Find all solutions for $z^3 = \overline{z}$

Question

I know that $z = a + ib$ and that $\overline{z} = a - ib$, but when I try and calculate the solutions I get an unsolvable equation.

Would appreciate any help.

Answer 1

Forget about real and imaginary parts and note that every solution $z$ is such that $|z|^3=|z^3|=|\bar z|=|z|$ hence $|z|=0$ or $|z|=1$. Furthermore, $z^4=z^3\cdot z=\bar z\cdot z=|z|^2$.

Can you finish in both cases?

Answer 2

First note that, if $|z|=r$, then $|z^3|=|z|^3=|\overline{z}|=|z|$, giving $r\in\{0,1\}$, so $z=0$ works and $z=1 \text{cis } \theta$ works, for some $\theta$. Plugging in gives $z^3=\text{cis } 3\theta$ and $\overline{z}=\text{cis } -\theta$, so we have $ 3 \theta \equiv -\theta \pmod {2\pi} $.

That means that $4\theta$ is a multiple of $2\pi$, which is true iff $\theta$ is a multiple of $\frac{\pi}{2}$.

Can you finish?