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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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    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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The formula is incorrect in general because the right hand side might be defined while the left hand side is not.

For example if $f_n(z)=(-1)^n$, then $\Pi f_n(z)$ is not defined and so $|\Pi f_n(z)|$ isn't either; but obviously $\Pi |f_n(z)|$ is defined and its value is $1$.

But under auspicious circumstances I cannot exclude that something could be salvaged...

Warning
In contrast to what happens for series, there is no naïve notion of absolute convergence for infinite products of complex numbers: else the example above shows you would have the disastrous terminology that some absolutely convergent products are divergent!
The best substitute is that the convergence of $\Pi (1+|a_n|) $implies the convergence of $\Pi (1+a_n) $.
Beware that some books' tratment of infinite products is not quite satisfactory. If you want to take the safe way, I cannot recommend warmly enough Remmert's Classical Topics in Complex Function Theory where the theme is handled right at the beginning of the book.

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    @ Georges: Concerning your example above, the value of $\Pi f_n(z)$ is $1$ or $-1$, so after taking $|.|$ we get $|\Pi f_n(z)|=1$ !! why is that not true?!2011-12-01