For which real $a$ does the series $\sum_{k=0}^{\infty}\frac{1}{(1+6k)^{ia}}-\frac{1}{(5+6k)^{ia}}$ vanish? The question is to find when will $L(\chi, s)$ of $\chi=\left(\frac{-12}{n}\right)$ vanish. So  $\left(\frac{-12}{n}\right)$ is defined as the Jacobi symbol for odd $n$ ($\chi($even$)$=0). Using quadratic reciprocity I got $$\left(\frac{-12}{n}\right)=\left(\frac{-1}{n}\right)\left(\frac{3}{n}\right)\left(\frac{4}{n}\right)=\left(\frac{n}{3}\right)$$ so we have $$L(\chi,s)=\sum_{n=1\pmod 3, n\,\text{odd}}\frac{1}{n^s}-\sum_{n=2\pmod 3, n\,\text{odd}}\frac{1}{n^s}.$$ And we are only interested in when $s=ia$. How can I find where $L(\chi, s)$ vanish for those $s$?
 A: Your character is $\chi(n)=(\frac{n}3)1_{2\ \nmid\ n}$, the Dirichlet series is $L(\chi,s)=\sum_{n\ge 1}\chi(n)n^{-s}$.
It doesn't converge anywhere on $\Re(s)=0$. You need the analytic continuation, for a non-trivial Dirichlet character it amounts to $L(\chi,it)=\lim_{\sigma\to 0^+}L(\chi,\sigma+it)$.
$$L(\chi,0)=\lim_{s\to 0} L(\chi,s)=\lim_{s\to 0} s\int_1^\infty (\sum_{n\le x} \chi(n))x^{-s-1}dx$$
The mean value of $\sum_{n\le x} \chi(n)$ is $\frac23$.
$\sum_{n\le x} \chi(n)-\frac23$ is periodic zero-mean implies that
$$\int_1^\infty (\sum_{n\le x} \chi(n)-\frac23)x^{-s-1}dx$$ converges for $\Re(s) > -1$. Whence
$$L(\chi,0)=\lim_{s\to 0} s\int_1^\infty \frac23 x^{-s-1}dx= \frac23$$
Next, following the same argument as for $\zeta(s)$ we can show that $L((\frac{n}3),s)\zeta(s)$ has no zeros on $\Re(s)=1$, whence $L((\frac{n}3),s)$ has no zeros on $\Re(s)=1$. The functional equation implies that $L((\frac{n}3),s)$ has no zeros  on $\Re(s)=0$, and $$L(\chi,s)=(1+2^{-s})L((\frac{n}3),s)$$ implies that its zeros on $\Re(s)=0$ are at $\frac{(2 \Bbb{Z}+1)i\pi}{\log 2}$.
