I'm interested in the isometries of the hyperbolic plane, i.e. the mappings which leave the geometric properties of objects invariant. I'm working with the Poincare disc model. My lecture notes suggest that there, the isometries are given by elements of SU(1,1). This is the subgroup of Möbius transformations of the form $\phi_A(z) = \frac{az + b}{\bar b z + \bar a}$.
The rotation by an angle $\theta$ in the complex plane is given by the mapping $f(z) = e^{i\theta} z$. This mapping is $not$ in SU(1,1) because $a$ is non-zero but $\bar a$ is not. But it seems clear to me that it should be an isometry because it doesn't change lengths and relative angles.
How does this fit together?
Transformations of the form $\frac{az+b}{\bar b z + \bar a}$ are assumed to be normalized: $a \bar a + b \bar b = |a|^2 + |b|^2 = 1$. But $f(z)=e^{i\theta} z$ is not normalized. When you normalize it you get $$f(z) = \frac{e^{i\theta/2} z + 0}{0 + e^{-i\theta/2}} $$ and the universe is saved.