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Math Help - Complex Algebra

  1. #1
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    Complex Algebra

    I'm a little rusty in complex algebra I was wondering how i would go about solving this equation for w

    \frac{w+i}{w-i} = \frac{-1+i}{1-i}\left(\frac{z+1}{z-1}\right)

    The correct answer being w = \frac{i-z}{i+z}
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  2. #2
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    Multiply both sides by (w-i). Then move both terms to one side. Expand the term with (w-i) and factorise w.
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  3. #3
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    I'm not sure exactly what you're getting at, could you write some of the steps out?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    The long way round?

    \dfrac{w+i}{w-i} = \dfrac{-1+i}{1-i}\left(\dfrac{z+1}{z-1}\right)

    \dfrac{w+i}{w-i} = \dfrac{-z-1+iz+i}{z-1-iz+i}

    (w+i)(z-1-iz+i) = (-z-1+iz+i)(w-i)

    w(z-1-iz+i) + i(z-1-iz+i) = w(-z-1+iz+i) - i(-z-1+iz+i)

    w(z-1-iz+i) - w(-z-1+iz+i) = - i(-z-1+iz+i) -  i(z-1-iz+i)

    w(z-1-iz+i +z+1-iz-i) = - i(-z-1+iz+i + z-1-iz+i)

    w(2z-2iz) = - i(-2+2i)

    2w(z-iz) = - 2i(-1+i)

    w(z-iz) = i +1

    w = \dfrac{i + 1}{z-iz} = \dfrac{i}{z}

    Are you sure this is the correct question?
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  5. #5
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    Thanks for taking the time to work it out, Unknown008!

    Mathematica agrees with your answer.

    Original poster, please check either your question or answer.
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  6. #6
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    Sorry I was off by -1 on the RHS. it should be -\frac{1+i}{1-i}\frac{z+1}{z-1}, but i have the general method now. thanks
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