Find all entire functions which have poles at \(\infty\)
Solution
1. Let\[ f(z)=\sum_{n=0}^{\infty}a_nz^n \]be entire. Near infinity, use the coordinate\[ w=\frac1z. \]Then\[ f(1/w) = \sum_{n=0}^{\infty}a_nw^{-n} \]for every \(w\neq0\).
Because \(f\) is entire, its Taylor series at \(0\) has infinite radius of convergence. Substituting \(z=1/w\) for \(w\neq0\) gives the displayed Laurent series about \(w=0\).
2. If \(f\) has a pole at \(\infty\), then \(f\) is a nonconstant polynomial.
By definition, \(f\) has a pole at \(\infty\) exactly when \[ w\longmapsto f(1/w) \] has a pole at \(w=0\). A Laurent series at a pole has only finitely many negative-power terms. By step 1, the coefficient of \(w^{-n}\) is \(a_n\). Hence there is an integer \(d\geq0\) such that \[ a_n=0 \] for every \(n>d\). Thus \[ f(z)=\sum_{n=0}^{d}a_nz^n \] is a polynomial.
Since the singularity at infinity is a pole rather than removable, the principal part of \(f(1/w)\) is nonzero. Therefore some \(a_n\) with \(n\geq1\) is nonzero, so the polynomial is nonconstant.
3. Every nonconstant polynomial has a pole at \(\infty\).
Let \[ p(z)=a_0+a_1z+\cdots+a_dz^d, \qquad d\geq1, \qquad a_d\neq0. \] Then \[ p(1/w) = a_0+a_1w^{-1}+\cdots+a_dw^{-d}. \] Its Laurent series at \(w=0\) has highest negative power \(w^{-d}\) with nonzero coefficient \(a_d\). Hence \(w=0\) is a pole of order \(d\) for \(p(1/w)\), so \(p\) has a pole of order \(d\) at \(\infty\).
4. The complete list is the set of all nonconstant polynomials.
Step 2 shows that every entire function with a pole at infinity is a nonconstant polynomial, and step 3 shows that every nonconstant polynomial has such a pole.
5. Q.E.D.
Step 4 is the requested classification.