<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] $$z^3+z^*=0.$$
Let $z=x+iy$. Then
$$(x+iy)^3+x-iy=0,$$ so that
$$x+x^3-3xy^2=0,\tag{1}$$ and $$-y+3x^2y-y^3=0.\tag{2}$$
We ignore the trivial solution $x=y=0$ and suppose that $x,y\neq 0$. Then using (1) and (2) we see that $$x^3+x=x\left(1+\frac{y^3+y}{3y}\right).$$ Hence, $$-3xy^2+x\left(1+\frac{y^3+y}{3y}\right)=0.$$ Since $x\neq 0$ this gives $$-3y^2+1+\frac{y^3+y}{3y}=0,$$ leading to $$y(1-2y^2)=0.$$ Therefore, we must have $$y=\pm\frac{1}{\sqrt{2}}.$$ Substituting into $(2)$ gives $$x=\pm\frac{1}{\sqrt{2}}.$$ Therefore,
$$z=x+iy=\pm\frac{1}{\sqrt{2}}\pm i\frac{1}{\sqrt{2}},$$ where each $\pm$ is chosen independently of the other.
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