1. Image of Mobius transformation

Hello! First time I ask for help, usually I'm the one giving it!

I need to find the image of the first quadrant $x>0, y>0$ under the Mobius transformation $z \mapsto \frac{z-i}{z+i}$. For some reason I'm having trouble doing that. Please explain how you would do it. Thanks!

2. Originally Posted by Bruno J.
Hello! First time I ask for help, usually I'm the one giving it!

I need to find the image of the first quadrant $x>0, y>0$ under the Mobius transformation $z \mapsto \frac{z-i}{z+i}$. For some reason I'm having trouble doing that. Please explain how you would do it. Thanks!

Let z = x + iy , x, y > 0, so z --> [x + (y-1)i]/[x + (y+1)i] = [(x^2 + y^2 -1) + xyi]/[x^2 + y^2 + 2y + 1]

It's easy to check that both the real and the inaginary parts of the above expression are positive ==> it looks like the in image is the positive cuadrant of the unit disk (since the Cayley transform z --> (z-i)/(z+i) maps conformally the WHOLE half plane Im(z) > 0 onto the unit disk.

Tonio

Hmmm...on a second thought it may well be that x^2 + y^2 - 1 is NOT positive: when we get z from the positive cuadrant of the unit disk! So perhaps we get the whole upper half unit disk?

3. Originally Posted by Bruno J.
Hello! First time I ask for help, usually I'm the one giving it!

I need to find the image of the first quadrant $x>0, y>0$ under the Mobius transformation $z \mapsto \frac{z-i}{z+i}$. For some reason I'm having trouble doing that. Please explain how you would do it. Thanks!
Here is my (nice?) way to find the answer.

Notice that $\frac{|z-i|}{|z+i|}=1$ iff $z$ is at the same distance from $i$ and $-i$, i.e. iff $z$ is real.

Thus, the real axis is mapped to the unit circle. $0$ is mapped to $-1$ and $\infty$ to $1$.

On the other hand, the positive imaginary half-axis $\mathbb{R}_+ i$ is mapped to $(-1,1]$ because $\frac{\lambda i - i}{\lambda i +i}=\frac{\lambda-1}{\lambda +1}=1-\frac{2}{\lambda+1}$.

A connexity argument implies that the first quadrant is either the top half unit disk ( $\{|z|< 1, \Im(z)>0\}$ or the bottom half unit disk.

It remains to compute the image of $1$ to decide: $\frac{1-i}{1+i}=-i$, so it's the bottom half of the unit disk.

This argument probably requires a few details (left to the reader) to be turned to a full proof, but that's how I would find the image.