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In Ahlfors' book, chapter $\textbf{V}-2.3$, we have a method to find a form to entire functions, that is: if $f$ is an entire function with a finite number of zeros $a_1, ..., a_N$, and a zero of multiplicity $m$ on the origin then $$f(z)=z^me^{g(z)}\prod_{n=1}^{N}\Big(1-\frac{z}{a_n}\Big)$$ after, generalize this for infinitely many zeros $$f(z)=z^me^{g(z)}\prod_{n=1}^{\infty}\Big(1-\frac{z}{a_n}\Big)$$

My question is: Why that generalization doesn't mean $f$ is a constant function? I said that because we know that an analytic (nonconstant) function has a finite number of zeros. My point is that there is a contradiction between this two facts.

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    $\begingroup$ Doesn't the sine function have infinitely many zeroes? $\endgroup$ Commented Dec 14, 2018 at 20:01

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You should mention that in the finite case multiple zeros are repeated. Ahlfors says that if there are infinitely many zeros one can try to obtain a similar representation by means of an infinite product. He states that this product converges absolutely if and only if $\sum_{n=1}^\infty 1/\lvert a_n \rvert$ is convergent.

A neccessary condition for the convergence of this series is that $a_n \to \infty$ as $n \to \infty$. This means that $(a_n)$ does not have a cluster point in $\mathbb{C}$. Therefore you cannot conclude that $f$ is constant. See Lord Shark the Unknown's comment.

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