Meromorphic function with finite isolated singularities Let $f$ be an entire function and let $a_1,\ldots,a_n$ be all zeroes of $f$ in $\mathbb{C}$. Suppose that there exist real numbers $r_0>0$ and $t>1$ such that $|f(z)|\geq |z|^t$ for all $|z|\geq r_0$. Prove that
$$\sum_{j=1}^n\text{Res}\left(\frac{1}{f},a_j\right)=0$$
I'm fairly sure that you would use Cauchy's integral formula for the computation of the sum of the residues, but I'm not sure how to go about doing so?
 A: Hint
Integrate over a circle $\gamma_R$ with radius $R$ around 0. Consider the case $R \to \infty$. Show that $\int _{\gamma_R} \frac{1}{f(z)} dz \to 0$ using the standard inequality for integrals and $\left\vert \frac{1}{f(z)} \right\vert \leq R^{-t}$ for $\vert z \vert =R$ and $R > r_0$. Then use the Cauchy integral formula to represent the same intregral via the residua.
A: First note that for all $i$, $|a_i| < r_0$, or else if such a $a_j$ is, then $0=|f(a_j)|\geq |a_j|^t$, which implies that $|a_j|=0$ (!). In other words, for all $i$, $a_i \in D(0,r_0)$. Now for any $\epsilon >0$, we integrate over $1/f$ over $\partial D(0,r_0+\epsilon)$. Then by Cauchy's residue theorem, we have $$\frac{1}{2\pi i}\int_{\partial D(0,r_0+\epsilon)} \frac{dz}{f}=\sum\limits_{i=1}^n \text{Res}(1/f,a_i)$$ However a simple estimation tells us that $$\bigl \lvert \int_{\partial D(0,r_0+\epsilon)} \frac{dz}{f} \bigr \rvert\leq {2\pi(r_0+\epsilon)}.\text{sup}_{\partial D(0,r_0+\epsilon)}|1/f|\leq \frac{2\pi}{(r_0+\epsilon)^{t-1}}$$ Since $t>1$ and $f$ is entire, we can let $\epsilon \rightarrow \infty$, the RHS of the above approximation will result to $0$. Therefore, $$\sum\limits_{i=1}^n \text{Res}(1/f,a_i)=0$$
EDIT: didn't know that someone already posted a hint! But I will just leave this up in case you need to check on the details :)
