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I'm reading Conway's complex functions of one variable, and in chapter 3 he goes over Cross-Ratios. He defines the cross ratio to be $(z,z_1,z_2,z_3)=\frac{(z-z_3)(z_2-z_4)}{(z-z_4)(z_2-z_3)}$, where notice that this sends $z_2$ to $0$, $z_3$ to $1$ and $z_4$ to $\infty$. The interesting fact he discusses is that the cross ratio maps points of the circle to the real line. He goes on to say that all Mobius Transformations can map Circles onto Circles, by simply composing the cross ratio of points on a circle with the inverse of the cross ratio of points on another circle.

However, Mathematica gives me the inverse of $\frac{(z-z_3)(z_2-z_4)}{(z-z_4)(z_2-z_3)}$ to be $\frac{-z_2 z_3+z z_2 z_4+z_3 z_4-z z_3 z_4}{-z_2+z z_2-zz_3+z_4}$ and if I plug in any real number to this, it will not be sent to a point on a circle because the y-value (the imaginary part) will be zero. What am I doing wrong? Mathematica could be giving me the wrong inverse, but I don't get anything better by hand.

Any help is much appreciated.

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  • $\begingroup$ Notice that you can write $z_1$ rather than $z1$. $\qquad$ $\endgroup$ – Michael Hardy Mar 12 '16 at 3:24
  • $\begingroup$ @MichaelHardy Thx, took care of it. $\endgroup$ – Mike Mar 12 '16 at 3:25
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Hmm. You say that $$ \frac{-z_2 z_3+z z_2 z_4+z_3 z_4-z z_3 z_4}{-z_2+z z_2-zz_3+z_4} $$

will not give you points of a circle because the imaginary part will be zero. But rememebr that $z_2, z_3, z_4$ will all be not REAL NUMBERS, but points on the circle you're mapping to. So picking them to be, say $z_2 = 1, z_3 = i, z_4 = -i$, I get

$$ \frac{-1 i+z (-i)+1-z 1}{-1+z -zi-i} = \frac{-i-iz+1-z}{-1+z -zi-i} $$ Plugging in, for instance, $z = 0$ gives $$ -\frac{1-i}{1+i} $$ which most ditinctly is NOT a real number.

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  • $\begingroup$ Ahh, this made everything fall into place, thx a bunch! $\endgroup$ – Mike Mar 12 '16 at 5:54
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This should all be done in $\mathbb C\cup\{\infty\}$ rather than just $\mathbb C$, and circles that pass through the point $\infty$ are straight lines.

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  • $\begingroup$ Yes, that's an underlying assumption. $\endgroup$ – Mike Mar 12 '16 at 3:27
  • $\begingroup$ So you're doing nothing wrong: if four points on a circle are mapped to four points on a straight line, then they're mapped to four points on a circle because that line is a "circle". $\qquad$ $\endgroup$ – Michael Hardy Mar 12 '16 at 3:28
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    $\begingroup$ I see what you mean, technically I have mapped on a circle. But there's this graphic in my professor's notes where the composition of a cross ratio with the inverse of another cross ratio will map points of one circle onto points of another. In the graphic the circles are aside from each other with a line segment between them, where the map Circle -> Line -> Circle is drawn, and I'm trying to do this in mathematica. I believe my error is in the fact that I am not composing the functions. I will try to post a picture if I can. Thx for ur help. $\endgroup$ – Mike Mar 12 '16 at 3:32

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