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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?

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# Entire functions and polynomials

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