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Presuming the Riemann Hypothesis, the non-trivial zeros of the zeta function occur when both $\Re\{\zeta (s)\}$ and $\Im\{\zeta(s)\}=0$, where $s=\frac{1}{2}+it$.

and since $$\zeta(s)=2^s\pi^{s-1}\sin\bigg(\frac{\pi s}{2}\bigg)\Gamma(1-s)\zeta(1-s),$$ when $$\Im\{2^s\pi^{s-1}\sin\bigg(\frac{\pi s}{2}\bigg)\Gamma(1-s)\}=0$$ then either $\Re\{\zeta (s)\}$ or $\Im\{\zeta(s)\}=0 $, but not both.

I don't really understand how some of the imaginary / real zeros of the zeta function can be so regular and predictable, when others of the same function can be so irregular.

share|improve this question
@DietrichBurde OP refers to so-called trivial zeros. martin: take any complicated function with weird zeros, like $\sin(e^{-x} + \cos (x^2))$, and multiply it by $\sin x$. You will have some predictable zeros, and some hard-to-find ones. Same with $\zeta$: the functional equation exhibits it as a product, in which one of the terms is just a trigonometric function. –  user127096 Apr 21 '14 at 16:09

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