If $u_1(z)u_2(z)\equiv 0$ on a simply connected domain then either $u_1\equiv 0$ or $u_2\equiv 0$

Question

Suppose $u_1(z)$ and $u_2(z)$ are harmonic in a simply connected domain $D$, with $u_1(z)u_2(z)\equiv 0$ in $D$. Prove that either $u_1(z)\equiv 0$ or $u_2(z)\equiv 0$ in $D$.

Since $D$ is simply connected, there are holomorphic functions $f,g\in\mathcal{H}(D)$ such that $\Re f = u_1$ and $\Re g = u_2$ (or $\Im$ if necessary). I don't know how to use this fact. I first thought maybe harmonic functions have discrete roots but $\log|z|$ has roots $|z| =1$ so this is not true. Any hint on this problem?

Hint: if $u_i$ is nonzero, its zero set has empty interior. – Aphelli Dec 06 '22 at 09:02 — Aphelli, Dec 06 '22 at 09:02
@Aphelli Thank you! – one potato two potato Dec 06 '22 at 09:07 — one potato two potato, Dec 06 '22 at 09:07

score 2 · Answer 1 · answered Dec 06 '22 at 09:07

If $u_2$ is not identically $0$ then there is an open disk contained in it. But then $u_1=0$ on this disk and a holomorhic function $f$ with real part $u_1$ is constant on this disk (by C-R equations). It follows that $f$ is a constant everywhere and the conclusion follows.

If $u_1(z)u_2(z)\equiv 0$ on a simply connected domain then either $u_1\equiv 0$ or $u_2\equiv 0$

1 Answers1