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A glimpse of figures in this preprint seems to suggest that curves $\Re{\zeta(s)}=0$ (or $\Im{\zeta(s)}=0$) do not touch each other in the half-plane $\Re{s}>1$.

Question: Is there any conditional(e.g. RH is true) argument for such an observation?

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    $\begingroup$ For any meromorphic function $f$ and any $\theta$, the curves comprising $\mathrm{Arg}f(s)=\theta$ can only touch (or rather, intersect) in a root or pole of $f$. $\endgroup$ Aug 7, 2014 at 11:59
  • $\begingroup$ @EmilJeřábek: I'm not certain whether your statement is correct. Let $s_0$ be a zero of $\zeta^{\prime}(s)$ and $Arg\zeta(s_0)=\theta$. Let $f(s)=e^{-i\theta}\zeta(s)$. Then $\Im{f}=0$ touch at $s_0$. $\endgroup$
    – Y. Zhao
    Aug 7, 2014 at 12:35
  • $\begingroup$ Ah yes, you are right. So, the question is really if/why $\zeta$ doesn’t take real or purely imaginary values at zeros of $\zeta'$ in the given region. I’d say that since $\zeta'$ has only countably many zeros that do not follow a particularly regular pattern, it would need a very good reason for $\zeta$ to attain a special argument value at them. $\endgroup$ Aug 7, 2014 at 13:55
  • $\begingroup$ @EmilJeřábek: I remember that the universality theorem of Voronin shows that fixing real part of s between 1/2 and 1, $(\log(\zeta(s)),\zeta^\prime(s)/\zeta(s))$ is everywhere dense in $\mathbb{C}^2$, so it might be a very hard problem even assuming RH. I have changed the half plane to $\Re{s}>1$. $\endgroup$
    – Y. Zhao
    Aug 7, 2014 at 14:06
  • $\begingroup$ It is intriguing looking at these curves since they remind me of this: link .The curves there are for real part of gamma function equal to 1 and imaginary part equal to 0. Interesting... $\endgroup$ Aug 7, 2014 at 16:33

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