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Math Help - Complex Number Problem

  1. #1
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    Complex Number Problem

    I am new to this forum.
    Please help me with this problem-

    Find the real values of x and y
    if
    ((1+i)x-2)/(3+i)+((2-3i)y+i)/(3-i)=i where i = imaginary number
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  2. #2
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    Re: Complex Number Problem

    The first step is to simplify the fraction so that there is no imaginary part below the line. To do this multiply the top and bottom of the fraction by the complex conjugate of the bottom.
    Thanks from varunkanpur
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  3. #3
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    Re: Complex Number Problem

    Quote Originally Posted by Shakarri View Post
    The first step is to simplify the fraction so that there is no imaginary part below the line. To do this multiply the top and bottom of the fraction by the complex conjugate of the bottom.
    On simplification, it comes

    4x + 9y + i(2x-13-7)=0

    What to do for getting real values of x and y ?
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  4. #4
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    Re: Complex Number Problem

    What I did is-

    \frac{(1+i)x-2}{3+i}+\frac{(2-3i)y+i}{3-i}= i

    \frac{4x +2xi-3i-3+9y-7yi}{10}=i

    4x +2xi-3i-3+9y-7yi=10i

    4x +9y+i(2x-13-7y)=0

    What to do to find real values of x and y
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  5. #5
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    Re: Complex Number Problem

    Quote Originally Posted by varunkanpur View Post
    What I did is-
    \frac{(1+i)x-2}{3+i}+\frac{(2-3i)y+i}{3-i}= i
    4x +9y+i(2x-13-7y)=0
    What to do to find real values of x and y
    Solve this system:
    \begin{cases}4x +9y=0\\2x-13-7y=0\end{cases}
    Thanks from varunkanpur
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