Let $$\space\space\space \eta(z) = \sum_{a=1}^{\infty} \frac{1}{a^{z}} \cdot (-1)^{a-1} $$
Now let $\space$$z = \sigma + it $ ,
$$\eta(\sigma + it) = \sum_{a=1}^{\infty} \frac{1}{a^{\sigma + it}} \cdot (-1)^{a-1} $$
May we split $\space\space$$\cfrac{1}{a^{\sigma + it}}$ $\space\space$ into its real and complex components:
$$\frac{1}{a^{\sigma + it}} = \frac{1}{a^{\sigma}} \cdot \left(\frac{1}{a^{it}}\right) = \frac{1}{a^{\sigma}} \cdot \left(\frac{1}{(e^{\ln{a}})^{it}}\right) = \frac{1}{a^{\sigma}} \cdot \left(\frac{1}{e^{(i\cdot t\ln{a})}}\right) = \frac{1}{{a^{\sigma}}} \cdot e^{-(i\cdot t\ln{a})}$$
$ \\ $ $ \\ $ $$ \qquad \qquad \quad = \frac{1}{{a^{\sigma}}} \cdot \bigg( \cos(-t\ln{a}) + i\cdot \sin(-t\ln{a}) \bigg) = \frac{1}{{a^{\sigma}}} \cdot \bigg( \cos(t\ln{a}) - i\cdot \sin(t\ln{a}) \bigg) $$
Therefore,
$$\space\space\space\Re \big(\eta(\sigma+it) \big) = \sum_{a=1}^{\infty} \frac{1}{a^{\sigma}}\cdot \cos(t\ln{a}) \cdot (-1)^{a-1}$$
$\\$
$$\Im \big(\eta(\sigma+it) \big) = \sum_{a=1}^{\infty} \frac{1}{a^{\sigma}}\cdot \sin(t\ln{a}) \cdot (-1)^{a} $$
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Where $\Re$ and $\Im$ respectively represent the real and complex components of its argument.
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For $z$ to be a root of $\eta$ we thus require that $\Re \big(\eta(z)\big) = 0 = \Im \big(\eta(z)\big) $.
Presumably there will be no such $z$ , with $\Re(z) = \sigma\neq \frac{1}{2} $ satisfying the condition above (for sure if Riemann Hypothesis turns to be true).
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So now I will focus on the numbers $z$, such that $\Re(z) = \frac{1}{2}$.
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Given that, the unknown of the problem will be $\space$ $t = \Im(z)$ $\space$ for which $\space$ $ \eta(\frac{1}{2} + it) = 0$.
Let me now reformulate what we're working on:
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$$ \eta(\tfrac{1}{2} + it) = \sum_{a=1}^{\infty} \frac{1}{a^{\frac{1}{2} + it}} \cdot (-1)^{a-1} $$
Which leads us to:
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$$ \space\space\space \Re \big(\eta(\tfrac{1}{2}+it) \big) = \sum_{a=1}^{\infty} \frac{1}{a^{\frac{1}{2}}}\cdot \cos(t\ln{a}) \cdot (-1)^{a-1}$$
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and
$$ \Im \big(\eta(\tfrac{1}{2}+it) \big) = \sum_{a=1}^{\infty} \frac{1}{a^{\frac{1}{2}}}\cdot \sin(t\ln{a}) \cdot (-1)^{a}$$
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Now letting $\space$ $a^{\frac{1}{2}} = \sqrt{a} $ $\space$, and writing it by extend:
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$$ \Re \big(\eta(\tfrac{1}{2}+it) \big) = \frac{1}{\sqrt{1}} \cdot \cos(t\ln{1}) - \frac{1}{\sqrt{2}} \cdot \cos(t\ln{2}) + \frac{1}{\sqrt{3}} \cdot \cos(t\ln{3}) - \dots $$
$ \\ $
$$ \Im \big(\eta(\tfrac{1}{2}+it) \big) = -\frac{1}{\sqrt{1}} \cdot \sin(t\ln{1}) + \frac{1}{\sqrt{2}} \cdot \sin(t\ln{2}) + \frac{1}{\sqrt{3}} \cdot \sin(t\ln{3}) + \dots $$
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Onward,
$ \\ $
$$ \space \space \Re \big(\eta(\tfrac{1}{2}+it) \big) \space = \space 1 - \frac{1}{\sqrt{2}} \cdot \cos(t\ln{2}) + \frac{1}{\sqrt{3}} \cdot \cos(t\ln{3}) - \dots $$
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$$ \space \space \space \Im \big(\eta(\tfrac{1}{2}+it) \big) \space = \space 0 \space + \frac{1}{\sqrt{2}} \cdot \sin(t\ln{2}) - \frac{1}{\sqrt{3}} \cdot \sin(t\ln{3}) + \dots $$
$ \\ $ $ \\ $ $ \\ $
For $\space$ $z = \frac{1}{2} + it$ $\space$ to be a root of $\eta$, we ought to require that:
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$$ \space \space 0 \space = \space 1 - \frac{1}{\sqrt{2}} \cdot \cos(t\ln{2}) + \frac{1}{\sqrt{3}} \cdot \cos(t\ln{3}) - \dots $$
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$$ \space \space 0 \space = \space 0 \space + \frac{1}{\sqrt{2}} \cdot \sin(t\ln{2}) - \frac{1}{\sqrt{3}} \cdot \sin(t\ln{3}) + \dots $$
$ \\ $ $ \\ $ $ \\ $
$$\therefore$$
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$$ \frac{1}{\sqrt{2}} \cdot \cos(t\ln{2}) - \frac{1}{\sqrt{3}} \cdot \cos(t\ln{3}) + \dots = 1 $$
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$$ \frac{1}{\sqrt{2}} \cdot \sin(t\ln{2}) - \frac{1}{\sqrt{3}} \cdot \sin(t\ln{3}) + \dots = 0 $$
$ \\ $ $ \\ $ $ \\ $
This system of equations doesn't look too alien, at least.
