# Moebius transformation in upper half plane model

I am studying Hyperbolic Geometry on my own. More especially, I am studying the Moebius transformation of the upper half plane model in hyperbolic geometry. But I got stuck studying it. I am looking for a solution of this problem. The problem is as follows:

Find the Moebius transformation that rotates the hyperbolic plane about $$i$$ through an angle of $$\frac { \pi}{4}$$.

• The stabilizer of $i$ is (the image in the group of Möbius transformations of) $SO_2(\Bbb{R})$ and the derivative of $f(z)=(az+b)/(cz+d),ad-bc=1$ is $1/(cz+d)^2$. – reuns Dec 18 '20 at 19:59
The stabilizer of $$i$$ in the upper half-plane consists of elliptic linear-fractional transformations $$g_A(z)=\frac{az+b}{cz+d}$$ corresponding to matrices $$A=\left[\begin{array}{cc} a & b\\ c & d \end{array}\right]= \left[\begin{array}{cc} \cos(\phi) & \sin(\phi)\\ -\sin(\phi) & \cos(\phi) \end{array}\right]$$ (which makes it easy to remember: Euclidean rotations correspond to hyperbolic rotations). The angle of rotation $$\phi$$ of the matrix $$A$$ acting on $${\mathbb R}^2$$ correspond to the angle of rotation $$2\phi$$ for the Moebius transformation $$g_A$$. (This is again easy to remember since for $$A=-I$$, rotation by $$\pi$$, $$g_A=id_{\mathbb H^2}$$, rotation by $$2\pi$$.) The rest you can figure out on your own.