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Let $f:U\subset\mathbb{C}^n\to\mathbb{C}$ be non-trivial and holomorphic with $U$ open and connected. Is the zero set $Z(f)=\{z\in U\mid f(z)=0\}$ a nowhere dense set (i.e. is the interior of the closure of $Z(f)$ empty?)?

My question comes from Huybrechts' Complex Geometry exercise 1.1.9 - I want to show that the set of meromorphic functions $K(U)$ on $U$ is a field. Here a meromorphic function is a function defined on the complement of a nowhere dense subset, expressible locally as a quotient of holomorphic functions.

The result of the previous exercise shows that for any non-trivial holomorphic function $f:U\to\mathbb{C}$, the complement $U\setminus Z(f)$ is connected and dense in $U$.

Edit: I understand that if the zeros were dense, then $f\equiv 0$. I don't understand how to relate this to the explicit definition of nowhere dense.

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If $Z(f)$ were dense at some point, then $f\equiv 0$ followed immediately from the power series expansion.

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  • $\begingroup$ Assume the interior of the closure of $Z(f)$ contains a point $z_0$. Then there is an $\epsilon > 0$ such that $\forall z \in B_\epsilon(z_0)\;:\;f(z) = 0$. $\endgroup$ Jun 9, 2014 at 15:31
  • $\begingroup$ Oh, of course! Well that was very silly of me -- thanks! $\endgroup$ Jun 9, 2014 at 16:11

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