回復 23# cut6997 的帖子
第 8 題
Z_1 = 2(cosα + isinα)，Z_ 2 = 3(cosβ + isinβ)
3Z_1 - 2Z_2 = 3/2 - i
6(cosα - cosβ) = 3/2
-12sin[(α + β)/2]sin[(α - β)/2] = 3/2
6(sinα - sinβ) = -1
12cos[(α + β)/2]sin[(α - β)/2] = -1
tan[(α + β)/2] = 3/2
sin(α + β) = 12/13，cos(α + β) = -5/13
Z_1Z_2 = 6[cos(α + β) + isin(α + β)]
= (-30/13) + (72/13)i
