TITLE: Taylor's results on zeros of the linear combination of Gamma-completed Riemann Zeta functions
QUESTION [9 upvotes]: Let $s,z$ be two complex variables, 
$\zeta(s)$ be the Riemann $\zeta$-function. Let
\begin{equation}
\zeta_1(s)=\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s),
\end{equation}
be the Gamma-completed Riemann zeta function, and let
\begin{equation}
f(s)=\zeta_1(s+1/2)-\zeta_1(s-1/2)
\end{equation}
It satisfies the functional relation $f(s)=-f(1-s)$
Let
\begin{equation}
F(z)=f(iz+1/2)
\end{equation}
Taylor proved in 1945 (On the Riemann Zeta function) that all the zeros of $F(z)$ are real.
We plotted $Re(F(x+I y))=0$ and $Im(F(x+I y))=0$ in Mathematica 9.0.


We found out that at $x=0,y=\pm 6.8246$, $F(x+I y)=0$.  Thus $F(z)$ seemed to have non-real zeros.
Has anyone notice this problem?
EDIT:
 Here is a snapshot of Taylor's formulas.

Here is a little more

REPLY [3 votes]: Here is a 3D plot of daniel's function $F(s)=F(\sigma+i t)=F(1/2+x+i y)$.  The dots that pops out of the figure are the zeros because I am plotting $-\log|F(s)|$ vs. $s$.

It seems to me that these two zeros that are not on the $\sigma=1/2$ line are the only exceptions. See two figures below.