Complex solutions of a quintic equation: $z^5+4\overline z^3=0$ 
Solve $$z^5+4\overline z^3=0$$

This is what I did. Let $z=r(\cos\theta+i\sin\theta)$ so, the equation is:
$$r^5(\cos5\theta+i\sin5\theta)+4r^3(\cos3\theta-i\sin3\theta)=0.$$
Suppose $r$ is not $0$ and divide both sides by $r^3$ to get:
$$(r^2\cos5\theta+4\cos3\theta)+i(r^2\sin5\theta-4\sin3\theta)=0+0i$$
so we get the following system of equations: 
$$r^2\cos5\theta+4\cos3\theta=0$$
$$r^2\sin5\theta-4\sin3\theta=0$$
I'm not sure how to continue.
I've tried subtract $4\cos3\theta$ from both sides in the first equation and in the second one add $4\sin3\theta$ to both sides, and then divide the equations, to get $\tan5\theta=-\tan3\theta$, but then i suppose im not dividing be zero and that's a problem.
What can I do from here? Is there a simpler way to solve this problem? Thanks!
 A: First of all, this is not polynomial. It looks like it, but it's not: the fundamental theorem of algebra doesn't apply (notice that we get nine solutions at the end, not five).
Notice that $z = 0$ is an obvious solution, so we will look for non-zero solutions. Multiply the equation by $z^3$ to get $$z^8 + 4|z^6| = 0\\ z^8 = -4|z|^6$$
so $z^8$ is negative real number, i.e. $8\arg z \equiv \pi \pmod {2\pi}$, by de Moivre's formula, and $\arg z = \frac{\pi + 2k\pi}{8}.$
On the other hand, $$z^8 = - 4|z|^6 \implies |z|^8 = 4|z|^6 \implies |z|^2 = 4 \implies |z|=2.$$
Thus, the solutions are $2e^{i\frac{\pi + 2k\pi}{8}},\ k\in\mathbb Z$ and $0$.
A: Note that as per Euler's Formula the complex number $z$ can be expressed as $z=re^{i\theta}$ where $|z|=r$ and $\theta$ is the argument of $z$. Likewise, the complex number $\bar z $ can be expressed as $\bar z=re^{-i\theta}$ where $|\bar z|=r$ and $-\theta$ is the argument of $\bar z$.
Hence, you are looking for solutions to $$r^5e^{5i\theta}+4r^3e^{-3i\theta}=0$$
Two "solution branches" are possible: $r=0$ and $r\ne0$
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For $r=0$ the solution is obviously the number $0$
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For $r\ne 0$, 
$$r^5e^{5i\theta}+4r^3e^{-3i\theta}=0$$
Dividing by $r^3$ (since $r\ne 0$)
$$r^2e^{5i\theta}+4e^{-3i\theta}=0$$
$$\Rightarrow r^2e^{8i\theta}+4e^{0i}=0$$
$$\Rightarrow r^2e^{8i\theta}=-4e^{0i}=4e^{i\pi}=4e^{i(\pi+2k\pi)}$$
$$\Rightarrow r^2e^{8i\theta}=4e^{i(\pi+2k\pi)}$$
$$\Rightarrow r=2,\theta=\left(\dfrac{\pi}{8}+\dfrac{2k\pi}{8}\right)$$ where $k=0,1,2,3,...,7$
Note that $-1=e^{i\pi}$
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Thus all the solutions to the equation are $$z=0 , 2e^{i(\frac\pi8+\frac{2k\pi}8)}$$ where $k=0,1,2,3,...,7$
