# Mobius transformation on the open upper half plane

Let $a, b, c, d$ be complex numbers with $ad - bc \ne 0$. Then $$f(z) = \frac{az + b}{cz + d}$$ is called the Mobius transformation.

If $H := \{ z \in \mathbb{C} : \Im(z) > 0 \}$ is the open upper half plane, show that any Mobius transformation from $H$ onto itself can be written with real coefficients $a, b, c, d$ with $ad - bc = 1$.

Can someone please show me how to do this problem?

I already showed that the Cayley transform $C(z) = \frac{z-i}{z+i}$ is a biholomorphic map (analytic, injective, and onto) from $H$ onto the unit disk $B(0,1)$. I also showed that any biholomorphic map from $H$ onto $H$ is a Mobius transformation. How can I go further and use this?

There is also another hint telling me that such transformation maps the real axis into the real axis, but I don't see how that would help me.

Thank you.

Here is a sketch of a possible proof.

Suppose $f$ is a Möbius transformation taking $\mathbb{R}_{\infty}$ to $\mathbb{R}_{\infty}$. Then I claim the coefficients $a,b,c,d$ of $f$ are all real. suppose $a_1,a_2,a_3\in\mathbb{R_{\infty}}$ are three distinct points that are mapped to $b_1,b_2,b_3\in\mathbb{R_{\infty}}$. Then by invariance of cross-ratio we have for all $z\in H$ $$\frac{z − a_1}{a_3 − a_1}:\frac{z − a_2}{a_3 − a_1}=\frac{f(z) − b_1}{b_3 − b_1}:\frac{f(z) − b_2}{b_3 − b_2}$$ with the appropriate conventions on $\infty$. It is easy to see this implies that $f$ can be written in the desired form.

Next, we have $$\Im(f(z))=\frac{ad-bc}{|cz+d|^2}\cdot\Im(z).$$ This implies $ad-bc>0$, which is what we wanted.

• Thank you for your answer. I used a different approach but were able to show that $ad - bc > 0$, but then how can I show it is, in fact, equal to $1$?
– dh16
Oct 22, 2016 at 5:14
• Since $ad-bc$ is nonzero you can just normalize by dividing all the coefficients by $\sqrt{ad-bd}$
– ArtW
Oct 22, 2016 at 13:03

$w=\frac{az+b}{cz+d}$ gives $z=\frac{b-wd}{cw-a}$. The upper half plane is $y\geq 0$ or $z-\overline z\geq 0$

$\implies \frac{b-wd}{cw-a}-\frac{b-\overline wd}{c\overline w-a}\geq 0$

$\implies (ad-bc)(w-\overline w)\geq 0$ (On simplifying)

You need $ad-bc=1$ so that $w-\overline w\geq 0$ is the upper half $w$-plane.

• This is invalid - you haven't shown that $a,b,c,d$ are real
– ArtW
Oct 25, 2016 at 9:21
• $z-\bar{z}$ is not even real necessarily, how can you write something like $z-\bar{z}\geq 0$? Feb 21, 2019 at 20:17