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Let $f,g$ be entire functions such that $g \not\equiv 0.$ If $fg\equiv0$ in $\mathbb{C},$ could anyone advise me how to show $f \equiv0$ in $\mathbb{C} \ ?$ Thank you.

Alexy Vincenzo
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2 Answers2

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Suppose there exists $z$ such that $f(z) \ne 0$. Then $f$ is non-zero in some neighbourhood of $z$, so $g$ must be zero in the same neighbourhood. And if an entire function is identically zero in the neighbourhood of any point, it is zero in the whole of $\mathbb C$.

TonyK
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  • why must $g$ be zero in the same neighborhood that $f$ is nonzero? So there exists some $\epsilon >0$ such that $f:B_\epsilon(z) \rightarrow \Bbb{C}$ is nonzero, how does this imply $g \equiv 0$? – homosapien Sep 04 '22 at 19:28
  • @MyMathYourMath: because $fg\equiv 0$ in $B_\epsilon(z)$. That means $f(z)g(z)=0$ for all $z\in B_\epsilon(z)$. – TonyK Sep 04 '22 at 19:41
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Since the Complexes are an integral domain, $f(z)g(z)=0$ implies either $f(z)=0$ or $g(z)=0$. But then one of $f,g$ , say $f$ must have an uncountable number of zeros in $\mathbb C$. But an uncountable subset of $\mathbb C$ has a limit point in $\mathbb C$ . Then the set of zeros of $f$ has a limit point in $\mathbb C$, so that, by the identity theorem, we must have $f==0$.

user172643
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