Solving $z^3+3i\overline{z}=0$ $$z^3+3i\overline{z}=0$$
$z=x+yi$
$$(x+yi)^3+3i(x-yi)=0$$
$$x^3+3x^2yi-3xy^2-y^3i+3ix+3y=0$$
$$x^3-3xy^2+3y=0\text{  and  }  3x^2y-y^3+3x=0$$
How to continue from here?

$$z^3+3i\overline{z}=0\to \overline{z}^3-3iz=0$$
sum both equation
$$(z^3+\overline{z}^3)+3i(z-\overline{z})=0\to Re(z^3)+3i(iIm(z))=0$$
$$Re(z^3)=3Im(z)$$
If we call $z=re^{i\theta}$ then
$$r^3\cos(3\theta)=3r\sin\theta$$
as already said before, it is easy to get that $r=0$ or $r=\sqrt{3}$. For $r=0$ then $z=0$ and for $r=\sqrt{3}$ then:
$$\cos(3\theta)=\sin\theta\to\cos(3\theta)-\cos(\theta-\pi/2)=0$$
$$-2\sin(2\theta-\pi/4)\sin(\theta+\pi/4)=0$$
finish the calculations and find all possibles values of $\theta$.
