Let $g(z)$ be entire analytic with the growth condition $|g(z)| \to \infty$, as $|z|$ $\to$ $\infty$. Then, how can I show that $g$ must be a polynomial?
Some thoughts:
Does the singularity at infinity have to be a pole? If so, why?
I think it does not, since there was no additional, stronger assumptions such as $g$ being one-to-one. If $g$ were 1:1, then by Big Picard's theorem, the singularity is obviously a pole.
I'm also thinking about the singularity of $g(z):=f(\frac{1}{z})$ at the origin. Is it a pole or an essential singularity ... or could it be either? (Again, since there was no one-to-one assumption of $g$, I don't see why the singularity must be a pole.)
EDIT: I am also concerned that the question is incorrect and perhaps needs additional assumptions - 1:1 or something else.
Thanks,