Saturday, November 18, 2017

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.


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