Sum of the roots of unity, $z_{1}^p+...+z_{n}^p$

Question

Let $z_1,...,z_n$ be the $n$ roots of unity. I am not able to find a value for the sum: $$z_{1}^p+...+z_{n}^p,\ p \in \Bbb N$$ I know that this sum can also be written as $$\sum_{k=0}^{n-1}e^{i(\frac{2\pi}{n}p\ k)}=1+\sum_{k=1}^{n-1}e^{i(\frac{2\pi}{n}p\ k)}$$ I have been trying with some examples and if I am correct this sum should be equal to $0$, so I think that $1+\sum_{k=1}^{n-1}e^{i(\frac{2\pi}{n}p\ k)}$ should converge to $-1$. That is the furthest I have reached, but I don't know which is the formal mathematical procedure here.

Answer 1

This isn't a formal answer but it's good foundation for intuition.

First of all, if $Z=\{\zeta\}$ is the set of $n$th roots of unity, the sum $S=\sum\zeta$ of these $n$th roots must be $0$ "by symmetry." That is, since the set $Z$ is symmetric (rotating by $1/n$th of a full rotation doesn't change the set), we can expect the sum to also be symmetric, but the only point that doesn't change under rotation is the origin itself. In other words, the sum ought'nt be biased in any particular direction. More algebraically, if we pick a particular root of unity $\xi\ne1$, multiplying-by-$\xi$ is a bijection from $Z$ to itself, so $\xi S$ is the same sum as $S$ but with the summands cyclically reordered. Writing $S=\xi S$, we can divide $(1-\xi)S=0$ to get $S=0$.

Define $f(x)=x^p$. If we consider $Z^p:=\{f(\zeta)\mid \zeta\in Z\}$ (that is, $z_1^p,\cdots,z_n^p$) as a multiset (that is, a set where elements can be members multiple times, i.e. "with multiplicity") then we can say $Z$ is the $m$th roots of unity, where $m=n/g$ with $g=\gcd(n,p)$, but each $m$th root has $g$ copies in $Z^p$. This is a fact from group theory: the function $f$ is a surjective homomorphism of with kernel the $g$th roots of unity, so we can apply the first isomorphism theorem and Lagrange's theorem.

Thus, the sum of the elements of $Z^p$ counted with multiplicity is actually $g\sum \eta$, where $\eta$ ranges over the set $W$ of $m$th roots of unity. As long as $m\ne 1$ (i.e. as long as $p$ is not a multiple of $n$), the sum $\sum\eta=0$, so the sum over elements of $Z^p$ is also $0$. That is, $z_1^p+\cdots+z_n^p=0$ unless $n\mid p$ in which case the sum is $n$.

Answer 2

Notice that this is a sum of a geometric sequence: $$ \sum_{k=0}^{n-1} e^{i\frac{2\pi}{n}p\,k} = \sum_{k=0}^{n-1} \left(e^{i\frac{2\pi}{n}p}\right)^k$$ Then you can use formula $$ \sum_{k=0}^{n-1} q^k = \frac{q^n-1}{q-1}$$ which can be applied if $q=e^{i\frac{2\pi}{n}p}\neq 1$, that is, if $\frac{p}{n} \notin \mathbb{Z}$.

If $\frac{p}{n}\in\mathbb{Z}$, that is $p=mn$ for some $m\in\mathbb{Z}$, note that $z_k^p=1$.