Does a Möbius transformation map the diameter points of a circle to diameter points?

Question

The Möbius transformation given by $(az+b)/(cz+d)$ maps a circle to circle or a line. Say we are given the unit circle centered at the origin and that the pole $-d/c$ does not touch the boundary of the circle. Under these assumptions, the Möbius transformation maps the unit circle to another circle. From all the examples I have seen, it looks like the points $-1$, $1$ which are diameter points on the unit circle gets mapped to another pair of diameter points on the image circle. Same for $-i$ and $i$. I was wondering if this is always the case?

Answer 1

Your conjecture is not correct. A simple counterexample is $T(z) = 1/(z+2)$, which maps the circle $|z| = 1$ onto the circle $|z-2/3|=1/3$.

But the diametral points $\pm i$ on the unit circle are mapped to $\frac 25 \mp \frac i 5$, which are not diametral on the image circle.

One can show that if $T(z) = \frac{az+b}{cz+d}$ has the property that diametral points on the unit circle (or any circle with center at the origin) are mapped to diametral points on the image circle, then $c=0$ or $d=0$, i.e. $T$ is of the form $T(z) = \alpha z+\beta $ or $T(z) = \alpha+\beta/z$.

Answer 2

The automorphism group of the unit circle consists of all elements of the form $e^{i\theta}\frac{\alpha-z}{1-\bar{\alpha}z}$ where $\alpha$ is an element of the open unit disc.Note that if you take a general Mobius transformation The unit disc may not always map to the unit disc. Now for the automorphisms as I've written down just take $z=e^{i\phi}$ and you'd see that it maps to a point on the unit disc.