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Complex Numbers

  1. Jul 27, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the locus of z where

    [tex]

    \arg(\frac{z-2i}{z-3}) = \frac{\pi}{6}

    [/tex]


    2. Relevant equations

    [tex]

    \arg(x+iy) = \arctan(\frac{y}{x})

    [/tex]

    3. The attempt at a solution

    I wrote z=x+iy in which case we have

    [tex]\frac{z-2i}{z-3} = \frac{x+iy-2i}{x-3+iy} = \frac{(x^3 - 3x + y^2 - 2y)+(6 - 2x - 3y)}{(x-3)^2+y^2}[/tex]

    I'm ommiting my working obviously, but i've checked it and i think it's correct. This means that

    [tex]

    \arctan(\frac{6-2x-3y}{x^2 -3x + y^2 - 2y}) = \frac{\pi}{6}

    [/tex]

    [tex]

    18 - 6x - 9y = \sqrt{3} (x^2 - 3x + y^2 - 2y)

    [/tex]

    [tex]

    (x + \frac{2\sqrt{3} - 3}{2})^2 + (y + \frac{3\sqrt{3} - 2}{2})^2 = 13

    [/tex]

    which is the equation of a circle. This seems correct, but I think the locus is only part of the circle not the whole thing. How would you guys go about tackling this problem?

    Thanks.
     
    Last edited: Jul 27, 2008
  2. jcsd
  3. Jul 27, 2008 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi kidsmoker! :smile:

    Personally, I'd use ordinary geometry.

    Let A = 2i, B = 3, then the equation is angle BXA = 30º (taking care not to get -30º or 390º).

    And the locus of points X with BXA constant is … ? :smile:

    And just check "by hand" the solutions you have to ignore because you get 390º. :smile:
     
  4. Jul 27, 2008 #3
    Okay so X lies on an arc of a circle passing through A and B (above to the right)?
     
  5. Jul 27, 2008 #4

    tiny-tim

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    Yes! :smile:

    (erm … forget what I said about 390º … I forgot exactly what the question was. :redface:)
     
  6. Jul 27, 2008 #5
    Cool thanks.

    Yeah I was wondering what all that 390º stuff was about lol.
     
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