Using the complex exponentiation (this is the MathWolrd's Page) one can deduce for $t>0$ $$2^{\frac{1}{2}+it}=\sqrt{2}e^0(\cos(t\log 2)+i\sin(t\log 2)),$$ since $a=2,b=0,c=\frac{1}{2}, d=t$ and $\arg 2=0$, thus from $$\zeta(s)= \left( 1-2^{1-s} \right)^{-1} \eta(s)$$ that holds for $0<\Re s<1$ where $\zeta(s)$ is the Riemann Zeta function and $\eta(s)$ is the Dirichlet Eta function, one can write by specialization the following

Lemma. For $t>0$ $$\eta(\frac{1}{2}+it)\sqrt{2}(\cos(t\log 2)+i\sin(t\log 2))$$ $$\qquad\qquad\qquad=\zeta(\frac{1}{2}+it) \left[ \sqrt{2}\left( \cos(t\log 2)+i\sin(t\log 2) \right) -2 \right]. $$


Question 1. Is previous Lemma right?

I believe that yes by my easy computations that is a direct deduction of previous identities but is appreciated a yes or a proof with more details to do a comparison with mine.

When I check previous with Wolfram Alpha *, I've asked to me if

Question 2. Can you convince to me, with a mathematical explanation that $$\zeta(\frac{1}{2}+it) \left[ \sqrt{2}\left( \cos(t\log 2)+i\sin(t\log 2) \right) -2 \right]$$ has a numerical root about $\approx-1.06^{-19} + 2.5i$? You can work also with the other expression, I say that involves the Dirichlet eta function if it is more easy to tell us why we've a root with more or less those real and imaginary part. Thanks in advance.

  • $\begingroup$ * My code for Wolfram Alpha online calculator were DirichletEta[1/2+it]*sqrt(2)*(cos(t log(2))+i sin(t log(2))) (next) Zeta[1/2+it]*(sqrt(2)*(cos(t log(2))+i sin(t log(2)))-2) $\endgroup$ – user243301 Jun 6 '16 at 11:11
  • $\begingroup$ I do not get your definition of root here. Do you mean your function vanishes for $t=2.5i?$ This would be trivial because $\zeta(1/2+5/2\times i\times i)=\zeta(-2)=0$ $\endgroup$ – gammatester Jun 6 '16 at 11:55
  • $\begingroup$ I say that Wolfram Alpha say that there is a root. My Question 2 is to clarify, on assumption that previous Lemma and my computaions with the online calculator are rights, if there is a such root and how to give a reasoning to find. Then you say that there is a root in $0+\frac{3}{2}i$, please add an answer with the proof, I then I understand that it is easy by yourself previous computations. Ver thanks much @gammatester Of course you can presume those trivial zeros of the Riemann zeta as you said with your qick answer $\endgroup$ – user243301 Jun 6 '16 at 13:03

Your function has roots for $t=(2n+\tfrac{1}{2})i, \; n \in \mathbb{Z}>0$. These are the well-known trivial zeroes of the $\zeta$ function, because $\frac{1}{2}+(2n+\tfrac{1}{2})i\times i= -2n.\;$ The trivial zeroes are a consequense of the $\sin$ term in the reflection formula.

PS: I guess you are interested in the zeroes on the critical line $\tfrac{1}{2} + ti\;$ with $t \in \mathbb{R},$ so $t=2.5i\;$ is not what you really want.

  • $\begingroup$ Very thanks much then for your answer, I understand that previous Lemma was right. Now I am studying your zeros. My problem is that I am asking without a good viewpoint of the theory of such functions. It is appreciated your answer because clarify my doubts when from the online calculator was computing the root, then thanks for your trick to get related with $\zeta(-2)$ to compute the right root (I don't know why Wolfram Alpha online calculator don't know your trick!). $\endgroup$ – user243301 Jun 6 '16 at 13:23
  • $\begingroup$ My main purpose is to obtain good computations and reasonings. Very thanks much also for the explanation that the place of trivial zeros is consequence of the reflection function. Thanks truly for your trick. $\endgroup$ – user243301 Jun 6 '16 at 13:28

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