Solve $z^3=\overline{z}$ in $\mathbb{C}$

Question

Solve an equation in $\mathbb{C}$: $$z^3=\overline{z}$$

Now clearly $$|z|^3= |z^3|=| \overline{z}| =|z|\implies |z|=0 \;\;\vee\;\;|z|=1 $$

So if $|z|=0$ we get $z_1=0$ and if $|z|=1$ we get $$ z^4 = z \cdot \overline{z} =1$$ and so we get $$0=z^4-1 = (z-1)(z+1)(z+i)(z-i)$$ so $z_{2,3}=\pm1$ and $z_{4,5}=\pm i$. So this equation has 5 solutions.

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Second approach: we have $$z^3=\overline{z}\implies \overline{z}^3=z \implies (z^3)^3=z$$ so we get the equation $z^9-z=0$ which has $9$ (different) solution!?

What is going wrong here?

Answer 1

In that way from $z^8=1$ we obtain four extra solutions which doesn’t satisfy the original equation, notably in the derived form $z^4=1$, as in the real case when we square equations in order to eliminate square roots.

Answer 2

if $|z|=1$ then let $z=\cos{\alpha}+i\sin{\alpha}=>z^3=\cos{3\alpha}+i\sin{3\alpha}=\overline{z}=\cos\alpha-i\sin\alpha=>\cos3\alpha=cos\alpha$ and $\sin3\alpha=-sin\alpha\cdots$