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.
 A: If $z^3=4\overline z$, then $z^4=4z\overline z=4|z|^2$. So, $|z|^4=|z^4|=4|z|^2$, and therefore $z=0$ or $|z|=2$. So, unless $z=0$, $z$ can be written as $2(\cos\theta+i\sin\theta)$, in which case$$z^3=4\overline z\iff8\bigl(\cos(3\theta)+i\sin(3\theta)\bigr)=8(\cos\theta-i\sin\theta).$$Can you take it from here?
A: hint
$z=0$ is a solution.
Put $$z=r(\cos(t)+i\sin(t))=re^{it}$$
with $ r>0$.
the equation becomes
$$r^3e^{3it}=4re^{-it}$$
or
$$r^2e^{4it}=4=4e^{2ik\pi}$$
thus
$$r^2=4$$
and
$$4t=2k\pi$$
The solutions are
$$z=2\Bigl(\cos(k\frac{\pi}{2})+i\sin(k\frac{\pi}{2})\Bigr)$$
with $ k\in\{0,1,2,3\}$
The set of solution is
$$S=\{0,2,2i,-2,-2i\}$$
A: OK how about this? Say $z = \cos(\theta) + i \sin (\theta)$ as you say. Then basically we know that $3 \theta = 2\pi - \theta$ so $\theta$ is going to be $\frac{\pi}{2}$ and $z$ is purely imaginary.
If the magnitude is $m$ then we have $m^{3}=4m$ so $m=2$. I think the answer is $2i$. Does it work?
$(2i)^{3}=-8i$ and $4\cdot \overline{2i}=-8i$. They are equal so...Yeah :)
But I didn't get all the possible values for $\theta$. If we write $3 \theta = 2n\pi - \theta$ for $n=0,1,2,3$ we can read off $\theta=0,\frac{\pi}{2},\pi$ and $\frac{3\pi}{2}$ so this gives values of $i=2,2i,-2$ and $-2i$ and they all check back in the equation.
Oh and I just saw somebody else pointed out that $z=0$ works. Of course!
A: 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 $z^3=4\overline{z}$.
