2

Let $z \in \mathbb{H}$, where $\mathbb{H}$ denotes the half plane $\mathbb{H}=\{z \in \mathbb{C}:Im(z)>0\}$. Let \begin{equation*} f(z)=\frac{az+b}{cz+d} \end{equation*} which is called a Mobius Transformation, and let $ad-bc>0$. I want to show that $Im(f(z))>0$.

Following this solution, I should use $Im(z)=\frac{z-\overline{z}}{2i}$. Applying this formula, I get \begin{align*} Im (f(z)) &=Im \left(\frac{az+b}{cz+d}\right) \\ &=\frac{\frac{az+b}{cz+d}-\overline{\left(\frac{az+b}{cz+d}\right)}}{2i} \end{align*} but I am not sure how to show that this equals $\frac{ad-bc}{c^2+d^2}$. Is there a nice and clean way to simplify $Im(f(z))$? Thank you!

Community
  • 1
EthanAlvaree
  • 3,412
  • http://math.stackexchange.com/questions/36542/real-and-imaginary-parts-of-the-m%C3%B6bius-transformation – tired Apr 29 '15 at 08:43
  • That looks really messy. I'm surprised how the solution I linked to concluded so easily that $\textrm{Im}(f(z)) = \frac{ad-bc}{c^2+d^2}$. – EthanAlvaree Apr 29 '15 at 08:45

2 Answers2

4

There is an error in the answer you linked. We cannot expect $Im (f(z))$ not to depend on $z$; otherwise, $f$ would map $\mathbb{H}$ to an horizontal line, which contradicts its analyticity.

I think the best way is to multiply by the conjugate of the denominator:

$$Im \left( \frac{az+b}{cz+d}\right) = Im \left( \frac{(az+b)(c \overline{z}+d)}{(cz+d)(c \overline{z}+d)}\right) = \frac{1}{|cz+d|^2} Im \left( acz\overline{z}+bc\overline{z}+adz+bd\right) = \frac{(ad-bc) Im(z)}{|cz+d|^2}.$$

D. Thomine
  • 10,870
1

$$\frac{az+b}{cz+d}=\frac{(az+b)(c\bar{z}+d)}{|cz+d|^2}$$

And one has

$$(az+b)(c\bar{z}+d)=acz\bar{z}+adz+bc\bar{z}+bd$$

$acz\bar{z}+bd$ is real and $\mathcal{I}(adz+bc\bar{z})=(ad-bc)\mathcal{I}(z)$

Putting the above together, one gets

$$\mathcal{I}\left(\frac{az+b}{cz+d}\right)=\frac{ad-bc}{|cz+d|^2}\mathcal{I}(z)$$

marwalix
  • 16,773