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Four complex points lie on a circle proof

  1. Dec 21, 2012 #1
    This isn't really homework. I'm studying What Is Mathematics by myself. But I'm very stuck on one of its exercises.

    1. The problem statement, all variables and given/known data

    Prove that if for four complex numbers [itex]z_{1}[/itex], [itex]z_{2}[/itex], [itex]z_{3}[/itex] and [itex]z_{4}[/itex] the angles of [itex]\frac{z_{3} - z_{1}}{z_{3} - z_{2}}[/itex] and [itex]\frac{z_{4} - z_{1}}{z_{4} - z_{2}}[/itex] are the same, then the four numbers lie on a circle or on a straight line, and conversely.


    2. Relevant equations

    [itex]x^{2} + y^{2} = r^{2}[/itex] and [itex]ax + b = c[/itex]?

    3. The attempt at a solution

    I can prove that the quotient between two complex numbers with the same angle is real, using the trigonometric form of the complex numbers (z = [itex]\rho(\cos(\phi) + i\sin(\phi))[/itex]. I have [itex]\frac{z_{3} - z_{1}}{z_{3} - z_{2}} = \rho_{123}(\cos(\phi) + i \sin(\phi))[/itex] and [itex]\frac{z_{4} - z_{1}}{z_{4} - z_{2}} = \rho_{124}(\cos(\phi) + i \sin(\phi))[/itex].

    Between two complex numbers, z and z', if the angle between the two are real, then:

    [itex]\frac{z}{z'} = \frac{\rho_{z}(cos(\phi) + i \sin(\phi))}{\rho_{z'}(\cos(\phi) + i \sin(\phi))} = \frac{\rho_{z}}{\rho_{z'}}(\cos(\phi - \phi) + i \sin(\phi - \phi)) = \frac{\rho_{z}}{\rho_{z'}}(\cos{0} + i \sin {0}) = \frac{\rho_{z}}{\rho_{z'}}[/itex]

    I have [itex]\frac{z_{3} - z_{1}}{z_{3} - z_{2}} = \rho_{123}(\cos(\phi) + i \sin(\phi))[/itex] and [itex]\frac{z_{4} - z_{1}}{z_{4} - z_{2}} = \rho_{124}(\cos(\phi) + i \sin(\phi))[/itex] so that [itex]\frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}}= \frac{\rho_{123}}{\rho_{124}}[/itex].

    What I am not understanding is how this means that all four complex points lie on a circle or a straight line? Should I reduce the [itex]\frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}}[/itex] to their [itex]a + bi[/itex] components to try to get them into the form for the equation of a circle? So that [itex]\frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}} = \frac{\frac{(a_{3} + ib_{3}) - (a_{1} + ib_{1})}{(a_{3} + ib{3}) - (a_{2} + ib{2})}}{\frac{(a_{4} + ib_{4}) - (a_{1} + ib_{1})}{(a_{4} + ib_{4}) - (a_{2} + ib_{2})}}[/itex]?

    Does anyone have a hint for this problem?
     
  2. jcsd
  3. Dec 21, 2012 #2

    Dick

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    Use the angle being the same to show the cross ratio is real. The cross ratio is invariant under Mobius transformations. Use ideas like that rather than trying to mess around with components.
     
  4. Dec 21, 2012 #3

    pasmith

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    I think this is a question about Mobius maps. I say this because
    [tex]\phi(z) = \frac{z - z_1}{z - z_2}[/tex]
    is a Mobius map, and one of the properties of Mobius maps is that they map straight lines and circles to straight lines and circles. Now
    [tex]\phi^{-1}(z) = \frac{zz_2 - z_1}{z - 1}[/tex]
    maps the points 0, [itex]\infty[/itex], [itex](z_3 - z_1)/(z_3 - z_2)[/itex] and [itex](z_4 - z_1)/(z_4 - z_2)[/itex] to the points [itex]z_1, z_2, z_3, z_4[/itex] respectively. Now if
    [tex]
    \arg \left( \frac{z_3 - z_1}{z_3 - z_2} \right) = \arg \left( \frac{z_4 - z_1}{z_4 - z_2} \right)
    [/tex]
    then the points [itex]0, \infty, (z_3 - z_1)/(z_3 - z_2)[/itex] and [itex](z_4 - z_1)/(z_4 - z_2)[/itex] lie on a straight line, so their images lie on a straight line or a circle.

    Conversely, if [itex]z_1, z_2, z_3[/itex] and [itex]z_4[/itex] lie on a straight line or a circle, then so do [itex]0, \infty[/itex], [itex](z_3 - z_1)/(z_3 - z_2)[/itex] and [itex](z_4 - z_1)/(z_4 - z_2)[/itex]. Since [itex]\infty[/itex] is in that set, it must be a straight line (so that the arguments of [itex](z_3 - z_1)/(z_3 - z_2)[/itex] and [itex](z_4 - z_1)/(z_4 - z_2)[/itex] are equal or differ by [itex]\pi[/itex]).
     
  5. Dec 21, 2012 #4

    Dick

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    There's an even simpler argument if you've already proven that given three distinct points there is some mobius transformation such that f(z1)=1, f(z2)=0 and f(z3)=∞. Then you don't really have to worry about the details of what the formula for f even is. Just use that the cross ratio is real.
     
    Last edited: Dec 21, 2012
  6. Dec 21, 2012 #5
    Thank you guys soooooooooo much!!
     
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