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Complex analysis prove

  1. Dec 1, 2011 #1
    Q:Let f be entire and suppose that Im f(z) ≥ M for all z. Prove that f must be a constant function.

    A: i suppose M is a constant. So Im f(z) is a constant which means the function is a constant.

    Am i doing this right ?
    but i dont think there will be such a stupid question in my assignment ....
    Or M is not even a constant? M can be anything?

    THanks
     
  2. jcsd
  3. Dec 1, 2011 #2

    micromass

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    M is a constant alright. But why is Im f(z) a constant?? It depends on z, so I don't find it trivial that it is a constant...

    Mod note: I moved the thread to calculus and beyond.
     
  4. Dec 1, 2011 #3
    Suppose I have the function

    [tex]f(z)=e^{ig(z)}[/tex]

    and I know that [itex]|f|<100[/itex] everywhere. Then that must mean f is a constant right? But for [itex]e^{ig}[/itex] to be a constant, that must mean g(z) is a constant. So, suppose I have the expression:

    [tex]e^{ig}=e^{i(u+iv)}[/tex]

    and [itex]|e^{ig}|[/itex] is always bounded, say less than 100. Then what must be the constraints on the function u+iv for that to work?
     
  5. Dec 1, 2011 #4
    am i doing this prove by liouville's theorem?
    can i do it like this:
    suppose f = 1/e^f(z)=1/e^ Im(f) < 1/e^M
    so it is bounded
    by liouville's theorem, it is constant
     
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