How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about the 0?
EDIT:
ok, following some of your comments/answers this is what I have done:
$$z^3 = - \overline z$$
$$\rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
So
$$
\begin{Bmatrix}
\rho^3 = \rho\\ 
3\theta = -\theta + 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho^3 = \rho\\ 
2\theta =  2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho = 0 or \rho = 1\\ 
\theta =  k\frac{\pi}{2}
\end{Bmatrix}$$
is this the right way?

HINT
If $z = a+bi$ note that
$$
z^3 = a^3 - 3ab^2 + 3a^2bi - b^3i = (a^3 - 3ab^2) + (3a^2b-b^3)i
$$
so
$$
0 = \bar{z} + z^3 = (a^3 - 3ab^2+a) + (3a^2b-b^3-b)i,
$$
which implies you have 2 equations in 2 unknowns.
Can you take it from here?
