<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] there is the trivial solution $z = 0$
or
$z^3 + \bar z 3i = 0\\
z(z^3 + \bar z 3i) = 0\\
z^4 + z\bar z 3i = 0\\
z^4 + |z|^2 3i = 0\\
\frac {z^4}{|z|^2} = -3i \\
\arg z^4 = \frac {-\pi}{2}+2n\pi\\
\arg z = \frac {-\pi}{8}, \frac {-5\pi}{8},\frac {3\pi}{8},\frac {7\pi}{8}\\
|\frac {z^4}{|z|^2}| = |z^2| = 3\\
|z| = \sqrt 3\\
z = \sqrt 3 e^{\frac {-\pi}{8}i},z = \sqrt 3 e^{\frac {-5\pi}{8}i},z = \sqrt 3 e^{\frac {3\pi}{8}i},z = \sqrt 3 e^{\frac {7\pi}{8}i}$
</s>