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Show that an analytic mapping is an open mapping

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data

    As in in title.

    2. Relevant equations

    Open mapping: maps open sets to open sets.

    3. The attempt at a solution

    Not sure.
     
  2. jcsd
  3. Mar 11, 2012 #2

    Dick

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    That's a pretty poor attempt. If you don't know how to prove it can you at least give us a few thoughts on why you think it might (or might not) be true?
     
  4. Mar 12, 2012 #3
    If a function f(z) is differentiable on a set G, then it is continuous on G. If it continuous, the image of every open set is an open set. Done(?).
     
  5. Mar 12, 2012 #4

    Dick

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    Nice that you are trying to think about it. But that's not true in the real numbers. f(x)=x^2 is continuous and differentiable, but it's not open. The proof is going to have to involve special properties of analytic functions.
     
  6. Mar 12, 2012 #5
    Anything to do with "Analytic continuation" that I see on Wikipedia?
     
  7. Mar 12, 2012 #6

    Dick

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    That's part of it. Rouche's theorem is probably the main ingredient.
     
  8. Mar 12, 2012 #7

    Bacle2

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    It makes sense to me this way:

    An open set in f(z) (say an open ball B(f(z),r), to simplify) is one in which every point in B(f(z),r) is in f(z), i.e., every point in B(z,r) is in the image of f(z). This makes every point in B(f(z),r) a solution to , or a zero of, a certain equation.
     
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