solving ${x + y + z = 4, x^2 + y^2 + z^2 = 4, x^3 + y^3 + z^3 = 4}$ Knowing that :
 $$x + y + z = 4$$
$$x^2 + x^2 + z^2 = 4$$ 
$$x^3 + y^3 + z^3 = 4 $$
Solving the equation with complex numbers : 
I know that: 
$$4 = x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy + xz + yz)$$
So:
$$ (xy + xz + yz) = 6$$
$$4 = x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x^2y + x^2z + xy^2 + xz^2 + y^2z + yz^2) - 6xyz$$
Trying to find :
$$(x^2y + x^2z + xy^2 + xz^2 + y^2z + yz^2) = 
x(xy + xz) + y(xy + yz) + z(xz + yz)$$
Knowing that $x+y+z = 4$, I'm using that info so:
$$x(xy + xz + yz) + y(xy + yz + xz) + z(xz + yz + xy) - xyz - xyz - xyz$$
we get:
$$(xy + xz + yz)(x+y+z) - 3xyz$$
Once I find what the product $xyz$ is, I'm getting stuck. How do I find the $x,y,z$ (that are allegedly complex numbers) ?

From just the first two equations and the QM-AM inequality:
$$\sqrt{\frac43}= \sqrt{\frac{x^2+y^2+z^2}3} \geqslant \frac{x+y+z}3 = \frac43$$
we have the fact that there are no real solutions...

If looking for complex solutions, note
we have $2(xy + yz + zx) = (x+y+z)^2-(x^2+y^2+z^2) \implies xy+yz+zx = 6 $
and $3xyz = (x^3+y^3+z^3)-(x+y+z)^3+3(x+y+z)(xy+yz+zx) \implies xyz = 4$
Thus $x, y, z$ are roots of $t^3-4t^2+6t-4=(t-2)(t^2-2t+2)=0$.
Solving the quadratic is easy and so the roots are $(2, 1+i, 1-i)$ in some order..
