# Inversion as hyperbolic isometry (Poincaré disk model).

I'm beginning to study hyperbolic geometry in the Poincaré disk model, which is described as $$D = \{z \in \mathbb{C} : |z|<1 \},$$ and I need to show that inversion about a circle orthogonal to $$\partial D$$ is a reflection about the corresponding hyperbolic line.

For this, I started by considering a circle $$C$$ orthogonal to $$\partial D$$, then take any point $$P$$ not on $$C$$, and call $$P'$$ its image under inversion with respect to $$C$$. What I want to show is that for any point $$Q$$ on the arc of $$C$$ that corresponds to the hyperbolic line in $$D$$, the hyperbolic distance from $$Q$$ to $$P$$ is equal to the hyperbolic distance from $$Q$$ to $$P'$$. This would imply that this hyperbolic line is in fact the perpendicular bisector of the segment $$\overline{PP'}$$, which would finish the proof. Here's a diagram that illustrates what notation I've used.

I've already shown that inversion about $$C$$ maps $$Q_1$$ to $$Q_2$$, $$P_1$$ to $$P_2$$, $$Q$$ to itself and $$P$$ to $$P'$$. We've also shown during class that isometries in the Poincaré disk model are of the form

$$f(z) = e^{i \theta} \frac{z - z_0}{-\overline{z_0}z + 1}.$$

My plan was then to try and explicitly write inversion with respect to $$C$$ as an isometry of the form above, which would imply that the hyperbolic distances from $$Q$$ to $$P$$ and from $$Q$$ to $$P'$$ are indeed equal. However, I've had no luck in doing so, and so I wanted to ask for help in this. How do I write an inversion about a circle orthogonal to $$D$$, so that it has the form of an isometry in Poincaré disk model? Would this be necessary for the approach I'm taking, or would it be easier to try and prove (by some other method) that said inversion is an isometry in this hyperbolic model?

Thanks in advance for the help/suggestions!