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Complex numbers / cartesian equations etc

  1. Jul 27, 2004 #1
    Using converse of alternate segment theorem (i think it is)

    i.e. this:
    "If the line joining two points A and B subtends equal magnitude angles at two other points on the same side of it, then the four points lie on a circle"

    establish the cartesian equation, range and domain of the locus of points:

    Arg(z - a) - Arg(z) = pi/2

    a = 1 + 0i

    and for

    Arg(z) - Arg(z-a) = pi/2

    a = 1 + 0i

    say z = x + iy ( i think )

    i get

    ( x - 1/2 ) ^2 + y^2 = 1/4

    But i'm not sure about the domain, range etc

    If anyone could help me, or point me in the right direction, that would be great.

    Thanks.
     
  2. jcsd
  3. Jul 27, 2004 #2

    AKG

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    I can't really tell what it is you're trying to do at all.
     
  4. Jul 28, 2004 #3
    I've got two complex numbers,

    z and a
    Let z = x + iy and a = 1 + 0i

    And I want to establish a cartesian equation using this theorem:

    "If the line joining two points A and B subtends equal magnitude angles at two other points on the same side of it, then the four points lie on a circle"

    Basically, I'd like to determine the equation, range and domain of the locus of points such that

    Arg (z-a) - Arg(z) = pi/2

    Sorry if i'm not being clear, i'm finding it hard to explain :p
     
  5. Jul 28, 2004 #4

    AKG

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    I don't see how the top and bottom relate. Do you want the cartesian equation of a circle (the locus of points I assume you're talking about)? Do you want to express your equation (with the "Arg"s) in Cartesian form?

    Anyways, I'm not sure exactly what you want, but if we have z = x + iy, then arg(z) = arctan(y/x). So:

    [tex]\arg (z - a) - \arg (z) = \pi /2[/tex]

    [tex]\arctan \left (\frac{y}{x - 1} \right ) - \arctan \left (\frac{y}{x} \right ) = \pi /2[/tex]

    [tex]\tan \left [ \arctan \left (\frac{y}{x - 1} \right ) - \arctan \left (\frac{y}{x} \right ) \right ] = \tan \frac{\pi}{2}[/tex]

    [tex]\frac{\frac{y}{x - 1} - \frac{y}{x}}{1 + \frac{y}{x - 1}\frac{y}{x}} = \tan \frac{\pi}{2}[/tex]

    [tex]\frac{y}{x(x - 1) + y^2} \times \frac{x(x - 1)}{x(x - 1)} = \tan \frac{\pi}{2}[/tex]

    Since I'm not exactly sure what you want, I won't go further.
     
  6. Jul 28, 2004 #5
    I used vectors and scalar product for the first one, I thought it might be relevant to this.

    I'm after the cartesian of a circle - the locus of points. (I don't need the "Arg"s in there :))

    thanks
     
  7. Jul 28, 2004 #6

    AKG

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    Your confusing me even further. Scalar product of what? The cartesian equation of a circle is quite simply:

    x² + y² - r² = 0

    Of course, if it has its center elsewhere, you'd have to find the center and modify the equation. Another thing you could do:

    |z| - r² = 0
     
  8. Jul 29, 2004 #7
    don't worry, i did it myself

    it's (x - 1/2)^2 + y^2 = 1/4

    with a domain of (0, 1)
    and a range of (0, 1/2]

    Scalar product of vectors.

    Complex numbers can be represented as vectors, and then i used the scalar (dot) product.

    Thanks anyway
     
  9. Jul 29, 2004 #8

    AKG

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    Yes, that's standard for any circle, although the domain is [0,1] and the range is [-1/2, 1/2]. A circle centered at (a,b) with radius r has the cartesian equation:

    [tex](x-a)^2 + (y-b)^2 - r^2 = 0[/tex]
    I know what a scalar product and complex numbers are, I just didn't know which specific vectors/complex numbers you were talking about. This is odd, it seems you're learning about vectors and complex numbers before learning basic stuff about circles? Anyways, I suppose there's nothing wrong with that.
     
  10. Aug 25, 2004 #9
    It's the domain for which the locus of points exists.... not the circle

    for any z below the x axis, Arg(z-a) - Arg(z) is not pi/2, it's -pi/2

    (Also z = 0 + 0i gives Arg(z) as undefined - this is the reason for excluding (0,0))

    Capische?
     
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