Math 4330, Homework 2, 1/31/2014
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Let K be a positive integer and ω = exp(2πi/K).
1. Show that each of ω 0 , ω 1 , ω 2 , . . . , ω K−1 is a K-th root of unity.
Solution:
Let j be an integer satisfying 0 ≤ j < K, and α = ω j . Then αK = ω Kj = exp(2Kjπi/K) =
(exp(2πi))j = 1j = 1, so α is a K-th root of unity.
2. Show that ω 0 + ω 1 + · · · + ω K−1 = 0.
Solution:
Since ω is a root of the polynomial xK − 1, and xK − 1 = (x − 1)(1 + x + x2 + · · · + xK−1 ),
we have that
0 = ω K − 1 = (ω − 1)(1 + ω + ω 2 + · · · + ω K−1 ).
Since ω − 1 6= 0, it follows that 1 + ω + ω 2 + · · · + ω K−1 = 0.
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Chris Monico, and may not be reproduced in any form without
written permission from the author.
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