Show that for every $z,w \in \mathbb{C}$ there exist $a,b,c,d \in \mathbb{R}$, s.t. $\frac{az+b}{cz+d}=w$ with $ad-bc=1$ and Im(z)>0 and Im(w)>0

Question

Show that for every $z,w \in \mathcal{H}$ there exist $a,b,c,d \in \mathbb{R}$, s.t. $\frac{az+b}{cz+d}=w$ with $ad-bc=1$.

This is just a small lemma in a bigger proof about a relation between bijections in the upper half-plane and $SL_2(\mathbb{R})$ regarding the mobius transform. The problem seems kinda obvious given the geometric interpretation of the mobius transform, but I can't figure out how to proof it "algebraically", meaning just shuffling the terms around / guessing explicit values for a,b,c,d to fulfill the equation...

Answer 1

Let $z=x+iy$, $w=u+iv$; note that $y,v>0$. Let's express $w$ through $z$, excluding the explicit mention of $i$ in the process:

$w = u+iv = u + v(z-x)/y = \frac{vz-vx+uy}{y} = \frac{az+b}{cz+d}$, where $a=v/\sqrt{vy}$, $d=y/\sqrt{vy}$, $c=0$, $b=(uy-vx)/\sqrt{vy}$.

Answer 2

If you calculate $Im(w)$ you get $$Im(w)=\frac{(ad-bc) Im(z)}{|cz+d|^2},$$ so you just need to choose $c, d$ so that $|cz+d|^2=\dfrac{Im(z)}{Im(w)}$ (for instance , $c=0$, $d^2=\dfrac{Im(z)}{Im(w)})$, then follow through to find $a, b$.