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Let's dive into the first equation, taken it's left side to be a function $f$ of $t$
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$$ f = \frac{1}{\sqrt{2}} \cdot \cos(t\ln{2}) - \frac{1}{\sqrt{3}} \cdot \cos(t\ln{3}) + \dots $$
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We will be then interested in finding the numbers $t$ for which $\space$ $f = 1$ $\space$ .
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Using the expansion
$$\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \space \space \cdots \space \space (-1)^{j} \cdot \frac{x^{2j}}{(2j)!} \space \space \cdots $$
$ \\ $ $ \\ $ We can rewrite $f$ as
$ \\ $
$$ f = \frac{1}{\sqrt{2}} \cdot \bigg( 1 - \frac{(t\ln{2})^{2}}{2} + \frac{(t\ln{2})^{4}}{24} - \cdots \bigg) - \frac{1}{\sqrt{3}} \cdot \bigg( 1 - \frac{(t\ln{3})^{2}}{2} + \frac{(t\ln{3})^{4}}{24} - \cdots \bigg) + \cdots $$
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which we can reassemble to be written as a infinite polynomial in $t$:
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$$ f = \bigg( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + \cdots \bigg) - \frac{t^{2}}{2} \bigg( \frac{(\ln{2})^{2}}{\sqrt{2}} - \frac{(\ln{3})^{2}}{\sqrt{3}} + \cdots \bigg) + \frac{t^{4}}{24} \bigg( \frac{(\ln{2})^{4}}{\sqrt{2}} - \frac{(\ln{3})^{4}}{\sqrt{3}} + \cdots \bigg)-\cdots $$
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Now we've reached a remarkable point.
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The above polynomial's coefficients are no more than just constant numbers. And those numbers are pretty familiar.
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The independent coefficient,
$$\bigg( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + \cdots \bigg)$$
is equal to
$$ 1 - \Re \big(\eta(\tfrac{1}{2}) \big)$$
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The quadratic coefficient,
$$ -\frac{1}{2}\bigg( \frac{(\ln{2})^{2}}{\sqrt{2}} - \frac{(\ln{3})^{2}}{\sqrt{3}} + \cdots \bigg) $$
is equal to
$$ -\frac{1}{2} \bigg( \frac{d^2 \Re \big(\eta(\tfrac{1}{2}+it) \big)}{dt^2} \bigg) _{t=0} $$
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And the quartic coefficient
$$ \frac{1}{24}\bigg( \frac{(\ln{2})^{4}}{\sqrt{2}} - \frac{(\ln{3})^{4}}{\sqrt{3}} + \cdots \bigg)$$
is equal to
$$ \frac{1}{24} \bigg( - \frac{d^4 \Re \big(\eta(\tfrac{1}{2}+it) \big)}{dt^4} \bigg) _{t=0} $$
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And so on with the subsequent coefficients.
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We can thus write $f$ as:
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$$ f = 1 - \Re \big(\eta(\tfrac{1}{2}) \big) + \frac{t^{2}}{2} \bigg( \frac{d^2 \Re \big(\eta(\tfrac{1}{2}+it) \big)}{dt^2} \bigg) _{t=0} - \frac{t^{4}}{24} \bigg(\frac{d^4 \Re \big(\eta(\tfrac{1}{2}+it) \big)}{dt^4} \bigg) _{t=0} + \cdots $$
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If we want $f = 1$, then $t$ must be the roots of $\space$ $\beth$ $\space$, where
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$$ \beth = \Re \big(\eta(\tfrac{1}{2}) \big) - \frac{t^{2}}{2} \bigg( \frac{d^2 \Re \big(\eta(\tfrac{1}{2}+it) \big)}{dt^2} \bigg) _{t=0} + \frac{t^{4}}{24} \bigg(\frac{d^4 \Re \big(\eta(\tfrac{1}{2}+it) \big)}{dt^4} \bigg) _{t=0} - \cdots $$
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Compactly written as
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$$ \beth = \sum_{l=0}^{\infty} (-1)^{l} \cdot \frac{t^{2l}}{(2l)!} \cdot \bigg( \frac{d^{(2l)} \Re \big(\eta(\tfrac{1}{2}+it) \big)}{dt^{(2l)}} \bigg) _{t=0}$$
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Let's go through an analogous procedure now with the second of our two main equations.
