Trouble computing a sum of Dirichlet characters.

Question

Let $\chi(n)$ be a character mod $m$, and let $\rho$ be an $h$th root of unity. I am trying to compute the following sum \begin{equation} \sum_{\chi}(\rho^{-1}\chi(a) + \rho^{-2}\chi(a^2) + \cdots + \rho^{-h}\chi(a^h))\end{equation} and then use the result to prove that the $\phi(m)$ characters mod $m$ at $a$ take all $h$th roots of unity with equal frequency. I know that $\chi(a)$ is an $h$th root of unity, but am not sure how to use this in the computation. Any tips would be appreciated.

Answer 1

Because $\chi(a)$ is an $h$th root of unity, $\chi(a)^h = 1$, and because $\chi$ is a multiplicative function, $\chi(a)^h = \chi(a^h) = 1$. We also know that $$ \sum_\chi \chi(b) = \begin{cases} \phi(m) & \text{if $b \equiv 1 \pmod m$, or}\\ 0 & \text{otherwise.} \end{cases}$$ Without loss of generality, we can assume that the order of $a$ is $h$. Then the sum is \begin{align} & \sum_\chi \left(\rho^{-1}\chi(a) + \rho^{-2}\chi(a^2) + \cdots + \rho^{-h}\chi(a^h)\right)\\ & = \rho^{-1} \sum_\chi \chi(a) + \rho^{-2} \sum_\chi \chi(a^2) + \cdots + \rho^{-h} \sum_\chi \chi(a^h)\\ & = \rho^{-1} \cdot 0 + \rho^{-2} \cdot 0 + \cdots + \rho^{-(h-1)} \cdot 0 + 1 \sum_\chi 1\\ & = \sum_\chi 1\\ & = \phi(m). \end{align}

That answers your question. Here is a hint to help you prove that the $\phi(m)$ characters modulo $m$ at $a$ take all $h$th roots of unity with equal frequency: Find the value of $$1 + z + z^2 + \cdots + z^{h-1}$$ if $z$ is an $h$th root of unity, and then take $z = \rho^{-1} \chi(a)$.