 Solution

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.