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?

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    $\begingroup$ Here is a counterexample to your lemma: Let $\omega=e^{2\pi i/3}$ and choose $z_1=4$, $z_2=\omega$, $z_3=3\omega$, $z_4=3\bar\omega$, $z_5=\bar\omega$. Then the $a_k(P)$'s coincide with the $a_k(T)$'s for the equilateral triangle $T$ with vertices $4$, $4\omega$, $4\bar\omega$. $\endgroup$ – Christian Blatter Feb 11 '11 at 15:20
  • $\begingroup$ @Christian: Interesting. However, your counter-example is consistent with the pseudo-proof, so I'm not worried yet. Assume for the lemma that the polygon is convex. I've edited the question ... but for non-convex polygons, maybe I should look at moments instead. $\endgroup$ – Eric Nitardy Feb 11 '11 at 16:04

Here is a proof of the lemma, where I now assume $0\leq \theta_1<\ldots<\theta_n<2\pi$ and all $r_j>0$; furthermore I replace $e^{i k\theta_j}$ by $e^{-i k\theta_j}$ in the definition of $a_k(P)$.

Let $b(t)$ be the $2\pi$-periodic box function which is $=1$ for $0\leq t<{2\pi\over m}$ and $=0$ for ${2\pi\over m}\leq t<2\pi$, and consider the function $f(t):=\sum_{j=1}^n r_j b(t-\theta_j)$. For $k\ne0$ the Fourier coefficients of $f$ compute to $$\hat f(k)={i\over 2\pi k}(e^{-2\pi i k/m}-1)\> a_k(P)=0,$$ so $f$ is a constant.

Consider a summand $r_j b(t-\theta_j)$ of $f$. As this summand jumps down to zero at the point $t=\theta_j+{2\pi\over m}$ there has to be another summand $r_l b(t-\theta_l)$ to compensate for this jump; in fact one necessarily has $\theta_l= \theta_j+{2\pi\over m}$ and $r_l=r_j$. It follows that $P$ is invariant under a rotation by ${2\pi\over m}$.

  • $\begingroup$ @ Christian: Cool! Your box function may provide a way to explore the second half of my question. $\endgroup$ – Eric Nitardy Feb 11 '11 at 19:29

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