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This is for homework, and I would really appreciate a hint. The question states

"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 $z \in \Omega$."

I tried contrapositive first. So suppose there exists points $z_0$ and $z_1$ in $\Omega$ such that $f(z_0) \neq 0$ and $g(z_1) \neq 0$. But I immediately get stuck here because I can't say anything about $f(z_0)g(z_0)$ or $f(z_1)g(z_1)$. Maybe a direct proof would be better? So by the Cauchy integral formula, I can say $$ \int_{\partial \Omega} \frac{f(w)}{w - z} dw \int_{\partial \Omega} \frac{g(w)}{w - z} dw = f(z)g(z) = \int_{\partial \Omega} \frac{f(w)g(w)}{w - z} dw = 0 $$ for all $z \in \Omega$. Again I get stuck here, because I don't see how this implies $f(z) = 0$ or $g(z) = 0$ for all $z \in \Omega$.

tylerc0816
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  • Do you know the identity theorem? – Daniel Fischer Sep 17 '13 at 20:47
  • @DanielFischer We have not covered that, no. We've gone over Cauchy's theorem, Morera's, Liuoville's. The basics. – tylerc0816 Sep 17 '13 at 20:49
  • Related: http://math.stackexchange.com/q/20403/ – Jonas Meyer Sep 17 '13 at 20:50
  • @tylerc0816 Identity theorem is the most basic of them :) You probably covered it, but maybe not its name. It says that if the set of zeroes of a holomorphic function has an acumulation point, the function must be zero on the domain. – N. S. Sep 17 '13 at 20:51
  • $\large% {\rm g}\left(z\right) = \sum_{,n}a_{n},\delta\left(z - z_{n}\right)\quad$ where $\large\quad z_{n}\quad$ is a root of $\large\quad{\rm f}\quad$ and $\large\quad\left\lbrace a_{n}\right\rbrace\quad$ are arbitrary constants. – Felix Marin Sep 17 '13 at 20:55

2 Answers2

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Let's assume that $f$ is not identically $0$. 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. But since $fg=0$ we have $g=0$ on $U$. Now use the identity principle to conclude $g=0$ in $\Omega$.

sranthrop
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  • So, just to clarify, there is no proof using the integral formulas? It seems odd to me because the answer relies on none of the material we've been going over, nor have covered. – tylerc0816 Sep 17 '13 at 21:12
  • Have you already learned that every holomorphic function has (locally) a power series expansion? Because then the identity principle is an easy consequence. – sranthrop Sep 17 '13 at 21:35
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    Love this solution. – Numbersandsoon Oct 15 '13 at 02:30
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This question is about the Uniqueness/Identity theorem.

Suppose $f(z)$ and $g(z)$ are not identically zero in $G$. Let $h(z) = f(z) g(z)$ which is analytic in $G$ and $h(z)= 0 \; \forall z \in G$. Then the set $\{ z \in G \,|\, h(z) = 0\}$ has a limit point in G.
(i.e. G is the closed set where $h(z)=0$)

By the Uniqueness/Identity theorem, $h(z)$ is identically $0$ and the only way for that to be true is if either $f(z)$ or $g(z)$ is identically $0$ in G.

user770687
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