= Solution
{c}
Since $f(0)=0$, the function $g(z)=f(z)/z$ extends analytically to zero by $g(0)=f'(0)$. It is bounded on the disk: outside $|z|<1/2$, $|g(z)|\leq2$, while on the closed half-radius disk its analytic extension is continuous and hence bounded.
Fix $z\ne0$ and choose $|z|<r<1$. Let $T_r$ be the first exit time of Brownian motion started at $z$ from the radius-$r$ disk. The bounded analytic martingale identity in part (a) gives
$$
g(z)=\mathbb E_z[g(Z_{T_r})].
$$
On this exit circle, $|g(Z_{T_r})|=|f(Z_{T_r})|/r\leq1/r$. The triangle inequality gives $|g(z)|\leq1/r$. Letting $r\uparrow1$ proves $|g(z)|\leq1$, and therefore
$$
\boxed{|f(z)|\leq|z|\quad\text{for all }z\in D.}
$$
At zero the claim holds by the assumed zero value. This is the <Schwarz lemma>, obtained from Brownian optional stopping rather than an analytic maximum principle.
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