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Find all entire functions $f$ such that for all $z\in \mathbb{C}$, $|f(z)|\ge \frac{1}{|z|+1}$

This is one of the past qualifying exams that I was working on and I think that I have to find the function that involved with $f$ that is bounded and use Louiville's theorem to say that the function that is found is constant and conclude something about $f$. I can only think of using $1/f$ so that $\frac{1}{|f(z)|} \le |z|+1$ but $|z|+1$ is not really bounded so I would like to ask you for some hint or idea.

Any hint/ idea would be appreciated.

Thank you in advance.

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up vote 3 down vote accepted

Hint: As you suggest, you can consider $g = 1/f$, which is itself an entire function (why?). Use the estimate $|g(z)|\leq 1 + |z|$ and the Cauchy estimates to show that $g$ is a (nonvanishing) polynomial of degree $\leq 1$. What can you conclude?

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does that mean , the required function is a rational function having pole or order at most 1? Is that how you conclude? or am I missing something? – Deepak Dec 31 '12 at 1:53
@Deepak: a nonvanishing polynomial is constant. – froggie Dec 31 '12 at 2:05

Suppose $f$ is not constant. As an entire non-constant function it must have some sort of singularity at infinity. It cannot be a pole (because then it would have a zero somewhere, cf the winding-number proof of the FTA), so it must be an essential singularity. Then $zf(z)$ also has an essential singularity at infinity. But $|zf(z)|\ge \frac{|z|}{|z|+1}$, which goes towards $1$, which contradicts Big Picard for $zf(z)$.

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