# Product of $|z^k - 1|$

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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Have you tried pulling the modulus out of the product and simplifying the resulting polynomial? Another approach might be to give up on the trig and try representing $z_k = e^{2\pi i k/N}$. – Neal Sep 19 '11 at 8:42
Here is an alternative write up of the proof I wrote a while back: artofproblemsolving.com/Forum/viewtopic.php?f=38&t=368139 – Ragib Zaman Sep 19 '11 at 11:29

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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Note that, as a byproduct, you now also have a proof of that sine product identity! :) – Hans Lundmark Sep 19 '11 at 15:16
Wait. Isn't this incorrect since the product is undefined at $x=1$ (since we divided by $x-1$)? – tskuzzy Sep 19 '11 at 18:45
If you're uncomfortable with letting $x=1$, you can instead consider the limit of both sides as $x\to 1$. – Hans Lundmark Sep 19 '11 at 18:56
Good point. Thanks :) – tskuzzy Sep 19 '11 at 19:03

The primitive N'th roots correspond to factors of $(X^n - 1)/(X-1)$.

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