Showing the Equivalence of (1-w)(1-w^2)...(1-w^{n-1}) and n

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SUMMARY

The discussion focuses on proving the equivalence of the product (1-w)(1-w^2)...(1-w^{n-1}) and the integer n, where w is defined as the nth root of unity, specifically w = exp(2πi/n). The solution involves recognizing that the roots of the polynomial x^n - 1 = 0 are 1, w, w^2, ..., w^{n-1}. By dividing the polynomial by (x - 1) and evaluating at x = 1, the left-hand side simplifies to the desired product, confirming the equivalence.

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Homework Statement


if w is the nth root of unity, i.e. w= exp(2pi/n i) show:
(1-w)(1-w^2)...(1-w^{n-1})=n


Homework Equations





The Attempt at a Solution


since w^(n-a)= complex congugate of w^a
terms on the left hand side are going to pair up to give |1-w|^2 |1-w^2|^2...
but I'm not sure what to do from here.
Thanks
 
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I wouldn't do it that way at all!

It should be sufficient to note that the n roots of xn- 1= 0 are 1, w, w2, ..., wn-1 and so xn-1= (x-1)(x-w)(x-w2)...(x- wn-1). Dividing both sides by x- 1 we get (x-w)(x-w2)...(x- wn-1) on the right and what on the left? Now set x= 1.
 
thanks
 

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