$
\mathbb E|\theta_{T_{\mathbb D}}|
=\frac{2b}{\pi}\int_0^\infty\frac{s}{s^2+b^2}ds
=\boxed{\infty}.
$$
There is no contradiction with the terminal point lying on the <compact> <unit circle>: reducing its argument modulo $2\pi$ loses the number and direction of turns accumulated before exit.
\Image[/past-exam-of-the-stochastic-differential-equations-and-brownian-motion-in-complex-analysis-course-of-the-university-of-oxford/2007-winding-cauchy.png]
{title=The logarithmic lift and the Cauchy tail of the accumulated winding angle}
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