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 $\mathbb{C}^n$?

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 nonconstant 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. 


It is equivalent to show that the complement, say $V$, has empty interior. In fact $V$ has measure zero (say, with respect to Lebesgue measure on $\mathbb{C}^n$; any measure which is absolutely continuous with respect to Lebesgue measure would serve as well). If $V$ is smooth, this is very standard: the Implicit Function Theorem (or something like that). In general, you can use the fact that $V$ has a relatively open subset which is a smooth manifold whose complement has positive codimension and finish off by induction. 


Here is a theorem from the "Complex Varieties" section of Mumford's "Red Book of Varieties and Schemes" (2nd edition): Theorem: Let $X$ be a complex variety and $U \subseteq X$ a nonempty (Zariski) open subvariety. Then $U$ is dense in $X$ for the analytic topology. (I think varieties for Mumford are always irreducible.) 

