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. WHAT IF $f_k$ IS CONVERGENT?
On
I think is false, correct me if I am wrong but I think the left side can be not defined while the right side is. Consider $$f_n(z)=(-1)^n,$$ then $\Pi f_n(z)$ is not defined therefore $|\Pi f_n(z)|$ isn't either, and in the right side is easy to see that $\Pi |f_n(z)|$ is defined and its value is $1$.
Your notation higly implies that the sequence is index by $\Bbb N$ and I'll adhere to that. The statement is true for the values of $z$ such that the product without modulus either diverges to $0$ or converges, because pointwise $\left\lvert \prod_{k=1}^n a_k\right\rvert=\prod_{k=1}^n\left\lvert a_k\right\rvert$ for all $n$ and $\lvert \bullet\rvert$ is continuous.