= Solution
Fix an outer radius $R>\rho=|z_0|$. By continuity, reaching zero before radius $R$ requires first reaching every radius $\varepsilon\in(0,\rho)$. The <planar Brownian annulus hitting probability> from (i) therefore implies
$$
0\leq\mathbb P(\tau_0<\tau_R)
\leq\mathbb P(\tau_\varepsilon<\tau_R)
=\frac{\log(R/\rho)}{\log(R/\varepsilon)}\longrightarrow0
\quad(\varepsilon\downarrow0).
$$
Thus $\mathbb P(\tau_0<\tau_R)=0$ for every fixed $R>\rho$.
On any continuous <sample path> that reaches zero in finite time, the radius has a finite maximum before that time. There is therefore some integer $m>\rho$ for which zero is reached before radius $m$. Consequently
$$
\{\tau_0<\infty\}=\bigcup_{m\in\mathbb N,\,m>\rho}\{\tau_0<\tau_m\}.
$$
Every event on the right has <probability> zero, so
$$
\boxed{\mathbb P(\tau_0=\infty)=1.}
$$
This proves that <planar Brownian motion avoids a fixed point> even though it can reach arbitrarily small positive radii.
Back to article page
