complex number

x+3i=5-yi
x=5,
-y=3
y=-3

x-2+(2y+6)i=0
x-2=0
x=2,
2y+6=0
y=-6/2
y=-3

(x-5)i+4+y=6
x-5=0
x=5,
4+y=6
y=2

x+2y-(x-2y)i=7+i
x+2y=7
x=7-2y......(1)
-(x-2y)=1
-x+2y=1........(2)

(1)+(2)
2y=7-2y+1
4y=8
y=2
x=7-2y
x=7-4
x=3


2010-12-01 11:30:55 補充：
其實很簡單, 只要 對番 coefficent of i 要兩邊一樣,
另外無 i 果part, 對番無 i 果堆數
e.g.第 1 題 :x+3i=5-yi
L.H.S =coefficient of i =3
R.H.S=coeffficent of i =-y
所以 -y=3 , y=-3
另外果堆數, 就要對番 無 i 果part
L.H.S =x
R.H.S =5
所以 x=5

2010-12-01 11:31:42 補充：
第3題：(x-5)i+4+y=6
L.H.S. =coefficent of i =x-5
R.H.S =coefficent of i =0
所以 x-5=0
x=5
另外果堆數, 就要對番無 i 果 part
L.H.S =4+y
R.h.S =6
4+y=6
y=2

其實其他都是這樣對出來!