Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 28 / 5

Write $f=u+iv$. The prescribed complex-linear differential is exactly the Cauchy-Riemann equationsA holomorphic function is smooth, as follows locally from its Cauchy integral formula, so its mixed second partial derivatives commute. Differentiating the two equations givesand, rewriting them as $v_x=-u_y$, $v_y=u_x$,Thus $\boxed{\Delta \Re f=\Delta \Im f=0}$: both components are harmonic functions. Here $\Delta ={\partial }_x^2+{\partial }_y^2$, the sign convention used for the Brownian generator $\Delta /2$.

Let $u=\Re f$, and suppose $u$ has a local maximum at $z_0\in U$. Choose $R>0$ such that the closed disk $\overline{D(z_0,R)}$ lies in $U$ and $u(z)\le u(z_0)$ throughout it. Start planar Brownian motion $B$ at $z_0$, and for $0<r<R$ let ${\tau }_r$ be its first exit from $D(z_0,r)$.

By part (a) and Itô formula, $u(B_{t\wedge {\tau }_r})$ is a local martingale; boundedness of $u$ on the compact disk makes it a true bounded martingale. The exit time is finite almost surely. One direct check applies the same formula to $|B_{t\wedge {\tau }_r}-z_0{|}^2$, giving $2\mathbb{E}(t\wedge {\tau }_r)=\mathbb{E}|B_{t\wedge {\tau }_r}-z_0{|}^2\le r^2$. Thus $\mathbb{E}{\tau }_r\le r^2/2$.

The optional stopping theorem, first at $t\wedge {\tau }_r$ and then by bounded convergence, givesThe exit point is uniformly distributed on the circle: rotational invariance of planar Brownian motion about its starting point makes its exit law rotation invariant, and normalized arc length is the unique rotation-invariant probability measure on a circle. Since every boundary value is at most $u(z_0)$, equality of the mean forces every boundary value to equal $u(z_0)$; continuity rules out a strict deficit at any boundary point. This holds for every $0<r<R$, so $u$ is constant on the entire disk.

The Cauchy-Riemann equations then give both partial derivatives of $v=\Im f$ equal to zero there. Thus $f$ is constant on that disk. The identity theorem for holomorphic functions propagates this constant to the connected component of $z_0$ in $U$. Applying the same Brownian argument to $v$ proves the assertion for the imaginary part as well:If $U$ is connected, this is constancy on all of $U$. For a disconnected open set the printed global wording needs this qualification: a function taking different constants on two disjoint disks is holomorphic and has local maxima on each disk while not being globally constant.

Suppose $|g|$ has a local maximum at $z_0$. If $g(z_0)=0$, the nonnegativity of the modulus forces $g=0$ throughout a neighbourhood, and the identity theorem for holomorphic functions makes it zero on that connected component.

If $g(z_0)\ne 0$, take $c=2i/g(z_0)$. Then $cg(z_0)=2i$, and continuity gives a disk $D$ around $z_0$ on which $\Im (cg)>1$. This lies within the given half-plane for a branch of the holomorphic logarithm. On $D$ defineComposition preserves holomorphy, and the real logarithm is strictly increasing, so $\Re f$ has a local maximum at $z_0$. Part (i) forces $f$ to be constant on the disk. Exponentiating gives $cg=e^f$, hence $g$ is constant there and, by the identity theorem for holomorphic functions, on the whole connected component. ThereforeThis proves the maximum modulus principle by the requested logarithm argument. Multiplication by $c$ is what puts the image in the supplied logarithm branch; no global logarithm of $g$ is assumed.

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.

Again suppose $P$ is a nonconstant polynomial without a complex zero, and put $g=1/P$. As proved in part (c), $g$ is an entire holomorphic function with $|g(z)|\to 0$ at infinity. Choose $z_0$ and then a disk large enough that $|g(z)|<|g(z_0)|/2$ outside it. On the closed disk, $|g|$ attains its maximum by the extreme value theorem, and that maximum is positive and greater than every value outside the disk. It is therefore a global, and in particular local, maximum on $\mathbb{C}$.

Part (b)(ii) forces $g$ to be constant. A nonzero constant reciprocal makes $P$ constant, a contradiction. Hence $\boxed{P\text{ has a complex zero}}$. Equivalently one can minimize $|P|$ globally and apply the maximum modulus principle to its reciprocal if that minimum is positive. This supplies the second proof of the fundamental theorem of algebra using part (b).