✔ 最佳答案
The answer is (-3+4i)/5 !
2010-01-05 01:15:49 補充:
設z1=a(cosx+isinx), z2=a(cosy+i siny)
由z1-z2=2得
a(cosx-cosy)= 2
a(sinx-siny)=-1
兩式相除再用和差化積得 tan[(x+y)/2]=2
so, cos(x+y)=-3/5, sin(x+y)= 4/5
z1*z2/|z1*z2|= cos(x+y)+i sin(x+y)= (-3+4i)/5
幾何意含?
(1)菱形對角線為分角線,so,z1+z2為角z1,O,z2的分角線,
(2)z1-z2=2-i, 可視為向量(2, -1)(菱形另一對角線), so, z1+z2 (向量)//(1, 2)
(3) z1+z2 // (cos[(x+y)/2, sin[(x+y)/2]) , so, cos(x+y), sin(x+y)可得
故 (z1*z2)/|z1*z2|= cos(x+y)+i sin(x+y)可得
計算技巧?
(1)三角函數兩倍角
(2)normalize(觀念上用到)
(3)複數極式相乘,角度相加