Proving that if $u,v$ are harmonic on $D$ and $uv = 0$ on an open subset of $D$, then $u$ or $v$ is 0 on $D$ Let $D$ be an open connected set. Suppose $u$ and $v$ are harmonic functions on $D$ and that $u(z) v(z) = 0$ on an open subset of $D$.
Obviously on this open subset at least one of $u$ or $v$ is 0, but I want to extend this property to all of $D$ to show that $u = 0$ or $v = 0$ identically on $D$.
What is a good way to approach this?
 A: Let $D'\subset D$ be an open disc. We have $D' = \{u|_{D'}=0\}\cup \{v|_{D'}=0\}.$ These are closed subsets of $D'.$ Now $D'$ is homeomorphic to $\mathbb R^2,$ so it has the Baire property. Hence one of $\{u|_{D'}=0\}, \{v|_{D'}=0\}$ has nonempty interior. Suppose WLOG it's $\{u|_{D'}=0\}$ that does. Let's now use the result that Elizabeth mentioned: Since $u$ is harmonic in $D,$ $f=u_x-iu_y$ is holomorphic in $D.$ From the above we then see $f=0$ on an open subset of $D.$ By the identity principle for holomorphic functions, $f=0$ on $D.$ Therefore $u_x,u_y$ vanish everywhere in $D.$ Since $D$ is connected, $u$ is constant in $D,$ and that constant must be $0.$
A more general result is this: Suppose $u,v$ are real analytic in $D$ and $uv=0$ in $D.$ Then at least one of $u,v$ vanish on $D.$ Since harmonic functions are real analytic, this solves the given problem.
A: Either $u = 0$ or $v = 0$ on the subset of $D$ due to continuity. Suppose $u = 0$ on the subset without loss of generality. Then let $f = u_x - i u_y$. We have $f = 0$ on the subset. By the identity theorem for holomorphic functions, $f = 0$ on $W$. So $u = 0$ on $W$.
