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Here, please assume $B(z_0,r)\subseteq \Omega$.

I would like to know if this is really true because I've seen many proofs using this statement as a true statement.

For example, if you look at the proof of TonyK at this link " If $f,g$ are entire functions and$\ fg\equiv 0$ then either $f \equiv 0$ or $g\equiv0. $ " we can see that he claimed that we can find a neighborhood of $z_0$ s.t $f(z) \neq 0$ for $z \in$ the neighborhood.

However, if we have the function $f(x)=x^2$ where it is defined on $\mathbb{R}$, we can only find $1$ zero ($x = 0$), so there is no such neighborhood of $0$.

I am pretty sure there should be some critical differences between real analysis and complex analysis defining differentiability, but I think I should ask what makes the statement true within the complex analysis so that I can use the fact with confidence in the future. It is really confusing to me.

Any answer would be really appreciated.

john
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    For any neighborhood of zero, $f(z)=z^2$ is not identically vanishing. $\equiv$ symbol is important here. – mathcounterexamples.net Oct 30 '21 at 07:19
  • @mathcounterexamples.net Thanks for your comment. So, basically, the statement has to be false? And, complex functions can be considered as the real functions (as long as I have to draw these things in my head)? – john Oct 30 '21 at 07:34
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    Not sure exactly what the OP means, but what is true is that if $f(z_0)=c$ AND $f$ analytic non-constant in a domain $D$ containing $z_0$, then there is a small $r>0$ st $f(w) \ne c$ for all $0<|w-z_0|<r$ (and of course the disc centered at $z_0$ of radius $r$ is contained in $D$ but that is less relevant as for $r$ small enough that's always the case); this is true for real analytic functions also with intervals instead of domains as it is a combination of power series and topological connectedness properties – Conrad Oct 30 '21 at 13:52

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