HMMT 十一月 2016 · 团队赛 · 第 3 题

HMMT November 2016 — Team Round — Problem 3

[ 3 ] Complex number ω satisfies ω = 2. Find the sum of all possible values of 4 3 2 ω + ω + ω + ω + 1 .

解析

[ 3 ] Complex number ω satisfies ω = 2. Find the sum of all possible values of 4 3 2 ω + ω + ω + ω + 1 . Proposed by: Henrik Boecken Answer: 5 5 ω − 1 1 4 3 2 The value of ω + ω + ω + ω + 1 = = . The sum of these values is therefore the sum of ω − 1 ω − 1 1 5 5 4 3 over the five roots ω . Substituting z = ω − 1, we have that ( z + 1) = 2, so z + 5 z + 10 z + ω − 1 5 2 10 z + 5 z − 1 = 0. The sum of the reciprocals of the roots of this equation is − = 5 by Vieta’s. − 1