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HMMT 十一月 2015 · 冲刺赛 · 第 9 题

HMMT November 2015 — Guts Round — Problem 9

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [ 7 ] Find the smallest positive integer n such that there exists a complex number z , with positive realn nand imaginary part, satisfying z = ( z ) .
    HMMT November 2015, November 14, 2015 — GUTS ROUND
    Organization Team Team ID#

英文原题

[ 7 ] Find the smallest positive integer n such that there exists a complex number z , with positive real
and imaginary part, satisfying z n = ( z ) n .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
HMMT November 2015, November 14, 2015 — GUTS ROUND
Organization Team Team ID#

解析

英文解析

  1. [ 7 ] Find the smallest positive integer n such that there exists a complex number z , with positive realn nand imaginary part, satisfying z = ( z ) .
    Proposed by: Alexander Katz
    Since | z | = | z | we may divide by | z | and assume that | z | = 1. Then z = , so we are looking for the 1
    smallest positive integer n such that there is a 2 n root of unity in the first quadrant. Clearly therethzis a sixth root of unity in the first quadrant but no fourth or second roots of unity, so n = 3 is thesmallest.