回覆 1# Superconan 的帖子
第 7 題
已知複數\(z_1,z_2,z_3\)滿足\(\cases{\displaystyle |\;z_1|\;=|\;z_2|\;=|\;z_3|\;=1\cr \frac{z_1}{z_2}+\frac{z_2}{z_3}+\frac{z_3}{z_1}=1}\),則\(|\;z_1+2z_+3z_3|\;\)最大可能的值為 。
[另解]
為打字方便,把 z_1、z_2、z_3 分別以 p、q、r 表示,其共軛複數分別是 p'、q'、r'
p/q + q/r + r/p = (p/q + q/r + r/p)' = (p/q)' + (q/r)' + (r/p)' = p'/q' + q'/r' + r'/p'
|p| = |q| = |r| = 1
p' = 1/p、q' = 1/q、r' = 1/r 代入上式可得
p/q + q/r + r/p = q/p + r/q + p/r
同乘以 pqr
p^2r + q^2p + r^2q = q^2r + r^2p + p^2q
(p - q)(q - r)(r - p) = 0
p = q 或 q = r 或 r = p
若 p = q
1 + p/r + r/p = 1
r/p = ±i
|p + 2q + 3r| = |p||1 + 2 ± 3i| = √[(1 + 2)^2 + 3^2] = √18
同理
若 q = r,|p + 2q + 3r| = √[(2 + 3)^2 + 1^2] = √26
若 r = p,|p + 2q + 3r| = √[(3 + 1)^2 + 2^2] = √20
|p + 2q + 3r| 的最大值為 √26