# 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.

-
$U \subset \mathbb{C}$ 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