# Canonical products and general form for entire functions

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.

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

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.