If the product of two holomorphic functions is identically zero, then one of the functions is zero

Question

I have to answer the following question:

If $f$ and $g$ are holomorphic on some domain $\Omega$ and $f(z)g(z)=0$ for every $z\in \Omega$, then $f(z)=0$ or $g(z)=0$ for every $z\in\Omega$.

Is this correct: Let's assume that $f$ is not identically $0$ on the domain. Then there is a point $z_0\in\Omega$ such that $f(z_0)\neq0$. Since $f$ is continuous we find a neighbourhood of $z_0$ on which $f$ is free of zeros, $U$ say (why can I say this?) But since $fg=0$ we have $g=0$ on $U$. From the identity principle we can conclude $g=0$ in $\Omega$.

How can I show that this isn't true if the domain isn't connected? — Roos Jansen, Nov 13 '13 at 14:36
Regarding “why can I say this”: If $f$ is continuous, then ${z\colon f(z)\ne0}=f^{-1}[\mathbb C\setminus{0}]$ is open. — Carsten S, Nov 13 '13 at 14:41
If you have $\Omega = \Omega_1 \cup \Omega_2$ with $\Omega_1 \cap \Omega_2 = \varnothing$ and $\Omega_i$ open and nonempty, let $f(z) = 1$ if $z\in \Omega_1$, and $f(z) = 0$ if $z\in\Omega_2$. Let $g = 1-f$. Then $f\cdot g \equiv 0$, but neither $f$ nor $g$ vanishes identically. — Daniel Fischer, Nov 13 '13 at 16:28
Can somebody explain why there is a neighbourhood of $z_0$ which is free of zeros? I don't understand it yet. — Roos Jansen, Jan 02 '14 at 15:53

score 1 · Answer 1 · answered Nov 13 '13 at 14:28

1

Hint: Put $\epsilon=\frac12|f(z_0)|.$

Everything else looks fine.

answered Nov 13 '13 at 14:28

Cameron Buie

102,994

If the product of two holomorphic functions is identically zero, then one of the functions is zero

1 Answers1