Solve Complex Equation $z^3 = 4\bar{z}$ I'm trying to solve for all z values where $z^3 = 4\bar{z}$.
I tried using $z^3 = |z|(\cos(3\theta)+i\sin(3\theta)$ and that $|z| = \sqrt{x^2+y^2}$ so:
$$z^3 = \sqrt{x^2+y^2}(\cos(3\theta)+\sin(3\theta))$$ and $$4\bar z = 4x-4iy = 4r\cos(\theta)-i4r\sin(\theta)$$
but I have no idea where to go from there.

Just for fun, let's do this by expanding $(x+iy)^3=4(x-iy)$ with $x,y\in\mathbb{R}$ and separate the real and imaginary parts. We wind up with
$$x(x^2-(3y^2+4))=0\quad\text{and}\quad y(y^2-(3x^2+4))=0$$
If $x=0$ then $y(y^2-4)=0$, so $y=0,\pm2$, hence $z=0$, $2i$ and $-2i$ are solutions.
If $x\not=0$, then we must have $x^2=3y^2+4$, which implies $y(y^2-(9y^2+12+4))=-8y(y^2+2)=0$. The only real solution is $y=0$, which leads to $x^2=3\cdot0^2+4=4$, or $x=\pm2$.  So $z=2$ and $-2$ are also solutions.
In all we have $z=0,2,2i,-2$, and $-2i$ as solutions of $z^3=4\overline{z}$.
