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If we consider the two functions $\zeta (-3+it)$ and $\zeta(4+it)$, then according to the argument equation derived from the functional equation of the Riemann Zeta function

$Arg(\zeta(s))=Arg(\chi(s))+Arg(\zeta(1-s))$

and the equation of arguments of conjugate functions

$Arg(\zeta(\overline{1-s}))=Arg(\overline{\zeta(1-s)})=-Arg(\zeta(1-s))$

we obtain

$Arg(\zeta(s))-Arg(\chi(s))/2=-(Arg(\zeta(\overline{1-s}))-Arg(\chi(s))/2)$

then

1) in the coordinate system rotated by angle $-Arg(\chi(s))/2$ the angles of the vectors $\zeta (-3+it)$ and $\zeta (4+it)$ will be symmetric with respect to the axis rotated by angle $-Arg (\chi(s))/2$

2) in a fixed coordinate system $Arg (\zeta(4+it))\leq\varepsilon$, therefore in a moving coordinate system, the vectors $\zeta (-3+it)$ and $\zeta(4+it)$ will rotate in opposite directions, the vector $\zeta(-3+it)$ is clockwise, and the vector $\zeta(4+it)$ is counterclockwise

can we say that this rule holds for any values $\sigma<1/2$ for two functions $\zeta(\sigma+it)$ and $\zeta(1-\sigma+it)$ at any interval between the zeros of these functions (if they exist, because at the point of zero, perhaps $Arg(\zeta(\sigma+it))$ and $Arg(\zeta(1-\sigma+it))$ will have gaps)?

added 28.10

Is it possible to say something definite in this case about changing the module $\Delta|\zeta(s)_L|$, for example, if in the moving coordinate system $\pi<Arg(\zeta(s))<3\pi/4$ when $\sigma<1/2$, then $ \Delta|\zeta(s)_L|<0$, i.e. the module decreases, and if $\pi<Arg(\zeta(s))<\pi/2$ when $\sigma<1/2$, then $\Delta|\zeta(s)_L|>0$, i.e. the module increases?

Where $L$ is a line rotated by angle $-Arg(\chi(s))/2$ and $\zeta(s)_L$ is a vector of projection of $\zeta(s)$ on this line.

Because in this case in the moving coordinate system the vector $\zeta (s)$ rotates clockwise.

kkapitonets
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  • Sure, the only thing you didn't clarify is how you define $arg\ \zeta(s)$ – reuns Oct 25 '19 at 10:55
  • Dear, reuns, in this formulation, the specific value of $Arg(\zeta(s))$ does not matter, it is only about the change of $Arg(\zeta(s))$ and the direction of rotation of the vectors $\zeta(\sigma+it)$ and $\zeta (1-\sigma+it)$ in the moving coordinate system. – kkapitonets Oct 25 '19 at 11:39
  • Can I insert such a Lemma into my paper:

    In a moving coordinate system rotated by an angle $-Arg(\chi(s))$ relative to the axes of the complex plane for any values $\sigma<1/2$ for two functions $\zeta(\sigma+it)$ and $\zeta(1-\sigma+it)$ at any interval between the zeros of these functions (if they exist) them vectors of value will rotate in opposite directions, the vector $\zeta(\sigma+it)$ is clockwise, and the vector $\zeta(1-\sigma+it)$ is counterclockwise. Will these arguments be enough to prove this Lemma? Will the presence of zeros significantly affect the rotation of the vectors?

     – kkapitonets Oct 25 '19 at 11:40
  • The variation of $arg\ \zeta(s)\in \Bbb{R/2\pi Z}$ in direction $v$ is precisely the imaginary part of $v\zeta'/\zeta(s)$, then $\zeta'/\zeta(s)$ is meromorphic on the whole complex plane and it satisfies $\zeta'/\zeta(s) =\chi'/\chi(s)-\zeta'/\zeta(1-s)$ where $\chi'/\chi(s)$ is meromorphic too with poles only at negative even integers and positive odd integers – reuns Oct 25 '19 at 11:50
  • The fact is that in a fixed coordinate system, the vector $\zeta(s)$ at $\sigma>1/2$ makes not rotational but oscillatory motion, i.e. it does not make a complete revolution around the zero of the complex plane.

    I don't quite understand what the direction $v$ means in this case-is it a vector?, then it is not clear what $v\zeta'/\zeta(s)$ is a scalar product? And how do I go from $\zeta'/\zeta (s)=\chi'/\chi(s)-\zeta'/\zeta(1-s)$ to the statement I need to prove?

     – kkapitonets Oct 25 '19 at 13:13

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