I was trying to apply Poincare's Polygon theorem, for that I had to give a pairing of sides, i.e., to have an isometry of the hyperbolic plane that will take a side of a polygon to another side (of the same length). So my question is that -
Given any two hyperbolic line segments of equal hyperbolic length, is there an isometry of the hyperbolic plane that sends one of the line segments to the other?
Its is easy to see that there will be mobius transformations that will take one line segment to the other, but I couldn't see why will there be one such transformation that will also fix the upper half-plane (i.e. will be an isometry of the hyperbolic plane).
A proof of existence will be enough for me.
A Möbius transformation is uniquely defined by 3 points and their images. If you have $z_1\mapsto z_1'$ and $z_2\mapsto z_2'$ mapping the endpoints of the line segments, then add $\overline{z_1}\mapsto \overline{z_1'}$, i.e. map the complex conjugates for one point and its image. If the segments $(z_1,z_2)$ and $(z_1',z_2')$ are indeed of equal length, then the map defined by these three points will also map $\overline{z_2}\mapsto \overline{z_2'}$ and it will have a representation using real coefficients only, so that it preserves the real axis.
If some other reader wants the same for the Poincaré disk, use inversion in the unit circle instead of complex conjugate i.e. reflection in the real axis. The idea is that in a way, the upper and the lower half plane in the half plane model, or the inside and the outside (including the point at infinity) of the disk in the disk model, are algebraically pretty much equivalent. It makes sense to think of a hyperbolic point in the half plane model not as a single point in the upper half plane, but as a pair of points reflected in the real axis. Using this helps adding additional constraints for the Möbius transformation.