I know that if $f(z)$ is a nonconstant entire function, then $e^{f(z)}$ has an essential singularity at $z = \infty$,
Now suppose $f(z)$ is an entire function that is not a polynomial. What kind of singularity can $f(z)$ have at $\infty$?
My attempt is since $f(z)$ is entire,
$f(z) = \sum_{i=0}^{\infty} a_nz^n$ around $0$,
since $f$ is not polynomial, the series has infinite terms.
Then $f(\frac{1}{w})= \sum_{i=0}^{\infty} a_n\frac{1}{w^n}$. But I don't know how to continue from there
Best Answer
As you observed, we can classify the singularity of $f$ at $\infty$ by classifying the singularity of $g(z)=f(1/z)$ at $0$. We know that if $g(z)$ has a removable singularity at $0$, then $f$ has a constant limit at $\infty$, meaning that $f$ is bounded, and therefore it is constant, so that cannot happen. Now suppose $g$ has a pole at the origin, then there is some $n$ such that $z^n g(z)$ is holomorphic in a neighborhood of the origin, so $$g(z)=\frac{a_{n}}{z^n}+\cdots+\frac{a_{1}}{z}+a_0.$$ But then $f(z)=a_0+a_1z+\cdots+z_nz^n$, a polynomial, so this cannot happen either. Therefore, the singularity must be essential.
This leads to a useful characterization of an essential singularity: a singularity at $z_0$ is essential iff the Laurent series about $z_0$ has infinitely many negative powers. The singularity of an entire function at $\infty$ is essential iff the Laurent series has infinitely many positive powers.