Polynomial growth forces a meromorphic function to be rational

The bound \[ \abs{f(z)}\leq C\abs z^n \] is finite for every \(\abs z>R\), so \(f\) has no pole outside the closed disk \(\abs z\leq R\). The poles of a meromorphic function are isolated. If there were infinitely many poles in the compact disk \(\abs z\leq R\), they would have an accumulation point in that disk, contradicting isolatedness of the poles. Hence there are only finitely many.

Let the distinct poles of \(f\) be \(a_1,\ldots,a_r\), with respective orders \(m_1,\ldots,m_r\), and set \[ P(z)\coloneqq\prod_{j=1}^r(z-a_j)^{m_j}. \] At each \(a_j\), multiplication by \((z-a_j)^{m_j}\) removes the pole of \(f\), while the remaining factors are holomorphic and nonzero there. Thus every singularity of \(Pf\) is removable. Away from the poles, \(Pf\) is already holomorphic, so after filling in the removable singularities, \(g=Pf\) is entire.

Let \[ m\coloneqq m_1+\cdots+m_r=\deg P. \] For sufficiently large \(\abs z\), there is a constant \(C_1>0\) such that \[ \abs{P(z)}\leq C_1\abs z^m. \] Hence the assumed growth bound gives another constant \(C_2>0\) such that \[ \abs{g(z)}\leq C_2\abs z^{m+n} \] for all sufficiently large \(\abs z\).

Choose an integer \(k>m+n\). For all sufficiently large radii \(\rho\), the Cauchy estimate on the circle \(\abs z=\rho\) gives \[ \abs{g^{(k)}(0)} \leq \frac{k!}{\rho^k} \max_{\abs z=\rho}\abs{g(z)} \leq k!C_2\rho^{m+n-k}. \] Because \(m+n-k<0\), the right-hand side tends to \(0\) as \(\rho\to\infty\). Therefore \(g^{(k)}(0)=0\) for every sufficiently large integer \(k\). The Taylor series of the entire function \(g\) consequently has only finitely many nonzero coefficients, so \(g\) is a polynomial.

By steps 2 and 3, both \(P\) and \(g=Pf\) are polynomials. Therefore \[ f=\frac{g}{P} \] as meromorphic functions on \(\CC\), so \(f\) is rational.

Step 4 is the required conclusion.