Assume $\alpha, \beta \in \Bbb C,$ such that $\alpha^m = \beta^n = 1$ show that $(\alpha+\beta)^{mn}\in \Bbb R$

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.

Solution 1:

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 $\pm 1$.

Solution 2:

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, $$ \color{blue}{(\alpha +\beta)^{mn } =\overline{ (\alpha +\beta)^{mn }}\Longleftrightarrow (\alpha +\beta)^{mn }\in\Bbb R}$$

Solution 3:

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

Solution 4:

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$.