# $(n-1)^{\text{th}}$ root of conjugate

Statement: If $n \in \mathbb{Z}, n \gt 2$, then the equation $\overline{z} = z^{n-1}$ has $n+1$ (distinct) solutions.

I understand that a solution is a pair $(z,\overline{z})$ that satisfies the equation. I also know how to derive the result by expressing $z,\overline{z}$ in polar coordinates and solving for the angles, I will not put the solution here.

Question: In order to solve the equation, why is it not allowed to take the $(n-1)^{th}$ root directly? i.e

Let $\overline{z}=r\operatorname{cis}\theta$, take the $(n-1)^{th}$ root, then $z= r^{\frac{1}{n-1}}\operatorname{cis}\left(\frac{\theta+2k\pi}{n-1}\right)$, with $k=0,1,2...n-2$.

Then we get $n-1$ solutions instead of the required $n+1$ solutions!

Your proposed method of solution does not work because $z$ and $\bar z$ are related complex variables. It fails for the same reason that when one is asked to solve, say, $$z^3 = z + 1,$$ you cannot simply just take the cube root of both sides, because $z$ appears on both sides of the equation.
Instead, a correct solution should take into account that the magnitude and argument of both sides must be equal: $$z^{n-1} = \bar z$$ implies that $$|z|^{n-1} = |z^{n-1}| = |\bar z| = |z|.$$ Hence $$|z|(|z|^{n-2} - 1) = 0,$$ which implies $|z| = 0$ or $|z| = 1$. If $|z| = 0$ then $z = 0$, which furnishes one solution, so we now focus on the case $|z| = 1$: this implies $z = e^{i\theta}$, hence $$\bar z = e^{-i\theta},$$ and the original equation becomes $$e^{i(n-1)\theta} = e^{-i\theta}.$$ It follows that $$(n-1)\theta = -\theta + 2\pi k, \quad k \in \mathbb Z,$$ and the solution of this is straightforward.
well $0$ is one solution. otherwise if $z \ne 0$ then multiplying $z^{n-1}=\bar z$ by $z$ gives: $$z^n = z\bar z = |z|^2 \gt 0$$ which, if $n \gt 2$ gives $|z|=1$ and $z=e^{\frac{2\pi ki}{n}}$ for the $n$ values $k=0,1,\dots,n-1$