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Thread: Circles in a complex plane

  1. #1
    May 2009

    Circles in a complex plane

    If z = x + iy show that the general equation for a circle in the complex plane is |z|^2 + 2Re(az) > - |a|^2

    where a is an arbitrary complex number.

    Not sure how to go about this one.
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  2. #2
    MHF Contributor

    Apr 2005
    Well, first, that's not correct. Was that supposed to be "= ", not ">"? I'm going to assume "="

    You know, I hope, that the equation of a circle with center (a,b) and radius r is (x-a)^2+ (y-b)^2= r^2.

    Now look at what you are given: |z|^2+ 2Re(az)= -|a|^2. Write z= x+ yi and a= a+ bi (I'm using a to mean to different things but we aren't going to use them in the same equation).

    |z|^2= (x+ yi)(x- yi)= x^2+ y^2

    az= (a+ bi)(x+ yi)= ax- by+ i(bx+ ay) so 2Re(az)= ax- by.

    Finally, |a|^2= (a+bi)(a-bi)= a^2+ b^2.

    Sp |z|^2+ 2Re(az)= -|a|^2 becomes x^2+ y^2+ ax- by= -a^2- b^2 which is the same as x^2+ ax+ a^2+ y^2- by+ b^2= 0

    Hmmm, (x+a)^2+ (y-b)^2= 0 is NOT the equation of a circle, it is a single point, (-a, b). And keeping the ">" also does not give a circle, it gives every point in the plane except (-a, b).
    Last edited by HallsofIvy; May 17th 2009 at 09:19 AM.
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  3. #3
    MHF Contributor

    Aug 2008
    Paris, France
    Indeed there's a mistake, and here is the way I would derive the correct equation:

    The equation of the circle of center a\in\mathbb{C} and radius r is obviously |z-a|=r.

    Squaring this equation gives r^2=|z-a|^2=(z-a)(\bar{z}-\bar{a})=|z|^2-a\bar{z}-z\bar{a}+|a|^2=|z|^2-2\Re(\bar{a}z)+|a|^2 (I used the formulas |z|^2=z\bar{z} and z+\bar{z}=2\Re(z)).

    Hence the equation is |z|^2-2\Re(\bar{a}z)=r^2-|a|^2.

    More generally, we deduce that for any a\in\mathbb{C} and M\geq -|a|^2,


    is the equation of a circle (of center -\bar{a} and radius \sqrt{M+|a|^2}), which is perhaps what you meant (but wrote unproperly).
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