Problem finding zeros of complex polynomial I'm trying to solve this problem
$$ z^2 + (\sqrt{3} + i)|z| \bar{z}^2 = 0 $$
So, I know $ |z^2| = |z|^2 = a^2 + b ^2 $ and $ \operatorname{Arg}(z^2) = 2 \operatorname{Arg} (z) - 2k \pi = 2 \arctan (\frac{b}{a} ) - 2 k\pi $ for a $ k \in \mathbb{Z} $.  Regarding the other term, I know $ |(\sqrt{3} + i)|z| \bar{z}^2 | = |z|^3 |\sqrt{3} + i| = 2 |z|^3 = 2(a^2 + b^2)^{3/2} $ and because of de Moivre's theorem, I have $ \operatorname{Arg} [(\sqrt{3} + i ) |z|\bar{z}^2] = \frac{\pi}{6} + 2 \operatorname{Arg} (z) - 2Q\pi $. 
Using all of this I can rewrite the equation as follows
$$\begin{align*}
&|z|^2 \Bigl[ \cos (2 \operatorname{Arg} (z) - 2k \pi) + i \sin (2 \operatorname{Arg}(z) - 2k \pi)\Bigr]\\
&\qquad \mathop{+} 2|z|^3 \Biggl[\cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right) + i \sin \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right)\Biggr] = 0
\end{align*} $$
Which, assuming $ z \neq 0 $, can be simplified as 
$$\begin{align*}
&\cos (2 \operatorname{Arg} (z) - 2k \pi) + i \sin (2 \operatorname{Arg} (z) - 2k \pi) \\
&\qquad\mathop{+} 2 |z|\Biggl[\cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q \pi \right) + i \sin \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right)\Biggr] = 0
\end{align*} $$
Now, from this I'm not sure how to go on. I tried a few things that got me nowhere like trying to solve
$$ \cos (2 \operatorname{Arg}(z) - 2k \pi) = 2 |z| \cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right) $$
I'm really lost here, I don't know how to keep going and I've looked for error but can't find them. Any help would be greatly appreciated.
 A: Here is an alternative to solving it using polar form.
Let $z=a+bi$, so that $\bar{z}=a-bi$ and $|z|=\sqrt{a^2+b^2}$. Then you want to solve
$$(a+bi)^2+(\sqrt{3}+i)\sqrt{a^2+b^2}(a-bi)^2=0,$$
which expands to
$$(a^2-b^2)+2abi+(\sqrt{3}+i)\sqrt{a^2+b^2}\left((a^2-b^2)-2abi\right)=0$$
Thus, we need both the real part and the imaginary part of the left side to be 0, i.e.
$$(a^2-b^2)+\sqrt{a^2+b^2}\left(\sqrt{3}\cdot (a^2-b^2)+2ab\right)=0$$
and
$$2ab+\sqrt{a^2+b^2}\left(-2ab\sqrt{3}+(a^2-b^2)\right)=0.$$
It should be possible to solve these equations by simple manipulations, though I haven't worked it out myself yet.
A: The relation is equivalent to $z^2=-(\sqrt{3}+i)|z|\overline{z}^2$. $z=0$ is a solution, so in the following $z \neq 0$. Take modulus of both sides and denote $r=|z|=|\overline{z}|$. Then $r^2=2r^3$, which means $r=\frac{1}{2}$. 
The relations turns to $z^2+\frac{1}{2}(\sqrt{3}+i)\overline{z}^2=0$. Multiply by $z^2$ and get $z^4+\frac{1}{2}(\sqrt{3}+i)\frac{1}{16}=0$. Write it in trigonometric form 
$$ z^4=\frac{1}{16}\left(-\frac{\sqrt{3}}{2}-i\frac{1}{2}\right)=\frac{1}{16}(\cos \frac{7\pi}{6}+\sin \frac{7\pi}{6})$$. 
From here on it is just the extraction of complex roots.
[edit] I did not answer your question, as to how to continue your calculations, but I can say from experience that in most complex numbers problems the substitution $z=a+bi$ gets you in more troubles in the end, than working with the properties of complex conjugate, modulus and trigonometric form. You can see that in my solution no great computational problems were encountered.
