<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] $$z^3+\overline z=0$$
But $z=a+bi$, $\overline z = a-bi$ so:
$$(a+bi)^3+a-bi=a^3-3ab^2+3a^2bi-b^3i+a-bi=i(3a^2b-b^3-b)+(a^3-3ab^2+a)$$
We know that $a+bi=0 \iff a=0 \land b=0$:
$$\begin{cases}
3a^2b-b^3-b=0\\
a^3-3ab^2+a=0
\end{cases}$$
$$\begin{cases}
3a^2-b^2-1=0\\
a^2-3b^2+1=0
\end{cases}$$
$$a^2=3b^2-1 \implies 3a^2-b^2-1=9b^2-3-b^2-1=8b^2-4=0 \implies b=\pm\frac{\sqrt2}{2}$$
$$a^2=3b^2-1=1.5-1=0.5\implies a=b=\pm\frac{\sqrt2}{2}$$
So finally:
$$z=\frac{\sqrt2}{2}\pm\frac{\sqrt2}{2}i$$
or
$$z=-\frac{\sqrt2}{2}\pm\frac{\sqrt2}{2}i$$

To be clear there may be other solutions. I have divided my equations by $a$ and $b$, but each may be equal to $0$.


*

*$a=0$ then $b^3+b=0$


*

*$b=0$ OK

*$b\not=0$ then $b^2+1=0$ and $b\not\in \mathbb{R}$ (but it's a contradiction)


*$b=0$ then $a^3+a=0$, same as above, only $(a;b)=(0;0)$ is valid.


The other solution is then $$z=0$$
</s>