 Solution

Write $Z_0=r_0{e}^{i{\alpha }_0}$ 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 liftFor $t<T_{\mathbb{D}}$ 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 clockThe 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 lawIts first absolute moment diverges: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.