HMMT 二月 2014 · 代数 · 第 6 题

HMMT February 2014 — Algebra — Problem 6

Given that w and z are complex numbers such that | w + z | = 1 and | w + z | = 14, find the smallest 3 3 possible value of | w + z | . Here, | · | denotes the absolute value of a complex number, given by p 2 2 | a + bi | = a + b whenever a and b are real numbers.

解析

Given that w and z are complex numbers such that | w + z | = 1 and | w + z | = 14, find the smallest 3 3 possible value of | w + z | . Here, | · | denotes the absolute value of a complex number, given by √ 2 2 | a + bi | = a + b whenever a and b are real numbers. 41 1 3 3 2 2 2 2 Answer: OR 20 . 5 OR 20 We can rewrite | w + z | = | w + z || w − wz + z | = | w − wz + z | = 2 2 3 1 2 2 2 | ( w + z ) − ( w + z ) | . 2 2 3 1 1 3 1 1 2 2 2 2 2 2 2 2 By the triangle inequality, | ( w + z ) − ( w + z ) + ( w + z ) | ≤ | ( w + z ) − ( w + z ) | + | ( w + z ) | . 2 2 2 2 2 2 3 1 3 1 3 3 2 2 2 2 2 2 By rearranging and simplifying, we get | w + z | = | ( w + z ) − ( w + z ) | ≥ | w + z | − | w + z | = 2 2 2 2 3 1 41