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

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SUMMARY

The discussion focuses on finding a Möbius transformation that maps the real line to the unit circle, specifically using the function f(z) = (z - i)/(z + i). Participants confirm that this transformation is valid and discuss the properties of Möbius transformations, emphasizing the importance of checking three key points: f(∞) = 1, f(0) = -1, and f(1) = -i. The conversation highlights the necessity of understanding inversion, translation, rotation, and dilation to effectively manipulate regions in complex analysis.

PREREQUISITES
  • Understanding of Möbius transformations
  • Familiarity with complex numbers and their properties
  • Knowledge of basic calculus, particularly limits
  • Concepts of geometric transformations: inversion, translation, rotation, dilation
NEXT STEPS
  • Study the properties of Möbius transformations in detail
  • Learn how to derive transformations using specific points on the complex plane
  • Explore the geometric interpretations of inversion and rotation in complex analysis
  • Practice mapping various regions using different Möbius transformations
USEFUL FOR

Mathematicians, students of complex analysis, and anyone interested in geometric transformations in the complex plane will benefit from this discussion.

Amer
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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
 

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