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About LinkBacksHello! First time I ask for help, usually I'm the one giving it!
I need to find the image of the first quadrantunder the Mobius transformation
. For some reason I'm having trouble doing that. Please explain how you would do it. Thanks!
Hello! First time I ask for help, usually I'm the one giving it!
I need to find the image of the first quadrant
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
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?
Here is my (nice?) way to find the answer.Hello! First time I ask for help, usually I'm the one giving it!
I need to find the image of the first quadrant
Notice thatiff
is at the same distance from
and
, i.e. iff
Thus, the real axis is mapped to the unit circle.is mapped to
and
to
.
On the other hand, the positive imaginary half-axisis mapped to
because
.
A connexity argument implies that the first quadrant is either the top half unit disk (or the bottom half unit disk.
It remains to compute the image of, 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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