Let $w: \mathbb{C} -> \mathbb{C}$ be a Möbius transformation such that $w \circ w = id$. Let C be a line or circle, and $B = w(C)$ and $C \cap B = \{y, x\}$ ($x \ne y$). Prove that in this case $w(x) = y$ and $w(y) = x$
It's obvious that $w(x) \in \{x, y\}$. Can you give me a hint how to prove that $w(x) = y$? I know that $w$ must be $w(z) = \frac{az + b}{cz - a}$. Also C can't be a line if $w(x) = x$, because in this case $w$ must send it to line ($\frac{a}{c} \in C$). I don't know how to prove that it can't be a circle also
Without loss of generality assume that $x=0$ and $y = \infty$, otherwise consider a “conjugate” map $T^{-1} \circ w \circ T$ with a suitable Möbius transformation $T$.
Then $C$ and $B$ are both lines through the origin. If $w(0) = 0$ and $w(\infty) = \infty$ then $w$ is a rotation, and $w \circ w = id$ implies that $w(z) = z$ or $w(z) = -z$. In both cases, lines through the origin are mapped onto themselves, so that the intersection $C \cap B$ is not just two points.