The function $\zeta(\frac{1}{2}+it) \left[ \sqrt{2}\left( \cos(t\log 2)+i\sin(t\log 2) \right) -2 \right]$ has a numerical root

Question

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

$\quad$

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.

Answer 1

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.