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Math Help - Complex Circle Geometry

  1. #1
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    Complex Circle Geometry

    I am trying to show that the points 1, a,−(a*), 1/a in the complex plane lie on a circle. Where a* is the conjugate of a.

    I know that given four points A,B,C,D then they lie on a circle iff

    angle(ADC)+angle(ABC)=pi.

    Any help would be greatly appreciated.

    Thanks.
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  2. #2
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    Quote Originally Posted by DeFacto View Post
    I am trying to show that the points 1, a,−(a*), 1/a in the complex plane lie on a circle. Where a* is the conjugate of a.

    I know that given four points A,B,C,D then they lie on a circle iff

    angle(ADC)+angle(ABC)=pi.
    This condition seems wrong. The right condition is: angle(ABC)=angle(ADC).

    In term of complex numbers, it means that \frac{z_B-z_A}{z_B-z_C} and \frac{z_D-z_A}{z_D-z_C} have the same argument, which is equivalent to the fact that their ratio is real:

    (A,B,C,D are either colinear or cocyclic) iff \frac{\tfrac{z_B-z_A}{z_B-z_C}}{\tfrac{z_D-z_A}{z_D-z_C}}\in\mathbb{R}.

    (it is called the cross-ratio of z_A,z_B,z_C,z_D) If you substitute with the complex numbers a,-\bar{a},1,\frac{1}{a} and simplify the ratio, you will see it is real indeed.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    Quote Originally Posted by DeFacto View Post
    I know that given four points A,B,C,D then they lie on a circle iff

    angle(ADC)+angle(ABC)=pi.
    This condition seems wrong. The right condition is: angle(ABC)=angle(ADC).
    In fact, the correct condition is a combination of those two. If B and D are on the same side of AC then angle(ABC)=angle(ADC). If they are on opposite sides then angle(ADC)+angle(ABC)=π. In either case, the cross-ratio is real. In one case it is positive and in the other case it is negative.
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  4. #4
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    Quote Originally Posted by Opalg View Post
    In fact, the correct condition is a combination of those two. If B and D are on the same side of AC then angle(ABC)=angle(ADC). If they are on opposite sides then angle(ADC)+angle(ABC)=π. In either case, the cross-ratio is real. In one case it is positive and in the other case it is negative.
    Thank you, Opalg, for this reminder.

    Now I remember, a unifying condition is (with oriented angles): \widehat{ABC}\equiv\widehat{ADC}\!\!\!\pmod{\pi} (i.e. (\overrightarrow{BA},\overrightarrow{BC})\equiv (\overrightarrow{DA},\overrightarrow{DC})\!\!\!\pm  od{\pi}.)
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