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I want to prove that if $\dfrac{z_1-z_4}{z_1-z_2} \times \dfrac{z_2-z_3}{z_4-z_3}$ is real, then the four complex numbers are concyclic.

Now I'm aware that this can be done by drawing them up arbitrarily and then observing that we can make use of the fact that a quadrilateral is concyclic iff opposite angles are supplementary.

My question is, how do I really do this in detail without hand waving anything? My issues arise when considering cases. I understand that if that cross ratio is real, then its argument can either be 0 or $\pi$. However, I'm having some trouble understanding how we can possibly have the 0 case.

I can see the outline of the proof by considering the arguments and making use of that geometric property, but beyond that I'm having trouble constructing a proper proof.

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  • $\begingroup$ That they are concyclic of the cross-ratio is real is true, provided that straight lines are considered circles passing through $\infty$. $\endgroup$ Nov 3 '14 at 1:34
  • $\begingroup$ I am puzzled by your statement that the argument can be either $0$ or $\pi$. The cross-ratio function has four arguments and one value. Any of the four arguments or the one value can be anything in $\mathbb C\cup\{\infty\}$. If the cross-ratio is real, it certainly need not be $0$ or $\pi$. ${}\qquad{}$ $\endgroup$ Nov 3 '14 at 1:45
  • $\begingroup$ @MichaelHardy The reason lies in the fact that a complex number has the polar form r(cos x+ i sin x). So if the sin x is 0 ( which happens at 0 or $\pi$), then the number is real. $\endgroup$
    – TheLeogend
    Apr 27 '18 at 15:17
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Lemma 1: Mappings of the form $z\mapsto w = \dfrac{az+b}{cz+d}$ preserve cross-ratios. I.e., if $w_k=f(z_k)$ for $k=1,2,3,4$; then

$$ \text{cross-ratio}=\frac{z_1-z_4}{z_1-z_2} \times \frac{z_2-z_3}{z_4-z_3}=\frac{w_1-w_4}{w_1-w_2} \times \frac{w_2-w_3}{w_4-w_3}. $$

Lemma 2: Every circle and every straight line can be mapped onto $\mathbb R\cup\{\infty\}$ by a mapping of the form considered in Lemma 1. (Here we consider every line to contain the point $\infty$, which may be mapped either to $\infty$ or to some (finite) real number.)

Lemma 3: The inverse of a mapping of the form considered in Lemma 1 is another mapping of that form; consequently every circle is also the image of $\mathbb R\cup\{\infty\}$ under such a mapping.

So see if you can prove the three lemmas and then use them to get your result.

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  • $\begingroup$ Mobius transformations moment $\endgroup$
    – Buraian
    Aug 22 at 11:59
  • $\begingroup$ : Every circle and every straight line can be mapped onto R∪{∞} by a mapping of the form considered in Lemma 1. Can you explain this a bit more? $\endgroup$
    – Buraian
    Aug 22 at 12:01
  • $\begingroup$ @Buraian : For example, consider the circle of radius $5$ centered at $10+12i = 10 + 12\sqrt{-1}.$ One of the points on that circle is $13+8i,$ since the distance from $10+12i$ to $13+8i$ is $5.$ Another point on that circle is $10+17i$ and another is $5+12i.$ Now suppose I want to map $13+8i$ to $\infty$ and $10+17i$ to $0$ and $5+12i$ to $1.$ Then there are some complex numbers $a,b,c,d$ such that the mapping $z\mapsto\dfrac{az+b}{cz+d}$ does exactly that, so that $\dfrac{a(13+8i)+b}{c(13+8i)+d} = \infty$ and $\dfrac{a(10+17i)+b}{c(10+17i)+d} = 0$ and$\,\ldots\qquad$ $\endgroup$ Aug 22 at 19:34
  • $\begingroup$ $\ldots\,$and $\dfrac{a(5+12i) +b}{c(12+5i)+d} = 1,$ and moreover every point on that circle is mapped to some point in $\mathbb R\cup\{\infty\}$ and every point not on that circle is mapped to some complex number that is not in $\mathbb R\cup\{\infty\}. \qquad$ $\endgroup$ Aug 22 at 19:35
  • $\begingroup$ I think I get what you are saying but R U{infty} is liek the real line right? So what is the significance of collapsing those sets onto the real line? $\endgroup$
    – Buraian
    Aug 23 at 13:46

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