0
$\begingroup$

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.

  • 0
    Isn't this an implication of the strong form of the identity theorem?2012-05-05

1 Answers 1

0

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

  • 0
    For the refined versions to which literature would you refer me?2012-05-06