The distance from 1 to the other $n$th roots of unity

Question

I want to prove that the sum of the fourth powers of the diagonals of a regular $n$-gon inscribed in the unit circle is equal to $6n$. I consider the distance from 1 to the other $n$th roots of unity given by $\omega^k$, $k=1,2,\dots, (n-1)$. So basically my working is $$\sum_1^{n-1}|1-\omega^k|^4=\sum_1^{n-1}(|1-\omega^k|^2)^2=\sum_1^{n-1}\left[(1-\omega^k)(1-\omega^{-k})\right]^2=\sum_1^{n-1}(1-\omega^k-\omega^{-k}+1)^2=\sum_1^{n-1}(6-4\omega^k-4\omega^{-k}+\omega^{2k}+\omega^{-2k}).$$ So basically I now split the sum and $$\sum_1^{n-1}\omega^k=\sum_1^{n-1}\omega^{-k}=-1,$$ right?

But what about $$\sum_1^{n-1}\omega^{2k}$$ and $$\sum_1^{n-1}\omega^{-2k}?$$

Thanks.

Answer 1

Let $\omega_k=\omega^k$. Recall that $(\omega_k)_{1\leqslant k\leqslant n}$ is the set of roots of the polynomial $x^n-1$. The sum $$ \sum_{k=1}^n\omega_k=\sum_{k=1}^{n}\omega^k=1+\sum_{k=1}^{n-1}\omega^k $$ is the sum of the roots of $x^n-1$, hence its opposite equals the coefficient of $x^{n-1}$ in this polynomial, that is, zero for every $n\geqslant2$. Thus, $$ \sum_{k=1}^{n-1}\omega^k=-1+\sum_{k=1}^n\omega_k=-1. $$ Likewise, $$ 1+\sum_{k=1}^{n-1}\omega^{2k}=\sum_{k=1}^n\omega_k^2=\left(\sum_{k=1}^n\omega_k\right)^2-\sum_{1\leqslant k\ne\ell\leqslant n}\omega_k\omega_\ell. $$ The double sum in the RHS is the coefficient of $x^{n-2}$ in the polynomial $x^n-1$ hence, for every $n\geqslant3$, zero. Thus, $$ \sum_{k=1}^{n-1}\omega^{2k}=-1. $$ Finally, $z\mapsto1/z$ is an involution of the set of roots of $x^n-1$ with a fiwed point $1=\omega^n$, for every $n\geqslant1$, hence $$ \sum_{k=1}^{n-1}\omega^{-k}=\sum_{k=1}^{n-1}\omega^{k},\qquad \sum_{k=1}^{n-1}\omega^{-2k}=\sum_{k=1}^{n-1}\omega^{2k}. $$

Answer 2

If $n$ is odd, then $\omega^2$ is also a primitive $n$th root of unity. If $n=2m$, then $\omega^2$ is a primitive $m$th root of unity. In either case, you can use formulas you already know.

Answer 3

HINT: If $n$ is odd, the integers $2k\bmod n$ for $k=1,\dots,n-1$ run through $1\dots,n-1$, so

$$\sum_{k=1}^{n-1}\omega^{2k}=\sum_{k-1}^{n-1}\omega^k\;.$$

If $n$ is even, the integers $2k\bmod n$ for $k=1,\dots,n-1$ run through $2,4,\dots,n-2$ twice with a $0$ in the middle. The powers $\omega^{2k}$ for $k=1,\dots,\frac{n}2-1$ are the $\frac{n}2$-nd roots of unity different from $1$.