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Homework Help: Complex Analysis - Finding the equation of a circle

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data
    If [itex]\frac{z}{z + 3}[/itex] is purely imaginary, show that z lies on a certain circle and find the equation of that circle.

    3. The attempt at a solution


    [itex]\frac{z}{z + 3}[/itex] = [itex]\frac{x + iy}{x + iy + 3}[/itex]

    Multiplying by the complex conjugate (and simplifying), we get,

    [itex]\frac{x^{2} + y^{2} + 3x + 3iy}{x^{2} + y^{2} + 6x + 9}[/itex]

    Since we're only interested in the imaginary part here, I take,

    [itex]\frac{3iy}{x^{2} + 6x + 9 + y^{2}}[/itex]

    I am not too sure what to do from here, also...does "z lies on a certain circle" mean on the boundary line or anywhere in that enclosed zone including the boundary?
  2. jcsd
  3. Mar 18, 2012 #2
    You're not "only interested in the imaginary part". You are told z/(z+3) is pure imaginary; this means the real part is zero. That's probably where you want to start.

    A circle is the set of points equidistant from its center; the circle does not include its interior. So "on a circle" means on the curvy line.
  4. Mar 18, 2012 #3
    So if the real part is going to be zero, it means the numerator can only be zero. From that we can get an equation for a circle, which I believe is (x + [itex]\frac{3}{2}[/itex])[itex]^{2}[/itex] + [itex]y^{2}[/itex] = [itex]\frac{9}{4}[/itex] - but is this the circle I'm looking for?

    If so, why am I getting an equation for a circle from the real part, when the expression in question is supposedly purely imaginary?
  5. Mar 18, 2012 #4
    It's _because_ the expression is pure imaginary that the real part is zero, which gives you the equation you are after.
  6. Mar 19, 2012 #5
    Ah I see, that makes it more clear now.
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