Are the imaginary parts of these Riemann zeta related numbers equal?

Question

While we can't say anything about the real part of these numbers below, maybe it could be possible to say something about the imaginary parts?

Let:

$$A=\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s\right)}$$

and:

$$B=\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s+\frac{1}{n}\right)}$$

Then compute:

$$\rho_1=\frac{1}{1-\frac{A}{B}}+\frac{1}{n}+s$$

$$\rho_2=\frac{1}{1-\frac{B}{A}}-\frac{1}{n}-s$$

where $s$ is a complex number.

Show that $\Im(\rho_1)=-\Im(\rho_2)$ for all positive integers $n>1$.

Associated Mathematica program:

"Mathematica start"
"Part 1:"
Clear[s, A, B, n, z, k, rho1, rho2];
n = 30;
s = (1/3 + 14*I);
A = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n], {k, 1, 
   n}]; B = 
 Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n + 1/n], {k, 1, 
   n}]; rho1 = N[(s + 1/n + 1/(1 - A/B)), 30]
rho2 = N[(-s - 1/n + 1/(1 - B/A)), 30]
"Part 2:"
Clear[s, A, B, n, z, k, rho];
n = Prime[10]; Reduce[
 rho == s + 1/n + 1/(1 - A/B) && 
  s + 1/n + 1/(1 - A/B) == Conjugate[-s - 1/n + 1/(1 - B/A)] && 
  B != 0, Re[rho]]
"Mathematica end"

Answer 1

Let $s$ be: $$s=c+i d$$

and let the ratio $$\frac{A}{B}=a+i b$$

then $$\frac{B}{A}=\frac{1}{a+i b}$$

The formulas then become:

$$\rho_1=\frac{1}{1-(a+i b)}+\frac{1}{n}+c+i d$$

$$\rho_2=\frac{1}{1-\frac{1}{a+i b}}-\frac{1}{n}-(c+i d)$$

Formulas for imaginary parts: $$\Im(\rho_1) = \frac{\left(\frac{1}{1-(a+i b)}+(c+i d)+\frac{1}{n}\right)-\left(\frac{1}{1-(a-i b)}+(c-i d)+\frac{1}{n}\right)}{i 2}$$ $$\Im(\rho_2) = \frac{\left(\frac{1}{1-\frac{1}{a+i b}}-(c+i d)-\frac{1}{n}\right)-\left(\frac{1}{1-\frac{1}{a-i b}}-(c-i d)-\frac{1}{n}\right)}{i 2}$$

which when simplifying becomes:

$$\Im(\rho_1) = \frac{b}{(a-1)^2+b^2}+d$$

$$\Im(\rho_2) =-\frac{b}{(a-1)^2+b^2}-d$$

From which one sees that: $$\Im(\rho_1) = -\Im(\rho_2)$$

---

One can repeat for the real parts:

Formulas for real parts: $$\Re(\rho_1) = \frac{\left(\frac{1}{1-(a+i b)}+(c+i d)+\frac{1}{n}\right)+\left(\frac{1}{1-(a-i b)}+(c-i d)+\frac{1}{n}\right)}{2}$$ $$\Re(\rho_2) = \frac{\left(\frac{1}{1-\frac{1}{a+i b}}-(c+i d)-\frac{1}{n}\right)+\left(\frac{1}{1-\frac{1}{a-i b}}-(c-i d)-\frac{1}{n}\right)}{2}$$

which simplifies to:

$$\Re(\rho_1) =\frac{1}{2} \left(\frac{2-2 a}{(a-1)^2+b^2}+2 c+\frac{2}{n}\right)$$

$$\Re(\rho_2) = \frac{1}{2} \left(\frac{2 (a-1)}{(a-1)^2+b^2}-2 c-\frac{2}{n}+2\right)$$

Adding together: $$\Re(\rho_1) + \Re(\rho_2)$$

$$\left(\frac{1}{2} \left(\frac{2-2 a}{(a-1)^2+b^2}+2 c+\frac{2}{n}\right)\right)+\left(\frac{1}{2} \left(\frac{2 (a-1)}{(a-1)^2+b^2}-2 c-\frac{2}{n}+2\right)\right) = 1$$

Therefore: $$\Re(\rho_1) + \Re(\rho_2) = 1$$