<s>[INST] 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?
 [/INST] One solution is $z=0$. When $z\ne0$, multiplying by $z$ will not alter the solutions of the equations; so we can look at $z^4+|z|^2=0$. If we write $z=re^{it}$, the equation becomes 
$$
r^4e^{4it}+r^2=0.
$$
As we are assuming $r\ne0$, this reduces to $-r^2e^{4it}=1$. So $r=1$ and $e^{4it}=-1$, that is (since $-1=e^{i\pi}$), $4t=\pi+2k\pi$. Thus
$$
t=\frac14\,\left(\pi+2k\pi\right)=\frac{\pi}4+\frac{k\pi}2,\ \ k=0,1,2,3.
$$
So we get four solutions, from $z=\cos t+ i\sin t$, which are 
$$
\frac{\sqrt2}2+i\,\frac{\sqrt2}2,\ \ -\frac{\sqrt2}2+i\,\frac{\sqrt2}2,\ \ \frac{\sqrt2}2-i\,\frac{\sqrt2}2,\ \ \text{ and }-\frac{\sqrt2}2-i\,\frac{\sqrt2}2.
$$
</s>