Find All Complex solutions for $z^3+3i\bar{z}=0$ Find All Complex solutions for $z^3+3i \bar z =0$.
I tried substituting $z=a+bi$ and simplifying as much as I can
and this is what I ended up with: $a^3-2b^2a+3b+i(3ba^2-b^3+3a)=0$.
I just did not understand how do I get the values of $z$ from this equation

You let $z=a+bi$ and the system of equations to solve is: $a^3-3ab^2+3b=0$ and $3ba^2-b^3+3a=0$. If you multiply the first equation by $-a$ and the second by $b$ and then add the resulting equations you get $a^4+b^4=6a^2b^2$. Let $m=\frac{b}{a}$. Then, $$m^4-6m^2+1=0.$$ The solution of this equation is: $m=\tan(\frac{3\pi}{8}+\frac{k\pi}{2})$ where $k=0,1,2,3$. Hence, we found the Arguments (angles) of the 4 solutions. The other 4 Arguments can be found by formula $\pi-\theta$ from these Arguments, since $\tan(\pi-\theta)=-\tan\theta$ satisfies the equation too.
To find the magnitudes of them, multiply the first equation by $a$ and the second by $b$ and then add the resulting equations you get $a^2+b^2=\frac{6ab}{b^2-a}$. Then, since $b=ma$, we have $|z|=\sqrt\frac{6m}{m^2-1}$. We find out that $\sqrt\frac{6m}{m^2-1}=\sqrt{3}$ for all $m$ values we found above. I checked by WolframAlpha trying to prove by hand...
The solution $z=0$ could be found by polar method.
