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?

ANSWER: 
If $A \in SL(2,\mathbb R)$, then
\begin{align*}
 A\cdot i&=\frac{ai+b}{ci+d}\\
 &=\frac{ai+b}{ci+d}\cdot \frac{(-ci+d)}{(-ci+d)}\\
 &=\frac{ac+bd+i(ad-bc)}{c^2+d^2}.
\end{align*}
We're in $SL(2, \mathbb R$), so $ad-bc=1$, and
$$\frac{ac+bd+i(ad-bc)}{c^2+d^2}=\frac{ac+bd}{c^2+d^2}+i\frac{1}{c^2+d^2}.$$
If $c=0$ then $a \ne 0$, $d\ne 0$ and $ad-bc=1$ becomes $ad=1$. So, $d=1/a$. Also,
$$\frac{ac+bd}{c^2+d^2}+i\frac{1}{c^2+d^2}=\frac{bd}{d^2}+i\frac{1}{d^2}=\frac{b}{d}+i\frac{1}{d^2}=ab+a^2i.$$
Since $z \in \mathbb H$, then $z=x+iy$ where $y>0$. Now $a^2=y \implies a=\sqrt y$. Therefore, $ab=x \implies b=\frac{x}{\sqrt y}$.
So, $\begin{pmatrix} \sqrt y & \frac{x}{\sqrt y} \\ 0 & \frac{1}{\sqrt y} \end{pmatrix} \in SL(2,\mathbb R)$ and 
$$\begin{pmatrix} \sqrt y & \frac{x}{\sqrt y} \\ 0 & \frac{1}{\sqrt y} \end{pmatrix} \cdot i =z$$
implies transitive.
