# Solving another equation

• Jan 18th 2010, 03:54 PM
Paymemoney
Solving another equation
Hi i need help on where i have gone wrong in the following equations:
1)Solve $(x+2i)(x-2i)=5-12i$ for real values x and y.
$x^2+2ix+2ix+4i^2=5-12i$

$x^2+4ix-4=5-12i$

$x^2+4ix=9-12i$

$4ix+12i=9-x^2$

$i(4x+12)=9-x^2$

$4x+12=9-x^2$

$x^2+4x+3$

$(x-1)(x+3)$
x=1 or x=-3

2)Solve $(x+iy)(x+iy)=-18i$ for real values x and y.
$x^2+ixy+ixy+i^2y^2=-18i$

$x^2+2ixy-y^2=-18i$

$2ixy-18i=x^2-y^2$

$i(2xy-18)=y^2-x^2$

Also another question does $-i^2=1$ and $i^2=-1$

P.S
• Jan 18th 2010, 04:09 PM
Calculus26
For 2

a complex number z = 0 if and only if Re(z) = 0 and Im(z) =0

(x^2 - y^2) + i (2xy -18) = 0

x^2 = y^2 x = + y

and 2xy = 18

If x= y 2x^2 = 18 x= +3

If x = -y -2x^2 =18 and there are no solutions

only solutions are (3,3) and (-3,-3)

For 1)

(x+2i)(x-2i) = x^2 + 2ix - 2ix - 4i^2 = x^2 + 4

also

Even if http://www.mathhelpforum.com/math-he...8dca7823-1.gif

how do you conclude 4x+12 = 9 -x^2 ?

you should have x^2 + 4 = 5 -12 i

x^2 = 1-12i

if x is real x^2 must be real so there are no solutions

yes i^2 =-1 and -i^2 = 1