In the study of prime number theorem I have the function
$A= x^{s-1} h(s) = x^{s-1}\frac{1}{s (s+1)}{ (B)}=x^{s-1} \frac{1}{s (s+1)} (-\frac{\zeta'(s)}{\zeta(s)}- \frac{1}{s-1})$
where B is analytic at s=1 for some proof given in precedence.
A should have same module respect coniugate value in the interval $1\leq \sigma \leq c$.
If I consider $x^{s-1},s, (s+1) , s-1$ they have the same real part and opposite immaginary so module respect coniugate values is the same but why also $\zeta(s)$ and $\zeta'(s)$
Let $f(s)=\zeta(s)$ or any meromorphic function which is real for $s \in [1,3]$, its Taylor series $f(2+s) = \sum_{k=0}^\infty c_k s^k, c_k = \frac{f^{(k)}(2)}{k!}$ has real coefficients, thus for small $|s|$, $\overline{f(2+\overline{s})} =f(2+s)$.
Analytic continuation implies $\overline{f(2+\overline{s})} =f(2+s)$ stays true for every $s$. And hence $|f(s)|= |f(\overline{s})|$.
The only subtlety is with non-meromorphic functions such as $\log s$ which is real on $[1,3]$ but whose analytic continuation is not real on $[-3,-1]$.