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.

1$\begingroup$ +1: this is a slicker argument than mine. $\endgroup$ – Pete L. Clark Feb 17 '10 at 16:32
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.

1$\begingroup$ Sard's theorem would show the measure zero part. $\endgroup$ – Anweshi Feb 17 '10 at 15:08

2$\begingroup$ Sard's theorem claims that the image of something has measure zero, not that the preimage of something does, Anweshi $\endgroup$ – Mariano SuárezÁlvarez Feb 17 '10 at 15:22

1$\begingroup$ Actually, my first draft of the answer included a reference to Sard's theorem. Upon reflection, I took it out. I do maintain that, whatever particular theorem you want to cite, a closed submanifold of Euclidean space has measure zero, and that this is not hard to show! $\endgroup$ – Pete L. Clark Feb 17 '10 at 16:30
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.)

$\begingroup$ If they weren't irreducible, then his theorem would be false for him :) $\endgroup$ – Mariano SuárezÁlvarez Feb 17 '10 at 14:13