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Let $f: \mathbb{C} \rightarrow \mathbb{C}$ an entire function and let $(a_n)_{n\geq 1} \subset \mathbb{C}^{*}$ the sequence of the zeros of $f$. Suppose that $z=0$ is a zero of $f$ of order $m\geq 1$ and that $f$ has infinitely many zeros. Show that:

\begin{align} f \not\equiv 0 \quad \Longrightarrow \quad \lim_{n\to\infty} = |a_n| = \infty \end{align}

I tried to prove it by contradiction using the identity principle for holomorphic functions, without success.

Any suggestions? Thanks in advance!

(It’s a step in the proof of Weierstrass factorization theorem)

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Otherwise, $(a_n)_{n\in\mathbb N}$ has a bounded subsequence. So, by the Bolzano-Weierstrass, it has a convergent subsequence. But, since $f$ is not the null function, this is impossible, by the identity theorem.

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Assume that $\lim_{n\to\infty} |a_n| = \infty$ does not hold. Then there is a $R> 0$ and a subsequence $(a_{n_k})$ such that $|a_{n_k}| \le R$ for all $k$. The closed disk $D = \{ z : |z| \le R \} $ is compact, therefore $(a_{n_k})$ has an accumulation point. So $f$ and the zero function have the same values on a set with an accumulation point in $\Bbb C$, and the identity theorem says that $f \equiv 0$, contrary to the assumption.

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