How can the following product be expressed as a complex number in the form $a + bi$?
$$p = (\sigma +it)(\sigma+1 +it) (\sigma+2 +it)\cdot \cdot \cdot (\sigma+j +it) \\ = \prod_{j=0}^{2k-2}(\sigma+j +it) $$
To provide context, this product comes from a particular algorithm for the Riemann Zeta function for $\text{Re}(s) = \sigma >1$:
$$\zeta(s)=\sum_{j=1}^{n-1}j^{-s} + \frac12 n^{-s} + \frac{n^{1-s}}{s-1} + \sum_{k=1}^m T_{k,n}(s) + E_{m,n}(s)$$
Where, letting $B_{2k}$ denote the indicated Bernoulli number,
$$T_{k,n}(s) = \frac{B_{2k}}{(2k)!}n^{1-s-2k}\prod_{j=0}^{2k-2}(s+j)$$ and the error satisfies $$|E_{m,n}(s)|< \left| \frac{s+2m+1}{\sigma +2m +1}T_{m+1,n}(s) \right|.$$
I want to express this entire algorithm as a complex number in the form $a+bi$ by substituting $s=\sigma +it$. This way, $\zeta(s)$ can be separated into its real and imaginary parts, and each part can presumably be computed efficiently.
Here is my progress so far:
$$\zeta(\sigma+it)= \\ \sum_{j=1}^{n-1}\frac{\cos{(t\log{j})}}{j^\sigma} + \frac{\cos{(t\log{n})}}{2n^\sigma} + \frac{n^{1-\sigma}}{(\sigma-1)^2+t^2}\Big((\sigma-1)\cos(t\log{n})- t\sin{(t\log{n})}\Big) \\ -i \Big(\sum_{j=1}^{n-1}\frac{\sin{(t\log{j})}}{j^\sigma} + \frac{\sin{(t\log{n})}}{2n^\sigma} + \frac{n^{1-\sigma}}{(\sigma-1)^2+t^2}\Big(t\cos(t\log{n}) + (\sigma -1)\sin{(t\log{n})}\Big) \Big) + r$$
Where $r$ is the rest that I haven’t gotten into $a+bi$ form yet: $$r = \sum_{k=1}^m T_{k,n}(\sigma + it) + E_{m,n}(\sigma + it)$$
The process of getting this representation was straightforward—although tedious—using Euler’s identity. However, I’m having trouble with the terms contained within $r$; particularly, the $T_{k,n}(\sigma + it)$ term.
$$T_{k,n}(\sigma + it) = \frac{B_{2k}}{(2k)!}n^{1-\sigma-2k}\Big( \cos{(t\log{n})} -i \sin{(t\log{n})} \Big)\prod_{j=0}^{2k-2}(\sigma+j +it)$$
Simplifying the exponent was simple, but I don’t know how to simplify the product. This is where the question above comes from.