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Homework Help: Conic Sections on the Complex Plane (circle)

  1. Jun 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Describe the locus and determine the Cartesian Equation of:
    [itex]\left|z-3-5i\right|= 2[/itex]

    2. Relevant equations
    [itex]\left|z-C\right|= r[/itex] -----> formula for a circle on complex plane
    C = the centre
    z = the moving point (locus)

    [itex](x-h)^{2}+(y-k)^{2}=r^{2}[/itex] -----> Formula for a circle on the cartesian plane
    3. The attempt at a solution
    Ok I think I've got the first section, describe the locus

    Well if -C = -3-5i
    that means C = 3+5i

    So the centre of the circle will be at 3+5i on the complex plane.

    But I get stuck when converting it into the cartesian form.

    [itex]z = x + yi[/itex]

    [itex]\left|(x + yi)-3-5i\right|= 2[/itex]
    [itex]\sqrt{(x-3)^{2}-i(y + 5)^{2}}[/itex]
    But I don't know how to proceede from there because I can't figure out how to get rid of the [itex]i[/itex]

    Anyone know how to?
  2. jcsd
  3. Jun 18, 2012 #2
    I'm not sure exactly what you have done, I don't do these questions this way.

    The easy way to do it is to let [itex]z=x + iy[/itex] and then sub into [itex]\left|z-3-5i\right|= 2[/itex]. Now you just find the modulus as if it were an complex number and you end up with an eqn for a circle.
  4. Jun 18, 2012 #3
    The modulus of a complex number [itex]z = x + iy[/itex] is defined by

    [itex]|x+iy| \equiv \sqrt{x^2 + y^2}[/itex].​

    Note the absence of [itex]i[/itex] on the right-hand side of the above equation.
  5. Jun 18, 2012 #4


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    Science Advisor

    3+5i, on the complex plane, corresponds to the point (3, 5) in the Cartesian plane. There is no "i" when writing points in the Cartesian plane. Yes, the equation |z- (a+ bi|= |(x- iy)- (a+ bi)|= r is a circle in the Complex plane with center at a+ bi which corresponds to (a, b) and radius r. It has equation [itex](x- a)^2+ (y- b)^2= r^2[/itex]. You do NOT include the "i" in the equation when converting to the Cartesian form.
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