Let \(f\) be entire. Discuss, with proofs and examples, the types of singularities \(f\) might have (removable, pole, or essential) at \(\infty\) in each of the following cases.
\(f\) has at most finitely zeros in \(\CC\).
\(f\) has infinitely many zeros in \(\CC\).
Solution
For the singularity at infinity, use the local coordinate \[ w=\frac1z \] and study \[ F(w)=f(1/w) \] near \(w=0\).
1. The singularity of an entire function \(f\) at \(\infty\) is removableif and only if \(f\) is constant.
Suppose first that the singularity at \(\infty\) is removable. Then \(F(w)=f(1/w)\) is bounded for sufficiently small nonzero \(w\). Equivalently, there are \(R,M>0\) such that \[ \abs{f(z)}\leq M \] whenever \(\abs{z}>R\). Since \(f\) is continuous on the compact disk \(\abs{z}\leq R\), it is bounded there as well. Thus \(f\) is bounded on all of \(\CC\), and Liouville’s theorem implies that \(f\) is constant.
Conversely, if \(f\) is constant, then \(F(w)\) is constant on the punctured neighborhood of \(0\) and extends holomorphically across \(0\). Hence the singularity at \(\infty\) is removable.
2. The singularity of \(f\) at \(\infty\) is a pole if and only if \(f\) isa nonconstant polynomial.
Write the Taylor expansion \[ f(z)=\sum_{n=0}^{\infty}a_nz^n. \] Then \[ F(w) = f(1/w) = \sum_{n=0}^{\infty}a_nw^{-n}. \] If \(F\) has a pole at \(0\), its Laurent series has only finitely many negative powers and at least one such term. Hence only finitely many \(a_n\) are nonzero and at least one \(a_n\) with \(n\geq1\) is nonzero. Thus \(f\) is a nonconstant polynomial.
Conversely, if \[ f(z)=a_0+a_1z+\cdots+a_dz^d, \qquad d\geq1, \qquad a_d\neq0, \] then \[ F(w) = a_0+a_1w^{-1}+\cdots+a_dw^{-d}, \] so \(F\) has a pole of order \(d\) at \(0\).
3. The singularity of \(f\) at \(\infty\) is essential if and only if \(f\)is a nonpolynomial entire function.
The function \(F(w)=f(1/w)\) is holomorphic on a punctured neighborhood of \(0\), so its isolated singularity there is removable, a pole, or essential. By steps 1 and 2, the first two possibilities are exactly constant functions and nonconstant polynomials. Therefore the remaining entire functions, precisely the nonpolynomial entire functions, have an essential singularity at \(\infty\).
4. In part (a), where \(f\) has at most finitely many zeros, all threetypes of singularity at \(\infty\) can occur.
A removable singularity occurs for \[ f(z)=1, \] which has no zeros, by step 1.
A pole occurs for \[ f(z)=z, \] which has one zero, by step 2.
3.Thus the assumption of finitely many zeros excludes none of the threesingularity types.
5. In part (b), a pole at \(\infty\) is impossible.
If \(f\) had a pole at \(\infty\), step 2 would make \(f\) a nonconstant polynomial. A nonzero polynomial has only finitely many zeros, contradicting the hypothesis that \(f\) has infinitely many zeros.
6. In part (b), a removable singularity occurs only for the zerofunction.
If the singularity at \(\infty\) is removable, step 1 says that \(f\) is constant. The only constant function with infinitely many zeros is \[ f\equiv0. \] Conversely, the zero function has every point as a zero and has a removable singularity at \(\infty\).
7. Every nonzero entire function with infinitely many zeros has anessential singularity at \(\infty\), and this case occurs.
Let \(f\) be nonzero entire with infinitely many zeros. By step 6 its singularity at \(\infty\) is not removable, and by step 5 it is not a pole. Therefore it is essential.
For example, \[ f(z)=\sin z \] has the infinitely many zeros \(n\pi\), \(n\in\ZZ\), and is not a polynomial. Hence step 3 shows that its singularity at \(\infty\) is essential.
8. The classification is therefore:
with at most finitely many zeros, removable, pole, and essential are all possible;
with infinitely many zeros, a pole is impossible, the removable case is exactly \(f\equiv0\), and every nonzero such function has an essential singularity at \(\infty\).
4. Part (b) follows from steps <1>5–<1>7.
9. Q.E.D.
Step 8 gives the requested discussion, proofs, and examples.