HMMT 二月 2005 · TEAM2 赛 · 第 8 题
HMMT February 2005 — TEAM2 Round — Problem 8
题目详情
- [25] Let x and y be two k th roots of unity. Prove that ( x + y ) is real.
解析
- [25] Let x and y be two k th roots of unity. Prove that ( x + y ) is real. Solution: Note that ( ) k ∑ k k i k − i ( x + y ) = x y i i =0 ( ) k ∑ 1 k i k − i k − i i = ( x y + x y ) 2 i i =0 k − i i i k − i − 1 by pairing the i th and ( k − i )th terms. But x y = ( x y ) since x and y are k th i k − i k − i i roots of unity. Moreover, since x and y have absolute value 1, so does x y , so x y k is in fact its complex conjugate. It follows that their sum is real, thus so is ( x + y ) . This can also be shown geometrically. The argument of x (the angle between the vector 2 π x and the positive x -axis) is an integer multiple of , as is the argument of y . Since k π x + y bisects the angle between x and y , its argument is an integer multiple of . k k Multiplying this angle by k gives a multiple of π , so ( x + y ) is real.