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Entire functions and polynomials

  1. Mar 18, 2009 #1


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    Hi: I am trying to show:

    If f is analytic in C (i.e., f is entire) and : |z|>1 implies |f(z)|>1.
    Prove that f(z) is a polynomial.

    I have tried using the fact that f(z)=Suma_nz^n (Taylor series) valid in the whole of C,

    and derive a contradiction assuming |f(z)|>1 for |z|>1 . I checked that the "standard"

    non-poly. functions, e.g., e^z, sinz , were not counterexamples (they're not).

    Obviously, if this result were false for f(z) non-polynomial, then f(z) would need to

    have all its zeros( if any) inside of S^1, which is not impossible), and any poly.

    satisfying above condition should also have its zeros inside of S^1.

    I have considered using the fact that any f analytic has a singularity at oo (follows

    from Liouville's), but this only works if the singularity is an essential singularity; is

    this true, that any non-poly. analytic function has an essential singularity at oo ?.

    Any other ideasa?
  2. jcsd
  3. Mar 19, 2009 #2
    The important fact in this case, which follows from Casorati-Weierstrass, is that the singularity at [tex]\infty[/tex] is not an essential one (so it is a pole or removable). A pole at [tex]\infty[/tex] can be "removed" by adding a suitable polynomial, just like a pole in [tex]\mathbb{C}[/tex] can be removed by substracting the principal part. The result then follows from Liouville.

    btw, the only meromorphic functions [tex]\bar{\mathbb{C}} \to \bar{\mathbb{C}}[/tex] are rational functions.
  4. Mar 19, 2009 #3


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    Yes, thanks, I realized it right after posting it.

    Are you refering to the Mobius maps (az+b)/(cz+d) ?. I don't see how that
    relates here.
  5. Mar 19, 2009 #4
    By "rational map" I mean any quotient of polynomials (the Möbius transformations are the bijective ones). My statement implies that the answer to

    is yes.
  6. Mar 19, 2009 #5

    matt grime

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    Please don't assume this will lead to the right answer, but the first thing that occurs to me is that it might be easier to think of a different function. Define

    g(z) = 1/f(1/z)

    this is a function that maps the unit disk into itself.
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