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?

Answer

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.