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Complex Numbers help

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Sketch the line described by the equation:
    |z − u| = |z|

    z = x+jy
    u = −1 + j√3





    3. The attempt at a solution

    (x+1)^2 + j(y-√3)^2 = (x+jy)^2



    I just don't quite get where to go with this
    please give me a headstart
     
  2. jcsd
  3. Feb 10, 2013 #2

    Dick

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    |z|^2=(x^2+y^2). There is no j in there. There shouldn't be any j in the left hand side either. It's an absolute value.
     
  4. Feb 10, 2013 #3
    oh yeah
    and um did you mean |z|^2=(x+y)^2?
     
  5. Feb 10, 2013 #4

    Dick

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    Definitely not! |z|=sqrt(x^2+y^2). Look it up.
     
  6. Feb 10, 2013 #5

    Mark44

    Staff: Mentor

    If z = x + iy, then |z|2 = x2 + y2, which is what Dick wrote. kiwi101, it looks like you need to review the definition of the absolute value or magnitude of a complex number.
     
  7. Feb 10, 2013 #6
    I was just about to write a long argument about how I was right and then I realized you're right. I misinterpreted something.

    So this is what I have done, I feel its right.

    (x+1)^2 + (y-√3)^2 = x^2 + y^2

    x^2 + 2x + 1 + y^2 -2√3y + 3 = x^2 + y^2

    2x + 1 -2√3y + 3 = 0

    (x+2)/√3 = y

    and then rationalize it and this is the equation of the line?
     
  8. Feb 10, 2013 #7

    Dick

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    That looks ok to me.
     
  9. Feb 10, 2013 #8
    Thanks!
    Out of curiousity this question:

    Sketch the line or curve described by the equation
    ℜe{z} + ℑm{z} = ℜe{u}

    would be x + jy = -1 ?

    so is this a line or what?
     
  10. Feb 10, 2013 #9
    Wait do I solve for y and then rationalize like

    y = (-1 -x)/j
     
  11. Feb 10, 2013 #10

    Dick

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    Im(z)=y, not jy. Check the definition again.
     
  12. Feb 10, 2013 #11
    I just assumed that since it says Im(z) it meant to include the imaginary iota.

    So then I guess it is just a line y = -1-x
     
  13. Feb 10, 2013 #12

    Dick

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    Yep!
     
  14. Feb 10, 2013 #13
    Thanks once again! :)
     
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