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    MHF Contributor Bruno J.'s Avatar
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    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!
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    Quote Originally Posted by Bruno J. View Post
    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?
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    Quote Originally Posted by Bruno J. View Post
    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.
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