When is ArcTan a rational multiple of pi?

Question

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

$x\in\mathbb{R}$ ?

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
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
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
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

score 10 · Answer 1 · edited Apr 13 '17 at 12:19

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 $\pi$ .

Philip Thomas · Answer 2 · 2017-09-29T16:34:38.377

-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.

edited Sep 29 '17 at 16:34

answered Sep 29 '17 at 14:49

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

When is ArcTan a rational multiple of pi?

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