Let $SL_2(\mathbb{R})\curvearrowright \mathbb{C}\cup\{\infty\}$ where $Az=\frac{az+b}{cz+d}$
Show that if $z=x+iy$ with $y>0$ (has positive imaginary part) then $Az$ does too.
Then, considering this same action, show that the orbit of 1 is $ R \cup\infty$
Progress So far I have tried plugging it in, attempting to work around with some clever "1's" (e.g. $i/i$) to change the form. With $SL_2(\mathbb{R})$ we know that $ad-bc=1$ so I've been attempting to see if that has anything to do with the answer as well.
We have: $$ \frac{a(x+iy)+b}{c(x+iy)+d} = \frac{(ax+b)+iay}{(cx+d)+icy} $$ Multiplying by $1$, $$ \frac{(ax+b)(cx+d-icy)+iay(cx+d-icy)}{(cx+d)^2+(cy)^2} $$ The denominator is clearly real, and the imaginary part of the numerator is $$ (ax+b)(-cy)+ay(cx+d) = y(ad-bc)=y, $$ which is positive if and only if $y$ is.
For the second part, it is clear that the orbit of $1$ is contained in $\mathbb{R} \cup \{\infty\}$, since the coefficients of the transformation are real.
Next, it is also clear that $$ \begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix} $$ is in $SL_2(\mathbb{R})$; this is equivalent to $1/(1-z)$, which clearly maps $1$ to $\infty$. Finally, for any real number $\alpha$, the matrix $$ \begin{pmatrix} 1 & \alpha-1 \\ 0 & 1 \end{pmatrix} $$ is also in $SL_2(\mathbb{R})$, and corresponds to the translation $z \mapsto z+(\alpha-1)$, which obviously maps $1$ to $\alpha$. Hence the orbit is the whole of the $\mathbb{R} \cup \{\infty\}$.