Where are the nonreal zero's of $\zeta(s) - \zeta(s + 1) + \zeta(s + 2) $ ??

Question

Let $\zeta(s)$ be the Riemann Zeta function.

Define the function $f(s) = \zeta(s) - \zeta(s+1) + \zeta(s+2)$

I know $f(s)$ is defined on the whole complex plane and has a simple pole at $-1$, $0$ and at $1$ and no other poles.

I also know the function $f(z)$ has infinitely many zero's on the negative real line.

I wonder where are the nonreal zero's of $\zeta(s) - \zeta(s + 1) + \zeta(s + 2) $ ??

$$\zeta(s) - \zeta(s + 1) + \zeta(s + 2) = 0 $$

I found $ s = -5.957 + 2.776 i$ and $ s = 0.380 + 0.770 i$ and their conjugates.

I did not find zero's with $Re(s) > 1/2 $ by hand. Do they exist ??

And how about zero's with $Re(s) > 1 $ ?

Is there a real number $a$ such that any zero satisfies $Re(s) < a $ ?

Can this be analyzed theoretically ( without computer search ) ?

Similar question for $g(s) = \zeta(s) - \zeta(s+1) + \zeta(s + \frac{5}{4})$

I found $0.193 + 0.541 i$ as solution to $g(s) = 0$.

Answer 1

To answer one of your questions: for $s=\sigma+it$ with $\sigma>1$, the bounds $$ \frac{\zeta(2\sigma)}{\zeta(\sigma)} \le |\zeta(s)| \le \zeta(\sigma) $$ are easy to verify from the Euler product for $\zeta(s)$. Therefore $$ |f(s)| \ge \frac{\zeta(2\sigma)}{\zeta(\sigma)} - \zeta(\sigma+1) + \frac{\zeta(2(\sigma+2))}{\zeta(\sigma+2)} $$ for $\sigma>1$, and the right-hand side is strictly positive for all $\sigma>1.4867$.