# 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?

• We can write it as $\displaystyle \frac{3}{z^6} = -\left(5|z|^3+2\right) \Rightarrow z^6 = -\frac{3}{5|z|^3+2}$. So $z^6$ must be an purely real no bcz right side is purely real no.. Aug 18, 2015 at 18:44
• as for the answer I got: that's a conjugate, not a minus…I'll write it clearer in the question :) Aug 18, 2015 at 18:46
• I am slightly confused by your last comment, the last line gives you that $\theta=\frac{k\pi}{6}$ for $k=0,\cdots,11$, but do any of those satisfy the first equation? Aug 18, 2015 at 18:52
• Just a general note: If you have a system of equations, let's say two, solutions of the second equation are not necessarily the solutions of the first automatically. You need to solve each and find intersection. So, for example, if $\theta = 0$ is the solution for the second equation, doesn't mean it also satisfies the fist. Aug 18, 2015 at 18:52

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.