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A Locus in the Complex Plane

  1. Jul 18, 2011 #1
    I am reading Visual Complex Analysis by Dr. Tristan Needham and am hung up on some of the geometrical concepts. In particular, I am having trouble with ideas involving the geometric properties of numbers like:

    [itex]\frac{z-a}{z-b}[/itex]

    Note: I am still in the first and second chapters, which deal with the basic geometry of complex numbers and functions.

    As an example problem, here is one I've been trying to figure out for a few days:

    attachment.php?attachmentid=37296&stc=1&d=1311039855.png

    As an attempted solution, I chose three arbitrary points a, b, z and constructed the perpendicular bisector of a and b and a three point circle through a, b and z. The chord through a and b divides the circle into two regions.

    Since z is a variable under a constraint, it may move freely about this circle. As long as z stays on the same side of chord [itex]\stackrel{\rightarrow}{ab}[/itex], the angle between the chords [itex]\stackrel{\rightarrow}{az}[/itex] and [itex]\stackrel{\rightarrow}{bz}[/itex] will be a constant. I interpreted (but have found no way to prove) the constant angle as the constant referred to in the problem.

    That being said, it looked to me like the tangent of this angle was equal to [itex]Im((z-a)/(z-b))/Re((z-a)/(z-b))[/itex],
    where Im(z) and Re(z) are the imaginary and real parts of the complex number z.

    This is as far as I got. I showed that the angle between chords was constant along the three point circular arc of a, b and z. I have no idea if this is even the right place to look, and this problem is only one of many dealing with complex ratios that I simply don't understand.

    Any help?
     

    Attached Files:

  2. jcsd
  3. Jul 19, 2011 #2

    Gib Z

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    The useful result from Euclidean Geometry is the following: Angles at the circumference standing on equal chords are equal.
     
  4. Jul 19, 2011 #3
    I used this result earlier to move z along the arc to the perpendicular bisector of [itex]\stackrel{\rightarrow}{ab}[/itex], yielding an isosceles triangle inscribed in the three point circle. That didn't seem to help me. Is there a more useful approach?
     
  5. Jul 19, 2011 #4

    Gib Z

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    Draw a new diagram. Also keep in mind the fact arg(p/q) = arg(p) - arg(q) . You should be able to see from the diagram why arg(z-a) - arg(z-b) is the angle <azb (on one side of the chord connecting ab at least). It then follows from the quoted theorem that the locus is the arc of a semi-circle.
     
  6. Jul 20, 2011 #5
    Thanks a lot. This was a big help.
     
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