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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 $\frac { \pi}{4}$.

Please help me. Thanking in advanced.

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    $\begingroup$ Do you know an isometry from the upper half-plane model to the disk model? Conjugate a rotation of the disk by this isometry. $\endgroup$ – Robert Bell Dec 18 '20 at 19:16
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    $\begingroup$ 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 $1/(cz+d)^2$. $\endgroup$ – reuns Dec 18 '20 at 19:59
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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.

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