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It is well known that if $x$ is a rational multiple of $\pi$ then $\cos x$, $\sin x$, etc, are algebraic numbers. What is known about the inverse problem?

That is, is there a set of conditions that if imposed on $\alpha$ imply that $\cos^{-1} \alpha$ is a rational multiple of $\pi$?

Another way of putting it is, given the complex number $z=a+ib$, is there a way of deciding whether there exists some $n$ such that $z^n=1$?

EDIT: 'factor' -> 'multiple'

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I think you mean a rational multiple of $\pi$, not factor? –  joriki Mar 5 '12 at 17:35
    
you're right. edited –  whistles Mar 5 '12 at 19:02
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A sylistic observation: Greek letters like $\alpha$ are traditionally used for angles when trigonometric functions are under discussions. The argument of $\cos^{-1}$ is best thought of as a length or as a ratio of lengths, not truly an angle. So I would recommend a roman letter for the argument of $\cos^{-1}$, not a Greek one. I often wonder if confusion regarding inverse trig functions could be avoided if it was clear that, say, cosine eats angles and spits out lengths, while $\cos^{-1}$ does the opposite. –  alex.jordan Mar 6 '12 at 0:56
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3 Answers

Consider the case of $\theta = \arcsin(1/3)$ We have $\sin(\theta) = 1/3$ and $\cos(\theta) = 2\sqrt{2}/3;$ both of these numbers is algebraic. However, $\theta/\pi$ has a decimal expansion with no repetitions after a couple of thousand digits. This is not definitive, but it seems likely to me it's not rational.

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$\arcsin(1/3)$ is not a rational multiple of $\pi$, in fact the only cases where $r$ and $\sin(r\pi)$ are both rational are when $\sin(r\pi) \in \{\pm 1,\pm 1/2,0 \}$. See e.g.math.stackexchange.com/questions/94478/… –  Robert Israel Mar 5 '12 at 19:04
    
which is precisely the question. what conditions must be set, if they exist, so that $\theta/\pi$ is rational. As Robert says above, this case is clear. For instance, are algebraic integers $z$ with $|z|=1$, integers roots of unity? My understanding is that it is not the case. What conditions should be added then? –  whistles Mar 5 '12 at 19:11
    
There are lots of algebraic integers with $|z|=1$. Take any algebraic integer $x$ with $-2 < x < 2$, and $z = (x + i \sqrt{4-x^2})/2$ is an algebraic integer with $|z|=1$. In fact, $x = z + 1/z$, so if $P(x) = 0$ where $P(t)$ is a monic integer polynomial of degree $n$, then $Q(t) = t^n P(t + 1/t)$ is a monic integer polynomial with $Q(z)=0$. –  Robert Israel Mar 6 '12 at 0:01
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If you are given a polynomial $P$ with integer coefficients, irreducible over the rationals, such that $P(z) = 0$, then $z$ is a root of unity if and only if $P$ is a multiple of the cyclotomic polynomial of the same degree. Is that what you're looking for? –  Robert Israel Mar 6 '12 at 0:10
    
so simple. yes. I believe that will do. –  whistles Mar 7 '12 at 1:44
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"Given the complex number $z=a+bi$, is there a way of deciding whether there exists some [positive integer] $n$ such that $z^n=1$?"

That would depend on how $z$ is given. If you are given polynomials $f,g$ with integer coefficients and you are told that $a$ is the real root of $f$ between (say) $.5$ and $.7$ and $b$ is the real root of $g$ between (say) $.7$ and $.9$, then from the degrees of $f,g$ you can derive an upper bound for $n$, and you can calculate $z^n$, exactly, for $n$ up to that upper bound, and you can determine whether you ever get $1$.

If you are given $a,b$ as decimal numbers, rounded to (say) $17$ places, then, provided $a^2+b^2=1$ to $17$ places, it is guaranteed that there is a positive integer $n$ such that $z^n=1$, to $17$ places.

If you have a black box that gives you any finite number of decimal places of $a$ that you ask for, and if $b=\pm\sqrt{1-a^2}$, then I don't think there's any way you can ever know for certain whether there's a positive integer $n$ such that $z^n=1$, since both roots and nonroots of unity are dense on the unit circle.

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Another way of putting it is, given the complex number $z=a+ib$, is there a way of deciding whether there exists some n such that $z^{n}=1$?

Calculate the minimal monic polynomial $p(x)\in\mathbb{Q}[x]$ for which $z$ is a root. That is, $p(x)$ is the lowest power possible for which $p(z)=0$ and the coefficients of $p(x)$ are all rational, and the coefficient of the highest powered term is 1. Then $n$ exists iff $p(x)$ exists and is cyclotomic.

This follows from the fact that $$z^n-1=\Pi_{d|n}\Phi_d(z)$$ where $\Phi_d(z)$ is the $d$th cyclotomic polynomial, and one other unique factorization theorem.

Comment: In practice, when I have applied this theorem to a given $z$ in quantum computing problems, $p(x)$ is readily calculated. In all cases of interest, $p(x)$ contained a fractional coefficient, which immediately proved the corresponding angle was an irrational multiple of $\pi$, since all cyclotomic polynomials have integer coefficients.

Edit: Applying this to $\theta=\arcsin(1/3)$, we find $z=e^{i\theta}=2\sqrt{2}/3+i/3$. The corresponding minimal monic polynomial in $\mathbb{Q}[x]$ is $$p(x)=x^4-14/9x^2+1.$$ Because of the $-14/9$ fractional coefficient, we can immediately conclude $p(x)$ is not cyclotomic, there is no $n\in\mathbb{N}:z^n=e^{in\theta}=1$, and thus $\theta$ is an irrational multiple of $\pi$.

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I just noticed that Robert Israel gave this answer in a comment back in March 2012. –  Matt Aug 30 '13 at 23:14
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