Proving that $\sum \cos p \theta = 0$ where $\theta$ is argument of the primitive roots of unity

Question

Prove that if $\theta_1, \theta_2, \theta_3 \ldots$ be the arguments of the primitive nth roots of unity, $\sum \cos p \theta = 0$ when $p$ is a positive integer less than $\frac n {abc\cdots k}$ where $a,b,c,\ldots k$ are the different constituent primes of $n$; and that, when $p = \frac n {abc\cdots k}$, $\sum \cos p \theta = \frac {(-1)^{\mu}n} {abc \cdots k}$, where $\mu$ is the number of the constituent primes.

I know that the $n$th roots of unity are $e^{2\pi ik/n}$ for $k=0,1,\ldots n-1$, and the primitive roots are those whose powers give all the roots, so the integers $k$ which are coprime to $n$. Any hints on how to go about finding that sum?

Answer 1

The precise condition on this is a bit bizarre.

Consider any number $p < n$ and let $\theta = \theta_1$. We have $$\sum_{k=0}^{n-1} cos(p \theta_k) = Re \sum_{k=0}^{n-1} e^{(i\theta^k)^p} = Re \sum_{k=0}^{n-1} e^{(p i\theta)^k} = Re\frac{1-e^{(ip\theta)^n}}{1-e^{ip\theta}} = 0 $$

Since since $e^{ipn\theta} = (e^{in\theta})^p = 1$

I believe your second question is incorrectly stated as your given $p$ is less than $n$, so we should still have what's done above.

Answer 2

It is possible to show that $p\theta_i$ are also arguments of same roots of unity as $p$ is integer. So since roots of unity sum to zero, real and imaginary parts are zero, which gives your sum as zero.