Problem 1. A meromorphic function on \(\mathbb { C } \cup \{ \infty \}\) is a meromorphic function \(f ( z )\) on C such that \(g ( z ) = f ( 1 / z )\) is also meromorphic. Show that a meromorphic function on \(\mathbb { C } \cup \{ \infty \}\) must be rational, i.e. one can express it as the quotient of two polynomials.
Solution
1. There are only finitely many finite poles.
By hypothesis \(g(w)=f(1/w)\) is meromorphic near \(w=0\). Hence for some \(R>0\), the function \(f\) is holomorphic on \(|z|>R\) except possibly for the behavior corresponding to \(w=0\) itself. Thus every finite pole of \(f\) lies in the compact disk \(|z|\le R\).
Poles of a meromorphic function are isolated. If there were infinitely many finite poles in that compact disk, compactness would give an accumulation point in \(\mathbb C\), contradicting isolatedness. Therefore the finite poles are \(a_1,\dots,a_m\) for some finite \(m\).
2. Subtracting all finite principal parts leaves an entire function with at most polynomial growth.
At each pole \(a_j\), let \[ P_j(z)=\sum_{k=1}^{N_j}\frac{c_{jk}}{(z-a_j)^k} \] be the principal part of the Laurent expansion of \(f\). Define \[ h(z)=f(z)-\sum_{j=1}^mP_j(z). \] Every finite pole is removable for \(h\), so \(h\) extends to an entire function. Each \(P_j(z)\) tends to zero as \(z\to\infty\).
Since \(f(1/w)\) is meromorphic at \(w=0\), it has either a removable singularity or a pole there. In the removable case, \(f(z)\) is bounded for large \(|z|\); in the pole case of order \(N\), one has \(|f(z)|\le C|z|^N\) for sufficiently large \(|z|\). The same corresponding bound holds for \(h\), because the subtracted principal parts tend to zero at infinity. Thus \(h\) has at most polynomial growth.
3. The entire remainder is a polynomial, so \(f\) is rational.
Write \[ h(z)=\sum_{n=0}^\infty b_nz^n. \] If \(|h(z)|\le C(1+|z|^N)\) for large \(|z|\), Cauchy’s coefficient estimate on large circles gives, for every \(n>N\), \[ |b_n|\le \frac{C(1+R^N)}{R^n}\longrightarrow0 \qquad(R\to\infty). \] Hence \(b_n=0\) for all \(n>N\), so \(h\) is a polynomial. In the bounded case this is simply Liouville’s theorem and \(h\) is constant.
Therefore \[ f(z)=h(z)+\sum_{j=1}^mP_j(z) \] is a finite sum of rational functions, hence itself a rational function. This proves the assertion.