Solve complex equation with $\overline z$

Question

I need help solving this task, if anyone had a similar problem, it would help me.

The task is: Solve the equation in a set of complex numbers.

$z^3=\overline z$

I tried this :

$z^3=\overline z\\\frac{z^3}{\overline z}=1\\z^2=1\\w_0=(\cos(0)+i\sin(0))\\w_0=1\\w_1=\cos(\pi)+i\sin(\pi)\\w_1=-1$

I only get results: $ 1,-1$

How to get complex solutions?

Thanks in advance !

Set $z=re^{i \theta}$ giving you $r^3e^{3i \theta}=\dfrac{1}{r}e^{-i \theta}$ then $r^3=\dfrac{1}{r}$ and $3i \theta=-i \theta +k 2\pi$. — Jean Marie, Sep 22 '20 at 07:31
Also: https://math.stackexchange.com/q/766599, https://math.stackexchange.com/q/1283252 – all found with Approach0 — Martin R, Sep 22 '20 at 07:34
How did you conclude $z^2=1$ from $\frac{z^3}{\overline z}=1$? — Martin R, Sep 22 '20 at 07:36
Thanks a lot ! I got it, the solutions are : $-1,1,i,-i,0$. $\frac{z^3}{\overline z}=z^2$ that can only be done if z is real. — LogicNotFound, Sep 22 '20 at 07:52

score 1 · Accepted Answer · answered Sep 22 '20 at 07:35

The simplification $z^3/\bar{z}=z^2$ can only be done if $z$ is real, so $z=\bar{z}$ and nonzero.

Instead you can notice that the given equation implies $|z^3|=|\bar{z}|=|z|$. From this you get $|z|^3=|z|$ and therefore either $|z|=0$ or $|z|=1$.

The first case is easy; in the second case, $\bar{z}=z^{-1}$ and you can finish.

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