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Is there any sequence $\{ Z_{\nu} \}_{\nu \in \mathbb{N}}$ in $\mathbb{C}^{n}$, $Z_{\nu} \rightarrow 0$, such that any holomorphic function in $\mathbb{C}^{n}$ which vanishes in $Z_{\nu}$ for all $\nu \in \mathbb{N}$ is identically zero?

Thank you!

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    Dear @DavideGiraudo, roughly speaking, the Identity Theorem states that two holomorphic functions which agree in a open subset are equals. I am very convinced that sequence exists, and the main idea is in my previous comment.2012-10-13

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Let $\{ Z_{\nu}' \}_{\nu \in \mathbb{N}}$ be a dense sequence in the unit ball $B(0;1)$. Define a new sequence $\{ Z_{\nu} \}_{\nu \in \mathbb{N}}$ by

$Z_{1}', \frac{Z_{1}'}{2}, \frac{Z_{2}'}{3}, \frac{Z_{1}'}{4}, \frac{Z_{2}'}{5}, \frac{Z_{3}'}{6}, \frac{Z_{1}'}{7}, \frac{Z_{2}'}{8}, \frac{Z_{3}'}{9}, \frac{Z_{4}'}{10} \cdots$

Then for each fixed $Z' \in \{ Z_{\nu}' \}$, we have $\nu Z_{\nu} = Z'$ for infinite number of $\nu \in \mathbb{N}$. It means that $Z_{\nu} = \frac{Z'}{\nu} \rightarrow 0$. Then, any holomorphic function in $\mathbb{C}^{n}$ that vanishes in the sequence $\{ Z_{\nu} \}$ has, in particular, a subsequence of zeros accumulating to the origin that belong to the complex line generated by any $Z'$. Therefore it vanishes identically on that complex line, by the principle of isolated zeros in one variable. Since the union of these lines is dense in $\mathbb{C}^{n}$, the function is identically zero by continuity.

Thanks to Pietro Majer.

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    Hi @Migu$e$mat$e$, I got th$e$ $e$xercicise from my professor, here in Brazil. I did not find it in any book.2014-09-05