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Let $\alpha$ be a complex number such that both $\alpha$ and $\alpha+1$ have modulus $1$. If for a positive integer $n$, $1+\alpha$ is an $n$-th root of unity, then show that $\alpha$ is also an $n$-th root of unity and $n$ is a multiple of $6$.

I tried solving this problem in the following way:

We consider $\alpha +1= \cos\theta +i\sin\theta.$ Now, as $1+\alpha$ is an $n$-th root of unity ,so, we can say, that $$(\alpha +1)^n= (\cos\theta +i\sin\theta)^n=\cos n\theta+i\sin n\theta=1\implies \theta=\pi,$$ so that $\text{Im}(\alpha+1)=0.$ But then, we have, $(\alpha+1)^n=\cos n\pi=(-1)^n=1$ and comparing both sides, we may conclude that $n$ must be even. So, we consider $n=2p.$

Also, we know that the set of $n-th$ roots of unity are $\{\cos(\frac{2k\pi+\pi}{n})+i\sin(\frac{2k\pi+\pi}{n})| k=0,1,2,...,n-1\}.$ We assume, $\alpha +1=\cos(\frac{2k\pi+\pi}{n})+i\sin(\frac{2k\pi+\pi}{n})$ for some $k.$

Thus, $$\alpha=(1+\alpha)-1=\cos(\frac{2k\pi+\pi}{n})+i\sin(\frac{2k\pi+\pi}{n})+\cos(\pi)+i\sin(\pi)=2\cos(\frac{\frac{2k\pi+2\pi}{n}-\pi}{2})\cos(\frac{\frac{2k\pi+2\pi}{n}+\pi}{2})+2i\sin(\frac{\frac{2k\pi+2\pi}{n}+\pi}{2})\cos(\frac{\frac{2k\pi+2\pi}{n}-\pi}{2})=2\cos(\frac{\frac{2k\pi+2\pi}{n}-\pi}{2})(\cos(\frac{\frac{2k\pi+2\pi}{n}+\pi}{2})+i\sin(\frac{\frac{2k\pi+2\pi}{n}+\pi}{2})).$$ Now, $$(\alpha)^n=2^n\cos^n(\frac{\frac{2k\pi+2\pi}{n}-\pi}{2})(\cos(\frac{2k\pi+2\pi+n\pi}{2})+i\sin(\frac{2k\pi+2\pi+n\pi}{2}))=2^n\sin^n(\frac{k\pi+\pi}{n})(\cos(\pi+\frac{2k\pi+n\pi}{2})+i\sin(\pi+\frac{2k\pi+n\pi}{2}))=-2^n\sin^n(\frac{k\pi+\pi}{n})(\cos(\frac{2k\pi+n\pi}{2})+i\sin(\frac{2k\pi+n\pi}{2})).$$

But, $n$ is even, so, $\sin(\frac{2k\pi+n\pi}{2})=\sin(\frac{2k\pi}{2}+p\pi)=0,$ so, $\alpha^n=-2^n\sin^n(\frac{k\pi+\pi}{n})(\cos(\frac{2k\pi+n\pi}{2})).$ If, $\alpha=\cos\phi+i\sin\phi, $ then

$\alpha^n=\cos n\phi+i\sin n\phi=-2^n\sin^n(\frac{k\pi+\pi}{n})(\cos(\frac{2k\pi+n\pi}{2})),$

and on comparing, both sides, we find that, $\sin n\phi=0$ which means, $\phi=\pi.$ Thus, $\alpha^n=\cos n\phi+i\sin n\phi=\cos n\pi=(-1)^n=(-1)^{2p}=1,$ thus, $\alpha$ is an $nth$ root of unity.

However, I dont understand, how to prove, $3|n$ so that we can conclude, $6|n.$ I want to know, the way I showed $\alpha$ is an $nth$ root of unity, can be considered as a valid approach or not ?

metamorphy
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Arthur
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    You are overcomplicating it. If $|\alpha|=1$ and $|\alpha+1|=1$, show that $\operatorname{Re}{\alpha}=-1/2$ then $\ldots$ – dxiv May 03 '23 at 06:11
  • @dxiv Yeah, I managed to show $\Re \alpha=-\frac 12$ as you said. I got $\alpha=-\frac 12+i\frac{\sqrt 3}{2}$ and $\alpha +1=-\frac 12-i\frac{\sqrt 3}{2}$ or the vice-versa (as that can also be a valid solution ). But then ? – Arthur May 03 '23 at 06:21
  • @FdstZfsy they are the principal cube roots of unity. – D S May 03 '23 at 06:26
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    Just draw the picture. It is so obvious geometrically. Which regular polygon inscribed in a circle of radius 1 also has side-length 1? – Michal Adamaszek May 03 '23 at 06:26
  • @FdstZfsy You actually get two possible $\alpha = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$. Then check what's the minimum $n$ such that $(1+\alpha)^n=1$. That's assuming you don't "see" it geometrically, as hinted in another comment. – dxiv May 03 '23 at 06:28

3 Answers3

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A totally different approach to this problem.

