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Math Help - Complex Number problem

  1. #1
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    Complex Number problem

    Question: Express z = \ \frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)} in the form x+iy where x and y are real numbers


    Solution :
    z = \ \frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)}

    =  \frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)} = ................= \frac{\sqrt{104}i - \sqrt{72}}{24}......Is this correct???
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  2. #2
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    To do this problem you need to expand the denominator to the form x+yi after this you need to mulitply the both numerator and denominator by the complex conjugate x-yi . Is this what you did?
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  3. #3
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    This is what i have done?

    Quote Originally Posted by pickslides View Post
    To do this problem you need to expand the denominator to the form x+yi after this you need to mulitply the both numerator and denominator by the complex conjugate x-yi . Is this what you did?
    \frac{\sqrt{2} + \sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}

    = \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i +  \sqrt{2}i -\sqrt{6}i^2}

    = \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{16}i + \sqrt{6}}

    = \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{12}- \sqrt{16}i}

    = \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{12}- \sqrt{16}i} \times \frac{\sqrt{12}+ \sqrt{16}i}{\sqrt{12}+ \sqrt{16}i}


    = \frac{\sqrt{24} + \sqrt{32}i + \sqrt{72}i + \sqrt{96}i^2}{12 + \sqrt{192}i - \sqrt{192}i - 16 i^2}

    = \frac{\sqrt{24} + \sqrt{104}i - \sqrt{96}}{12  + 16}

    = \frac{ \sqrt{104}i - \sqrt{72}}{28}.............Could any body please tell me what i have done wrong and how should i rectify it ???
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  4. #4
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    Smile

    \frac{\sqrt{2} + \sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}

    = \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i +  \sqrt{2}i -\sqrt{6}i^2}

    = \frac{\sqrt{2} + \sqrt{6}i}{2\sqrt{6}- \sqrt{8}i}


    = \frac{\sqrt{2} + \sqrt{6}i}{2\sqrt{6}- \sqrt{8}i} \times \frac{2\sqrt{6} + \sqrt{8}i}{2\sqrt{6} + \sqrt{8}i}


    . . .
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  5. #5
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    Quote Originally Posted by zorro View Post
    [tex]

    = \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2}

    = \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{16}i + \sqrt{6}}
     - \sqrt{18}i + \sqrt{2}i \neq - \sqrt{16}i

     - \sqrt{18}i + \sqrt{2}i = - \sqrt{9\times 2}i + \sqrt{2}i =- 3\sqrt{ 2}i + \sqrt{2}i =-2\sqrt{ 2}i

    also

     \sqrt{6}+\sqrt{6} = 2\sqrt{6}\neq \sqrt{12}
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  6. #6
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    Quote Originally Posted by dedust View Post

    = \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2}

    = \frac{\sqrt{2} + \sqrt{6}i}{2\sqrt{6}- \sqrt{8}i}


    [/tex]


    . . .
    This can be simplified further - \sqrt{8}i = - 2\sqrt{2}i
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  7. #7
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    Thanks Pickslide and Mark Goodstein

    Thanks u for helping me
    pickslides and Mark Goodstein

    cheers
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  8. #8
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    Is this correct?

    Quote Originally Posted by zorro View Post
    Thanks u for helping me
    pickslides and Mark Goodstein

    cheers


    The result which i am getting is - \frac{1}{2}
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  9. #9
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    Is the answer correct ?

    Quote Originally Posted by pickslides View Post
    This can be simplified further - \sqrt{8}i = - 2\sqrt{2}i

    Mite it the answer correct?
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