Complex Analysis Question
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Complex Analysis Question

[From: ] [author: ] [Date: 12-04-30] [Hit: ]
-To prove this, let z = w^k = e^{2πik/n}. Since k is not a multiple of n, k/n is not an integer, so z ≠ 1.since z is an nth root of unity.......
This is a homework question that I'm not really sure how to approach.
Let k and n be positive integers with n > 1 and k not a multiple of n. Let
w:= cos(2pi/n) + i sin(2pi/n)
Show that
1 +w^k +w^(2k) +...+ w^(n-1)k = 0


I think that this has to do with roots of unity, more specifically the nth roots of k. I'm not really sure how to get started though, so any help would be appreciated!

-
To prove this, let z = w^k = e^{2πik/n}. Since k is not a multiple of n, k/n is not an integer, so z ≠ 1. Then

1 + w^k + w^{2k} + ··· + w^{(n - 1)k}

= 1 + z + z^2 + ··· + z^{n - 1}

= (1 - z^n)/(1 - z)

= 0

since z is an nth root of unity. QED
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