In Ch. 8 (p. 142) of the Opera de Cribro (Friedlander, Iwaniec) it is written:
To this end we apply the sieve to remove from each $p$ in a set $\mathcal{P}$ residue classes which are not squares modulo $p$. Since we are looking for an upper bound the sifting range $\mathcal{P}$ does not need to include every prime. Therefore, we choose $\mathcal{P}$ to be a large finite set which does not contain a few inconvenient primes, and we denote by $P$ the product of all primes in $\mathcal{P}$. Since the sieve weights are supported on divisors of $P$ we do not need to display the condition $d \mid P$ in various summations to follow.
In our case, the number of residue classes to be removed is given by $$\omega(p)=\frac{1}{2}(p-1),\qquad\text{for every $p \mid P$}.$$ We have $$S^\square(\mathcal{A})\leq S^+(\mathcal{A},\Omega),$$ where $S^+(\mathcal{A},\Omega)$ is given by (8.3). Our goal is to find good weights $ρ_d$ (these will be almost optimal) and to evaluate $S^+(A,\Omega)$ asymptotically for these weights.
We find that, for a non-principal character $\chi \pmod{p}$, $$\Omega_p(\chi)=\sum_{\nu\in\Omega_p}\chi(\nu)=-\omega(p)=-\frac12(p-1)$$ if $\chi = \chi_p$ is real and $\Omega_p(\chi) = 0$ if $\chi$ is complex.
My question is the assertion in last line above. I think for the real case it uses the fact that real characters come from Jacobi symbol and the fact that the summation of a character (real or complex) over the $p-1$ residues is zero, i.e. $\sum_{\nu=1}^{p-1}\chi(\nu)=0$, but the assertion in complex case in the last line above is not clear to me?