HMMT 二月 2014 · 代数 · 第 6 题
HMMT February 2014 — Algebra — Problem 6
题目详情
英文原题
- 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 byp
2 2
| a + bi | = a + b whenever a and b are real numbers.
解析
英文解析
- 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