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Fix an $n$th primitive root of unity $\xi$. I need to understand if we can characterize in an easy way all the solutions $k \in \mathbb{Z}$ of the equation $\left|1-\left(-\frac{\xi^k - 1}{\xi-1}\right)^n\right| = 1$ (note the complex modulus). Actually, I think that the only solutions are the trivial $k=an$, with $a \in \mathbb{Z}$. I know that similar problems can be really difficults, however I am not sure if this is the case.

Thank you very much for any suggestion. :-)

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The expression $(- (\xi^k - 1)/(\xi - 1))^n$ is real: it equals the $n$'th power of $\sin(k \pi/n)/\sin(\pi/n)$ times $(-1)^{n+k-1}$, if I calculated correctly. So the thing you're taking absolute value of must be $\pm 1$. If it's $1$, you get the trivial solution $k = an$. If it's $-1$, that tells you an $n$'th root of $\pm 2$ must be in the cyclotomic field, and you can probably work from there to get a contradiction and rule it out. –  Abhinav Kumar May 15 '13 at 14:36
    
@Abhinav: Why don't you put it as an answer? –  Mark Sapir May 15 '13 at 20:38
    
@Abhinav Kumar: Thanks a lot! You are right, so the problem now is if it is possible that $(-(\chi^k-1)/(\chi-1))^n = \pm 2$ (I think not) and, as you tell, this implies $\sqrt[n]{\pm 2} \in \mathbb{Q}(\chi)$. –  Richard Bonne May 15 '13 at 20:42
    
I mean $\chi = \xi$. –  Richard Bonne May 15 '13 at 20:43
    
@Richard:Abhinav reduced your question to a simple exercise in Galois theory. –  Mark Sapir May 15 '13 at 20:50

1 Answer 1

up vote 8 down vote accepted

The expression $(- (\xi^k - 1)/(\xi - 1))^n$ is real: it equals the $n$'th power of $\sin(k \pi/n)/\sin(\pi/n)$ times $(-1)^{n+k-1}$, if I calculated correctly. So the thing you're taking absolute value of must be $\pm 1$. If it's $1$, you get the trivial solution. If it's $-1$, that tells you an $n$'th root of $\pm 2$ must be in the cyclotomic field $\mathbb{Q}(\xi_n).$ But the degree of the field obtained by adjoining a root of the polynomial $x^n \pm 2$ is $n$ (since the poly is irreducible by Eisenstein), and so this leads to $n \leq \phi(n)$, which is a contradiction for any $n > 1$.

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