How Do Mobius Transformations Map Complex Domains?

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The discussion focuses on finding a Mobius transformation that maps the points 1, i, and 0 to ∞, 0, and 1, respectively. The transformation identified is (z-i)/i(z-1). The user is attempting to determine the image of the domain defined by { z : 0 < x < t} under this transformation but is struggling to visualize the resulting shape. Another participant confirms the mapping of specific points and suggests that the transformation should produce a line where the real part is equal to 1. The conversation highlights the complexities involved in visualizing conformal mappings through Mobius transformations.
sgcbell2
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Hi, I am currently working on problem to do with conformal mappings and Mobius Transformations.

My problem is:
Find the Mobius Transformation which carries the points 1, i, 0 to the points ∞, 0, 1 (in precisely this order). Find the image of the domain { z : 0 < x < t} under this Mobius Transformation, where t > 0 is some fixed positive real number.

I have found this Mobius transformation to be (z-i)/i(z-1).

I thought that in order to find the image of the domain given above, I should input different values along the boundaries x=0 and x=t into this mobius transformations to see where they are mapped to. I have done this but I'm finding it hard to see the shape of the image.

Maybe I am doing something wrong?
 
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Hello sg and welcome to PF.
I'm not really qualified re conformal mapping, but from what I know from complex numbers, I see the function you found transforms 1+0i to ##\infty##, 0+i to 0 and 0 + 0i to 1+0i.

I see x+0i is mapped to 1/(1-x) + ix/(1-x) which satisfies Re - I am = 1 , so I would expect a line I am = Re - 1
 
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That makes a lot of sense! Thank you
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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