Today I encountered the transformation $f(z) = \frac{z}{z-1}$. It has the following property:
- As the point $z$ makes a counter-clockwise revolution around the unit circle beginning at $1$, the point $f(z)$ parametrizes the line with real part $1/2$, travelling upward.
This is not a difficult thing to verify with a bit of fiddling around. I used this fact to show that a certain operator had a gap in its spectrum. But, if I hadn't already expected this was going to happen for another reason, I would never have noticed the above property of this transformation.
Another example: the Cayley transform $f(z) = \frac{z -i}{z+i}$.
- As $z$ travels the real line from left to right, $f(z)$ undergoes a counterclockwise revolution of the circle, beginning and ending at $1$.
Again, this is easy to check. I'm aware of the phenomenon because the transform is relatively famous, but, if I had encountered this transformation "in the wild", I would never have expected the above property. Which leads to my question:
Is there some more systematic way to see where lines and circles go under fractional linear transformations $f(z) = \frac{az+b}{cz+d}$? I can verify expected facts by doing things like computing the norm, or real and imaginary parts of the right hand side, but I cannot help but feel like there is some gap in my knowledge that prevents me from really understanding what is going on.
Best Answer
Since $$\frac{az+b}{cz+d} = \frac{a}{c} + \frac{bc-ad}{c}\frac{1}{cz+d} \tag1$$ every Möbius map is a composition of linear maps and $z\mapsto z^{-1}$. Linear maps are simple enough: lines clearly go to lines and circles to circles (with center of circle mapped to the center of the new circle). It remains to figure out the map $z\mapsto z^{-1}$. Geometrically it is easier to work with inversion map $z\mapsto \bar z^{-1}$ and compose it with conjugation (in either order). The reason is that in polar coordinates $z\mapsto \bar z^{-1}$ is written as $$re^{i\theta}\to r^{-1}e^{i\theta}\tag2$$ That is, the polar angle remains but the distance to the origin is inverted.
Inversion of a circle centered at $0$ is another circle centered at $0$. Inversion of a circle centered at $a\ne 0$ can be computed by focusing on the line $L$ through $0$ and $a$. Let $p$ and $q$ be the points where $L$ crosses the circle. Their images will be on the same line. Most importantly, the image of the circle will still be symmetric about $L$, since (2) preserves such symmetry. Thus, $\bar p^{-1}$ and $\bar q^{-1}$ are endpoints of a diameter. Thus the image circle has diameter $|p^{-1}-q^{-1}|$ and center $\frac12(\bar p^{-1}+\bar q^{-1})$. That is, ...
unless one of $p$ and $q$ is $0$; say $p=0$. Then we get a line passing through $\bar q^{-1}$. Since it's symmetric about $L$, it must be perpendicular to $L$. This determines the line.