If a number is an $n$-th root of unity, why is its conjugate also an $n$-th root of unity?

Question

If a number is an $n$-th root of unity, why is its conjugate also an $n$-th root of unity? I have read this property from various sources but I was wondering why it is so. For instance if $ z = e^{i 2k\pi/10} $ where $k$ can be $0,1,2,\dots, 9$ then why is $ \bar z $ a 1oth root of unity? Is it because if $a = e^{ix} $ then $\overline{a} = e^{-ix}$? But for roots the value of $k$ is always greater than or equal to zero. Would someone please help me get an understanding of this?

Answer 1

The complex conjugation operation respects multiplication: $\overline{zw}=\overline z\,\overline w$. By induction: $\overline{z^n}=\overline z^n$. If $z^n=1$ then $\overline z^n=\overline{z^n}=\overline 1=1$.

Answer 2

Assume $e^{in\theta} = 1$, that is $e^{i\theta}$ $n$-th root of unity. Then $$\overline{e^{in\theta}} = e^{-in\theta} = \frac{1}{e^{in\theta}}=1.$$

Answer 3

For a geometric interpretation, consider that the complex conjugate is the reflection of the point over the real axis (if drawn on a graph that has real and complex axis).

Also consider that multiplication of complex numbers is achieved by adding their angles towards the real axis together. (Nice pics) (Their absolute lengths multiply, but if we're dealing with roots of unity, that is always unity, so it's irrelevant here)

So an nth root has the property of having an angle a towards the real axis that will give 360° if multiplied by n. The conjugate c will have an angle of 360°-a, therefore c^n will have n*360° -n*a = (n-1)*360 -> unity