How to show the action of $SL_2(\mathbb R)$ on the complex upper half plane is transitive? I've seen other answers here and here but I still am not understanding how to show that this action is transitive. I want to show that for all $z \in \mathbb H$ there exists a matrix $g \in SL_2(\mathbb R)$ such that $gi=z$.
If I start with $\frac{ai+b}{ci+d}=z=x+iy$, then I get
$$\frac{ai+b}{ci+d}=z \implies ai+b=z(ci+d) \implies i=\frac{dz-b}{a-cz}$$ 
I don't know what to do with this or what this tells me.
If I do $\frac{az+b}{cz+d}=i$ I don't get anywhere either.
How can I go about this?

Here's a hint. You can write 
$$ z= a+bi = \frac{b\cdot i + a}{0\cdot i +1} ,$$
which is the transformation corresponding to the matrix
$$\begin{bmatrix} b & a \\ 0 & 1 \end{bmatrix} $$
evaluated on the point $i$ in $\mathbb{H}$. This matrix may not be in $SL_2 \mathbb{R} $, but you can easily fix that. 
