Consider an $n$-sided convex polygon $P$ that contains the origin in the complex plane. Let the $j$-th vertex be denoted $z_j = r_j e^{i\theta_j}$ ($0 \leq \theta_j < 2 \pi$) for $j= 1 \dots n$. I'm interested in non-zero values of
$ a_k(P)= \sum_{j=1}^{n} \frac{z_{j}^{k}}{|z_{j}|^{k-1}}=\sum_{j=1}^{n} r_j e^{ik\theta_j} \textrm{ for } k \geq 1.$
Lemma: Given a integer $m \geq 2$, if, for every $k \geq 1$ where $m$ does not divide $k$, $a_k(P)=0$, then the polygon $P$ is $m$-fold rotationally symmetric, that is, a rotation of $e^{i\frac{2 \pi}{m}}$ rotates the polygon into itself.
Pseudo-Proof: Re-imagine the $n$-sided polygon $P$ as a $2 \pi$-periodic function $f(\theta)$ of the angle $\theta$ where each vertex $z_j$ is represented as a Dirac delta function at $\theta_j$ with integral $r_j$, that is, $f(\theta)=\sum_{j=1}^{n} r_j \delta (\theta - \theta_j).$ The calculation $a_k(P)$ is then just $2 \pi$ times the $k$-th Fourier coefficient for $f(\theta)$. If, for all $k$ where $m$ does not divide $k$, $a_k(P)=0$, then the corresponding Fourier coefficients of $f(\theta)$ are all zero, implying that $f(\theta)$ is $\frac{2 \pi}{m}$-periodic. Hence the polygon will be $m$-fold rotationally symmetric. $\square$
Firstly, is there a good way to prove this lemma without resorting to non-converging Fourier series?
Then, in the same vein, the lemma implies that, if $P$ is not rotationally symmetric, then for every $m$, there are values of $k$, that are not multiples of $m$ for which $a_k(P) \neq 0$. But I believe much more is true, namely, that for 'almost' all $k$, $a_k(P) \neq 0$. In particular, if $k$ is the smallest so that $a_k(P) \neq 0$, I'd like to show that there is a k' relatively prime to $k$ so that a_{k'}(P) \neq 0, but I'm not sure how to approach the issue. Any thoughts?