Solve complex equation $5|z|^3+2+3 (\bar z) ^6=0$ I'm stuck in trying to solve this complex equation 
$$ 5|z|^3+2+3 (\bar z)^6=0$$
where $\bar z$ is the complex conjugate. 
Here's my reasoning: using $z= \rho e^{i \theta}$ I would write
$$ 5\rho^3+ 2 + 3 \rho^6 e^{-i6\theta} = 0 \\ 5\rho^3+ 2 + 3 \rho^6 (\cos(6 \theta) -  i \cdot \sin(6 \theta)) = 0 \\$$
from where I would write the system 
$$\begin{cases} 5\rho^3+ 2 + 3 \rho^6 \cos(6 \theta) = 0 \\ 3 \rho^6 \sin(6 \theta) = 0\end{cases}$$
But here I get an error, since, from the second equation, I would claim $ \theta = \frac{k \pi}{6}$ for $ k=0…5$, but $\theta = 0$ means the solution is real and the above equation doesn't have real solutions…where am I mistaken?

Let $w = \overline{z}^3$.  Then we have
$$
5|w|+2+3w^2 = 0
$$
As you point out, this constrains $w = k$ or $w = ki$ for real $k$.
Case 1. $w = k$
$$
3k^2+5|k|+2 = 0
$$
which yields no solutions since the left-hand-size is always positive.
Case 2. $w = ki$
$$
-3k^2+5|k|+2 = 0
$$
which yields $k = \pm 2$, so $w = \pm 2i$.
The rest is left as an exercise.
