Before telling me that this question maybe has already been done I would like to ask you for something more concrete-clear as an answer and not just comments please!
So, I do know that every nonsingular algebraic variety over $\mathbb{C}$ has the structure of a smooth complex manifold. And although sounds quite reasonable in my head, since a complex variety is a subset of the form $X \subseteq \mathbb{C}^{n}$, I can't see how does this smooth structure exists. $\mathbf{What}$ $\mathbf{charts}$ $\mathbf{do}$ $\mathbf{we}$ $\mathbf{use}$ $\mathbf{to}$ $\mathbf{do}$ $\mathbf{that}$? Apparently we have to change the topology of that space, since this topological space (equipped with the Zariski topology) is not even Hausdorff in the usual sense. So, strictly speaking not every nonsingular complex variety has a smooth structure doing it a complex manifold unless its dimension is zero. However, there must be an associated complex or evenmore analytic space.
I did learn somewhere, something more sophisticated and more algebraic, that Serre in the so-called GAGA paper he constructed a functor doing that job, but the idea I guess is pretty much the same (I mean it is based on the fact that nonsingular complex varieties admit a topology-structure of a complex manifold).
Also, one more question, I expect the converse of the above is not true at all. That is, not every complex manifold is given by a complex variety. But how do we give a formal proof of that or a counterexample?
Thank you for patience!
Best Answer
A) A smooth algebraic subvariety $X\subset \mathbb C^n$ is locally on X given by the set $V(f_1,\cdots,f_k)\subset \mathbb C^n$ of common zeros of a list $f_1,\cdots,f_k$ of polynomials with Jacobian matrix of rank $k$.
The implicit function theorem then shows that if we consider these polynomials as holomorphic functions, the variety $X$ is a holomorphic submanifold of $\mathbb C^n$.
Beware that in general it is impossible to find polynomials $f_1,\cdots,f_k$ such that $X=V(f_1,\cdots,f_k)$ and which satisfy the Jacobian condition at every $x\in X$ : the polynomials $f_i$ a priori depend on the choice of $x\in X$.
(If however one can find such polynomials working everywhere independently of $x\in X$, then $X$ is said to be a complete intersection in $\mathbb C^n$.)
B1) The simplest complex holomorphic manifold $X$ with no algebraic structure is $X=\mathbb C\setminus \mathbb Z$.
The precise statement is that there does not exist an algebraic variety $Y$ whose analytification is (isomorphic to) $X$ i.e. $Y^{an}=X$is impossible. We then say that $X$ is not algebraizable
The reason why $\mathbb C\setminus \mathbb Z$ is not algebraizable is that any complex non compact smooth algebraic variety of dimension $1$ is obtained by deleting finitely many points from a compact one-dimensional smooth variety.
So there is a purely topological obstruction to $\mathbb C\setminus \mathbb Z$ being algebraizable.
B2) Every one-dimensional compact complex manifold is algebraizable: this is celebrated theorem of Riemann.
B3) There exist compact complex tori $X$ of any dimension $n\geq 2$ which are not algebraizable.
A complex torus is a manifold of the form $X=\mathbb C^n/\Gamma$ where $\Gamma =\oplus _{j=1}^{2n}\mathbb Z\cdot v_j$ is the lattice obtained from some basis $(v_j)_{j=1\cdots 2n}$ of the real vector space underlying $\mathbb C^n$.
The subtle relations between the $v_j$'s dictating whether $X$ is algebraizable or not were discovered by Riemann (him again!) and are referred to as Riemann bilinear relations.
Edit (September 2, 2016): the technical condition for algebraizability of a torus.
The Riemann criterion for algebraicity of the torus $\mathbb C^n/\Gamma$ is that there exist a hermitian form $H:\mathbb C^n\times \mathbb C^n\to \mathbb C$ whose imaginary part is integral on the lattice:$$(Im H)(\Gamma\times \Gamma)\subset \mathbb Z$$ For example every complex torus of dimension $1$ is obtained by dividing out $\mathbb C$ by a lattice of the form $\Gamma =\mathbb Z \oplus \mathbb Z\tau$ where $\tau =a+ib$ with $b\gt0$.
Such a torus is algebraizable as witnessed by the hermitian form $H(z,w)=\frac {z\overline w}{b}$.
This is of course in line with the result mentioned in 2.
Bibliography
A nice introduction to these ideas is Shafarevich's Basic Algebraic Geometry 2: Chapter 8, page 153.