Let \(f\) be a meromorphic function in the plane such that \[ \lim_{|z|\to\infty}|f(z)|=\infty. \]
Show that \(f\) has only finitely many poles.
Show that \(f\) is a rational function.
Solution
1. There is \(R>0\) such that \(f\) has no poles in \(\{|z|>R\}\).
Choose \(R_0>0\) with \(|f(z)|\ge1\) for \(|z|>R_0\) away from the poles. Then \(g=1/f\) is holomorphic on \(\{|z|>R_0\}\): it is holomorphic away from the poles of \(f\), and it has a removable singularity with value \(0\) at each pole. Put \(G(w)=g(1/w)\) for \(0<|w|<1/R_0\). Since \(g(z)\to0\) as \(|z|\to\infty\), \(G\) is bounded near \(0\), so by Riemann’s removable singularity theorem it extends holomorphically to \(|w|<1/R_0\) with \(G(0)=0\). \(G\) is not identically zero, because \(g\) is nonzero away from the poles of \(f\). By the identity theorem the zero of \(G\) at \(0\) is isolated, so there is \(\delta>0\) with \(G(w)\neq0\) for \(0<|w|<\delta\). The poles of \(f\) in \(\{|z|>R_0\}\) are exactly the zeros of \(g\) there, so \(f\) has no poles in \(\{|z|>R\}\) with \(R=\max(R_0,1/\delta)\).
2. \(f\) has only finitely many poles.
By 1 every pole lies in the compact disc \(\{|z|\le R\}\). The poles of a meromorphic function have no accumulation point in \(\mathbb C\), so a compact set contains only finitely many of them.
3. \(f\) is a rational function.
Let \(p_1,\dots,p_k\) be the poles, with principal parts \(P_j(z)=\sum_{m=1}^{d_j}c_{j,m}(z-p_j)^{-m}\). Then \(h=f-\sum_{j}P_j\) has removable singularities at every \(p_j\), so it extends to an entire function. Each \(P_j(z)\to0\) as \(|z|\to\infty\), hence \(|h(z)|\to\infty\). An entire function with \(|h(z)|\to\infty\) has a pole at \(\infty\) and is therefore a polynomial. Thus \(f=h+\sum_jP_j\) is a sum of a polynomial and finitely many rational functions, hence rational.