7

Is there a characterisation for which $x\in\mathbb{R}$ the value $\arctan(x)$ is a rational multiple of $\pi$?

Or reformulated: What is the "structure" of the subset $A\subseteq\mathbb{R}$ which fulfils $$ \arctan(x) \in \pi\mathbb{Q} \Leftrightarrow x\in A$$ for all $x\in\mathbb{R}$?

J. M. isn't a mathematician
  • 3,985
Daniel Krenn
  • 567
  • 4
    I can't see that one can say anything much more than $x=\tan q\pi$ with $q$ rational. – Robin Chapman Aug 21 '10 at 12:04
  • 5
    All the elements of $A$ are real algebraic numbers, with all their Galois conjugates real as well. Other than that, of course, we can define polynomials $P_n$ and $Q_n$ for every $n\in\mathbb N$ such that $\tan\left(nx\right)=\frac{P_n\left(\tan x\right)}{Q_n\left(\tan x_\right)}$, and then (if we take these polynomials coprime) your set $A$ will be the union of the sets of roots of all $P_n$. – darij grinberg Aug 21 '10 at 13:06
  • 1
    The malformed equation should mean $\tan\left(nx\right)=\dfrac{P_n\left(\tan x\right)}{Q_n\left(\tan x\right)}$. – darij grinberg Aug 21 '10 at 13:06
  • 1
    A reference of possible interest, even though it only deals with the case where $x$ is rational: http://www.ma.utexas.edu/users/jack/gausspi.pdf – Doug Chatham Aug 22 '10 at 20:18
  • The article @Doug linked to is also at http://www.oberlin.edu/faculty/jcalcut/gausspi.pdf ; see also http://www.oberlin.edu/faculty/jcalcut/arctan.pdf . – J. M. isn't a mathematician Dec 18 '11 at 10:47
  • Note that darij's $P_n$ can be taken to be $P_n(t) = ((1+it)^n - (1-it)^n)/(2i)$. The only members of $Z[t]$ with degree $1$ that can be factors of a $P_n$ are $t$, $t+1$ and $t-1$.. Next question: what about irreducible quadratic factors? These will include $t^2-3$, $3t^2-1$, $t^2 \pm 2t - 1$ and $t^2 \pm 4t + 1$. – Robert Israel Dec 18 '11 at 18:11
  • Surely the tidiest thing to say is that $\arctan(x)\in\mathbb{Q}\pi$ iff $((1+ix)/(1-ix))^n=1$ for some $n>0$. – Neil Strickland Sep 29 '17 at 17:22

2 Answers2

10

A partial answer was provided in response to my MSE question, "ArcTan(2) a rational multiple of $\pi$?"

There Thomas Andrews showed that $\arctan(x)$ is not a rational multiple of $\pi$ for any $x$ rational, except for $-1,0,1$. More specifically:

$\arctan(x)$ is a rational multiple of $\pi$ if and only if the complex number $1+xi$ has the property that $(1+xi)^n$ is a real number for some positive integer $n$. This is not possible if $x$ is a rational, $|x|\neq 1$, because $(q+pi)^n$ cannot be real for any $n$ if $(q,p)=1$ and $|qp|> 1$. So $\arctan(\frac{p}q)$ cannot be a rational multiple of $\pi$.

Community
  • 1
Joseph O'Rourke
  • 149,182
  • 34
  • 342
  • 933
-2

It is easy to show that for $$\frac{\pi}{2^{n+1}}=\arctan x,$$ where $n=\left\{0,1,2,3,\ldots\right\}$, the argument $x$ is always an irrational number. Therefore $\arctan x$ cannot be a rational multiple of $\pi$ at $x\in\mathbb{R}$ for this specific case.

Philip Thomas
  • 35
  • 1
    So far, you have performed 9 edits of your own post in the span of less than 4 hours and the current version shows that you still don't understand the question. Please calm down! – Alex M. Sep 29 '17 at 19:21
  • I understood the question and replied properly. I stated that for this specific case $\arctan x$ cannot be equal to the rational multiplier $\frac{1}{2^{n+1}}$ times $\pi$ if the argument $x$ is a rational number. However I did not know that history of my corrections is recorded. Thanks for this information! – Philip Thomas Sep 29 '17 at 20:12
  • @ Joseph O'Rourke. I did not read your reply before posting my answer above. I completely agree with your statement. – Philip Thomas Sep 29 '17 at 20:30