How does one show that if $U \subseteq \mathbb{C}^n$ is nonempty and Zariski open then $U$ is also dense in the analytic topology on $\mathbb{C}^n$?



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 $z\in Z$ 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.


