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Math Help - [Solving Complex Numbers] Division legal?

  1. #1
    Junior Member Cthul's Avatar
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    [Solving Complex Numbers] Division legal?

    This is what I have done.
    (-4x+i)+(2-3yi)=6-2i
    Group real numbers and imaginary numbers
    (-4x+2)+(i-3yi)=6-2i
    \Rightarrow Solve for real number
    -4x+2=6

    -4x=4

    x=-1

    \Rightarrow Solve for imaginary number
    i-3yi=-2i

    3yi=3i

    .. Can someone verify if it's legal to divide both sides by 3i?
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    You can divide both sides by 3i but y=1 by inspection
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  3. #3
    Junior Member Cthul's Avatar
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    Oh, okay. Appreciated your verification.
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  4. #4
    MHF Contributor
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    Quote Originally Posted by Cthul View Post
    This is what I have done.
    (-4x+i)+(2-3yi)=6-2i
    Group real numbers and imaginary numbers
    (-4x+2)+(i-3yi)=6-2i
    \Rightarrow Solve for real number
    -4x+2=6

    -4x=4

    x=-1

    \Rightarrow Solve for imaginary number
    i-3yi=-2i

    3yi=3i

    .. Can someone verify if it's legal to divide both sides by 3i?
    Yes, the division by i is legal. However, these sorts of problems are easiest if you realise that for two complex expressions to be equal, their real parts have to be equal and their imaginary parts have to be equal.

    So if you write each side in terms of its real and imaginary parts, then you can equate real and imaginary parts.


    In your case:

    (-4x+i)+(2-3yi)=6-2i

    -4x + 2 + (1 - 3y)i = 6 - 2i

    Now equate real and imaginary parts:

    -4x + 2 = 6 and 1 - 3y = -2.
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