= Solution
{c}
Assume that a nonconstant polynomial $f$ has no zero. Then $g=1/f$ is entire. The leading polynomial term makes $|f(z)|\to\infty$ as $|z|\to\infty$, so $g(z)\to0$ there. On any remaining compact disk, continuity and absence of zeros bound $1/|f|$. Consequently $g$ is bounded on the entire plane.
Part (c)(i) makes $u(Z_t)=\operatorname{Re}g(Z_t)$ a bounded <local martingale>, hence a true bounded martingale. The <martingale convergence theorem> says it converges almost surely as $t\to\infty$.
The harmonic function $u$ cannot be constant. Otherwise the <Cauchy-Riemann equations> would make both derivatives of $\operatorname{Im}g$ zero, so $g$ and then $f$ would be constant. Choose points $z_1,z_2$ with $u(z_1)<u(z_2)$. Continuity supplies two disjoint disks on which the ranges of $u$ are separated by a positive gap. By part (b), the Brownian path visits these disks alternately at times tending to infinity. The corresponding subsequences of $u(Z_t)$ then remain separated, contradicting its almost sure convergence.
The zero-free assumption is impossible. Hence
\b[Every nonconstant complex polynomial has a complex root.]
This proves the <fundamental theorem of algebra> using recurrence and martingale convergence, without invoking Liouville's theorem as a prior algebraic shortcut.
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