= Solution
{c}
Write $Z_t=z+B_t^1+iB_t^2$, where the coordinates are independent standard <Brownian motions>, and write $f=u+iv$. The <Cauchy-Riemann equations> imply $\Delta u=\Delta v=0$. <Itô's formula> therefore makes
$$
M_t=u(Z_t)-u(z),\qquad N_t=v(Z_t)-v(z)
$$
<continuous local martingales>. Their <quadratic variations> and <quadratic covariation> are
$$
\langle M\rangle_t=\langle N\rangle_t=A_t:=\int_0^t|f'(Z_s)|^2ds,\qquad
\langle M,N\rangle_t=0.
$$
Indeed the two gradient norms are equal to $|f'|$, and their dot product vanishes by the Cauchy-Riemann equations.
The clock is finite at every finite time, since a continuous Brownian path has compact range there and $f'$ is continuous. To show its lifetime is infinite, choose a small disc on which $|f'|^2\geq c>0$; such a disc exists because the <entire function> is nonconstant. A supporting recurrence result is that planar Brownian motion spends infinite total time in every nonempty open disc almost surely. One proof combines recurrent returns to a smaller disc with the strong Markov property: each visit has a uniformly positive chance of lasting a fixed positive time in the larger disc. Equivalently, use the <divergence of a positive planar Brownian occupation integral>. Therefore
$$
\boxed{A_t\longrightarrow\infty\quad\text{almost surely}.}
$$
Finally the <common-clock time change of orthogonal local martingales> states that two continuous local martingales starting at zero with common bracket $A$, zero cross bracket, and $A_\infty=\infty$ are the coordinates of a planar Brownian motion evaluated at $A_t$. Applying it gives a <planar Brownian motion> $\widetilde Z$ starting at zero with
$$
\boxed{f(Z_t)=f(z)+\widetilde Z_{A_t}.}
$$
This proves <conformal invariance of planar Brownian motion> in the entire-function setting, including the <divergence of the conformal Brownian clock for an entire function>.
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