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Let $z_1,\ldots,z_n$ be complex numbers of modulus one. Does it exist an increasing sequence $k_j\in\mathbb{N}$ such that $\lim_{j\to\infty}z_i^{k_j}=1$ for all i?

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up vote 7 down vote accepted

Yes. Let $z=(z_1,...,z_n)$, a point on the torus $(S^1)^n$. Since the torus is compact, the sequence $z^m$ has a convergent subsequence $z^{m_1}$, $z^{m_2}$,..., and we may choose it so that $k_j=m_{j+1}-m_j$ are nonnegative and increasing. Then ${\rm lim} z_i^{k_j}=1$ for all $i$.

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Pavel, you can also delete your comments :P –  Mariano Suárez-Alvarez Feb 3 '10 at 18:31
    
Ah, thanks! I did. –  Pavel Etingof Feb 3 '10 at 18:33
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Yes.

A standard lemma is that some element of $\{\alpha, 2\alpha, ..., m\alpha\}$ is within $1/(m+1)$ of 0 mod 1, since otherwise, there would have to be two multiples of $\alpha$ between some $k/(m+1)$ and $(k+1)/(m+1)$ which means their difference would be close to 1.

The same pigeonhole argument works on $(S^1)^n$. Consider the first $m^n+1$ multiples of $(\alpha_1,...\alpha_n)$. Two must be in the same $n$-dimensional box $[k_1/m,(k_1+1)/m]\times...[k_n/m,(k_n+1)/m]$ which means their difference is a multiple $t_m\times(\alpha_1,..,\alpha_n)$ within $1/m$ of 0 on each coordinate.

If you want more details and better approximations, then there are some multidimensional versions of simple continued fractions which might work, but this suffices to show that a sequence of integers $\{t_m\}$ exists so that $z_i^{t_m}$ is within $2\pi/m$ of 1.

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