# Thread: Complex variables: Imaginary part of the Mobius transform

1. ## Complex variables: Imaginary part of the Mobius transform

If $L(z)=\frac {az+b}{cz+d}$ where $a, b, c, d \in \mathbb{R}$ and $ad-bc=1$
Show that:
$Im(L(z))=\frac {Im(z)}{|cz+d|^2}$ where $Im(z)=\frac {z-z^{*}}{2i}$.

I think that $Im(L(z))=\frac {L(z)-L(z)^{*}}{2i}$ can be written as the above statement but I am stuck on how to get the conjugate of L(z). I am a little rusty with complex variables. Thanks for your help.

2. Originally Posted by ordinalhigh
If $L(z)=\frac {az+b}{cz+d}$ where $a, b, c, d \in \mathbb{R}$ and $ad-bc=1$
Show that:
$Im(L(z))=\frac {Im(z)}{|cz+d|^2}$ where $Im(z)=\frac {z-z^{*}}{2i}$.

I think that $Im(L(z))=\frac {L(z)-L(z)^{*}}{2i}$ can be written as the above statement but I am stuck on how to get the conjugate of L(z). I am a little rusty with complex variables. Thanks for your help.

$L(z)=\frac{az+b}{cz+d}\cdot\frac{\overline{cz+d}}{ \overline{cz+d}}=\frac{(az+b)(c\bar{z}+d)}{|cz+d|^ 2}$ $=\frac{ab|z|^2+adz+bc\bar{z}+bd}{|cz+d|^2}$ .

But $adz+bc\bar{z}=z+2Re(z)bc=Re(z)+Im(z)\,i+2Re(z)bc$ , since $ad=1+bc$ , so $L(z)=\frac{ab|z|^2+Re(z)+2Re(z)bc+bc}{|cz+d|^2}+\f rac{Im(z)}{|cz+d|^2}\,i$ , and you're done.

BTW, the formula $Im(z)=\frac{z-\bar{z}}{2i}$ is pretty weird...it's true, but too fancy, imo.

Tonio