1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Where
    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]
    [itex]\uparrow[/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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Conic Sections on the Complex Plane (circle)
  1. Conical Section (Replies: 11)

Loading...