# Show that an entire function $f$ s.t. $|f(z)|>1$ for $|z|>1$ is a polynomial

I have been struggling on the following problem.

Suppose $f$ is an entire analytic function such that $|f(z)|>1$ if $|z|>1$. Show that $f$ is a polynomial.

My idea is as followed: all zeros of $|f(z)|$ lie inside $|z|\leq 1$. Applying Argument Principle, we can show that number of zeros of $f$ is bounded. So we can assume

$f(z)=(z-z_1)...(z-z_M)g(z)$

where g is entire analytic without any zeros. Then I would like to apply Liouville's Theorem: the point is that it isn't too clear to me why $|\dfrac{1}{g(z)}|$ is a bounded function.

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Hint: if $f$ is entire, then either $f$ is a polynomial, or $f$ has an essential singularity at $\infty$. – Chris Eagle May 20 '12 at 0:34
Hello Warwicker. Do you know about accepting answers to your questions? – Antonio Vargas May 20 '12 at 0:34
Oh dear. I have just known it. – iloveinna May 20 '12 at 0:45

If the Taylor series about 0 does not terminate, $f(1/z)$ has an essential singularity at $0$ (why?)

Then from the Casorati–Weierstrass theorem (have you learnt this?) you know $f(1/z)$ cannot be bounded from below near 0. Contradiction!

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This is great. I didn't approach in this way because the above exercise was taken from the chapter on "Argument Principle, Rouche's Thm and Liouville's Thm". – iloveinna May 20 '12 at 0:53
@Montez, can you explain the line "$f(z)$" cannot be bounded from below near 0" – Deepak Dec 21 '12 at 20:40
The Casorati-Weierstrass theorem says that if $p$ is an essential singularity of an analytic function $g$, the image of a deleted neighbourhood of $p$ is dense in the complex plane. If it's dense, it takes values arbitrarily close to $0$. – Robert Israel Dec 23 '12 at 7:05
@Deepak I didn't say $f(z)$ is bounded from below near $0$. I said $f(1/z)$ bounded from below near zero, which is the same as $f(z)$ bounded from below away from $0$, which is given, which contradicts the Casorati-Weierstrass theorem if the $f(1/z)$ has an essential singularity at zero. – Montez Dec 23 '12 at 16:33

The function $1/f$ is bounded at infinity, hence has a fake singularity there and can be holomorphically extended through infinity.
Hence $1/f$ is actually rational ($\iff$ meromorphic on the whole Riemann sphere) and so also $f$ is rational.
But since $f$ does not have any pole at finite distance (since it is an entire function) it is a polynomial (and has a pole at infinity if it isn't a constant).

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The number of zeros of $f$ can be finite as the set of zeros of $f$ will have a accumulation point and $f$ will be equivalent to $0$ (not possible) then $$g=\frac{f}{(z-a)\cdots(z-x)}$$ where $a$ to $x$ are the finite number of zeros of $f$ then $g$ is analytic consider $h=1/g$ see that $h(z)$ is never zero as we have deleted the zeros of $f$ and also $f$ tends to infinity as $z$ tends to infinity (by the given condition) which implies $$|h(z)| \le a+|z|^n,$$ where $n$ is the number of zeros of $f$, but by fundamental theorem of algebra we get $h(z)=c$ (constant) thus $$f=\frac{(z-a)\cdots(z-x)}{k}.$$

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