= Solution
{c}
First, <planar Brownian motion> started away from zero almost surely never hits zero. To see this without confusing it with neighborhood recurrence, stop in an <annulus> with inner radius $\epsilon$ and fixed outer radius $R>|Z_0|$. The preceding logarithmic calculation gives probability $\log(R/|Z_0|)/\log(R/\epsilon)$ of reaching the inner circle first. A hit of zero before the outer exit would require all these inner hits, so its probability is zero as $\epsilon\downarrow0$. On every finite path interval the continuous path is bounded; a countable union over integer outer radii and integer times proves the all-time assertion.
Consequently the <integrand> $1/Z$ is locally square integrable: on every <compact> time interval the nonvanishing continuous path has a positive minimum modulus. Write $Z=X+iY$ with independent standard real coordinates. Complex <quadratic covariation> satisfies $d[Z,Z]=dt-dt+2i\,d[X,Y]=0$, so analytic <Itô's lemma> has no second-derivative drift. For $L=\int Z^{-1}dZ$ one likewise has $d[L,L]=0$, and $d[e^{-L},Z]=0$. The product rule therefore gives
$$
d(Z_te^{-L_t})=e^{-L_t}dZ_t-Z_te^{-L_t}dL_t=0.
$$
Its initial value is $Z_0$, proving, simultaneously for all times by <continuity>,
$$
\boxed{Z_t=Z_0e^{L_t}.}
$$
Thus $\Re L_t=\log(|Z_t|/|Z_0|)$ and $\Im L_t$ is the \b[continuous accumulated change of <complex argument>] of the path about zero, measured in radians. It is a winding angle, without reduction modulo $2\pi$. A single-valued <logarithm> on the whole punctured plane is unnecessary; the path has a <continuous stochastic logarithm of planar Brownian motion>.
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