# Solve the complex equation $w^3=9\overline{w}$

by Jonathan S.   Last Updated July 16, 2019 16:20 PM - source

Apparently I'm having a hard time solving the complex equation $$w^3=9\overline{w}$$.

Here's what I did: $$r^3[\cos(\alpha)+i\sin(\alpha)]^3=9[\cos(-\alpha)+i\sin(-\alpha)]$$ $$r^3[\cos(3\alpha)+i\sin(3\alpha)]=9[\cos(-\alpha)+i\sin(-\alpha)]$$ $$\begin{cases}r^3=9\rightarrow r=\sqrt[3]{9}\\ 3\alpha=-\alpha+2k\pi\rightarrow \alpha=\frac{1}{2}k\pi\end{cases}$$ $$w_0=\sqrt[3]{9}[\cos(0)+i\sin(0)], w_1=\sqrt[3]{9}[\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})], w_2=\sqrt[3]{9}[\cos(\pi)+i\sin(\pi)], w_3=\sqrt[3]{9}[\cos(\frac{3}{2}\pi)+i\sin(\frac{3}{2}\pi)]$$

But when I compute Mathematica

$$(z)^3=9(Conjugate[z])$$

It gives me the solutions: $$z_0=-3; z_1=0; z_2=3; z_3=-3i; z_4=3i$$

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Let $$w=r(\cos\theta+i\sin\theta),$$ where $$r\geq0$$ and $$0^{\circ}\leq\theta<360^{\circ}.$$

Thus, $$r^3(\cos3\theta+i\sin3\theta)=9r(\cos(360^{\circ}-\theta)+i\sin(360^{\circ}-\theta)),$$ which gives $$r^3=9r$$ and $$3\theta=360^{\circ}-\theta+360^{\circ}k,$$ where $$k\in\mathbb Z$$.

Can you end it now?

I got: $$\{0,\pm3,\pm3i\}$$

Michael Rozenberg
July 16, 2019 15:22 PM

You are missing an $$r$$... If you write $$w = r e^{i \theta}$$ your equation becomes $$r^3 e^{3i \theta} = 9 r e^{-i \theta}$$

This means that $$r^3 = 9 r, \quad 3 \theta = -\theta + 2 k \pi, \quad z \in \mathbb{Z}.$$

so, you can have $$r = 0$$, or $$r=3$$ with $$\theta = k \frac{\pi}{2}$$, hence the solutions $$z_1=0$$, $$z_2=3$$, $$z_3=3 i$$, $$z_3=-3$$, $$z_4 = -3i$$.

PierreCarre
July 16, 2019 15:23 PM

Note that\begin{align}w^3=9\overline w&\implies\lvert w\rvert^3=9\left\lvert\overline w\right\rvert=\lvert w\rvert\\&\iff w=0\vee\lvert w\rvert=3.\end{align}And, yes, $$0$$ is a solution. Otherwise, $$w=3\bigl(\cos(\theta)+\sin(\theta)i\bigr)$$, for some $$\theta\in[0,2\pi)$$, and then\begin{align}w^3=9\overline w&\iff27\bigl(\cos(3\theta)+\sin(3\theta)i\bigr)=27\bigl(\cos(-\theta)+\sin(-\theta)i\bigr)\\&\iff\cos(3\theta)+\sin(3\theta)i=\cos(-\theta)+\sin(-\theta)i.\end{align}Can you take it from here?

José Carlos Santos
July 16, 2019 15:28 PM