# Modulus of infinite product of complex functions

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? – S.D. Nov 30 '11 at 13:45
$f_{k}$'s are all functions, in fact entire functions. – Jennifer Ad Nov 30 '11 at 13:49
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? – S.D. Nov 30 '11 at 13:55
Dear Jennifer, it is nice that you realize that the question is not so trivial as it looks at first sight : +1 – Georges Elencwajg Nov 30 '11 at 15:08
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? – Jennifer Ad Nov 30 '11 at 15:59

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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Won't it be worth to mention that if the RHS exists then the equality holds? – S.D. Nov 30 '11 at 14:50
Dear @S.D., I have just proved the opposite of what you say, so I guess you meant something else. However, as I said, there is something to be salvaged but for pedagogical reasons I'd rather it were the OP or another student (you, maybe?) who gave a precise formulation and proof of the correct statement. I would then be very glad to upvote them. – Georges Elencwajg Nov 30 '11 at 15:02
Dear Georges, thanks for pointing it out: indeed, I meant that the existence of LHS implies existence of RHS and the fact that the equality holds. – S.D. Nov 30 '11 at 15:11
@ 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?! – Jennifer Ad Dec 1 '11 at 0:15