/sci/ - Science & Math

Consider a number [math]z_0\in\mathbb{C}[/math] such that
[eqn]
z_0=-(\aleph_\mathcal{X}+b)+iy~~,\quad\text{where}\qquad b,y\in\mathbb{R}_0~~.
[/eqn]

Observe that the Euler product form of [math]\zeta[/math]takes [math]z_0[/math] as
[eqn]
\zeta(z_0)=\prod_{p}\cfrac{1}{1-p^{(\aleph_\mathcal{X}+b)-iy}}\nonumber\\
\qquad= \left(\cfrac{1}{1-P^{(\mathcal{X}\aleph_1+b)-iy}} \right) \prod_{p\neq P}\cfrac{1}{1-p^{(\aleph_\mathcal{X}+b)-iy}} \nonumber\\
\qquad=\left(\cfrac{1}{1-\cfrac{\vphantom{\hat1}1}{P^b}\big(P^{\mathcal{X}}\big)^{\!\widehat\infty}\left[ \cos(y\ln P)-i\sin(y\ln P)\right]} \right) \prod_{p\neq P}\cfrac{1}{1-p^{(\aleph_\mathcal{X}+b)-iy}} \nonumber~~.
[/eqn]

Let [math]y\ln P=2n\pi[/math] for some prime [math]P[/math] and [math]n\in\mathbb{N}[/math] or [math]n=0[/math]. Then
[eqn]
\zeta(z_0)=\cfrac{1}{\infty} \prod_{p\neq P}\cfrac{1}{1-p^{(\aleph_\mathcal{X}+b)-iy}}=0~~.
[/eqn]