non-trivial zeros of partial zeta functions  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 $\zeta$?
 A: Micah's answer answers your Q1. My reply gives some information on your Q2. Let us assume that $1 \leq a\leq N$ for simplicity (this condition can be removed).  Your zeta-functions $\zeta$ and $\Psi$ can be expressed by the Hurwitz zeta-function http://en.wikipedia.org/wiki/Hurwitz_zeta_function. For your first function
$$\zeta(s;N,a)=N^{-s} \zeta \left(s,\frac a N \right).$$
Davenport and Heilbronn proved that for any $c>0$, $a/N \neq 1/2,1$  there exists $\gg T$ zeros of this function in the strip $1<\Re(s)<1+c, | \Im (s)| < T $. By Voronin universality for the Hurwitz zeta-function the same can be said in any strip $1/2<\sigma_0< \Re (s)<\sigma_1<1$ (this can be found for example in Jörn Steuding's SLN or Garunkstis-Laurincikas book on the Lerch zeta-function).
Your second function can also be expressed in terms of the Hurwitz zeta-function 
$$\Psi(s,\omega;N,a)=N^{-s} \left(\zeta\left(s,\frac a N\right)+\omega(-1)\zeta \left( s,1-\frac a N \right) \right).$$
In this case joint universality of the Hurwitz zeta-function (see same references as above) can be used to show that except when $\Psi$ has an Euler-product (which only occurs for some small special cases when $ a / N$ equals 1/6, 1/4, 1/3, 1/2, 2/3, 3/4, 5/6, or  1.) there are $\gg T $ zeros in any strip $1/2<\sigma_0< \Re (s) < \sigma_1 < 1$ and $|\Im(s)| < T$. Davenport-Heilbronn's result for $\Re(s)>1$ can be obtained in this case as well.
For the special cases where your function has an Euler-product it will essentially be a Dirichlet L-function or the Riemann zeta-function, and its zeros will be the same.
A: EDIT Experimentally the zeros of $\zeta(s;5,1)$ and $\zeta(s;6,1)$ are those of zeta (the previous revision incorrectly included wrong zeros caused by insufficient precision).
The zeros of $\zeta(s;6,2)$ (if computed correctly) don't appear related to those of zeta.


sage code:
import mpmath
mpmath.mp.pretty=True
mpmath.mp.dps=100
N=5
a=1
an=[0]*N
an[a-1]=1
def L(x):
    return mpmath.dirichlet(x,an)

def search1():
    cac={}
    rr=list(range(-20,0))
    P=[]
    for k in xrange(1,40):
        rr += [0.1 + mpmath.j*k]
    for x in rr:
        try:
            r=mpmath.findroot(L,[x],maxsteps=1000)

        except:  continue
        print r
        ks=str(r)
        if ks in cac:  continue
        cac[ks]=1
        P += [(RR(r.real),RR(r.imag))]
    return P

P=search1()
pt=points(P)
pt.show()

A: 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 $\Re(s)>C$.
A: This is only a comment regarding the functional equation of the $\Psi$-function. So define
$$
\Psi^*(s,\omega;N,a):=\sum_{0\neq n\in\mathbf{Z}}\frac{\omega(n)}{|n|^s}e^{-2\pi i\frac{a}{N}n}
$$ 
Then one can show using Riemann's classical idea (see for example 1) that
$$
(-1)^p N^s\cdot\Gamma_{\infty}(s)\Psi(s,\omega;N,a)=\Gamma_{\infty}(1-s)\Psi^*(1-s,\omega;N,a),
$$
where $p=0$ (resp. p=$1$) if $\omega=1$ (resp. $\omega=sign$) and 
$\Gamma_{\infty}(s):=\pi^{-s/2}\Gamma(\frac{s+p}{2})$.
So it seems that many of the non-trivial zeros of the $\Psi$-functions are located along the symmetry axis of the functional equation. Is there some heuristic argument that could explain this phenomenon? 
