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

Question

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: $z_0=-3; z_1=0; z_2=3; z_3=-3i; z_4=3i$

Answer 1

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

Answer 2

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 $z_1=0$, $z_2=3$, $z_3=3 i$, $z_3=-3$, $z_4 = -3i$.

Answer 3

Note that\begin{align}w^3=9\overline w&\implies\lvert w\rvert^3=\lvert w^3\rvert=9\left\lvert\overline w\right\rvert=9\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?

Answer 4

Here is another approach, just for the value of looking at a different way to solve the problem. If $w = a+bi$ then $\bar{w} = a-bi$ and we are saying that $$ 9a-9bi = 9\bar{w} = w^3 = a^3-3ab^2 - i\left(b^3 - 3a^2b\right) $$ which implies the system of equations $$ \begin{split} a^3-3ab^2 &=9a\\ b^3-3a^2b &= 3ab \end{split} $$

Case I. $b=0$ then the second equation is trivially true and the first implies $a^3=9a \iff a\in \{0, \pm 3\}$, so we have the solutions $w \in \{0,\pm 3\}$.

Case II $b \ne 0$ and we can cancel a $b$ in the second equation, getting $$b^2-3a^2 = 3a \iff b^2 = 3a^2+3a,$$ which we plug into the first to get $$ a^3 - 3a\left(3a^2+3a\right) = 9a $$ which either implies $a=0, b=0$ or results in $$ a^2 - 3a(3a+3) = 9 \iff 8a^2+9a+9=0 $$ and solving the quadratic will yield the final answers for $w$.