Let \(f\) be entire, and discuss (with proofs and examples) the types of singularities \(f\) might have (removable, pole, or essential) at \(z=\infty\) in the following cases:
\(f\) has at most finitely many zeros in \(\CC\).
\(f\) has infinitely many zeros in \(\CC\).
Solution
Write \(f(z) = \sum_{k\geq 0} c_k z^k\) since it is entire.
If \(f\) has finitely many zeros, \(f\) is nonconstant and entire, and thus unbounded by Liouville. If \(f\) is nonconstant, \(z=\infty\) can not be removable, since this would force \(f\) to be constant. So \(z=\infty\) can be a pole or an essential singularity. Both possibilities can occur: if \(f\) is a polynomial, it is entire with finitely many zeros and a pole at \(z=\infty\). Taking \(f(z)= e^z\) has no zeros and an essential singularity at \(z=\infty\).
If \(f\) has infinitely many zeros, if \(f\) is nonconstant then infinitely many \(c_k\) are nonzero – otherwise \(f\) is a polynomial and can only have finitely many zeros. Then \(g(z) \da f(1/z) = \sum_{k\geq 0}{c_k\over z^k}\) has infinitely many nonzero terms, making \(z=0\) an essential singularity for \(g\) and \(z=\infty\) essential for \(f\).