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All:

Are any zeros of Riemann zeta function and the zeros of the derivatives of Riemann zeta function same ?

They shall be all different, right ? Is there a proof of this statement ?

Thank you.

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At the moment, this is completely unknown. If $\zeta(s) = \zeta'(s) = 0$, then $s$ is (at least) a double zero. We do not know whether $\zeta(s)$ should have multiple-zeroes or not. Further, multiple zeroes are somewhat independent of the Riemann Hypothesis. There doesn't seem to be any particular reason in particular for them to, or not to, occur.

In terms of what is currently known, in 1974 Montgomery showed that the Riemann Hypothesis implies that about 67 percent of zeroes of the Riemann zeta function are simple zeroes. [Montgomery, Distribution of zeros of the Riemann zeta function, Proc. Internat. Congr. Math. Vancouver, 1974, pp. 379-381].

In much greater generality, one might consider the same question for general $L$-functions. In short, many $L$-functions do have zeroes of higher multiplicity. The Birch and Swinnerton-Dyer Conjecture indicates that the multiplicity of a particular zero of an $L$-function relates to the rank of an associated elliptic curve, for instance. Some Dedekind zeta functions over number fields also have multiple zeroes.

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Following on from the post by @davidlowryduda, the zeros of the derivative $\zeta'(s)$ of the Riemann zeta-function are intimately connected with the behavior of the zeros of $\zeta(s)$ itself. Indeed, a theorem by Speiser (Speiser, A., Geometrisches zur Riemannschen Zetafunktion Math. Ann. 110 514–21 (1934)) states that the Riemann Hypothesis is equivalent to $\zeta '(s)$ having no zeros to the left of the critical line. Thus, understanding the properties the zeros of $\zeta '(s)$ can provide important tools and insight into the study of RH.

As mentioned already, and Montgomery proved a quantitative refinement of Speiser’s theorem. He was able to show that showed that $\zeta(s)$ and $\zeta '(s)$ have essentially the same number of zeros to the left of the critical line $\sigma = \Re(s) =\frac{1}{2}$, and proved that as $T \rightarrow \infty$, where $T$ is the height on the critical line, a positive proportion of the zeros of $\zeta'(s)$ lie in the region

$$\sigma < \frac{1}{2}+(1+ \epsilon) \frac{\log \log T}{\log T},\qquad \epsilon >0$$

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