Solving $z^3+3i\overline{z}=0$

Question

$$z^3+3i\overline{z}=0$$

$z=x+yi$ $$(x+yi)^3+3i(x-yi)=0$$ $$x^3+3x^2yi-3xy^2-y^3i+3ix+3y=0$$

$$x^3-3xy^2+3y=0\text{ and } 3x^2y-y^3+3x=0$$

How to continue from here?

Answer 1

Multiplying by $z$,

$$z^4+3i|z|^2=0$$ so that $z^4$ is purely imaginary. Then with $\omega$ a fourth root of $-i$,

$$-ir^4+3ir^2=0.$$

We have $r=0\lor r=\sqrt3$ and

$$z=0\lor z=\sqrt 3\,\omega$$ (five solutions in total).

Answer 2

$$z^3+3i\overline{z}=0\to \overline{z}^3-3iz=0$$

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 $\theta$.