Entire functions and polynomials

[WWGD]

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?

 

[yyat]

The important fact in this case, which follows from Casorati-Weierstrass, is that the singularity at $$\infty$$ is not an essential one (so it is a pole or removable). A pole at $$\infty$$ can be "removed" by adding a suitable polynomial, just like a pole in $$\mathbb{C}$$ can be removed by substracting the principal part. The result then follows from Liouville.

btw, the only meromorphic functions $$\bar{\mathbb{C}} \to \bar{\mathbb{C}}$$ are rational functions.

 

[WWGD]

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

Are you referring to the Mobius maps (az+b)/(cz+d) ?. I don't see how that
relates here.

 

[yyat]

Are you referring to the Mobius maps (az+b)/(cz+d) ?. I don't see how that
relates here.

By "rational map" I mean any quotient of polynomials (the Möbius transformations are the bijective ones). My statement implies that the answer to

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

is yes.

 

[matt grime]

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.