<s>[INST] Finding every solution for equation of complex numbers I need to find every solution for: 
$\ z^{3} + 3i \overline z = 0  $ 
So I tried was just to compare imaginary and complex part of $\ z^{3} $ and $\ 3i\overline z$
Ill spare you the alegbra, here is the result:
$$\ a^{3} - 3ab^{2} + i(3a^{2}b-b^{3}) = -3b -3ai \\a^{3} - 3ab^{2} = -3b \\3a^{2}b - b^{3} = -3a$$ and so $$\  a^{3} -3ab^{2} + 3b = 0  \\ b^{3} -3a^{2}b-3a=0
$$
but I'm pretty stuck here. not sure what do next. I also tried using eulers rule so
$$\ z^{3} = -3i\overline z \\ r^{3}e^{{i\theta}^{3}} = -3i \times re^{-i\theta} \\ r^{3}e^{{i\theta}^{3}} = -3i \times re^{2\pi-i\theta} \\ 3\theta = 2\pi - \theta +2\pi k \\ 4\theta = 2\pi + 2\pi k \\ \theta = \frac{\pi}{2} + \frac{\pi k }{2}$$
and
$$\ r^{3} = -3ir \\ r^{2} = -3i$$
 [/INST] Write your equation in the form: $$z(z^2+3i)=0$$
and then you will get: $$x^2-y^2+i(2xy+3)=0$$ or $$z=0$$
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