= Solution
{c}
Let $d=|z_0|$, the initial distance from the annulus center, and $T=T_{A_{r,R}}$. The function $z\mapsto\log|z-z_0|$ is harmonic on the annulus, so its value along stopped planar <Brownian motion> is a <local martingale> by <Itô's formula>. It is bounded between $\log r$ and $\log R$ before exit, hence is a true bounded martingale.
By the assumed almost sure finiteness of $T$, continuity, and <dominated convergence>,
$$
\log d=p\log r+(1-p)\log R,
$$
where $p$ is the inner-boundary exit probability. Solving gives the <harmonic measure of an annulus>:
$$
\boxed{\mathbb P(|Z_T-z_0|=r)=\frac{\log(R/d)}{\log(R/r)},\qquad
\mathbb P(|Z_T-z_0|=R)=\frac{\log(d/r)}{\log(R/r)}.}
$$
The two probabilities sum to one, as the exit must lie on one of the two circles.
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