Structural differences between closed forms of two related infinite products?

Question

In this question, I went quite a bit over the top, so I now tried to rephrase it in a much simpler way.

Take $a \in \mathbb{R}, s \in \mathbb{C}$ and:

$$\displaystyle C(s,a) := \prod_{n=1}^\infty \left(1- \frac{s}{a + n i} \right) \left(1- \frac{s}{{a - n i}} \right) = \frac{\xi_{c}(0 -a+s)}{\xi_{c}(0-a)}$$

with $\xi_{c}(z) = \frac{\sinh(\pi z)}{z}$.

And then also:

$$\displaystyle R(s,a) := \prod_{n=1}^\infty \left(1- \frac{s}{a + n i} \right) \left(1- \frac{s}{{1-(a + n i)}} \right) = \frac{\xi_{r}(0 -a+s)}{\xi_{r}(0-a)}\frac{\xi_{r}(1 -a-s)}{\xi_{r}(1-a)}$$

with $\xi_{r}(z) = \frac{1}{z\, \Gamma(z \, i)}$.

Questions:

For the infinite product of the conjugated zeros $C(s,a)$, the closed form has only two factors, however the closed form for the product of the reflexive zeros $R(s,a)$ does require four factors.

1) Is there a way to also reduce the closed form for $R(s,a)$ to two $\xi$-factors by altering $\xi_r(z)$?

Additional clarification:

I do believe that it is impossible to reduce these four factors:

$$\displaystyle \frac{\xi_{r}(0 -a+s)}{\xi_{r}(0-a)}\frac{\xi_{r}(1 -a-s)}{\xi_{r}(1-a)}$$

into a division of only two factors.

The most logical extension of $\xi_{r}(z)$ would be to assume: $\xi_{r}(z) = \frac{1}{z\, \Gamma(z \, i)} \, \, \frac{1}{(1-z)\, \Gamma((1-z) \, i)}$

and then to reduce the division to: $\displaystyle \frac{\xi_{r}(0-a+s)}{\xi_{r}(0-a)}$.

However this only correctly induces the original four factors for the nominators only and it gives the wrong results for the two denominators. Any tweak I tried to make to $\xi_r(z)$ fails on this apparently fundamental conflict, hence my conjecture that it is impossible. Keen to learn if this could be proven and/or which branch of Mathematics deals with this type of problems.

2) If not (per my conjecture), is this four factor closed form with $\xi_r(z)$ then dependent on the choice of $n$ or does it simply originate from the reflexive nature of the zeros in the product? To illustrate what I am after with an example; the relations $C(s,a) = C(2a-s,a)$ and $R(s,a) = R(1-s,a)$ are fully independent of the choice of $n$ and the product being finite or infinite, and just stem from the conjugated/reflexive relation between the zeros.

Thanks.

Answer 1

Too long for comment Think I made some progress, but not sure how to formalize this further. It seems that also for the individual factors $C(s,a,n)$ and $R(s,a,n)$, the closed forms are structured in exactly the same way as the full infinite product:

$C(s,a,n) := \left(1- \frac{s}{a + n i} \right) \left(1- \frac{s}{{a - n i}} \right) = \frac{\xi_{c n}(0 -a+s,n)}{\xi_{c n}(0-a,n)}$

with $\xi_{c n}(z,n) = z^2 + n^2$.

$R(s,a,n) := \left(1- \frac{s}{a + n i} \right) \left(1- \frac{s}{{1-(a + n i)}} \right) = \frac{\xi_{r n}(0 -a+s,n)}{\xi_{r n}(0-a,n)}\frac{\xi_{r n}(1 -a-s,n)}{\xi_{r n}(1-a,n)}$

with $\xi_{r n}(z,n) = z - n i$.

You can now obviously pick any $n$ you like (as long as $a \pm n i \ne 0, 1- (a+n i) \ne 0$), the structure remains the same. So does this imply that the structure of the closed forms is independent of $n$?