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The well-known Liouville theorem asserts that an irrational algebraic number $\alpha$ cannot have too good rational approximations, namely $|\alpha-p/q|\ge C(\alpha)/q^k$ where $k$ is the degree of $\alpha$. I wonder whether a similar result holds for the argument of an algebraic number that happens to lie on the unit circle.

For example, consider $\theta=\arccos(1/3)$. It is the argument of a root $\alpha$ of the polynomial $x^2-\frac23x+1$. Is there a similar lower bound for $|\theta/\pi-p/q|$, or equivalently, for $|\alpha^q-1|$? (Note that $|\alpha^q-1|\approx q\cdot |\theta/2\pi-p/q|$).

More generally, let $\alpha\in\mathbb C$ be an algebraic number, $|\alpha|=1$ and $\alpha$ is not a root of unity. Is it always true that $|\alpha^q-1|\ge C(\alpha)/q^k$ where $k$ depends only on the degree of $\alpha$ (or perhaps equals the degree minus one)?

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Sergei, Shouldn't both of your $\le$ signs be $\ge$'s? – SJR Apr 19 '11 at 10:57
Thanks, corrected it. – Sergei Ivanov Apr 19 '11 at 11:04
@Sergei: I asked Győry who asked Waldschmidt. They think your question is hard, and they don't know the answer. So it is very likely that your question is an open problem. – GH from MO Apr 26 '11 at 22:38

The answer is affirmative by a result of Fel'dman: An improvement of the estimate of a linear form in the logarithms of algebraic numbers (Russian), Mat. Sb. (N.S.) 77 (119) 1968, 423–436, MR0232736.

See also Theorem 3.1 in Baker's Transcendental Number Theory.

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Can you elaborate on how to apply Feldman's result here? I could only get something like $C(\alpha)/q^{\log q}$ from the first theorem cited on the MR page, and an estimate exponential in $q$ from the second one. – Sergei Ivanov Apr 19 '11 at 16:42
@Sergei: In the second theorem cited on the MR page, $m=2$, $H_0=q$, while $h$ and $n$ are positive constants depending on $\alpha$, so the bound is of the form $>\exp(-c_1-c_2*\ln q)=c_3 q^{-c_2}$ with constants $c_i>0$ depending on $\alpha$. – GH from MO Apr 19 '11 at 17:11
Thanks, I think I got it now. Do you know if the exponent can be bounded in terms of degree of $\alpha$ (not involving height or whatever)? – Sergei Ivanov Apr 19 '11 at 18:32
@Sergei: Unfortunately I don't know, so I was not completely right saying "the answer is affirmative". The state of the art seems to be set by Matveev (MR1817252) improving on the previous strong result by Baker-Wüstholz (MR1234835). The latter is available online:… I would probably ask the authors of these papers how significant is the dependence on the heights. Probably this dependence is substantial even for $\alpha$ on the unit circle, but I am no expert. – GH from MO Apr 19 '11 at 19:40

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