二次方程若 5-2i /3+i -(2-i)(3i-5)=(

(5 - 2i) / (3 + i) - (2 - i)(3i - 5) = (4 + xi) -i(2 + 5yi)
(5 - 2i)(3 - i) / (3 - i)(3 + i) - (2 - i)(3i - 5) = (4 + xi) -i(2 + 5yi)
(13 - 11i) / 10 + (7 - 11i) = 4 + xi - 2i - 5y
(83 / 10) - (121 / 10) i = (4 - 5y) + (x - 2)i

4 - 5y = 83 / 10 and x - 2 = - 121 / 10
y = - 43 / 50 and x = - 101 / 10

2014-01-03 14:24:59 補充：
(5 - 2i) / (3 + i) - (2 - i)(3i - 5) = (4 + xi) -i(2 + 5yi)
(5 - 2i)(3 - i) / (3 - i)(3 + i) - (2 - i)(3i - 5) = (4 + xi) -i(2 + 5yi)
(13 - 11i) / 10 + (7 - 11i) = 4 + xi - 2i + 5y
(83 / 10) - (121 / 10) i = (4 + 5y) + (x - 2)i

4 + 5y = 83 / 10 and x - 2 = - 121 / 10
y = 43 / 50 and x = - 101 / 10