A complex algebraic variety which is connected in the usual topology Hartshorne wrote in his book's Appendix B that it can be easily proved that a complex algebraic variety is connected in the usual topology if and only if it is connected in Zariski topology.
How can it be proved?
 A: If $X$ is a  complex projective  algebraic variety, the canonical morphism $$ H^i(X, F)\stackrel {\cong}{\to} H^i(X^{an},F^{an})$$ is an isomorphism of complex finite-dimensional vector spaces for all coherent sheaves on the algebraic variety $X$.
In particular for $i=0$ and $F=\mathcal O_X$ we get an isomorphism $$\Gamma(X, \mathcal O_X) \stackrel {\cong}{\to}  \Gamma(X^{an}, \mathcal O_X^{an}) \quad (\bigstar)$$  Now $X$ is connected in the Zariski topology iff the left-hand side has dimension $1$,  and similarly $X^{an}$ is connected in the classical topology iff the right-hand side has dimension $1$.
The isomorphism $(\bigstar)$ then implies   that$X$ is connected   in its Zariski topology iff $X^{an}$ is connected in its classical topology.  
The proof  only works for  projective (or slightly more generally for complete) varieties and uses the full power of GAGA but , hey, who can resist the pleasure of shooting at flies with ballistic missiles?  
Edit
As an answer to Makoto's comment, in order to prove that on a connected reduced compact analytic space all holomorphic functions $f:X\to \mathbb C$ are constant, it may be amusing in the spirit of nuking mosquitoes to invoke Remmert's theorem according to which $f(X)$ is  analytic in $\mathbb C$ , since $f$ is proper. Hence it is a point since it is compact and connected! 
A: Fredrik already explained why the connectedness for the complex topology implies the connectedness for the Zariski topology. 
The converse is harder. In the projective case, see Georges's wonderful answer. Let me give a reference for the general case. Let $X_1, \dots, X_n$ be the (Zariski) irreducible components of $X$. By Shafarevich, Basic Algebraic Geometry 2, Chapter VII, § 2.2, Theorem 1, $X_i(\mathbb C)$ is connected in complex topology for all $i\le n$. Let $Y$ be a connected component of $X(\mathbb C)$. Then $Y$ is the union of some $X_i(\mathbb C)$ and $X(\mathbb C)\setminus Y$ is the union of the other $X_i(\mathbb C)$'s. So $Y$ and $X\setminus Y$ are both Zariski closed and disjoint. This implies that $X$ is not Zariski connected if $X(\mathbb C)$ is not connected for the complex topology. 
A: It is easy at least in the affine case. (so the variety $X=\text{Spec }\mathbb{C}[x_1,\cdots,x_n]$ for some $n$.
If $X$ is disconnected (say, in two components for simplicity) in the usual topology, then its global section decomposes into direct products: $\Gamma(X,\mathcal{O}_X)= \Gamma(U,\mathcal{O}_X) \times \Gamma(V,\mathcal{O}_X)$. Taking $\text{Spec }$ of this gives $X$ back, and it is a standard result (Hartshorne, 2.19, Chapter II, for example), that $\text{Spec } A$ is disconnected if and only if $A$ can be written as a direct product of nonzero rings. Much of this part is wrong. See Zhen Lin's comment below.
On the other hand, if assume $X$ is disconnected in the Zariski topology. Since the Zariski topology is contained in the usual topology, every clopen (closed and open) set is also clopen in the usual topology. A topological space is disconnected if and only if there exists a non-trivial clopen set.
