Denote by $\zeta$ the Riemann zeta function. Does $\Re\zeta(s)$ ever vanish for $\frac{1}{2}<\Re(s)\leq 1$ ?
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4$\begingroup$ Why do you care about that region? The real part can vanish at numbers where ${\rm Re}(s) > 1$. See pdfs.semanticscholar.org/2e18/…. $\endgroup$– KConradCommented Jun 30, 2018 at 11:17
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1$\begingroup$ The are quantitative results for what is called the "argument of the Riemann zeta function", which essentially tells you how often the graph of the zeta function circles around the origin. Clearly this also gives lower bound on how often the real part vanishes. Try to google for "argument of the Riemann zeta function". $\endgroup$– Kurisuto AsutoraCommented Jul 3, 2018 at 7:07
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4 Answers
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Maybe like this. For $0<x<1$ we have $\mathrm{Re}\;\zeta(x)<0$ and $\mathrm{Re}\;\zeta(x+i)>0$, so for some $y = y(x)$ between $0$ and $1$ we have $\mathrm{Re}\;\zeta(x+iy) = 0$.
answered Jun 30, 2018 at 13:39
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Actually, much more is true by Theorem 11.10 in Titchmarsh: The theory of the Riemann zeta function (see also the remarks after the theorem in the book).
Let $\frac{1}{2}<a<b<1$. Then, by the quoted result (which is probably due to Bohr), for any nonzero complex number $c$ and for any sufficiently large positive number $T$, there are $\gg T$ points in the rectangle $a<\Re(s)<b$ and $0<\Im(s)<T$ such that $\zeta(s)=c$. The implied constant here depends on $a,b,c$.
answered Jul 2, 2018 at 20:01
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Mathematica seems to think the answer is yes:
Plot[Re[Zeta[9/16 + I t]], {t, 0, 100}, PlotStyle -> Thick]
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where the authors investigate the behavior of the real and imaginary parts of the Riemann zeta function and especially the vanishing thereof.
answered Jun 30, 2018 at 12:10