Let $\alpha=\cos(x)+i\sin(x)$ and $\alpha +1=\cos(y)+i\sin(y)$, where $0\le x,y<2\pi$ since $\alpha$ and $(\alpha +1)$ have modulus $1$

$\alpha +1=(\cos(x)+1)+i\sin(x)=\cos(y)+i\sin(y)$ Since $\cos^2(y)+\sin^2(y)=1$, we get $(\cos(x)+1)^2+\sin^2(x)=1\\\Rightarrow 1+2\cos(x)+\cos^2(x)+\sin^2(x)=1\\\Rightarrow 2+2\cos(x)=1\\\Rightarrow \cos(x)=-\dfrac{1}{2}$

Now, using $\cos(x)=-\dfrac{1}{2}$ and $0\le x<2\pi$, we get $x=\dfrac{2\pi}{3}$ or $x=\dfrac{4\pi}{3}$
$\Rightarrow\alpha=\dfrac{-1+i\sqrt{3}}{2}$ or $\alpha=\dfrac{-1-i\sqrt{3}}{2}$ respectively
$\Rightarrow\alpha+1=\dfrac{1+i\sqrt{3}}{2}$ or $\alpha+1=\dfrac{1-i\sqrt{3}}{2}$ respectively
$\Rightarrow y=\dfrac{\pi}{3}$ or $y=\dfrac{5\pi}{3}$ since $0\le y<2\pi$

If $(\alpha +1)$ is an $n^{th}$ root of unity, then $(\alpha +1)^n=1$ where $(\alpha +1)=e^{iy}$
$\Rightarrow ny$ is an integral multiple of $2\pi$
$\Rightarrow \dfrac{n\pi}{3}$ or $\dfrac{5n\pi}{3}$ is an integral multiple of $2\pi$
$\Rightarrow n$ is an integral multiple of $6$ or $\dfrac{6}{5}$
Since $n$ is an integer, both of the above conditions imply that $6|n$.

We also get that if $6|n$, then $nx=\dfrac{2n\pi}{3}$ or $nx=\dfrac{4n\pi}{3}$ where both are integral multiples of $2\pi$.
Hence, if $\alpha +1$ is an $n^{th}$ root of unity, then so is $\alpha$ and $6|n$.

Kendall
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Note that $z\bar{z} = |z|^2$, so the facts that $\alpha, 1+\alpha$ have modulus $1$ can be written $\alpha\bar{\alpha} = 1,$ so $\bar{\alpha} = 1/\alpha,$ and $$1 = (1+\alpha)\overline{(1+\alpha)} = (1+\alpha)(1+\bar{\alpha}) = (1+\alpha)^2/\alpha$$ so that $\alpha = (1+\alpha)^2$.

Therefore, if $(1+\alpha)^n = 1$, then also $$\alpha^n = \left((1+\alpha)^2\right)^n = \left((1+\alpha)^n\right)^2 = 1^2 = 1.$$

To find the order of $1+\alpha$, we can turn back to $\alpha = (1+\alpha)^2 = 1+2\alpha + \alpha^2,$ which implies $\alpha^2 = -1-\alpha$. We use this to compute $$(1+\alpha)^3 = (1+\alpha)(1+\alpha)^2 = (1+\alpha)\alpha = \alpha + \alpha^2 = \alpha + (-1-\alpha) = -1$$ and therefore $(1+\alpha)^6 = (-1)^2 = 1$. The divisors of $6$ are $1,2,3,6$, and we've seen $(1+\alpha)^k \neq 1$ for $k=1,2,3$, so the order of $1+\alpha$ is $6$.

Brian Moehring
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Forgetting about geometric intuition, and proving it directly (same as dxiv's first hint) $$|\alpha + 1| = 1 \implies \alpha+1 = \cos \theta + i\sin \theta \implies \alpha = \underbrace{(\cos \theta - 1) + i(\sin \theta)}_\text{a+ib}$$for some $\theta$. Now, $$|\alpha| = 1 \implies (\cos \theta - 1)^2 + \sin^2 \theta = 1 \implies \cos \theta = 1/2$$Hence $\sin \theta = \pm \sqrt{3}/2$. Note that $(\alpha+1)^2 = \frac 14 - \frac 34 \pm \frac{\sqrt{3}}{2} i = \alpha$. Since you have already shown that $n$ is even: $$1 = (1+\alpha)^n = (1+\alpha)^{2k} = \alpha^k$$ $$\boxed{\alpha^k = 1 \implies \alpha^{2k} = \alpha^n = 1}$$

D S
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