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!
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$\begingroup$ Of course you mean a nonempty open set ... This has nothing to do with holomorphicity, just continuity. $\endgroup$ – Ted Shifrin Oct 13 at 19:11
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Because $f$ is continuous and $\Bbb C\setminus\{0\}$ is open.
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$\begingroup$ i don't understand why do we need continuity...how does it work? $\endgroup$ – beigecamellia Oct 13 at 19:18
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$\begingroup$ 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. $\endgroup$ – David Jaramillo Oct 13 at 20:11
