I am reading a paper where the authors state that there exist infinite solutions to
$$ \sum_{n=1}^N\frac{\sin(m\pi a_n)}{m\pi}e^{-im\pi(2b_n+a_n)}=0, m\in\mathbb{Z}, m\neq0 $$
where $a_n$ and $b_n$ are between $0$ and $1$. I have spent lots of time figuring out how the solutions, i.e., $a_n$'s and $b_n$'s, look like for this problem but I have failed to proceed. I was hoping I could find some advice on this here.
Not an answer but too long for a comment.
Suppose $a_1=\ldots = a_N=\frac{1}{2}$, then we see that the sum is \begin{align} \frac{\sin\frac{\pi m}{2}}{m\pi}e^{-im\pi/2}\sum^N_{n=1}e^{-i2\pi mb_n}. \end{align} If $m$ is even then any thing will work, i.e. the sum will be zero (not in a very interesting way). If $m$ is odd then, we have that \begin{align} \frac{\sin\frac{\pi (2k+1)}{2}}{(2k+1)\pi}e^{-i(2k+1)\pi/2} = \frac{1}{(2k+1)\pi i}\ne 0 \end{align} which means that the sum is zero if \begin{align} \sum^N_{n=1}e^{-i2\pi mb_n}=0 \end{align} which means that $e^{-i2\pi mb_n}$ are the $N$th roots of unity (and any rotation of them).