# Holomorphic functions

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? – m_l May 5 '12 at 10:28
Meromorphic functions in one variable have isolated zeroes and singularities – Andrea Mori May 5 '12 at 10:28
@AndreaMori: nonconstant. – Chris Eagle May 5 '12 at 10:29
@ChrisEagle: non zero, actually :) – Andrea Mori May 5 '12 at 10:30
Isn't this an implication of the strong form of the identity theorem? – Johannes Kloos May 5 '12 at 13:12

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.
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 $f: \mathbb{C}^{n} \rightarrow \mathbb{C}^n$? – Novo May 6 '12 at 17:45
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 $\mathbb{C} \times \{ 0 \}$. – mrf May 6 '12 at 17:59
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. – mrf May 6 '12 at 19:00