三角函數.複數.向量

[happy520]

令Z=cos(2π/7)+isin(2π/7),求z/(1+zz) + zz/(1+z^4) + zzz/(1+z^6)之值

求[(1+ √3 i)/2]^20+[(1- √3 i)/2]^20之值

設A(2,1),B(7,X),C(y,3),D(2,9),若｜AB向量｜=｜CD向量｜,且AB向量與CD向量方向相反,則(X,Y)

[thepiano]

第 1 題
z^7 = 1 且 1 + z + z^2 + z^3 + z^4 + z^5 + z^6 = 0

(1 + z^2) + (z^3 + z^5) + (z^4 + z^6) = -z
(1 + z^2)(1 + z^3 + z^4) = -z
1/(1 + z^2) = (-1 - z^3 - z^4)/z

(z + z^4) + (z^2 + z^5) + (z^3 + z^6) = -1
(1 + z^3)(z + z^2 + z^3) = -1
1/(1 + z^3) = -z - z^2 - z^3

(z + z^2) + (z^3 + z^4) + (z^5 + z^6) = -1
(1 + z)(z + z^3 + z^5) = -1
1/(1 + z) = -z - z^3 - z^5

z^2/(1 + z^4)
= z^2/(1 + 1/z^3)
= z^2/[(1 + z^3)/z^3]
= z^5/(1 + z^3)

z^3/(1 + z^6)
= z^3/(1 + 1/z)
= z^3/[(1 + z)/z]
= z^4/(1 + z)

原求值式 = z/(1 + z^2) + z^5/(1 + z^3) + z^4/(1 + z) = -2


第 2 題
原求值式 = (-ω^2)^20 + (-ω)^20 = -1
其中 ω^3 = 1 且 ω^2 + ω + 1 = 0


第 3 題
向量 AB = (5，x - 1)
向量 CD = (2 - y，6)

由於向量 AB 和 向量 CD 方向相反
令 (5，x - 1) = t(2 - y，6)，t ＜ 0
t = 5/(2 - y) = (x - 1)/6
(x - 1)(2 - y) = 30 ...... (1)

又 5^2 + (x - 1)^2 = (2 - y)^2 + 6^2 ...... (2)

解聯立方程可得 x = -5，y = 7