pole at infinity iff f is a polynomial

Question

I need to show that if $f$ is an entire function it has a pole at infinity if and only if it is a polynomial. If I start with a polynomial, it is easy to show that it has a pole at infinity, but the other implication is harder.

That is, I want to prove that if $f$ has a pole at infinity it has to be a polynomial:

$f$ entire, pole at infinity $\rightarrow$ $f$ is polynomial

I do know that if $f$ is a polynomial than around zero we have that:

$f(1/z)=z^{-n}\cdot h(z)$, where $h$ is holomorphic and non-vanishing around zero. But how does this help?

I could try contrapositive. That is, if $f$ is not a polynomial, then showing that it can't have a pole at infinity. That is:

$f$ is not polynomial $\rightarrow$ $f(1/z)$ does not have a pole at zero

But this also seems hard, any tips?

Answer 1

Since $f$ has a pole at $\infty$ there exists an $R > 0$ such that $|f(z)| \geq 1$ for all $|z| > R$. Now let $z_1, ..., z_k$ be the zeros of $f$, counted with multiplicity, in $\overline{D(0,R)}$. There are only finitely many inside $\{ z: |z| \leq R \}$ or else the zeros would have an accumulation point in this compact set and then $f$ would be identically zero by the uniqueness theorem which is not possible since $f$ has a pole at $\infty$.

Consider the function $h(z) = \frac{(z-z_1)...(z-z_k)}{f(z)}$ then $h$ is an entire function. On the set $\{z: |z| > R \}$, $|h(z)| \leq C R^k$ by the Cauchy estimates since $h$ is an entire function of polynomial growth and hence it must be a polynomial. On the compact set $\overline{D(0,R)}$, we have $h(z) \leq M$ for some $M>0$ because $h$ is a continuous function on the compact set an hence it attains its maximum there. So $h$ is an entire function that is bounded since $|h(z)| \leq \max \{CR^k, M\}$ on all of $\mathbb{C}$, hence $h$ is a constant function.

But then, $f(z) = h(z)(z-z_1)..(z-z_k)=C(z-z_1)...(z-z_k)$ a polynomial.

Answer 2

Suppose $f$ has a pole at $\infty$ of order $n$. Use the integral formula for the $n$th derivative of $f$, which is as follows: $f^{(n)}(a) = \frac{n!}{2 \pi i} \int_C \frac{f(z)}{(z-a)^{n+1}}dz$, where $C$ is a positively oriented circle enclosing $a$ with radius.

To employ the formula here, suppose $|a|< R$ and let $C$ be the circle of radius $2R$ centered at zero. Then $|f^{(n+1)}(a)| \le \frac{(n+1)!}{2\pi} \int_C |\frac{f(z)}{(z-a)^{n+2}}||dz|$. Note that $|z-a| > R$ on the domain of integration, so we can upper bound further by $\frac{(n+1)!}{2\pi R^{n+2}} \int_C |f(z)||dz|$. The fact that $f$ has a pole at $\infty$ of order $n$ means that there exists a positive constant $c_1$ such that $ |f(z)| < |c_1z^n|$ for $|z|$ sufficiently large. Therefore as $R \to \infty$, $|f^{(n+1)}(a)| \le \frac{c_1 (n+1)! (2R)^n}{2 \pi R^{n+2}} \int_C |dz| = \frac{c_1 (n+1)! 2^n R^{n+1}}{R^{n+2}}$, which clearly approaches zero as $R \to \infty$. It follows that $f^{(n+1)}(a) = 0$. Since $f$ has a derivative which is identically zero, it is a polynomial.