TITLE: When is a singular point of a variety ($\mathcal{C}^\infty$-) smooth?
QUESTION [73 upvotes]: If $X$ is a nonsingular  algebraic (or analytic) variety over $\mathbb C$ or $\mathbb R$ then it is certainly $C^\infty$ over the reals.
The converse is  false for a silly reason : in the real or complex affine plane with coordinates $x,y$ the variety $x^2=0$ is  singular since it is not reduced, but set-theoretically it is the $y$-axis, a $C^\infty$ submanifold of the plane.
So let me concentrate on reduced varieties.  
Over the reals you have the disturbing phenomenon that the  plane curve $C$ defined by $y^3+2x^2y-x^4=0$ is algebraically singular at the origin, but your favourite computer graphics system  won't show you that, because the curve is actually $C^\infty$.
Worse: it is the graph of a real analytic function! (I learned this example in Milnor's Singular Points of Complex Hypersurfaces)  
I think this pathology is impossible over $\mathbb C$ but I cannot find a reference. So let me ask the official question:  
Can a reduced algebraic (or analytic) variety which has a singularity at a point be $C^\infty$ at that point? 
(Information on the more complicated real case welcome, of course)  
Edit Francesco's answer shows that indeed a singular complex variety  cannot be smooth.
This is  quite interesting, because many books will give you the Jacobian criterion for a  map $\mathbb C^n\to \mathbb C^k$ to define a submanifold but none (to my knowledge) adds that this Jacobian criterion also allows to prove nonsmoothness (modulo some technicalities), which is  paradoxically a more delicate question.    
Second Edit (June 29th,2012))
Here is a  proof  more geometric than Milnor's that the curve $X$ defined by $y^3+2x^2y-x^4=0$ is an analytic submanifold of $\mathbb R^2$.
The curve $X$ is actually rational: it is the image of $\mathbb R$ under $\phi (t)=(t(t^2+2),t^2(t^2+2))$.
The analytic morphism $\phi$ is injective, proper and immersive, hence an embedding into $\mathbb R^2$ with closed image $X$. [It  boils down to the fact that the polynomial $t^3+2t$ has positive  derivative $3t^2+2$ and is thus strictly increasing!]

REPLY [3 votes]: For the real case, here's another
example: the graph in ${\mathbb R}^2$ of
  $ y = x (1 + x^2)^{1/3} $
is real analytic and is the
${\mathbb R}$-points of a
variety defined over ${\mathbb R}$,
but the origin is 
singular; the defining equation is
 $  y^3 = x^3 + x^5 $.