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

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In summary, to show that (1-w)(1-w^2)...(1-w^{n-1})=n when w is the nth root of unity, it is sufficient to note that the n roots of xn-1=0 are 1, w, w^2, ..., w^{n-1}, and dividing by x-1 on both sides gives (x-w)(x-w^2)...(x-w^{n-1}) on the right. Setting x=1, we get n on the left side.
  • #1
plmokn2
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Homework Statement


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


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 [itex]|1-w|^2 |1-w^2|^2...[/itex]
but I'm not sure what to do from here.
Thanks
 
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  • #2
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.
 
  • #3
thanks
 

1. What does the equation (1-w)(1-w^2)...(1-w^{n-1}) represent?

The equation represents the product of n terms, where each term is equal to (1-w^k) and k ranges from 1 to n-1.

2. How is the equivalence of (1-w)(1-w^2)...(1-w^{n-1}) and n shown?

The equivalence can be shown using mathematical induction, where the base case n=2 is assumed to be true and then it is proven for n+1 by using the fact that (1-w)(1-w^2)...(1-w^{n-1}) can be expanded to (1-w^n) + (1-w^n)(1-w^{n+1}) and then simplified to (1-w^{n+1}).

3. Why is the equivalence of (1-w)(1-w^2)...(1-w^{n-1}) and n important?

This equivalence is important because it allows us to simplify complex expressions involving products of (1-w^k) terms into a single term, making it easier to work with and manipulate in calculations and proofs.

4. How can the equivalence of (1-w)(1-w^2)...(1-w^{n-1}) and n be applied in real-world situations?

This equivalence has applications in various fields, including mathematics, physics, and engineering. For example, it can be used in solving problems involving complex numbers and in circuit analysis.

5. Are there any limitations to the equivalence of (1-w)(1-w^2)...(1-w^{n-1}) and n?

Yes, the equivalence only holds true for values of w that are not equal to 1. If w=1, then the product would equal 0 and the equivalence would not hold. Additionally, the equivalence is only valid for whole number values of n.

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