HMMT 十一月 2017 · 冲刺赛 · 第 30 题

HMMT November 2017 — Guts Round — Problem 30

[ 19 ] Given complex number z , define sequence z , z , z , . . . as z = z and z = 2 z + 2 z for n ≥ 0. 0 1 2 0 n +1 n n Given that z = 2017, find the minimum possible value of | z | . 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2017, November 11, 2017 — GUTS ROUND Organization Team Team ID#

解析

[ 19 ] Given complex number z , define sequence z , z , z , . . . as z = z and z = 2 z + 2 z for n ≥ 0. 0 1 2 0 n +1 n n Given that z = 2017, find the minimum possible value of | z | . 10 Proposed by: Yuan Yao √ 1024 4035 − 1 Answer: 2 1 1 Define w = z + , so z = w − , and the original equation becomes n n n n 2 2 1 1 1 1 2 2 w − = 2( w − ) + 2( w − ) = 2 w − , n +1 n n n 2 2 2 2 2 which reduces to w = 2 w . it is not difficult to show that n +1 n 1 1 4035 1023 1024 z + = 2017 + = = w = 2 w , 10 10 0 2 2 2 √ √ 1024 1024 4035 4035 1 th and thus w = ω , where ω is one of the 1024 roots of unity. Since | w | = > , 0 1024 1024 0 2 2 2 √ 1024 1 4035 − 1 to minimize the magnitude of z = w − , we need ω = − 1, which gives | z | = . 0 1024 2 2