How small can a sum of a few roots of unity be?

In this paper they talk about this problem for 5 instead of 10 roots.

http://www.jstor.org/stable/2323469

EDIT: In view of Todd Trimble's comment, here's a summary of what's in the paper.

Let $f(k,N)$ be the least absolute value of a nonzero sum of $k$ (not necessarily distinct) $N$-th roots of unity. Then

$f(2,N)$ is asymptotic to $cN^{-1}$, where $c$ is $2\pi$ for even $N$, $\pi$ for odd $N$,

$f(3,N)$ is asymptotic to $cN^{-1}$, where $c$ is $2\pi\sqrt3$ for $N$ divisible by 3, $2\pi\sqrt3/3$ otherwise,

$f(4,N)$ is asymptotic to $cN^{-2}$, where $c$ is $4\pi^2$ for even $N$, $\pi^2$ for $N$ odd,

$f(k,N)>k^{-N}$ for all $k,N$,

$f(2s,N)<c_sN^{-s}$ for $N$ even and $s\le10$,

$f(k,N)<c_kN^{-[\sqrt{k-6}]-1}$ for $N$ even and $k>5$, and

If $N$ is twice a prime, and $k<N/2$, then there exists $k'<2k$ such that $f(k',N)\le2k2^{k/2}\sqrt{k!}N^{-k/2}$.

The only result in the paper for 5 roots of unity is (the trivial) $f(5,N)>5^{-N}$, but it is suggested that maybe $f(5,N)>cN^{-d}$ for some $d$, $2\le d\le3$, and some $c>0$.

From a computational point of view one can probably use the LLL algorithm for getting fairly good solutions: Indeed consider the sublattice of $\mathbb Z^{n+2}$ spanned by integral vectors of the form $(0,\dots,0,1,0,\dots,\lfloor A\cos(2\pi k/n)\rfloor,\lfloor A\sin(2\pi k/n)\rfloor)$. Fine-tuning of the the real number $A$ (which has to be choosen not too small) and searching for a short vector in this lattice yields solutions. Using known bounds for lattice packings yields perhaps some useful upper bounds (but the computations are probably a little tricky).

To expand on the Prouhet-Tarry-Escott problem, this is to find (multi)-sets of integers $A$ and $B$ with $\sum_Aa^k=\sum_Bb^k$ for $0 \le k \le m-1$. Then $|A|=|B|$ and perhaps one can always get $|A|=m$ although no-one really knows. This translates into ways to choose $n=2|A|$ Nth roots of unity (at least for even N): take the set $S$ consisting of the $n$ roots $\alpha^a$ and $-\alpha^b$ where $\alpha=e^\frac{2\pi i}{N}$. Note that -1 is a power of $\alpha$. I'm not sure what to do when $n$ and/or $N$ is odd but other people probably do. Fast forwarding over some details, one ends up with a polynomial of the form $(\alpha-1)^kg(\alpha)$ and that first factor gives the whole thing a size $O(\cos(\frac{2\pi}{N})^k)=O(N^{-k})$ The constant is easily computable although kind of large and requiring a fairly large $N$ to be accurate (for n=10 I got multi-digit accuracy by N=1000 although maybe N=100 was ok too). A reference I like is P. Borwein, C. Ingalls, The Prouhet-Tarry-Escott Problem revisited.

The referenced article by G. Myerson says (if my quick read is correct) that an approximately equal spacing around the unit circle can be $O(N^{-1})$ but not better but that no one knows a general construction which is better. It is intriguing that the solution sketched above has no special use of the number theoretic properties of $N$ except parity. Perhaps (some of) the best solutions (for an even number of roots) involve roots from 2 thin wedges which are nearly antipodal. For 4 roots the optimum is to take 1 twice and two other roots one on each side of -1.