Let $N,a\in\mathbf{Z}\_{\geq 1}$. Define a partial $\zeta$-function as
$$
\zeta(s;N,a):=\sum\_{\substack{n\geq 1\newline n\equiv a\pmod{N}}} \frac{1}{n^s}
$$
where $Re(s)>1$. Let $\omega$ be either the trivial character or the sign character i.e. $x\mapsto sign(x)$. Define a partial $\Psi$-function as
$$
\Psi(s,\omega;N,a):=\sum\_{\substack{0\neq n\in\mathbf{Z}\newline n\equiv a\pmod{N}}}\frac{\omega(n)}{|n|^s},
$$
for $Re(s)>1$. Then it is well known that $\zeta(s;N,a)$ and $\Psi(s,\omega;N,a)$ admit a meromorphic continuation to $\mathbf{C}$ with at most of pole of order $1$ at $s=1$. If $\omega$ is the sign character then
$$
\Psi(s,\omega;N,a)+\Psi(s,1;N,a)=2\zeta(s;N,a)
$$
and 
$$
\zeta(s;N,a)-\zeta(s;N,-a)=\Psi(s,\omega;N,a).
$$
Note that when $N>1$, the functions $\Psi$ and $\zeta$ do not have an Euler product. So here are 2 natural questions:


Q1: For a fixed $N$ do we know if there exists a constant $C\_N>1$ such that if $Re(s)>C\_N$ then $\Psi$ and $\zeta$ do not vanish (if the answer is yes then how to prove it)?


Q2: What do we know in general about the nontrivial zeros of $\Psi$ and $\zeta$?



non-trivial zeros of partial zeta functions 
The answer to question 1 is classical: Any Dirichlet series which has a finite abscissa of absolute convergence has a zero-free half-plane.


Suppose the Dirichlet series
$$ A(s)=\sum\_{n=1}^\infty \frac{a\_n}{n^s}$$
has an abscissa of absolute convergence $\sigma\_a$. If $a\_m$ is the first non-zero coefficient, then for $\Re(s)>\sigma\_a$
$$ m^s A(s) = a\_m + a\_{m+1} \Big(\frac{m}{m+1}\Big)^s + a\_{m+2} \Big(\frac{m}{m+2}\Big)^s +\cdots \to a\_m \ \text{ as } \ \Re(s) \to +\infty. $$
Hence, there exists an absolute constant $C$ such that for $\Re(s)>C$ we have
$$ \left|a\_{m+1} \Big(\frac{m}{m+1}\Big)^s + a\_{m+2} \Big(\frac{m}{m+2}\Big)^s +\cdots \right| \leq \frac{|a\_m|}{2}.$$
Consequently, $m^s A(s)$ has no zeros in the half-plane $\Re(s)>C$.


