This is a quick question about when the sum of a product is the product of a sum. We have the definition of the zeta function as
\begin{equation} \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}. \end{equation} Now, if we write $s=\sigma+i\tau$, this becomes
\begin{equation} \zeta(\sigma+i\tau)=\sum_{n=1}^{\infty}\frac{1}{n^{\sigma+i\tau}}. \end{equation}
My question is, can we do the following:
\begin{align} \zeta(\sigma+i\tau)&=\sum_{n=1}^{\infty}\frac{1}{n^{\sigma}n^{i\tau}}\\&=\sum_{n=1}^{\infty}\frac{1}{n^{\sigma}}\frac{1}{n^{i\tau}}\\&=\sum_{n=1}^{\infty}\frac{1}{n^{\sigma}}\sum_{n=1}^{\infty}\frac{1}{n^{i\tau}}\\&=\zeta(\sigma)\zeta(i\tau) \end{align}
The reason I am unsure about this, is I'm not fully sure under what conditions we can simply take the product and split it into the sum of two smaller products. Can anyone comment on this?
Of course you cannot do this. Consider the following counterexample:
$$ \zeta(8+9i) \neq \zeta(8)\zeta(9i) $$
You can verify that the numerical values differ.
Notice that the multiplication of sums you introduced means that every number summed by the left-hand sum is multiplied by the entire sum on the right, whereas what you want is a one-to-one multiplication.