I am reading notes on complex analysis and it states that if $f$ is holomorphic in $\Omega$ and $f(z_0) = 0$ for some $z_0 \in \Omega,$ then there exists an open neighborhood $U$ of $z_0$ such that $f(z) \neq 0$ for all $z \in U - \{z_0\}.$ Why is this true? So I understand that if $f$ is holomorphic then it can be written as a power series. Does it have something to do with the continuity of its derivative?

  • $\begingroup$ If $f$ is holomorphic then it is analytic. And it becomes obvious that its derivatives are continuous (and analytic..) $\endgroup$
    – reuns
    May 6, 2017 at 7:30

1 Answer 1


The claim is not true as stated: $f$ might be identically zero. But if it is not, you have $f(z)=(z-z_0)^kg(z)$ for some $k$ and holomorphic $g$ with $g(z_0)\ne 0$. By continuity of $g$, $g(z)\ne0$ in some neighbourhood $U$ of $z_0$.

  • $\begingroup$ By the holomorphic $\implies$ analytic theorem $\endgroup$
    – reuns
    May 6, 2017 at 7:26
  • $\begingroup$ Oh wow. I guess my notes were alluding to this theorem. They talk about this right after. I should have just kept reading... $\endgroup$
    – green frog
    May 6, 2017 at 7:29

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