ag.algebraic geometry - Zariski open sets are dense in analytic topology

It is enough to show that the complement of $U$ has empty interior. Also, that complement is contained in the zero set $Z$ of a non-constant polynomial $f$, so it is enough to show that $Z$ does not contain open sets.

If $zin Z$ is a point in the interior of $Z$, then the Taylor series of $f$ at $z$ is of course zero. Since $f$ is an entire function, this is absurd.

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