# When do three points determine the same orientation?

I'm trying to understand the following claim: If $z_1,z_2,z_3,z_4$ are points (as complex numbers) on a circle, then $z_1,z_3,z_4$ and $z_2,z_3,z_4$ determine the same orientation iff $CR(z_1,z_2,z_3,z_4)>0$.

Why is this?

This was the explanation I tried to explain to myself, but I don't know if it's fully correct, since I make a lot of assumptions to simplify the work.

Since the cross ratio is invariant under transformation, we can assume that $z_1,z_2,z_3,z_4$ lie on the real axis. Moreover, we can use a transformation to assume that $z_2=0$, $z_3=1$, and $z_4=2$. Now $$(z_1,z_2,z_3,z_4)=\frac{2(z_1-z_3)}{z_1-z_4}$$ and so $(z_1,z_2,z_3,z_4)>0$ if any only if $\frac{z_1-z_3}{z_1-z_4}>0$. Now note that for any $z$, $$(z,z_1,z_3,z_4)=\frac{(z_1-z_4)z-(z_1-z_4)}{(z_1-z_3)z-2(z_1-z_3)}$$ and $$(z,z_2,z_3,z_4)=\frac{2z-2}{z-2}.$$ So the determinant of the first transformation is $-(z_1-z_3)(z_1-z_4)$, and that of the latter is $-2$. But $\frac{z_1-1}{z_1-2}>0$ when numerator and denominator have the same sign, that is, either when $z_1>z_3$ and $z_1>z_4$, or when $z_1<z_3$ or $z_1<z_4$, and in either case the determinant is negative. It follows that $\Im(z,z_1,z_3,z_4)$ and $\Im(z,z_2,z_3,z_4)$ always have the same sign, and thus determine the same orientation. Thanks.

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## 2 Answers

For four different points $z_k\in{\mathbb C}$ $\ (1\leq k\leq 4)$ the cross ratio of these points is defined by $${\rm CR}(z_1,z_2;z_3,z_4):={z_4-z_2\over z_4-z_1}\Big/{z_3-z_2\over z_3-z_1}\ .$$ If, e.g., $z_4=\infty$, then $${\rm CR}(z_1,z_2;z_3,z_4):={z_3-z_1\over z_3-z_2}\ .$$ The (complex) value of ${\rm CR}$ is $\ne0,1,\infty$, and this value is real iff the four given points lie on a (generalized) circle. The ${\rm CR}$ is invariant under Moebius transformations, i.e. $${\rm CR}\bigl(T(z_1),T(z_2);T(z_3),T(z_4)\bigr)={\rm CR}(z_1,z_2;z_3,z_4)\qquad (T\in{\cal M})\ .$$

Addressing the question of the OP we may assume that the four points $z_k$ lie on the real axis (a generalized circle), and even that $z_3=0$, $z_4=\infty$. In this case the cross ratio $${\rm CR}(z_1,z_2,z_3,z_4)={\rm CR}(z_1,z_2,0,\infty)={z_1\over z_2}$$ is real to begin with, and it makes sense to consider its sign: This sign is positive iff $z_1$ and $z_2$ are on the same side of $0=z_3$, and this means geometrically that the triple $(z_1,z_3,z_4)$ induces the same orientation of the "circle" ${\mathbb R}$ as does the triple $(z_2,z_3,z_4)$.

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Hi Christian, thanks for your answer. I am wondering how one would go about showing that the cross ratio is real iff the four points lie on a generalized circle. I posted a question about that here: math.stackexchange.com/questions/1257294/… – EthanAlvaree Apr 29 '15 at 9:59

The answer is easier and more general. Just check that $\begin{split}\mbox{Im}\left(z,z_{1},z_{3},z_{4}\right)>0 & \Leftrightarrow\mbox{Im}\left(\frac{z-z_{3}}{z-z_{4}}\frac{z_{1}-z_{4}}{z_{1}-z_{3}}\right)>0\\ & \Leftrightarrow\frac{z_{1}-z_{3}}{z_{1}-z_{4}}\frac{z_{2}-z_{4}}{z_{2}-z_{3}}\mbox{Im}\left(\frac{z-z_{3}}{z-z_{4}}\frac{z_{1}-z_{4}}{z_{1}-z_{3}}\right)>0\\ & \Leftrightarrow\mbox{Im}\left(\frac{z-z_{3}}{z-z_{4}}\frac{z_{1}-z_{4}}{z_{1}-z_{3}}\frac{z_{1}-z_{3}}{z_{1}-z_{4}}\frac{z_{2}-z_{4}}{z_{2}-z_{3}}\right)>0\\ & \Leftrightarrow\mbox{Im}\left(\frac{z-z_{3}}{z-z_{4}}\frac{z_{2}-z_{4}}{z_{2}-z_{3}}\right)>0\\ & \Leftrightarrow\mbox{Im}\left(z,z_{2},z_{3},z_{4}\right)>0.\end{split}$

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