Did I do this complex analysis proof right?

  • Thread starter tylerc1991
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  • #1
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Homework Statement


Show that if c is any nth root of unity other than unity itself that:

1 + c + c^2 + ... + c^(n-1) = 0

Homework Equations


1 + z + z^2 + ... + z^n = (1 - z^(n+1)) / (1 - z)

The Attempt at a Solution


c is an nth root of unity other than unity itself => (1-c) =/= 0.
so,
1 + c + c^2 + ... + c^(n-1) = (1 - c^n) / (1 - c) (= 0 by assumption)
hence,
(1 - c)(1 + c + c^2 + ... + c^(n-1)) = 0
so either (1 - c) = 0 or (1 + c + c^2 + ... + c^(n-1)) = 0

but (1 - c) =/= 0 by definition

so (1 + c + c^2 + ... + c^(n-1)) = 0
 

Answers and Replies

  • #2
Dick
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Sure. That looks fine to me. Except instead of saying (= 0 by assumption) I would say (= 0 since c^n=1).
 
  • #3
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but since c is not unity itself doesn't that mean that c^n =/= 1? or is it that just c =/= 1?
 
  • #4
Dick
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but since c is not unity itself doesn't that mean that c^n =/= 1? or is it that just c =/= 1?

Wasn't the 'assumption' you were talking about that c is an nth root of unity? Doesn't that mean c^n=1??
 
  • #5
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oh right, the roots of unity raised to any power = 1, i got it now, thank you!
 
Last edited:

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