Didn't see any like this in the similar questions, so hopefully it isn't a repeat. There was a question on a past qual that asked if $f$ is entire, can $e^f$ have a pole at infinity. I think the answer is going to be no. But would like confirmation. If possible it would be much appreciate if someone could tell me why my logic is correct or incorrect.
Suppose $e^f$ has a pole at infinity. Clearly we cannot have $f$ be constant, hence as a consequence of Louisville's theorem we must have $\infty$ be either a pole or an essential singularity of $f$. If $\infty$ were essential, then we could find a sequence $z_n\to\infty$ such that $f(z_n)\to c$ for some constant $c$. Then $e^{f(z_n)}\to e^c$ and so $\lim_{z\to \infty}e^{f(z)}\not=\infty$ and $e^f$ does not have a pole at $\infty$.
The last case we have to consider is were $f$ has a pole at infinity. Let $g(w)=f(1/w)$. Let $D$ an exterior domain, and define $1/D=\{1/z:z\in D\}$. Then as $g$ has a pole inside $1/D$ it follows that either $$\int_{\partial 1/D} d\arg(g)\not=0\text{ }(*)$$ or $$\int_{\partial 1/D}d\arg(g)=0\text{ }(**)$$ If $(*)$ then by the argument principle modulo $2\pi$ it follows that $\arg(g)$ takes on every value in $[0,2\pi)$, in particular we can choose $w\in 1/D$ with $g(w)\in i\mathbb{R}$.
If $(**)$ then as $g$ has a pole in $1/D$ at $0=1/\infty$ it follows that $g$ must have a zero in $1/D$ as $$\int_{\partial1/D}d\arg(g)=\#\{\text{ zeros of }g\text{ in }1/D\}-\#\{\text{ poles of }g\text{ in }1/D\}$$ In this case we have $g(w)=0\in i\mathbb{R}$ for some $w\in 1/D$.
We conclude that regardless of our situation we can choose $w\in 1/D$ with $g(w)\in i\mathbb{R}$. As a consequence we can choose $z\in D$ with $f(z)\in i\mathbb{R}$. Well then as $D$ was an arbitrary exterior domain we can choose $z_n\to\infty$ with $f(z)\in i\mathbb{R}$, so $|e^{f(z_n)}|=1$, and we cannot have $$\lim_{z\to\infty}e^{f(z)}=\infty$$ Consequently $e^f$ does not have a pole at infinity.
