HMMT 二月 2021 · 冲刺赛 · 第 29 题

HMMT February 2021 — Guts Round — Problem 29

[18] Compute the number of complex numbers z with | z | = 1 that satisfy 5 10 15 18 21 24 27 1 + z + z + z + z + z + z + z = 0 . 2

解析

[18] Compute the number of complex numbers z with | z | = 1 that satisfy 5 10 15 18 21 24 27 1 + z + z + z + z + z + z + z = 0 . Proposed by: Daniel Zhu Answer: 11 Solution: Let the polynomial be f ( z ). One can observe that 15 15 20 12 1 − z 1 − z 1 − z 1 − z 15 18 f ( z ) = + z = + z , 5 3 5 3 1 − z 1 − z 1 − z 1 − z so all primitive 15th roots of unity are roots, along with − 1 and ± i . To show that there are no more, we can try to find gcd( f ( z ) , f (1 /z )). One can show that there exist a b a, b so that z f ( z ) − z f (1 /z ) can be either of these four polynomials: 5 10 32 5 10 15 30 (1 + z + z )(1 − z ) , (1 + z + z + z )(1 − z ) , 3 6 9 12 32 3 6 9 30 (1 + z + z + z + z )( z − 1) , (1 + z + z + z )( z − 1) . 15 32 5 Thus any unit circle root of f ( z ) must divide the four polynomials (1 − z )(1 − z ) / (1 − z ), 20 30 5 15 32 3 12 30 3 (1 − z )(1 − z ) / (1 − z ), (1 − z )(1 − z ) / (1 − z ), (1 − z )(1 − z ) / (1 − z ). This implies that z must be a primitive k th root of unity, where k ∈ { 1 , 2 , 4 , 15 } . The case k = 1 is clearly extrane- ous, so we are done. 2