What's wrong with the way I solved the complex equation $z^5=-16\overline{z} (z \in \mathbb{C})$ 
Solve the fallowing complex equation($z \in \mathbb{C})$: $z^5=-16\overline{z}$.

Here's how I tried to solve the equation:
$$ z^5=-16\overline{z} \rightarrow z^6=-16(x^2+y^2)$$
$$z= \varepsilon(\cos{\theta} +i\sin{\theta}) \rightarrow z^6=\varepsilon^6(\cos{6\theta} +i\sin{6\theta})$$
$$w=r(\cos{\varphi} +i\sin{\varphi})=-16(x^2+y^2) $$
$$r=\sqrt{(-16)^2(x^2+y^2)^2} = 16(x^2+y^2)$$
$$ \varphi = \arctan{0} = \pi$$
$$ 6\theta = \varphi + 2k\pi, k\in \mathbb{Z}\rightarrow \theta_0 = \frac{\pi}{6}; \theta_1 = \frac{\pi}{2}; \theta_2 = \frac{5\pi}{6} \theta_3 = \frac{7\pi}{6}; \theta_4 = \frac{3\pi}{2}; \theta_5 = \frac{11\pi}{6}$$
$$ z_0=16(x^2+y^2)e^{i\frac{\pi}{6}};z_1=16(x^2+y^2)e^{i\frac{\pi}{2}};z_2=16(x^2+y^2)e^{i\frac{5\pi}{6}};z_3=16(x^2+y^2)e^{i\frac{7\pi}{6}};z_4=16(x^2+y^2)e^{i\frac{3\pi}{2}};z_5=16(x^2+y^2)e^{i\frac{11\pi}{6}} $$
But my teacher told me that I didn't solve it correctly. Where am I wrong?

$x$ and $y$ are unknown, so it makes no sense saying $z_0=16(x^2+y^2)e^{i\pi/6}$. Recall that $z=x+iy$, according to what you wrote.
A proper way would be first noting that $z=0$ is a solution, so we can henceforth assume $z\ne0$.
Write $z=re^{i\varphi}$; then the equation becomes
$$
r^5e^{5i\varphi}=-16re^{-i\varphi}
$$
and therefore, being $r\ne0$,
$$
r^4e^{6i\varphi}=-16
$$
Then, as $r>0$, we get $r=2$ and $e^{6i\varphi}=-1=e^{i\pi}$, so
$$
6i\varphi=\pi+2k\pi
$$
You get six distinct solutions for $k=0,1,\dots,5$.
Alternatively, take the modulus of both sides:
$$
|z|^5=16|\bar{z}|=16|z|
$$
so $|z|^4=16$ and $|z|=2$ (we're still assuming $z\ne0$). Write $z=2u$; then the equation becomes $u^5=-\bar{u}=-u^{-1}$, since $|u|=1$. Thus we get $u^6=-1$ and so $u$ is one of the sixth roots of $-1$.
