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The "Littlewood Problem" in the title asks for a characterization of finite sequences

n1< ...< nk of integers such that zn1+zn2+...+znk≠0 for any complex number z of unit modulus.

Does anybody know about the current status of this problem?

Some Background

1)I came to know this Littlewood Problem through the paper of Casazza & Kalton, http://www.jstor.org/pss/2699467.

2)For k=2,3,4, by some simple geometric argument, a complete characterization can be easily obtained. I wonder if such a result has already appeared in the literature.

3)Furthermore, I wonder if at least for the case of k=5, (or indeed, similarly for any k),the following is true? And if it is, whether it is in the literature somewhere.

Suppose that for some complex number z of unit modulus and some integers n1< ...< n5,

zn1+zn2+...+zn5=0

then either zn1,zn2,..,zn5 are evenly distributed on the unite circle (i.e., they look like the 5th roots of unit after a certain rotation is applied to each) or three pounts among zn1,zn2,..,zn5 are evenly distributed on the unite circle.

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1 Answer 1

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For every $\alpha$, $\pi/3\le\alpha\le\pi$, there is a $\beta$ such that $1+e(\alpha)+e(-\alpha)+e(\beta)+e(-\beta)=0$, where I write $e(x)$ for $e^{2\pi ix}$. Moreover, $\beta$ depends continuously on $\alpha$, so there will be infinitely many $\alpha$ such that $\beta/\alpha$ is rational, say, $\beta/\alpha=m/n$. Let $z=e(\alpha/n)$. Then $z^0+z^n+z^{-n}+z^m+z^{-m}=0$, and in general these numbers will not be vertices of a regular pentagon, nor include vertices of an equilateral triangle. This is in (negative) answer to part 3) of the question, case $k=5$, and surely the answer is negative as well for any $k\ge5$.

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  • $\begingroup$ Thanks a lot for Gerry's answer. Sorry I do not know how to make this a comment(in stead of an answer), nor how to formally "accept" Gerry's answer. In view of Gerry's answer, the complete solution of the problem looks much less likely. I am still(even more eagerly) looking for any possible reference about the status of the said problem. Thanks again! $\endgroup$
    – user8760
    Commented Aug 24, 2010 at 15:37

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