The alternating zeta function and functional equation

Question

The Dirichlet eta function (the alternating zeta function) is given by

$$η(s)=∑_{n=1}^{∞}(-1)ⁿ⁻¹/n^{s}$$

The functional equation for $η(s)$ is given by

$$η(s)=ϕ(s)η(1-s)$$

where $$ϕ(s)=2(((1-2^{-1+s}))/(1-2^{s}))π^{s-1}sin(((πs)/2))Γ(1-s)$$

Apply it twice we get

$$(1-ϕ(1-s)ϕ(s))η(1-s)=0$$

Assume that $η(s)≠0$, then my question is: Can we deduce that $$1-ϕ(1-s)ϕ(s)=0$$ since $η(1-s)≠0$ or we get $$1-ϕ(1-s)ϕ(s)=0$$ for all $s∈ℂ$. Maybe I missing some thing!

Answer 1

We have

$$ϕ(s)=2(((1-2^{-1+s}))/(1-2^{s}))π^{s-1}sin(((πs)/2))Γ(1-s)$$

$$ϕ(1-s)=2(((1-2^{-1+(1-s)}))/(1-2^{(1-s)}))π^{(1-s)-1}sin(((π(1-s))/2))Γ(s)$$

Let $s=α+iβ$ such that $ϕ(s)≠0$ and $ϕ(1-s)≠0$

then the equation

$$1-ϕ(1-s)ϕ(s)=0$$ imply

$$sin(πα)cosh(πβ)-2cos((1/2)πα)sin((1/2)πα)cosh²((1/2)πβ)-2cos((1/2)πα)sin((1/2)πα)sinh²((1/2)πβ)=0$$ and $$cos(πα)sinh(πβ)-2cos²((1/2)πα)cosh((1/2)πβ)sinh((1/2)πβ)+2sin²((1/2)πα)cosh((1/2)πβ)sinh((1/2)πβ)=0$$ since the partial derivatives are all zero with respect to $α$ and $β$.