At the page fractional calculus, completed Riemann zeta
it is claimed that the symmetric functional equation for the Dirichlet eta function is:
Formula 7.3.2 $$\Gamma\left(\frac{z}{2}\right)\pi^{\Large-\frac{z}{2}}(1-2^z)\eta(z)=\Gamma\left(\frac{1-z}{2}\right)\pi^{\Large-\frac{1-z}{2}}(1-2^{1-z})\eta(1-z) \;\;\;\;\;\;\;(1)$$
Where $z \neq 0,1 $
and the following equation that I have not checked:
$$\pi^{\Large-\frac{z}{2}}\Gamma\left\{\frac{1}{2}\left(\frac{1}{2}+z\right)\right\}\left(1-2^{\Large\frac{1}{2}+z}\right)\eta\left(\frac{1}{2}+z\right)=\pi^{\Large\frac{z}{2}}\Gamma\left\{\frac{1}{2}\left(\frac{1}{2}-z\right)\right\}\left(1-2^{\Large\frac{1}{2}-z}\right)\eta\left(\frac{1}{2}-z\right) \;\;\;\;\;\;\;\;\;(2)$$
Where $z \neq \pm \frac{1}{2} $
But is the first equation really correct? Is the Dirichlet eta function completed in this way? I am new to the symmetric functional equation for the Riemann zeta function but to my mind this seems to be equivalent to completing the zeta function as:
$$\zeta(1-z)\zeta(z)$$
which other mathematicians would say is not the way to complete the zeta function.
What is the correct way to complete the Dirichlet eta function?
$$\eta(s)=\zeta(s)\left(1-1/2^{s-1}\right)$$
I don't understand your objection; there's nothing like $\zeta(1 - s) \zeta(s)$ appearing here. (Also, it is customary to use $s$ to denote a complex variable, not $z$, when referring to zeta functions.)
The completed zeta function \[\Lambda(s) = \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s)\] satisfies the functional equation \[\Lambda(1 - s) = \Lambda(s)\] for all $s \in \mathbb{C} \setminus \{0,1\}$. As \[\eta(s) = (1 - 2^{1 - s}) \zeta(s),\] it follows that \[\Lambda(1 - s) = \pi^{-(1-s)/2} \Gamma\left(\frac{1 - s}{2}\right) \zeta(1 - s) = \pi^{-(1-s)/2} \Gamma\left(\frac{1 - s}{2}\right) (1 - 2^s)^{-1} \eta(1 - s) \] and that \[\Lambda(s) = \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s) = \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) (1 - 2^{1 - s})^{-1} \eta(s),\] and these are both equal by the functional equation.
If you want the symmetric functional equation (though I can't imagine why you would), then you just replace $s$ with $1/2 + s$. The result, after some simple rearranging, is \[\pi^{-\frac{1}{2} \left(\frac{1}{2} - s\right)} \Gamma\left(\frac{1}{2} \left(\frac{1}{2} - s\right)\right) \left(1 - 2^{\frac{1}{2} - s}\right) \eta\left(\frac{1}{2} - s\right) = \pi^{-\frac{1}{2} \left(\frac{1}{2} + s\right)} \Gamma\left(\frac{1}{2} \left(\frac{1}{2} + s\right)\right) \left(1 - 2^{\frac{1}{2} + s}\right) \eta\left(\frac{1}{2} + s\right).\]
Again, this is all follows trivially from the functional equation for the Riemann zeta function and some trivial rearranging.