HMMT 二月 2005 · TEAM2 赛 · 第 8 题

英文解析

8. [25] Let x and y be two k th roots of unity. Prove that ( x + y ) is real.
   Solution: Note that
   ( )
   ∑kkk i k − i
   ( x + y ) = x yii =0
   ( )
   ∑k
   1 ki k − i k − i i = ( x y + x y )
   2 ii =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 thi k − i k − i iroots of unity. Moreover, since x and y have absolute value 1, so does x y , so x ykis 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 πx and the positive x -axis) is an integer multiple of , as is the argument of y . Sincex + y bisects the angle between x and y , its argument is an integer multiple of .πkk
   Multiplying this angle by k gives a multiple of π , so ( x + y ) is real.k