# a holomorphic function that is not identically zero

Let $$f: \Omega \to \mathbb{C}$$ be a holomorphic function, if $$f$$ is not identically zero, then we can find an open set $$U$$ on which $$f$$ is nonzero. why is that?

Thank you!

• Of course you mean a nonempty open set ... This has nothing to do with holomorphicity, just continuity. – Ted Shifrin Oct 13 at 19:11

Because $$f$$ is continuous and $$\Bbb C\setminus\{0\}$$ is open.
• Just recall the definition of continuity. $f$ is continuous if the pre-image of every open set is open. what @Gae. S. is pointing out is that $f^{-1}(\mathbb C-{0})$ is open and $f$ is non-zero there. if $f$ is not continous the result of course doesn't hold, consider for instance $f:\mathbb C\to \mathbb C$ given by $f(x)=0$ for $x\neq 0$ and $f(0)=1$. By the way, it seems that you think he is putting extra assumptions. This is not true, holomorphicity implies continuity. – David Jaramillo Oct 13 at 20:11