Results 1 to 4 of 4

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

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    157

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

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

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

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,742
    Thanks
    2814
    Awards
    1
    Quote Originally Posted by db5vry View Post
    The complex number z and its complex conjugate Z satisfy the equation $\displaystyle 2z + Z = \frac{11 + 7i}{1 + i}$
    Find z in the form x + iy. [6]
    $\displaystyle \begin{gathered}
    2z + \overline z = 3x + yi \hfill \\
    \frac{{11 + 7i}}
    {{1 + i}} = \frac{{\left( {11 + 7i} \right)\left( {1 - i} \right)}}
    {2} \hfill \\
    \end{gathered} $

    Now carry on.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    First of all $\displaystyle \frac{11+7i}{1+i}=\frac{(11+7i)(1-i)}{2}=9-2i$

    $\displaystyle 2(x+yi)+x-yi=9-2i$

    $\displaystyle 3x+yi=9-2i$

    Then $\displaystyle 3x=9\Rightarrow x=3$

    $\displaystyle y=-2$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member mathemanyak's Avatar
    Joined
    Jul 2008
    From
    TURKEY
    Posts
    39
    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; Jul 12th 2009 at 06:40 PM. Reason: Fixed a minor typo (in red)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex number in exponential form
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 17th 2010, 06:59 PM
  2. Complex number(exponential form)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Dec 8th 2009, 02:35 AM
  3. complex number polar form
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Feb 23rd 2009, 10:26 AM
  4. complex number exponential form
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Feb 21st 2009, 09:42 AM
  5. Trig. Form Of A Complex Number
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jun 5th 2008, 08:41 PM

Search Tags


/mathhelpforum @mathhelpforum