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A simple observation of the behavior of $\zeta(s), s=\sigma +it$ that I wonder if there's a explanation for:

Take $t_k$ as the height of the $kth$ non-trivial zero $z_k$ on the critical line ($\sigma=0.5$), i.e.,

$z_k=0.5+t_k i$

$\zeta(z_k)=0$.

Now consider $z'_k$ a short distance $\epsilon$ away parallel to the real axis, i.e.,

$z'_k= z_k+\epsilon$

$\epsilon \in \mathbb{R}, 0 < \epsilon \ll 1$

The observation has to do with the orientation of $\zeta(z'_k)$ and $\zeta(1-z'_k)$ in the complex plane.

E.g., for $t_{100}=236.524229...$ and $\epsilon=.001$ (origin of complex plane at top center): enter image description here

Looking at a number of such cases, it would appear that in the horizontal vicinity of a zero on the critical line ($\epsilon$ small), $\zeta(s)$ and $\zeta(1-s)$ approach being equal in magnitude and reflected around the imaginary axis, i.e., the bisector between them is oriented either at $+\pi/2$ or $-\pi/2$.

Does this reflection around the imaginary axis follow from the functional equation?

EDIT:

As an illustration of what's happening, here's an animation:

https://www.youtube.com/watch?v=H3t-0zYrs2w

Joe Knapp
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  • No functional equation here and you are confusing the real and imaginary axis. For $\Re(s) = 1/2$ then $s = \overline{1-s}$ so $\zeta(s) = \overline{\zeta(1-s)}$. For $\Re(s) \approx 1/2$ then $s \approx \overline{1-s}$ so $\zeta(s) \approx \overline{\zeta(1-s)}$. – reuns Apr 08 '19 at 18:15
  • But does that explain the reflection around the imaginary axis? They could be approximately equal at another angle, no? For $\sigma=0.5$ the bisector is $0$ or $-\pi$ (the real axis). – Joe Knapp Apr 08 '19 at 18:23
  • I believe the axis of reflection in this case as stated is the imaginary one. – Joe Knapp Apr 08 '19 at 18:30
  • The complex conjugaison is $a+ib \mapsto \overline{a+ib}=a-ib$ – reuns Apr 08 '19 at 18:42
  • The chart above shows the values for $s=0.501+i*236.52422967...$ which can be checked, per MATLAB, zeta(s)=0.0022438 - 0.0032966i and zeta(1-s)=-0.0022474 - 0.00331172i, close to a reflection around the imaginary axis as shown (slight difference in magnitude). – Joe Knapp Apr 08 '19 at 18:53
  • Nope, it's consistent. For all zeros $t_k$'s tested, for $\sigma$ offset by a small amount, the axis of reflection is always the imaginary axis. Even for $\sigma$ pretty far off, the bisector angle is close to vertical. For example, for s=0.6+i∗236.52422967..., the angle is $-1.05(\pi/2)$. As you go through the zero from left to right, the angle flips by $\pi$. – Joe Knapp Apr 08 '19 at 19:43
  • come on $\zeta(1/2+x+it) =\zeta(1/2+it)+x\zeta'(1/2+it)+O(x^2)$ if $\zeta(1/2+it)=0$ then .. – reuns Apr 10 '19 at 07:03
  • It is what it is, which anyone can check. That the axis of reflection as $\epsilon$ gets small (not $0$!) approaches the imaginary axis is hardly a coincidence or approximation, any more than any numerical calculation is an approximation--in this case about 14 significant figures using MATLAB zeta(). It may not be that profound, just an observation that I wonder if there's an explanation for. – Joe Knapp Apr 10 '19 at 10:13

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This equality of angles holds for any values of a complex variable in a movable coordinate system that is rotated by an angle of $-Arg(\chi(s))/2$.

It is easy to explain if we consider the argument equation of the functional equation of the Riemann zeta function:

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

If we substitute

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

and subtract from both parts of the equation $Arg(\chi(s))/2$ (which corresponds to a rotation by an angle of $-Arg (\chi(s))/2$), we obtain:

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

By the way it is the key to the Riemann hypothesis,

because when $\sigma<1/2$ the vectors $\zeta(s)$ and $\zeta(\overline{1-s})$ rotate in a moving coordinate system, and is alternately aligned with the axes of this moving coordinate system, and therefore the projection of the vectors $\zeta(s)$ and $\zeta(\overline{1-s})$ (each of them separately) on these axes cannot be zero simultaneously, since these projections are conjugate harmonic functions, that define the Riemann zeta function, then Riemann zeta function cannot have zeros with $\sigma<1/2$ and $\sigma>1/2$.

kkapitonets
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