Thread: 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.

Originally Posted by DeFacto

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 and 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 .

(it is called the cross-ratio of ) If you substitute with the complex numbers and simplify the ratio, you will see it is real indeed.

Originally Posted by Laurent

Originally Posted by DeFacto

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.

Originally Posted by Opalg

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): (i.e. .)