Let $f_1, \ldots, f_m$ be the characters of a finite group $G$ with order $m$, and let $a$ be an element of $G$ with order $n$. Theorem 6.7 shows us that each number $f_r(a)$ is an $n$th root of unity. Prove that every $n$th root of unity occurs equally often among the numbers $f_1(a), \ldots, f_m(a)$.
They also give the hint: Evaluate the sum $$\sum_{r=1}^m \sum_{k=1}^n f_r(a^k)e^{-\frac{2\pi i k}{n}}$$ in two ways to determine the number of times $e^{\frac{2\pi i}{n}}$ occurs.
I think evaluating in the order that it is currently written gives me 0, but I'm not sure how that helps me answer the problem.
We need the orthogonality relations for characters, specifically $$\sum_{r = 1}^m f_r(b) = \begin{cases} m &\text{if } b = 1 \\ 0 &\text{if } b \neq 1 \end{cases}$$ and the formula for a geometric sum, specifically $$\sum_{k = 1}^n \rho^k = \begin{cases} n &\text{if } \rho = 1 \\ 0 &\text{if } \rho \neq 1 \end{cases}$$ for an $n^{\text{th}}$ root of unity $\rho$.
Thus, since $f_r(a^k) = f_r(a)^k$ and $f_r(a)e^{-\frac{2\pi i}{n}}$ is an $n^{\text{th}}$ root of unity, for each $r$ we have $$\sum_{k = 1}^n f_r(a^k)e^{-\frac{2\pi ik}{n}} = \begin{cases} n &\text{if } f_r(a) = e^{\frac{2\pi i}{n}} \\ 0 &\text{otherwise}\end{cases}$$ and consequently $$\sum_{r = 1}^m \sum_{k = 1}^n f_r(a^k)e^{-\frac{2\pi ik}{n}} = n\cdot \#\bigl\{r : f_r(a) = e^{\frac{2\pi i}{n}}\bigr\}.$$
Changing the order of summation, we have $$\sum_{r = 1}^m f_r(a^k) = \begin{cases} m &\text{if } k = n\\ 0 &\text{if } k < n \end{cases}$$ and therefore $$\sum_{k = 1}^n \sum_{r = 1}^m f_r(a^k)e^{-\frac{2\pi ik}{n}} = \sum_{r = 1}^m f_r(a^n)e^{-\frac{2\pi in}{n}} = m.$$
It follows that $$\#\bigl\{r : f_r(a) = e^{\frac{2\pi i}{n}}\bigr\} = \frac{m}{n}.$$
No special properties of $e^{\frac{2\pi i}{n}}$ were used, only that it is an $n^{\text{th}}$ root of unity, thus the same argument works for $$\sum_{r = 1}^m \sum_{k = 1}^n f_r(a^k)\rho^{-k}$$ where $\rho$ is any $n^{\text{th}}$ root of unity.