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Math Help - Complex Analysis-Isomorphisms

  1. #1
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    Complex Analysis-Isomorphisms

    Hey there,

    I'll be delighted to get some help in these two questions:

    1. Find an Isomorphism of D=C-[-1,1] on (z:Imz>0).

    2. Find an Isomorphism of (z:Rez>0,Imz>0,|z|>1) on (z:Imz>0) such as  i \to 0 ,  1 \to 1 , \infty \to \infty .


    Thanks a lot !
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  2. #2
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    By isomorphism, do you mean conformal mapping (functions that preserve angles)?
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  3. #3
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    The correct definition I've got is:
    Let R and S be two regions. If there exists an analytic,injective function in R that satisfies: f(R)=S then the two regions R,S are isomorphic...
    So I don't think it has to be conformal...

    About the second one:
    I know that if we take the composition of the functions z^2,\frac{1}{z} then we'll get a mapping from the mentioned region to the upper half plane. But this composition doesn't preserve the points as needed...

    Hope you'll be able to help me

    Thanks !
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  4. #4
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    So you don't care if the function is onto, then? That's interesting...

    I'd fiddle around with fractional linear transformations, or Mobius transformations, as they're also called. Find out which points you want to go to which points. The key there is that Mobius transformations preserve cross-ratios. And that's the information you can use to construct your transformation.
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  5. #5
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    Ofcourse it should be onto... In the definition I've written: f(R)=S...It means that the map should be onto...
    Unfortunately,I've already tried solving these two questions using the guidance you mentioned- but without any success... I can't figure out how to continue the second question , from the part I mentioned above...
    About the first one-I think that if we take the composition of the functions:
     \frac{1}{z} , iz, \sqrt(z) - we'll get the needed isomorphism...Am I right?

    Hope you'll be able to continue your guidance...

    Thanks !
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  6. #6
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    My bad. I saw " \to" where you had put equality. So you have to have analytic bijections from one region to another. Such functions are conformal, at least if their derivative is everywhere non-zero.

    Hmm. Initial thoughts: #2 you're guaranteed to be able to do, since the Reimann Mapping Theorem says you can go from the starting region to the open unit disk, as well as from the the open unit disk to the ending region. All the regions there are simply connected, nonempty, open, proper subsets of the complex plane.

    I'll have to get back to this another time.
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  7. #7
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    Intuitively, with #2, one approach would be to rotate the complex plane in a counter-clockwise direction about the point 1+0i through an angle \pi/2, and then to dilate all points towards the point 1+0i, such that the point 1-\sqrt{2}+0i maps to 1+0i.
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  8. #8
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    Your candidate for #1 won't work, because the point z=-2, which is in the desired domain, gets mapped to 0.5-0.5i, which is not in the desired range.
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  9. #9
    MHF Contributor Bruno J.'s Avatar
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    For #1, apply a Mobius transformation which maps -1 to 0 and 1 to \infty, while mapping the real line to itself. Then you have the plane cut along the half-line [0, \infty]. Then use a fixed branch of the square root to map this to the upper-half plane.

    By the way, any analytic mapping is automatically conformal at any point where the derivative does not vanish.
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