1. ## 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

Any help would be greatly appreciated.

Thanks.

2. 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

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

In term of complex numbers, it means that $\displaystyle \frac{z_B-z_A}{z_B-z_C}$ and $\displaystyle \frac{z_D-z_A}{z_D-z_C}$ 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 $\displaystyle \frac{\tfrac{z_B-z_A}{z_B-z_C}}{\tfrac{z_D-z_A}{z_D-z_C}}\in\mathbb{R}$.

(it is called the cross-ratio of $\displaystyle z_A,z_B,z_C,z_D$) If you substitute with the complex numbers $\displaystyle a,-\bar{a},1,\frac{1}{a}$ and simplify the ratio, you will see it is real indeed.

3. 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

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.

4. 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): $\displaystyle \widehat{ABC}\equiv\widehat{ADC}\!\!\!\pmod{\pi}$ (i.e. $\displaystyle (\overrightarrow{BA},\overrightarrow{BC})\equiv (\overrightarrow{DA},\overrightarrow{DC})\!\!\!\pm od{\pi}$.)