TITLE: Powers of roots of unity
QUESTION [5 upvotes]: Let $\omega_1,\omega_2,...,\omega_k$ be $n$th roots of unity (not necessarily primitive) and let $q$ be coprime to $n$. If elements of $\omega_j$ have a vanishing sum is it true that their $q$th powers also have a vanishing sum? 
I assume not so the next question is "what is the smallest $n$ for which there is a counterexample?" ($q$ and $n$ are of course natural numbers.)

REPLY [11 votes]: There's no counterexample; the claim is true, as shown below . . .

Fix positive integers $n,q$, with $\gcd(q,n)=1$.

Let $\omega=\exp\bigl(\frac{2i\pi}{n}\bigr)$.

Claim:

If $e_1,...,e_k$ is a sequence of nonnegative integers such that $$\omega^{e_1}+\cdots+\omega^{e_k}=0$$
then the equality
$$\omega^{qe_1}+\cdots+\omega^{qe_k}=0$$
also holds.

Proof:

Let $f(x) = x^{e_1}+\cdots+x^{e_k}$, and let $\phi(x)$ be the $n$-th cyclotomic polynomial.

Since $\gcd(q,n)=1$, it follows that $\omega^q$ is a primitive $n$-th root of unity, hence $\phi(\omega^q)=0$.
\begin{align*}
\text{Then}\;\;&\omega^{e_1}+\cdots+\omega^{e_k}=0\\[4pt]
\implies\;&f(\omega)=0\\[4pt]
\implies\;&f(x)=\phi(x)g(x),\;\text{for some}\;g\in\mathbb{Q}[x]\\[4pt]
\implies\;&f(x^q)=\phi(x^q)g(x^q)\\[4pt]
\implies\;&f(\omega^q)=\phi(\omega^q)g(\omega^q)\\[4pt]
\implies\;&f(\omega^q)=0\\[4pt]
\implies\;&\omega^{qe_1}+\cdots+\omega^{qe_k}=0\\[4pt]
\end{align*}
as was to be shown.