0
$\begingroup$

I have the notes on the proof but I cannot fully understand the proof.

Let $C$ be a hyperbolic straight line through $z_o\in \mathbb{D}$ and $z^*_o$ the point symmetric to $z_o$ wrt the unit circle $S^1$. Then we have $z^*_o$ is outside $S^1$.

Since C is perpendicular to $S^1$, we have $z^*_0$ lies on $C$.

I can't understand that how the last statement will happen. Thanks.

$\endgroup$
0
$\begingroup$

$z_o^*$ is the inverse of $z_o$. The inverse of $C$ is $C$ itself, since circle inversion is conformal and knowing the two points of intersection with $S^1$ already uniquely defines the single orthogonal circle through these two points. Inversion preserves incidence, so if $z_o$ lies on $C$, then $z_o^*$ lies on the image of $C$, i.e. on $C$ itself as well.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.