HMMT 二月 2013 · 冲刺赛 · 第 25 题

HMMT February 2013 — Guts Round — Problem 25

[ 17 ] The sequence ( z ) of complex numbers satisfies the following properties: n • z and z are not real. 1 2 2 • z = z z for all integers n ≥ 1. n +2 n n +1 z n +3 • is real for all integers n ≥ 1. 2 z n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ z z 3 4 ∣ ∣ ∣ ∣ • = = 2. ∣ ∣ ∣ ∣ z z 4 5 Find the product of all possible values of z . 1

解析

[ 17 ] The sequence ( z ) of complex numbers satisfies the following properties: n • z and z are not real. 1 2 2 • z = z z for all integers n ≥ 1. n +2 n n +1 z n +3 • is real for all integers n ≥ 1. 2 z n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ z z 3 4 ∣ ∣ ∣ ∣ • = = 2. ∣ ∣ ∣ ∣ z z 4 5 Find the product of all possible values of z . 1 iθ Answer: 65536 All complex numbers can be expressed as r (cos θ + i sin θ ) = re . Let z be n iθ n r e . n 2 5 2 z z z z z πk n +3 n +1 n n +2 n +1 n 5 = = = z is real for all n ≥ 1, so θ = for all n ≥ 2, where k is an n n n +1 2 2 2 z z z 5 n n n πk 1 integer. θ + 2 θ = θ , so we may write θ = with k an integer. 1 2 3 1 1 5 2 r r r r 1 1 3 4 3 4 2 2 2 = ⇒ r = = r r , so r = 1. = 2 ⇒ r = , r = r r ⇒ r = , and r = r r ⇒ r = 4. 5 3 3 4 4 2 2 3 1 1 4 3 2 r r r r 2 2 4 5 3 4 10 10 Therefore, the possible values of z are the nonreal roots of the equation x − 4 = 0, and the product 1 10 4 8 of the eight possible values is = 4 = 65536. For these values of z , it is not difficult to construct 1 2 4 1 a sequence which works, by choosing z nonreal so that | z | = . 2 2 2