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How does one get the form of a circle out of this equation?

  1. Aug 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Sketch modulus((z+1)/(2z+3))=1 on the complex plane where z=x+iy

    2. Relevant equations



    3. The attempt at a solution
    I know it is a circle but i need help simplifying the equation into the form of a circle.

    i'm stuck at
    0= 3x^2 + 3y^2 + 10x + 8

    I usually complete the square on these type of problems but it has a 3x^2. What can i do to get it to the form of a circle?
     
  2. jcsd
  3. Aug 11, 2012 #2
    If you use Latex, this looks neater:
    [tex]\left| \frac{z+1}{2z+3} \right| = 1[/tex]
    Play around with the absolute value:
    [tex]\frac{|z+1|}{|2z+3|} = 1[/tex]
    This means that [itex]|z+1|=|2z+3|[/itex].
    Now substituting a+bi for z, we get [itex]|(a+1)+bi|=|(2a+3)+2bi|[/itex].
    Using the definition of the absolute value of a complex number, we then obtain
    [tex]a^2+2a+b^2+1=4a^2+12a+9+4b^2[/tex]
    Rearranging yields
    [tex]3a^2+10a+3b^2+8=0[/tex]
    which you already obtained.
    This is a quadratic equation which you can solve to get the imaginary part of z in terms of its real part. Then, plug in values for the real part and get the imaginary part values!
     
  4. Aug 11, 2012 #3
    If you want to complete the square but can't because of the 3x2 there, what can you do to remove the 3? Divide by it!
     
  5. Aug 11, 2012 #4

    vela

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    That's what you want to do. What specifically is it about completing the square that is a problem for you?
     
  6. Aug 11, 2012 #5
    ofcause!! oh how foolish i am :)

    answer:

    1/9 = (x+ 5/3)^2 + y^2

    thanks guys
     
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