請參考附件
官方於 112 年 4 月 26 日更正答案如下：
填充第 8 題 28
填充第 12 題 0 ≦ X ≦ 13
112 師大附中
版主: thepiano
Re: 112 師大附中
第 7 題
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
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