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First, taken it's left side to be a function $g$ of $t$
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$$ g = \frac{1}{\sqrt{2}} \cdot \sin(t\ln{2}) - \frac{1}{\sqrt{3}} \cdot \sin(t\ln{3}) + \dots $$
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We will be then interested in finding the numbers $t$ for which $\space$ $g = 0$ $\space$ .
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Using the expansion
$$\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \space \space \cdots \space \space (-1)^{j} \cdot \frac{x^{2j+1}}{(2j+1)!} \space \space \cdots $$
$ \\ $ $ \\ $ We can rewrite $g$ as
$ \\ $
$$ g = \frac{1}{\sqrt{2}} \cdot \bigg( (t\ln{2}) - \frac{(t\ln{2})^{3}}{6} + \frac{(t\ln{2})^{5}}{120} \cdots \bigg) - \frac{1}{\sqrt{3}} \cdot \bigg( (t\ln{3}) - \frac{(t\ln{3})^{3}}{6} + \frac{(t\ln{3})^{5}}{120} \cdots \bigg) + \cdots $$
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which we can reassemble to be written as an infinite polynomial in $t$:
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$$ g = t \bigg( \frac{(\ln{2})}{\sqrt{2}} - \frac{(\ln{3})}{\sqrt{3}} + \cdots \bigg) - \frac{t^{3}}{6} \bigg( \frac{(\ln{2})^{3}}{\sqrt{2}} - \frac{(\ln{3})^{3}}{\sqrt{3}} + \cdots \bigg) + \frac{t^{5}}{120} \bigg( \frac{(\ln{2})^{5}}{\sqrt{2}} - \frac{(\ln{3})^{5}}{\sqrt{3}} + \cdots \bigg)$$
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$$ - \cdots $$
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Those polynomial's coefficients are also familiar.
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The linear coefficient,
$$ \bigg( \frac{(\ln{2})}{\sqrt{2}} - \frac{(\ln{3})}{\sqrt{3}} + \cdots \bigg) $$
is equal to
$$ \bigg( \frac{d \Im \big(\eta(\tfrac{1}{2}+it) \big)}{dt} \bigg) _{t=0} $$
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The cubic coefficient,
$$ - \frac{1}{6} \bigg( \frac{(\ln{2})^{3}}{\sqrt{2}} - \frac{(\ln{3})^{3}}{\sqrt{3}} + \cdots \bigg) $$
is equal to
$$ \frac{1}{6} \bigg( \frac{d^3 \Im \big(\eta(\tfrac{1}{2}+it) \big)}{dt^3} \bigg) _{t=0} $$
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And the quintic coefficient,
$$ \frac{1}{120} \bigg( \frac{(\ln{2})^{5}}{\sqrt{2}} - \frac{(\ln{3})^{5}}{\sqrt{3}} + \cdots \bigg) $$
is equal to
$$ \frac{1}{120} \bigg( \frac{d^5 \Im \big(\eta(\tfrac{1}{2}+it) \big)}{dt^5} \bigg) _{t=0} $$
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So on with the subsequent coefficients we can write $g$ as:
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$$ g = t \bigg( \frac{d \Im \big(\eta(\tfrac{1}{2}+it) \big)}{dt} \bigg) _{t=0} + \frac{t^{3}}{6} \bigg(\frac{d^3 \Im \big(\eta(\tfrac{1}{2}+it) \big)}{dt^3} \bigg) _{t=0} + \frac{t^{5}}{120} \bigg(\frac{d^5 \Im \big(\eta(\tfrac{1}{2}+it) \big)}{dt^5} \bigg) _{t=0} + \cdots $$
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As we want $g = 0$, then $t$ must be the roots of $\space$ $\gimel$ $\space$, where
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$$ \gimel = t \bigg( \frac{d \Im \big(\eta(\tfrac{1}{2}+it) \big)}{dt} \bigg) _{t=0} + \frac{t^{3}}{6} \bigg(\frac{d^3 \Im \big(\eta(\tfrac{1}{2}+it) \big)}{dt^3} \bigg) _{t=0} + \frac{t^{5}}{120} \bigg(\frac{d^5 \Im \big(\eta(\tfrac{1}{2}+it) \big)}{dt^5} \bigg) _{t=0} + \cdots $$
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Compactly written as
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$$ \gimel = \sum_{l=0}^{\infty} \frac{t^{2l+1}}{(2l+1)!} \cdot \bigg( \frac{d^{(2l+1)} \Im \big(\eta(\tfrac{1}{2}+it) \big)}{dt^{(2l+1)}} \bigg) _{t=0}$$
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So far this is the most explicit representation of $t$, which is the imaginary component of a non-trivial root of $\eta$(or $\zeta$), I could reach.
$ \\ $ To partially conclude this post, and to state what has been accomplished: $ \\ $ $ \\ $ There are two infinite polynomials, $\beth$ and $\gimel$, defined as mentioned, such that their common roots happen to be the imaginary component of the (some, probably all) non-trivial roots of the Riemann Zeta function.
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I'd like to thank you if you went along all through this point, and to say that any comments or suggestions will be greatly welcomed.