How do you evaluate this trigonometric sum?

Question

I have strong reason$^{\dagger}$ to believe that the following equation is true:

$$\sum_{m=0}^{n} \left[\left(e^{i\pi\frac{k+k'}{n}}\right)^m+\left(e^{i\pi\frac{k-k'}{n}}\right)^m+\left(e^{i\pi\frac{-k+k'}{n}}\right)^m+\left(e^{i\pi\frac{-k-k'}{n}}\right)^m\right]=2[1+\cos \left[(k+k')\pi\right]$$

Where $0<k,k'<n$ are positive integers, but with $k\neq k'$ (for simplicity we can take $k>k'$). I have already verified this for various values of $k$, $k'$, and $n$, but I can't seem to prove it in general. Any suggestions?

$^{\dagger}$I came to this formula by writing out the character orthogonality relation for different 2-dimensional representations of the dicyclic group $Q_{2n}$.

Answer 1

$$ \begin{aligned} \sum_{m=0}^{n} &\left[\left(e^{i\pi\frac{k+k'}{n}}\right)^m+\left(e^{i\pi\frac{k-k'}{n}}\right)^m+\left(e^{i\pi\frac{-k+k'}{n}}\right)^m+\left(e^{i\pi\frac{-k-k'}{n}}\right)^m\right]\\ &= \sum_{m=0}^{n} \left[\left(e^{i\pi\frac{k+k'}{n}}\right)^m+\left(e^{i\pi\frac{k-k'}{n}}\right)^m\right] + \sum_{m=-n}^{0} \left[\left(e^{i\pi\frac{k+k'}{n}}\right)^m+\left(e^{i\pi\frac{k-k'}{n}}\right)^m\right]\\ &= 2 + \sum_{m=-n}^{n} \left[\left(e^{i\pi\frac{k+k'}{n}}\right)^m+\left(e^{i\pi\frac{k-k'}{n}}\right)^m\right]\\ &= 2 + e^{i\pi(k+k')} + e^{i\pi(k-k')} + \sum_{m=-n}^{n-1} \left[\left(e^{i\pi\frac{k+k'}{n}}\right)^m+\left(e^{i\pi\frac{k-k'}{n}}\right)^m\right]\\ &= 2 + e^{i\pi(k+k')} + e^{i\pi(-k-k')}\\ &\text{(By periodicity of $e$; and note that the sum comes from geometric series)}\\ &= 2(1+\cos(k+k')\pi). \end{aligned} $$

Answer 2

Using the formula for the sum of a geometric series, $$ \sum_{m=-n}^{n-1}e^{i\pi\frac knm}= \left\{\begin{array}{} \frac{e^{i\pi k}-e^{-i\pi k}}{e^{i\pi\frac kn}-1}=0&\text{if $2n\nmid k$}\\ 2n&\text{if $2n\mid k$} \end{array}\right. $$ Therefore, $$ \begin{align} \sum_{m=0}^n\left[e^{i\pi\frac knm}+e^{-i\pi\frac k nm}\right] &=\overbrace{\ \ 1\ \ \vphantom{e^{i\pi m}}}^{m=0}+\overbrace{e^{i\pi k}}^{m=n}+\sum_{m=-n}^{n-1}e^{i\pi\frac knm}\\ &=\left\{\begin{array}{} 1+(-1)^k&\text{if $2n\nmid k$}\\ 2n+2&\text{if $2n\mid k$} \end{array}\right.\\[9pt] &=1+(-1)^k+2n\,[2n\mid k] \end{align} $$ using Iverson Brackets.

Thus, $$ \begin{align} &\sum_{m=0}^n\left[e^{i\pi\frac{k+k'}nm}+e^{-i\pi\frac{k+k'}nm}+e^{i\pi\frac{k-k'}nm}+e^{-i\pi\frac{k-k'}nm}\right]\\ &=2+(-1)^{k+k'}+(-1)^{k-k'}+2n\,[2n\mid k+k']+2n\,[2n\mid k-k']\\ &=2+2\cos((k+k')\pi)+2n\,[2n\mid k+k']+2n\,[2n\mid k-k'] \end{align} $$ So the formula in the question will fail only in the case where $2n\mid k+k'$ or $2n\mid k-k'$. However, if $0\lt k,k'\lt n$ and $k\ne k'$, these divisibility conditions cannot be satisfied.