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Complex Numbers Involving a Circle

  1. Sep 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the equation [tex]|z - z_0| = R[/tex] of a circle, centered at [tex]z_0[/tex] with radius R, can be written

    [tex]|z|^2 - 2Re(z\bar{z_0}) + |z_0|^2 = R^2[/tex].

    2. Relevant equations

    3. The attempt at a solution

    Honestly, I have no clue where to start with this problem. I know that I need to reduce the given equation to the basic equation of a circle but I do not know where to start.
    I also know that the two equations are almost exact except for the [tex]- 2Re(z\bar{z_0})[/tex] which should reduce to zero somehow I just do not know where to start.

    I know [tex]Re(z) = Re(\bar{z}) = Re\frac{(z + \bar{z})}{2} = x[/tex]. Is this where I start?
  2. jcsd
  3. Sep 1, 2009 #2


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    Homework Helper

    Start with z=x+iy and z0=x0+iy0
  4. Sep 1, 2009 #3
    Thank you for that tidbit it really helped me almost solve it. So here is what I got and I am stumped:

    [tex]2Re(z\bar{z_0}) = 2\frac{z\bar{z_0} + \overline{z\bar{z_0}}}{2} = z\bar{z_0} + \bar{z}z_0[/tex]

    [tex]z\bar{z_0} = (x+iy)(x_0 - iy_0) = (xx_0 - yy_0 + iyx_0 - xiy_0)[/tex]

    [tex]\bar{z}z_0 = (x - iy)(x_0 + iy_0) = (xx_0 - yy_0 - iyx_0 + xiy_0)[/tex]

    So [tex]z\bar{z_0} + \bar{z}z_0 = 2xx_0 - 2yy_0[/tex]

    Is there a method where I can simplify this anymore? because I'm clueless.
  5. Sep 1, 2009 #4


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    [itex]|z- z_0|= \sqrt{(z- z_0)(\overline{z}- \overline{z_0})}[/itex] so [itex]z-z_0|= R[/itex] is the same as [itex]|z- z_0|= \sqrt{(z- z_0)(\overline{z}- \overline{z_0})}= R[/itex]. square both sides of that. I would NOT go to "x+ iy".

    Sorry about the nessed up Latex!
    Last edited by a moderator: Sep 1, 2009
  6. Sep 1, 2009 #5
    Thank you Halls but that seriously confuses me. I do not understand how I should apply it to my problem. Anyone else got anything in mind?
  7. Sep 1, 2009 #6
    Neither would I use the "x+iy" route.
    Recall the identity [tex]z\overline{z} = |z|^{2}[/tex]
  8. Sep 1, 2009 #7
    Thank you so much Fightfish. It was so easy to solve once I used that identity. Goes to show I need to learn the properties better.
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