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We know that for complex (entire) functions $f,g$ we have $|f(z)g(z)|=|f(z)||g(z)|$, where $|.|$ means complex modulus.

What about if we have an infinite product? Is it true that $$\bigg| \prod_{k=1}^{\infty} f_{k}(z)\bigg|= \prod_{k=1}^{\infty} |f_{k}(z)| $$ where $\{f_{k}\}$ is any set of entire functions.

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    Is it crucial that $f_k$ are function, or you are only interested in a pointwise convergence?2011-11-30
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    $f_{k}$'s are all functions, in fact entire functions.2011-11-30
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    Have you considered the following argument: $$\left|\prod_{k=1}^\infty f_k\right| =\left|\lim\limits_{n\to\infty}\prod_{k=1}^n f_k\right| = \lim\limits_{n\to\infty}\prod_{k=1}^n |f_k|$$ because $|\cdot|$ is a continuous function?2011-11-30
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    Dear Jennifer, it is nice that you realize that the question is not so trivial as it looks at first sight : +12011-11-30
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    Thank you all for your coments. I don't have any other assumptions about the functions $f_{k}$, but what I realy need to know is that if this statement is not true in general (in this case why!) when it could be true? Do we need a condition on the limit to be exist, for example, or something else?2011-11-30

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