 Solution

Let $\rho =|z_0|$ and $f(x,y)=\log \sqrt{x^2+y^2}$. On the annulus $r<\sqrt{x^2+y^2}<R$, its gradient and Laplace operator areThus $f$ is a harmonic function there. The stopped Itô formula givesThe sum of the squared coefficients is $|B_s{|}^{-2}\le r^{-2}$ before $\tau$. The Itô isometry therefore makes the stochastic term a true square-integrable martingale on every finite interval. Continuity prevents overshoot at exit, so $\log r\le \log |B_{t\wedge \tau }|\le \log R$. This proves the required bounded martingale property.

Before taking the terminal expectation, prove that $\tau$ is finite. Set ${\theta }_R=\inf \{t\ge 0:|B_t|=R\}$. The Itô formula for $|B{|}^2$ shows thatis a martingale, since its stopped stochastic integrands are bounded. Taking expectations yieldsThe monotone convergence theorem gives $\mathbb{E}{\theta }_R\le (R^2-{\rho }^2)/2<\infty$. Hence $\tau \le {\theta }_R<\infty$ almost surely.

The bounded convergence theorem, equivalently the optional stopping theorem applied to this bounded martingale, gives $\mathbb{E}\log |B_{\tau }|=\log \rho$. The two exit events partition the sample space up to a null set. Writing $p=\mathbb{P}({\tau }_r<{\tau }_R)$,Solving gives the planar Brownian annulus hitting probability