Complex Roots of Unity occur in Conjugate Pairs

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $U_n$ denote the complex $n$th roots of unity:
 * $U_n = \set {z \in \C: z^n = 1}$

Let $\alpha \in U_n$ be the first complex $n$th root of unity.

Then:
 * $\forall k \in \Z_{>0}, k < \dfrac n 2: \overline {\alpha^k} = \alpha^{n - k}$

That is, each of the complex $n$th roots of unity occur in conjugate pairs:


 * $\tuple {\alpha, \alpha^{n - 1} }; \tuple {\alpha^2, \alpha^{n - 2} }; \ldots; \tuple {\alpha^s, \alpha^{n - s} }$

where:
 * $s = \dfrac {n - 1} 2$ for odd $n$
 * $s = \dfrac {n - 2} 2$ for even $n$.

Proof
Consider the polynomial equation:
 * $(1): \quad z^n - 1 = 0$

The complex $n$th roots of unity are:
 * $1, \alpha, \alpha^2, \ldots, \alpha^{n - 1}$

From Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs, the roots of $(1)$ occur in conjugate pairs.

Let $k \in \Z$ such that $1 \le k \le n$.

Then:

That is, the complex $n$th root of unity which is the other half of the conjugate pair with $\alpha^k$ is $\alpha^{n - k}$.

When $n$ is odd, these pair up as:
 * $\tuple {\alpha, \alpha^{n - 1} }; \tuple {\alpha^2, \alpha^{n - 2} }; \ldots; \tuple {\alpha^s, \alpha^{n - s} }$

where $s$ is the largest integer less than $\dfrac n 2$; that last pair can be expressed:
 * $\tuple {\alpha^s, \alpha^{s + 1} }$

When $n$ is even:
 * $\alpha^s = \alpha^{n - s}$ when $s = \dfrac n 2$.

and in fact $\alpha^{n / 2}$

and so is wholly real.

Thus from Complex Number equals Conjugate iff Wholly Real:
 * $\alpha^{n / 2} = \alpha^{n - n / 2}$

Hence the complex $n$th roots of unity pair up as:
 * $\tuple {\alpha, \alpha^{n - 1} }; \tuple {\alpha^2, \alpha^{n - 2} }; \ldots; \tuple {\alpha^s, \alpha^{n - s} }$

where $s$ is the largest integer less than $\dfrac n 2$; that last pair can be expressed:
 * $\tuple {\alpha^s, \alpha^{s + 2} }$

The result follows;