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Suppose a nonconstant polynomial $P$ has no zero in $\mathbb{C}$. Its reciprocal $g=1/P$ is then an entire holomorphic function. If $d\ge 1$ is the degree and $a_d\ne 0$ 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 $g(z)\to 0$. 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, andFor $\epsilon >0$, choose $R$ such that $|g(w)|\le \epsilon$ whenever $|w|>R$, and write $K={\sup }_w|g(w)|$. The two-dimensional Gaussian density of $z+B_t$ is bounded by $1/(2\pi t)$, soConsequentlyLet $t\to \infty$, then $\epsilon \downarrow 0$, to obtain $g(z)=0$. This is impossible for $1/P$. ThusThis 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.