f.4 maths 10mark

(x+3yi)+2(5yi)-(7-2xi)=0?

2010-09-05 16:56:56 補充：
(x+3yi)+2(y-5i)-(7-2xi)=0
x+3yi+2y-10i-7+2xi=0
(x+2y-7)+(3y+2x-10)i=0

For a complex number a+bi=0, then a=b=0
{x+2y-7=0...(1)
{3y+2x-10=0...(2)

Sub (1) into (2):
3y+2(7-2y)-10=0
3y+14-4y-10=0
-y+4=0
y=4 and x=7-2(4)=-1

(x+3yi)+2(Y5i)-(7-2xi)=0

x +3yi +10yi -7 +2xi=0
(x - 7) +13yi+2xi = 0
(x – 7) + (13y + 2x)i = 0

The real part is (x -7)
The imaginary part is (13y + 2x)

For any real number, the imaginary part must be zero.
If the real number is zero, the real part of complex number must be zero.

(x – 7) = 0 --------------- (1)
(13 y + 2x) = 0 --------  (2)

From equation (1)
x – 7 = 0
x = +7

From equation (2)
13 y + 2x = 0
If x = 7,
13y + 2(7) = 0
13y + 14 = 0
13y = -14
y = -14/13

x = 7 and y = -14/13