What is the sum of the squares of the 10th roots of unity?

Question

Obviously the sum of the roots of unity is 0, but is there a way to calculate this other than calculating them all individually and squaring them?

Answer 1

square of 10th root of unity is a fifth root of unity. can you prove the sum is 0?

You have $$ \sum_{k=0}^9 e^{-ik\pi/10} = 0, $$ and you are asked to compute $$ \sum_{k=0}^9 \left(e^{-ik\pi/10}\right)^2 = \sum_{k=0}^9 e^{-ik\pi/5}, $$ which is zero as well because it is just traversing the 5-roots twice.

Answer 2

The square of a tenth root is a fifth root. From any fifth root you get two tenth roots. Different fifth roots give rise to two different tenth roots.

Thus you're summing twice the fifth roots.

Answer 3

We have $q^n-1 =(q-1)(1+q+\ldots+q^{n-1})$.

Thus if $\xi$ is a primitive $n$-th root of unity, $\xi^n=1$ and so $1+\xi+\ldots+\xi^{n-1}=0$ as required.