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Math Help - Complex number. How is this expressed in the form x + iy?

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    Complex number. How is this expressed in the form x + iy?

    The complex number z and its complex conjugate Z satisfy the equation

    2z + Z = \frac{11 + 7i}{1 + i}

    Find z in the form x + iy. [6]

    This is a new topic I am doing in maths and I would like to know how this is done because I don't know the basics in answering it. I would greatly appreciate any help if you could tell me what methods are used. Thanks

    note: "Z" actually symbolizes a little z with the line on top of it - z's conjugate, to clarify.
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  2. #2
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    Quote Originally Posted by db5vry View Post
    The complex number z and its complex conjugate Z satisfy the equation 2z + Z = \frac{11 + 7i}{1 + i}
    Find z in the form x + iy. [6]
    \begin{gathered}<br />
  2z + \overline z  = 3x + yi \hfill \\<br />
  \frac{{11 + 7i}}<br />
{{1 + i}} = \frac{{\left( {11 + 7i} \right)\left( {1 - i} \right)}}<br />
{2} \hfill \\ <br />
\end{gathered}

    Now carry on.
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  3. #3
    MHF Contributor red_dog's Avatar
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    First of all \frac{11+7i}{1+i}=\frac{(11+7i)(1-i)}{2}=9-2i

    2(x+yi)+x-yi=9-2i

    3x+yi=9-2i

    Then 3x=9\Rightarrow x=3

    y=-2
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  4. #4
    Junior Member mathemanyak's Avatar
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    we know z=x+iy and Z=x-iy implies 2x+2iy+x-iy=11+7i/1+i implies 3x+iy=
    (11+7i)(1-i)/2 = (11+7+7i-11i)/2 = 9-2i => x=3 and y=-2 => z=3-2i
    Last edited by mr fantastic; July 12th 2009 at 06:40 PM. Reason: Fixed a minor typo (in red)
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