Let $K=\mathbb{Q}(\sqrt{D})$ with discriminant $D<0$ and $\mathfrak{a}$ be an ideal in $K$. For $Q \in \mathbb{N},\rho \in \mathfrak{a},Norm(\mathfrak{a})=A,$ we define the theta series $$\vartheta(\tau;\rho,\mathfrak{a},Q\sqrt{D})=\sum_{\mu \equiv\rho(\mathfrak{a}Q\sqrt{D})}e^{2\pi i \tau\frac{\mu \overline \mu}{AQ|D|}}. $$ One can also write this theta series as $$\vartheta(\tau;\rho,\mathfrak{a},Q\sqrt{D})=\sum_{n=0}^\infty c\Bigl(\frac{n}{|QD|}\Bigr)e^{2\pi i \tau \frac{n}{A|DQ|}}.$$ Why do the coefficients vanish if $n$ contains a prime with prime power $1$ that is not split in $K$?
If $n=p$ that is inert in $K$ then there are no ideals in $K$ of norm $p$. Now I have to lead the equation $N(\mu)=p$ to a contradiction but I do not know how.
$$\zeta_K(s)=\sum_{c\in C_K} \zeta_{K,c}(s)$$ we know the Euler product of the LHS in term of split/inert/ramified primes
and the coefficients of $\zeta_{K,(1)}$ are those of your theta series.