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Math Help - Complex number problem

  1. #1
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    Complex number problem

    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

    ----------------------
    i am getting the answer as

    x+iy = -\frac{\sqrt{72}}{28} + \frac{\sqrt{104}}{28}i

    is it right or wrong if wrong can u give the right one .
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  2. #2
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    Quote Originally Posted by varunnayudu View Post
    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

    ----------------------
    i am getting the answer as

    x+iy = -\frac{\sqrt{72}}{28} + \frac{\sqrt{104}}{28}i

    is it right or wrong if wrong can u give the right one .
    I get i/2.

    If you show your working your error(s) can be pointed out.
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  3. #3
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    My Work....

    My work:

    z = \frac{\sqrt{2}+\sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}=\frac{\sqrt{2}+\sqrt{6}i}{\sqrt{12}-\sqrt{16}i}

    = <br />
\frac{\sqrt{2}+\sqrt{6}i}{\sqrt{12}-\sqrt{16}i}.\frac{\sqrt{12}+\sqrt{16}i}{\sqrt{12}+  \sqrt{16}i}

    = <br />
\frac{\sqrt{24}+ \sqrt{32}i + \sqrt{72}i - \sqrt{96}}{(\sqrt{12}-\sqrt{16}i)(\sqrt{12}+ \sqrt{16}i)}<br />

    = <br />
\frac{- \sqrt{72}+\sqrt{104}i}{12+16}<br />
=  \frac{- \sqrt{72}+\sqrt{104}i}{28}<br />
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  4. #4
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    Hello, varunnayudu!

    My work: . z \:= \:\frac{\sqrt{2}+\sqrt{6}\,i}{(\sqrt{3}+i)(\sqrt{2  }-\sqrt{6}\,i)}=\frac{\sqrt{2}+\sqrt{6}\,i}{\sqrt{12  }-\sqrt{16}\,i} . . . . How? .**
    I did some factoring first:

    . . \frac{\sqrt{2} + \sqrt{6}\,i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}\,i)} \;=\;\frac{{\color{red}\rlap{///}}\sqrt{2}\,(1+\sqrt{3}\,i)}{(\sqrt{3}+i){\color{r  ed}\rlap{///}}\sqrt{2}\,(1-\sqrt{3}\,i)} . =\;\frac{1+\sqrt{3}\,i}{(\sqrt{3}+i)(1-\sqrt{3}\,i)}


    The denominator is: . (\sqrt{3}+i)(1-\sqrt{3}\,i) \:=\:\sqrt{3} - 3i + i + \sqrt{3} \:=\:2\sqrt{3} - 2i \:=\:2(\sqrt{3}-i)

    The fraction becomes: . \frac{1+\sqrt{3}\,i}{2(\sqrt{3}-i)}

    Rationalize: . \frac{1+\sqrt{3}\,i}{2(\sqrt{3}-i)}\cdot{\color{blue}\frac{\sqrt{3}+i}{\sqrt{3}+i}  } \;=\;\frac{\sqrt{3} + i + 3i - \sqrt{3}}{2(3 + 1)} \;=\;\frac{4i}{8} \;=\;\frac{i}{2} . . . as promised by mr. f


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    I see what he did to the denominator!

    . . . This is depressing . . .


    \left(\sqrt{3} + i\right)\left(\sqrt{2} - \sqrt{6}\,i\right) \;=\;\left(\sqrt{3}\right)\left(\sqrt{2}\right) + \left(\sqrt{3}\right)\left(\text{-}\sqrt{6}\,i\right) + (i)\left(\sqrt{2}\right) + (i)\left(\text{-}\sqrt{6}\,i\right)

    . . . . . . . . . . . . . . = \;\sqrt{6} - \sqrt{18}\,i + \sqrt{2}\,i + \sqrt{6}


    and, of course: . \begin{array}{ccc}\sqrt{6} + \sqrt{6} &=& \sqrt{12} \\ \\[-4mm] \text{-}\sqrt{18}\,i + \sqrt{2}\,i &=& \text{-}\sqrt{16}\,i \end{array}\quad\hdots\quad NOT\:!

    Last edited by Soroban; December 29th 2008 at 01:35 PM.
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  5. #5
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    Quote Originally Posted by varunnayudu View Post
    My work:

    z = \frac{\sqrt{2}+\sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}=\frac{\sqrt{2}+\sqrt{6}i}{\sqrt{12}-\sqrt{16}i} Mr F says: The first mistake is in the first line. This denominator is not correct. Try expanding again.

    = <br />
\frac{\sqrt{2}+\sqrt{6}i}{\sqrt{12}-\sqrt{16}i}.\frac{\sqrt{12}+\sqrt{16}i}{\sqrt{12}+  \sqrt{16}i}

    = <br />
\frac{\sqrt{24}+ \sqrt{32}i + \sqrt{72}i - \sqrt{96}}{(\sqrt{12}-\sqrt{16}i)(\sqrt{12}+ \sqrt{16}i)}<br />

    = <br />
\frac{- \sqrt{72}+\sqrt{104}i}{12+16}<br />
=  \frac{- \sqrt{72}+\sqrt{104}i}{28}<br />
    ..
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