Let \(f\) be entire and let \(a_1,\ldots,a_n\) be all of its zeros in \(\mathbb C\). Suppose there exist \(R>0\) and \(\alpha>1\) such that \[ |f(z)|\ge |z|^\alpha \] for every \(|z|\ge R\). Prove that \[ \sum_{j=1}^n \operatorname{Res}_{z=a_j}\frac1{f(z)}=0. \]
Solution
Set \[ g(z)\coloneqq\frac1{f(z)}. \] Choose \(R_0>R\) so large that every zero \(a_1,\ldots,a_n\) lies in the disk \(\abs{z}<R_0\).
1. For every \(r\ge R_0\),\[ \int_{\abs{z}=r}g(z)\,dz =2\pi i\sum_{j=1}^n\operatorname{Res}_{z=a_j}g(z). \]
The function \(g=1/f\) is meromorphic on \(\CC\), with poles precisely at the zeros \(a_1,\ldots,a_n\) of \(f\). For \(r\ge R_0\), all of these poles lie inside the circle \(\abs{z}=r\), and there are no poles on or outside that circle. The residue theorem therefore gives the displayed identity.
2. For every \(r\ge R_0\),\[ \left|\int_{\abs{z}=r}g(z)\,dz\right| \le 2\pi r^{1-\alpha}. \]
On the circle \(\abs{z}=r\), the hypothesis gives \[ \abs{g(z)} =\frac1{\abs{f(z)}} \le\frac1{r^\alpha}. \] The circle has length \(2\pi r\), so the ML estimate yields \[ \left|\int_{\abs{z}=r}g(z)\,dz\right| \le 2\pi r\cdot r^{-\alpha} =2\pi r^{1-\alpha}. \]
3. The contour integrals in step 1 tend to \(0\) as\(r\to\infty\).
Since \(\alpha>1\), the exponent \(1-\alpha\) is negative. Hence \(2\pi r^{1-\alpha}\to0\), and step 2 gives the claim.
4. The sum of residues satisfies\[ \sum_{j=1}^n\operatorname{Res}_{z=a_j}\frac1{f(z)} =\boxed{0}. \]
By step 1, for every \(r\ge R_0\) the contour integral equals \(2\pi i\) times the displayed sum, which is independent of \(r\). Step 3 shows that the left-hand side tends to \(0\) as \(r\to\infty\). Therefore the constant residue sum must be \(0\).
5. Q.E.D.
Step 4 is the required identity.