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



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