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Here's a random result I came up as a side effect of something I was fiddling around with, which has a nice proof:

Assume $\alpha, \beta \in \mathbb{C}$ satisfy $\alpha^m = 1$, $\beta^n = 1$ for some $m, n \in \mathbb{N}^+$. Prove that $(\alpha + \beta)^{mn} \in \mathbb{R}$.

So, I decided to post the problem here just in case somebody might find it interesting.

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Probably the easiest approach to follow is to first render $\alpha$ and $\beta$ as vectors in the complex plane, each with unit magnitude. Then the resultant will have the form

$\alpha+\beta=r\exp(i\theta)$

where $\theta$ is halfway between the arguments of $\alpha$ and $\beta$. Now, the argument of $\alpha$ is $(2k\pi)/m$ another of $\beta$ is $(2l\pi)/n$ with $k,l$ both integers, thus $\theta=((kn+lm)\pi)/(mn))$. Put that in for $\theta$ in the equation above, raise to the power of $mn$ and observe that the "complex" exponential factor becomes $\pm 1$.

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    $\begingroup$ Makes it transparent. $\endgroup$ – Orest Bucicovschi Oct 20 '17 at 1:03
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    $\begingroup$ Yup, this is pretty much the solution I had in mind. My writeup would probably have mentioned something about "a diagonal of a rhombus bisects the incident angles". Also, there's a small issue that $\arg(\alpha + \beta)$ could also be $\frac{1}{2} (\arg(\alpha) + \arg(\beta)) \pm \pi$ depending on the positions of $\alpha,\beta$ and the branch cut you use for $\arg$. $\endgroup$ – Daniel Schepler Oct 20 '17 at 1:40
  • $\begingroup$ Or allow negative $r$ in the resultant depending on said branch cut, because the proposed exponent $mn$ is a whole number. $\endgroup$ – Oscar Lanzi Oct 20 '17 at 9:15
  • $\begingroup$ @OscarLanzi you may have look to my answer $\endgroup$ – Guy Fsone Jan 31 '18 at 8:38
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This easily derive from Binomial formula Since $|\alpha| = |\beta| =1$

Clearly we have the following $$ \color{red}{\overline{\alpha} =\alpha^{-1},\overline{\beta} =\beta^{-1}~~~\text{and}~~~\beta^{mn} = \alpha^{mn}= 1}$$

\begin{split}(\alpha +\beta)^{mn } &=& \sum_{j= 0}^{mn}{mn\choose j}\alpha^j\beta^{mn-j}=\sum_{j= 0}^{mn}{mn\choose j}\alpha^j\beta^{-j}\\&=& \sum_{j= 0}^{mn}{mn\choose j}\alpha^j\overline {\beta^{j}}=\overline{ \sum_{j= 0}^{mn}{mn\choose j}\overline{\alpha^j}\beta^{j}}\\ &=&\overline{ \sum_{j= 0}^{mn}{mn\choose j}\alpha^{-j}\beta^{j}}=\overline{ \sum_{j= 0}^{mn}{mn\choose j}\alpha^{mn-j}\beta^{j}}\\&=& \color{blue}{\overline{ (\alpha +\beta)^{mn }}} \end{split} Finally, $$ \color{blue}{(\alpha +\beta)^{mn } =\overline{ (\alpha +\beta)^{mn }}\Longleftrightarrow (\alpha +\beta)^{mn }\in\Bbb R}$$

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Very nice, thanks. However I guess this fact need to show up in the early step of the proof of constructible polygons. After all you are showing that the middlepoint (on the circle) of vertices from a $n$-agon and a $m$-agon is ideed vertex of a $nm$-agon. Moreover a little stronger version could give a construction of the generated (in $\mathbb{S}^1$) of two cyclic groups as a sort of (normalized) sum of the two groups. This isn't an answer but I did want to share these geometric and algebraic interpretations. Thanks

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You can start in this way:

$$\alpha = e^{\displaystyle i\frac{2a\pi}{m}}, \beta = e^{\displaystyle i\frac{2b\pi}{n}},$$

with $a, b \in \mathbb{Z}$.

Also, notice that:

$$\begin{align} \alpha + \beta & = e^{ i\left(\frac{a\pi}{m}+ \frac{b\pi}{n}\right)}\left(e^{ i\left(\frac{a\pi}{m}- \frac{b\pi}{n}\right)}+e^{ -i\left(\frac{a\pi}{m}- \frac{b\pi}{n}\right)}\right) \\ & = 2e^{ i\left(\frac{a\pi}{m}+ \frac{b\pi}{n}\right)}\cos\left(\frac{a\pi}{m}- \frac{b\pi}{n}\right) \end{align},$$

and then:

$$(\alpha + \beta)^{nm} = 2^{nm}e^{ i\left(\frac{a\pi}{m}+ \frac{b\pi}{n}\right)nm}\left(\cos\left(\frac{a\pi}{m}- \frac{b\pi}{n}\right)\right)^{nm}.$$

We can forget about $2\cos(\ldots)$, it is real! Moreover...

$$e^{ i\left(\frac{a\pi}{m}+ \frac{b\pi}{n}\right)nm} = e^{ i(an + bm)\pi}.$$

Since $an+bm$ is integer, then $ e^{ i(an + bm)\pi} = \pm 1$.

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A geometric proof is done as follows. On the $2$-dimensional Euclidean plane, let $O$ denote the origin $(0,0)$, $A$ the point whose coordinates correspond to the complex number $\alpha$, $B$ the point whose coordinates correspond to the complex number $\beta$, and $C$ the point on the plane such that $ACBO$ is a rhombus. Since $OA$ makes an angle $\frac{2\pi a}{m}$ with the horizontal axis and $OB$ makes an angle $\frac{2\pi b}{n}$ with the horizontal axis, where $a,b\in\mathbb{Z}$, we conclude that $OC$, being the internal angular bisector of the angle $AOB$, makes an angle $$\theta:=\frac{1}{2}\left(\frac{2\pi a}{m}+\frac{2\pi b}{n}\right)=\frac{\pi(an+bm)}{mn}$$ with the horizontal axis.

We further note that the point $C$ is given by the complex number $\alpha+\beta$. If $D$ denote the point corresponding to the complex number $(\alpha+\beta)^{mn}$, then $OD$ makes the angle $mn\theta$ with the horizontal axis. However, $$mn\theta =\pi(an+bm)\equiv 0\pmod{\pi}\,,$$ whence $D$ lies on the horizontal axis, making $(\alpha+\beta)^{mn}$ a real number.

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