MHB Find a Mobius Transformation to Map Real Line to Unit Circle

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Hey mobius transformation defined as
f(z) = \frac{az+b}{cz+d}
and ad \ne bc
it is a one to one function how i can find a mobius transformation that take the real line into the unit circle
I read it in the net
f(z) = \frac{z - i}{z+i}
and i checked it, it takes the real line into the unit circle, but there is a properties of the mobius transformation as the book said it is a combination of translation, inversion, rotation, dilation.

My question is how to find such map, or if we have the real line what first we have to do inversion,rotation,translation, ? to get the circle.

Thanks
 
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Well, Mobius transformations take lines or circles to lines or circles. All you have to do is check three points.

$$f( \infty)=1, \quad f(0) = -1, \quad f(1)= \frac{1-i}{1+i}= \frac{1-i}{1+i} \cdot \frac{1-i}{1-i}
= \frac{1-2i-1}{1+1} = -i.$$

Therefore, you seem to have done it. This method you can use to do most any of these transformations. Take a line or circle into an appropriate line or circle by making sure your $a,b,c,d$ are chosen correctly. Then, if you must map a region, pick a point in the origin region, and make sure it winds up in the destination region.
 
still not clear, how did you determine f(\infty) = 1 , f(0) = -1
what I was thinking about I said
\mid f(0) \mid = 1 \\ \frac{\mid b \mid}{\mid d\mid } = 1 \\ \mid b \mid = \mid d\mid
thats one

then I found that \mid a \mid = \mid c \mid by mapping infinity
after that guessing ?

what I was looking for is to master the inversion,translation, dilation, rotation
so I can imagine what i have to use to take a region to another
 
Amer said:
still not clear, how did you determine f(\infty) = 1 , f(0) = -1

Technically, the $f( \infty)$ is the limit:
$$ \lim_{x \to \infty}f(z)= \lim_{z \to \infty} \frac{z-i}{z+i}
= \lim_{z \to \infty} \frac{1-i/z}{1+i/z}=1.$$

I determined to check $0, 1, \infty$, because those are easy values to check on the real line.

what I was thinking about I said
\mid f(0) \mid = 1 \\ \frac{\mid b \mid}{\mid d\mid } = 1 \\ \mid b \mid = \mid d\mid
thats one

then I found that \mid a \mid = \mid c \mid by mapping infinity
after that guessing ?

what I was looking for is to master the inversion,translation, dilation, rotation
so I can imagine what i have to use to take a region to another

I've always just transformed the boundaries of regions, and made sure the inside of the region gets mapped correctly. You can check out inversions, translations, etc., here.
 
Thanks very :D
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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