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Math Help - Solving another equation

  1. #1
    Super Member
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    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
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  2. #2
    MHF Contributor Calculus26's Avatar
    Joined
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    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

    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
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