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?

Given
$$
z^3 = - \bar{z}.
$$
First we note that
$$
|z^3| = |-\bar{z}| \Longrightarrow |z|^3 = |z|.
$$
Therefore
$$
z = 0 \vee z = \exp(\zeta \pi \mathbf{i}).
$$
The case $z \ne 0$
We obtain
$$
\exp(3 \zeta \pi \mathbf{i}) = \exp( \pi \mathbf{i} - \zeta \pi \mathbf{i}).
$$
Whence
$$
3 \zeta = 2 k + 1 - \zeta \Longrightarrow \zeta = \frac{1}{4} + k.
$$

The general solution can be written as
$$
z = 0 \vee z = \exp\Big( \big[ \tfrac{1}{4} + k \big] \pi \mathbf{i} \Big)
= \exp\big( \pi \mathbf{i} / 4 \big) \exp\big( k \pi \mathbf{i} \big).
$$
Or as
$$
z_0 = 0 \vee k \in {1,2,3,4} : z_k = \exp\big( \pi \mathbf{i} / 4 \big) \mathbf{i}^k.
$$

Note that
$$
\exp\big( \pi \mathbf{i} / 4 \big) = \frac{1+\mathbf{i}}{\sqrt{2}},
$$
so
$$
z_1 = \frac{-1+\mathbf{i}}{\sqrt{2}}\\
z_2 = \frac{-1-\mathbf{i}}{\sqrt{2}}\\
z_3 = \frac{1-\mathbf{i}}{\sqrt{2}}\\
z_4 = \frac{1+\mathbf{i}}{\sqrt{2}}
$$
