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Show that if $a$ is a zero of the zeta function in the critical strip, then so are $\bar{a}$, $1-a$, and $1-\bar{a}$.
The definition of $\zeta$ is $$\dfrac{1}{\zeta(s)}=\prod_p\left(1-\frac{1}{p^s}\right)$$
I don't see how to get the desired fact from this. Or perhaps we should use the definition as sum of $1/n^s$?
That is only the definition of $\zeta(s)$ for ${\rm Re}(s)>1$. The Euler product does not converge if the real part of $s$ is $\le1$*. Rather, $\zeta(s)$ is defined by the analytic continuation of the $p$-series $\sum n^{-s}$ (or I guess equivalently the Euler product $\prod (1-p^{-s})^{-1}$ if you want) to the rest of the complex plane.
This analytic continuation is achieved explicitly via the functional equation. You can conclude that $s$ is a zero iff $\bar{s}$ is a zero since $\zeta(\bar{s})=\overline{\zeta(s)}$ (as I pointed out in the comments, this follows from the series definition $\sum n^{-s}$; how does conjugation affect each term of this?) and conclude that $s$ is a zero iff $1-s$ is a zero using the functional equation.
*Actually I think the Euler product might converge for some $s$ with real part $1$, IIRC.
