= Solution
The clock $A_t=\int_0^te^{2\beta_s}ds$ is continuous, strictly increasing and unbounded. Define $\tau_r=\inf\{t:A_t\geq r\}$. Part (a), with the initial vector retained, proves that
$$
\boxed{\mathcal B_r=Z_{\tau_r}\text{ is planar Brownian motion started at }(1,0)\text{ in }\mathcal F_{\tau_r}.}
$$
Every finite $r$ has finite $\tau_r$, and $|\mathcal B_r|=e^{\beta_{\tau_r}}>0$. On the common event where the clock is a continuous bijection of $[0,\infty)$, this holds simultaneously for all times. Thus the origin is a <polar point for planar Brownian motion> started at $(1,0)$:
$$
\boxed{\mathbb P(\mathcal B_r\ne0\text{ for all }r\geq0)=1.}
$$
To prove unbounded-time visits near zero, let $t_n=\inf\{t\geq n:\beta_t=-n\}$. Brownian recurrence makes each $t_n$ finite almost surely, and $t_n\geq n\to\infty$. Set $r_n=A_{t_n}$. Since $A_t\to\infty$, $r_n\to\infty$, while
$$
|\mathcal B_{r_n}|=|Z_{t_n}|=e^{-n}\longrightarrow0.
$$
Consequently
$$
\boxed{\liminf_{r\to\infty}|\mathcal B_r|=0\quad\text{almost surely}.}
$$
For every fixed neighbourhood of zero there are visits at arbitrarily large times. Since $\mathcal B$ has the law of planar <Brownian motion> with this initial value, both conclusions hold for that Brownian law. This <complex exponential construction of planar Brownian motion> proves the two properties together without confusing recurrence near a point with hitting the point.
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