Prove $f$ entire and with a pole at infinity to be polynomial

Solution 1:

Well, $g(0)=f\left(\frac10\right)=f(\infty)=\infty$.

Solution 2:

Note that when $z\rightarrow \infty, w=\frac{1}{z}\rightarrow 0$. And since $f(z)\rightarrow \infty$ as $z\rightarrow \infty$, there is no essential singularity. So $g(w)$ must have a pole of finite order at $\infty$.

Solution 3:

To show that 0 is a pole for f(1/z), try this:

Define $h(z)=\frac{1}{f(\frac{1}{z})}$, then $h(z)$ is analytic for $f(\frac{1}{z})$ non-zero. And note that $\lim_{z\to 0} |h(z)|=0 \implies lim_{z\to 0}h(z)=0$.

As the limit exists, we can take $h^*$ to be an analytic extension. Knowing that $h^*(z)$ is not identically zero, gives us that zeros are separated, so we can find a ball of radius $R>0$ such that $h^*(z)$ is analytic on $B_R(0)$.

Thus $h^*$ has a Taylor expansion centered about 0 on that region. As the first coefficient of that is 0 (Looking at the limit), but not all coefficients are non-zero or $h^*$ would be identically zero.

We have that $h^*(z)=z^N*\sum_{n=N}^\infty a_n z^{n-N}=z^N*k(z)$, where k is analytic and doesn't vanish on the disk.

$$lim_{z\to 0}Z^{N+1}f\left(\frac{1}{z}\right)=lim_{z\to 0} \frac{z}{k(z)}=0$$

Therefore $f(1/z)$ has a pole of order N at 0 $\implies$ f(z) has a pole of order N at infinity. Given this, dietervdf proof follows easily, but proving that the singularity at infinity is a pole and not an essential singularity is, I think, non-trivial.