= Solution
Suppose a nonconstant <polynomial> $P$ has no zero in $\mathbb C$. Its reciprocal $g=1/P$ is then an entire <holomorphic function>. If $d\geq1$ is the degree and $a_d\ne0$ its leading coefficient, division by $z^d$ shows $P(z)/z^d\to a_d$ as $|z|\to\infty$. Hence $|P(z)|\to\infty$, so $g(z)\to0$. Continuity on compact disks now makes $g$ bounded on all of $\mathbb C$.
Fix $z\in\mathbb C$ and let $B$ be standard <planar Brownian motion> started at zero. By part (a) and <Itô formula>, both components of $g(z+B_t)$ are <local martingales>. Since $g$ is bounded, they are true <martingales>, and
$$
g(z)=\mathbb E[g(z+B_t)]\qquad(t\geq0).
$$
For $\varepsilon>0$, choose $R$ such that $|g(w)|\leq\varepsilon$ whenever $|w|>R$, and write $K=\sup_w|g(w)|$. The two-dimensional Gaussian density of $z+B_t$ is bounded by $1/(2\pi t)$, so
$$
\mathbb P(|z+B_t|\leq R)\leq\frac{\pi R^2}{2\pi t}=\frac{R^2}{2t}.
$$
Consequently
$$
|g(z)|\leq\mathbb E|g(z+B_t)|\leq\varepsilon+\frac{KR^2}{2t}.
$$
Let $t\to\infty$, then $\varepsilon\downarrow0$, to obtain $g(z)=0$. This is impossible for $1/P$. Thus
$$
\boxed{\text{Every nonconstant complex polynomial has a complex zero.}}
$$
This <Brownian proof of the fundamental theorem of algebra> uses only harmonicity, the bounded-martingale property and the spreading Gaussian transition density. It does not use either maximum principle from part (b), nor does it incorrectly assume that planar Brownian motion hits a specified point.
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