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Problem: Prove the following identity about the product involving the nth roots of unity:

$ \prod_{k=1}^{N-1}|z^k-1| = N $

where $ z^k $ is the primitive nth root of unity.


Attempt:

$ \begin{align} \prod_{k=1}^{N-1}|z^k-1| &= \prod_{k=1}^{N-1}\left|(\cos(\frac{2\pi k}{N})-1)+i\sin(\frac{2\pi k}{N})\right| \\ &=\prod_{k=1}^{N-1}\sqrt{\cos^2(\frac{2\pi k}{N})-2\cos(\frac{2\pi k}{N})+1+\sin^2(\frac{2\pi k}{N})} \\ &=\prod_{k=1}^{N-1}\sqrt{2-2\cos(\frac{2\pi k}{N})} \\ &=\prod_{k=1}^{N-1}2\sqrt{\frac{1}{2}-\frac{1}{2}\cos(\frac{2\pi k}{N}))} \\ &=2^{N-1}\prod_{k=1}^{N-1}\sin(\frac{k\pi}{N}) \end{align} $

I found on Wikipedia that there is an identity for the last product: $ \prod_{k=1}^{N-1}\sin(\frac{k\pi}{N}) = N/2^{N-1} $. However I do not know how to prove it.

Could someone help me prove the last identity or perhaps suggest a different approach to the problem?

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    Here is an alternative write up of the proof I wrote a while back: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=3681392011-09-19

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Aha, I just solved it:

First consider the polynomial $ \prod_{k=0}^{N-1}(x-z^k) $ The roots of the polynomial are the nth roots of unity, which are precisely the roots of the polynomial $ x^n-1 $ and so the two are equal.

Dividing both sides by $ x-1 $, we get $ \prod_{k=1}^{N-1}(x-z^k) = 1+x+\dots+x^{N-1} $ Substituting $ x=1 $, we get that the product equals $ N $.

The product of the magnitudes is simply the magnitude of the product, so we get the desired result $ \prod_{k=1}^{N-1}|1-z^k| = N $

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    Good point. Thanks :)2011-09-19
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The primitive N'th roots correspond to factors of $(X^n - 1)/(X-1)$.