Prove that $\alpha_{1} ^k+ \alpha_{2} ^k +.....+ \alpha_{n} ^k = n$ for $k=0$ and $0$ for $k = 1,2,....,n-1$?

Question

For $n\geq 2$, let $\alpha_{1},\alpha_{2},\dots,\alpha_{n} $ be all the $n$th roots of unity over a field and the roots are not necessarily distinct. We have to prove that $\alpha_{1} ^k+ \alpha_{2} ^k +\dots+ \alpha_{n} ^k = n$ for $k=0$ and $\alpha_{1} ^k+ \alpha_{2} ^k +\dots+ \alpha_{n} ^k = 0$ for $k = 1,2,\dots,n-1$.

For $k=0$ the case is trivial and I have become mentally exhausted but still no idea how to prove for $k = 1,2,...,n-1$.

Answer 1

You probably mean that $\alpha_1,\alpha_2,\dots,\alpha_n$ are the $n$-th roots of unity. The case $k=0$ is indeed trivial.

For the other cases, recall that those elements are the roots of $X^n-1$, so by Viète's formulas, you know that \begin{gather} S_1(\alpha_1,\dots,\alpha_n)=0\\ S_2(\alpha_1,\dots,\alpha_n)=0\\ \vdots\\ S_{n-1}(\alpha_1,\dots,\alpha_n)=0\\ S_n(\alpha_1,\dots,\alpha_n)=-1 \end{gather} where $S_k$ is the $k$-th degree basic symmetric polynomial: \begin{gather} S_1(X_1,\dots,X_n)=X_1+X_2+\dots+X_n\\ S_2(X_1,\dots,X_n)=X_1X_2+X_1X_3+\dots\\ \vdots\\ S_n(X_1,\dots,X_n)=X_1X_2\dots X_n \end{gather} Since $$ X_1^k+X_2^k+\dots+X_n^k $$ is a symmetric polynomial of degree $k$, it can be expressed as a polynomial in the basic symmetric polynomial of degree at most $k$.

Answer 2

You're still just summing $n$th roots of unity, but potentially with repetitions and different $n$...

Consider the image of $x\mapsto x^k$ on the roots of unity, and the multiset $\{a_1^k,a_2^k,\cdots,a_n^k\}$.

Answer 3

The roots of unity are the roots of the equation $x^n-1=0$ the sum $\alpha_1+ \cdots +\alpha_n$ is the coefficient of $x^{n-1}$ which is zero. By taking the $k$ th power we either get the roots again in a different order in case $(n,k)=1$ or if $(n,k)=l$ we are getting the $n/l$ th roots $l$ times. But we know they sum to zero by the first argument.