How to show that $SL_2(\Bbb R)/SO_2(\Bbb R) \cong \Bbb H$?

Question

I've already shown that $SL_2(\Bbb R)$ acts on $\Bbb H$ on the left : $$SL_2(\Bbb R) \times \Bbb H \rightarrow \Bbb H$$ $$\gamma*z \mapsto \frac{az + b}{cz + d}$$ where $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\Bbb R)$ and that the stabilizer of $i$ is $SO_2(\Bbb R)$

Therefore, there is a bijection between $SL_2(\Bbb R)/SO_2(\Bbb R)$ and $\mathcal{O}_i$, where $\mathcal{O}_i$ is the orbit of $i$. So I think I have to show that $\mathcal{O}_i = \Bbb H$ but so far, I haven't succeed. We have that $\mathcal{O}_i \subseteq \Bbb H$ but I didn't manage to show the other inclusion.

Answer 1

Yes exactly ! $$x+iy=\frac{ai+b}{ci+d}\implies x+iy=\frac{(ai+b)(d-ci)}{c^2+d^2} =\frac{ac+bd}{c^2+d^2}+\frac{i}{c^2+d^2}.$$ Now, take $c=0$, $a=\sqrt y$, $d=\frac{1}{\sqrt y}$ and $b=\frac{x}{\sqrt y}$ and you'll have the result.

Answer 2

Hint: Any element of $\mathbb{H}$ can be written (suggestively) as $a(i)+b$. Do you know how to scale and translate points in $\mathbb{H}$ via Mobius transformations?