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.
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.