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Given four points $a, b, c, d \in \mathbb{C}$ on a circle I want to show that the ratio $ \frac{(a-c)(b-d)}{(a-b)(c-d)}$ is a positive real number. I have shown that it is real by expanding $ \frac{(a-c)(b-d)\overline{(a-b)(c-d)}}{(a-b)(c-d)\overline{(a-b)(c-d)}} $ and showing that the numerator (and denominator) is a sum of a complex number and its conjugate. I have no idea how to show this real number is positive. Is this true and if so how would I begin to show this?

Edit: a,b,c,d are in order as you traverse the circle

SchwarzWithNoT
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    If you've already seen that cross-ratios are preserved by Möbius transformations, maybe it would be easier to move your points to the real line. – Josh Keneda Sep 15 '14 at 21:12
  • If you mean that for any four points on a circle, the cross-ratio is positive, I don't think that's true. – Michael Hardy Sep 15 '14 at 21:19
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    Yeah, it's true if you can see $a, b, c, d$ in that order as you traverse the circle, but it's not necessarily true otherwise. I assumed that Theo just omitted that part. :) – Josh Keneda Sep 15 '14 at 21:26

3 Answers3

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Send three of the points to $0,1,\infty$ via a Mobius transform. The image of the fourth point will then be equal to the cross ratio. Further the image of the circle through the first three points will be the real line. You claim now follows.

Note that by changing the order of two points, say $a$ and $c$ the sign of the cross ratio changes. Make of that what you will.

Rene Schipperus
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I think what you mean is that it is a real number, not necessarily positive (which is most definitely not true in general). Since each point is on a circle, we can write each as $p = z_0+re^{i\theta}$ for some $z_0$, $r$ and $\theta$, where $z_0$ is the center of the circle, $r$ is the radius of the circle and $\theta$ is the angle corresponding to the point. Thus $a-b = r(e^{i\theta_1}-e^{i\theta_2})$ and so on. So we can - without loss of generality - assume that the circle is centered at the origin and has radius $1$. Thus we want to show that

$$\frac{(e^{i\theta_1}-e^{i\theta_3})(e^{i\theta_2}-e^{i\theta_4})}{(e^{i\theta_1}-e^{i\theta_2})(e^{i\theta_3}-e^{i\theta_4})}$$

is a real number. Taking a complex conjugate gives

$$\frac{(e^{-i\theta_1}-e^{-i\theta_3})(e^{-i\theta_2}-e^{-i\theta_4})}{(e^{-i\theta_1}-e^{-i\theta_2})(e^{-i\theta_3}-e^{-i\theta_4})} = \frac{\left(\dfrac{1}{e^{i\theta_1}}-\dfrac{1}{e^{i\theta_3}}\right)\left(\dfrac{1}{e^{i\theta_2}}-\dfrac{1}{e^{i\theta_4}}\right)}{\left(\dfrac{1}{e^{i\theta_1}}-\dfrac{1}{e^{i\theta_2}}\right)\left(\dfrac{1}{e^{i\theta_3}}-\dfrac{1}{e^{i\theta_4}}\right)}.$$

Making common denominators gives

$$\frac{\left(\dfrac{e^{i\theta_3}-e^{i\theta_1}}{e^{i\theta_1}e^{i\theta_3}}\right)\left(\dfrac{e^{i\theta_4}-e^{i\theta_2}}{e^{i\theta_2}e^{i\theta_4}}\right)}{\left(\dfrac{e^{i\theta_2}-e^{i\theta_1}}{e^{i\theta_1}e^{i\theta_2}}\right)\left(\dfrac{e^{i\theta_4}-e^{i\theta_3}}{e^{i\theta_3}e^{i\theta_4}}\right)} = \frac{(e^{i\theta_3}-e^{i\theta_1})(e^{i\theta_4}-e^{i\theta_2})}{(e^{i\theta_2}-e^{i\theta_1})(e^{i\theta_4}-e^{i\theta_2})}.$$

And after slight rearranging, you can easily see how this is equal to our original expression. Since the number is equal to its complex conjugate, it is real.

Cameron Williams
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If the point points are real, the cross ratio is clearly real.

Now the cross ratio is invariant under all Moebius transformations, and all circles are images of the real line under such transformations, so the general case follows at once.

Mariano Suárez-Álvarez
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