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Can someone solve this complex number equation for x and y

  1. Jan 20, 2012 #1
    If x and y are real quantities what are the solutions for x and y

    (ix)(1+iy)=(3x+i4)/(x+3y)

    I have tried grouping the equation into two equal complex numbers but have failed to find a solution which isolates x and y from the two quite long polynomial equations.

    Does anybody know how to do this.
     
  2. jcsd
  3. Jan 20, 2012 #2

    vela

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    Your approach sounds fine. Show us what you did so far.
     
  4. Jan 20, 2012 #3
    (xi)/(1 + yi) = (3x + 4i)/(x + 3y)

    First rationalized the left hand fraction.

    (xi)/(1 + yi) * (1 - yi)/(1 - yi) = (xi -yxi^2)/(1 + y^2)

    = (yx + xi)/(1 + y^2)
    Grouped real and imaginary terms

    [ yx/(1 + y^2) ] + [ x/(1 + y^2) ]i

    Then grouped real and imaginary terms on the right hand fraction

    (3x + 4i)/(x + 3y) = [ 3x/(x + 3y) ] + [ 4/(x + 3y) ]i

    Set the two complex numbers equal

    [ yx/(1 + y^2) ] + [ x/(1 +y^2) ]i = [ 3x/(x + 3y) ] + [ 4/(x +3y) ]i

    Take the two imaginary parts to be identical
    Take the two real parts to be identical

    [ yx/(1 + y^2) ] = [ 3x/(x +3y) ]
    [ x/(1 + y^2) ] = [ 4/(x + 3y) ]

    Expressing both fractions as polynomials

    (yx)(x + 3y) = (3x)(1 + y^2)
    (x)(x + 3y) = (4)(1 + y^2)

    Then multiplying out both equations in x and y

    1. yx^2 + 3y^2x = 3x + 3xy^2

    2. x^2 + 3yx = 4 +4y^2

    Grouping terms on one side to form two equations that equal zero.

    1. yx^2 + 3y^2x - 3x - 3xy^2 = 0

    2. x^2 +3yx - 4 - 4y^2 = 0

    Now trying to find some way of dealing with these terms.
     
    Last edited by a moderator: Jan 20, 2012
  5. Jan 20, 2012 #4

    I like Serena

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    Hmm, let's just cross multiply and simplify:

    (xi)/(1 + yi) = (3x + 4i)/(x + 3y)
    (xi)(x + 3y) = (3x + 4i)(1 + yi)
    x²i + 3xyi = 3x + 4i + 3xyi - 4y
    x²i = 3x + 4i - 4y
    (4y-3x) + (x² - 4)i = 0

    That doesn't look so bad does it?
    I suspect you made a mistake somewhere, but I haven't figured out where yet.
     
  6. Jan 20, 2012 #5

    vela

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    Note that you can write these two equations as
    \begin{align*}
    \frac{x(x+3y)}{1+y^2} &= \frac{3x}{y} \\
    \frac{x(x+3y)}{1+y^2} &= 4
    \end{align*}So you have y=(3/4)x. You can plug use that to eliminate y in one of the equations and solve for x.

    Or you can do it the way ILS did it.
     
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