Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2009 / B / 6

Let $d=|z_0|$, the initial distance from the annulus center, and $T=T_{A_{r,R}}$. The function $z\mapsto \log |z-z_0|$ is harmonic on the annulus, so its value along stopped planar Brownian motion is a local martingale by Itô's formula. It is bounded between $\log r$ and $\log R$ before exit, hence is a true bounded martingale.

By the assumed almost sure finiteness of $T$, continuity, and dominated convergence,where $p$ is the inner-boundary exit probability. Solving gives the harmonic measure of an annulus:The two probabilities sum to one, as the exit must lie on one of the two circles.

Suppose first that $d=|z_0|>ϵ$. Apply part (a) with inner radius $ϵ$ and outer radius $R>d$:Each event implies finite hitting time of the $ϵ$ circle, so that hitting probability is one. If $d=ϵ$, the hitting time is zero. If $d<ϵ$, it is the exit time from a bounded disk. The martingale $|Z_t-z_0{|}^2-2t$, stopped before that exit, gives a finite mean exit-time bound $({ϵ}^2-d^2)/2$. Thus the asserted hitting time is almost surely finite in this case too.

For alternating visits, choose two closed smaller disks $K_i$ strictly inside the given open disks $D_i$, so their distance is positive. Define successive first hitting times of $K_1,K_2,K_1,K_2,\dots$. Translation invariance and the Strong Markov property apply the just-proved hitting result from every successive starting point. A countable intersection makes all these times finite almost surely, with the required disk memberships.

These increasing times cannot accumulate at a finite time. On a compact time interval a continuous Brownian path is uniformly continuous, whereas every consecutive visit must cross the fixed positive distance between $K_1$ and $K_2$. ThereforeThis is the recurrence of planar Brownian motion in neighborhoods; it does not assert that every individual point is hit.

Write $Z_t=X_t+i{Y}_t$ and $u=\mathrm{Re}g$. Since $g$ is entire, $u$ is a harmonic function, so $u_{xx}+u_{yy}=0$. The two-dimensional Itô formula givesThe drift vanishes. Stop on successively larger disks to bound the derivatives and define the stochastic integrals without global growth assumptions. ThusThis is the harmonic functions of Brownian motion principle.

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 $g(z)\to 0$ 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 $u(Z_t)=\mathrm{Re}g(Z_t)$ 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 $\mathrm{Im}g$ 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
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.