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Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be a holomorphic function such that for an open interval $V \subset \mathbb{R}$ the following holds: $f(V)=0$. Does there exist an open set $U \subset \mathbb{C}$ such that $f(U)=0$.

Intuitively I would say it is true. I haven't been able to construct a counter example yet.

Any help is welcome.

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    $U \subset \mathbb{C}$ an open subset?2012-05-05
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    Meromorphic functions in one variable have isolated zeroes and singularities2012-05-05
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    @AndreaMori: *nonconstant*.2012-05-05
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    @ChrisEagle: non zero, actually :)2012-05-05
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    Isn't this an implication of the strong form of the identity theorem?2012-05-05

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This is certainly true. In fact, if $f$ and $g$ are entire holomorphic functions and $f = g$ on a set $E$ with an accumulation point, then $f = g$ everywhere, so your set $U$ can be taken as the whole complex plane.

See for example Wikipedia: Identity theorem

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    Thank you for your answer. Does the identity theorem also hold for a holomorphic function of the form $f: \mathbb{C}^{n} \rightarrow \mathbb{C} $ or $f: \mathbb{C}^{n} \rightarrow \mathbb{C}^n $?2012-05-06
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    Variants hold in higher dimension, but the sets of uniqueness are larger. For example $f(z,w) = z$ and $g(z,w) = 0$ agree on $\mathbb{C} \times \{ 0 \}$.2012-05-06
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    Ah, that is nice. Does the proof for the extension of the multivariable case directly follow from the identity theorem in one complex variable?2012-05-06
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    That $f = g$ on an open set implies $f = g$ everywhere is usually proven the same way as in the one variable case (via power series), but the more refined versions are more difficult.2012-05-06
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    For the refined versions to which literature would you refer me?2012-05-06