Finding every solution for equation of complex numbers

Question

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

$$\ r^{3} = -3ir \\ r^{2} = -3i$$

Answer 1

there is the trivial solution $z = 0$

or

$z^3 + \bar z 3i = 0\\ z(z^3 + \bar z 3i) = 0\\ z^4 + z\bar z 3i = 0\\ z^4 + |z|^2 3i = 0\\ \frac {z^4}{|z|^2} = -3i \\ \arg z^4 = \frac {-\pi}{2}+2n\pi\\ \arg z = \frac {-\pi}{8}, \frac {-5\pi}{8},\frac {3\pi}{8},\frac {7\pi}{8}\\ |\frac {z^4}{|z|^2}| = |z^2| = 3\\ |z| = \sqrt 3\\ z = \sqrt 3 e^{\frac {-\pi}{8}i},z = \sqrt 3 e^{\frac {-5\pi}{8}i},z = \sqrt 3 e^{\frac {3\pi}{8}i},z = \sqrt 3 e^{\frac {7\pi}{8}i}$

Answer 2

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 $$z=0$$

Answer 3

$HINT:$

You can take conjugate on both sides. This will give you one more equation. Now you have two equations involving $z$ and $\bar z$ . Eliminate $\bar z$, and solve for $z$.