The non-trivial zeros of $L(s, \chi)$ are symmetrically placed with respect to the critical line

Question

Let $\chi$ be a primitive Dirichlet character modulo $q$. My book says that "The non-trivial zeros of $L(s, \chi)$ are symmetrically placed with respect to the critical line $\sigma=1/2$". Now, if $\rho$ is a non-trivial zero of $L(s, \chi)$, then, by the functional equation, one has that $1-\rho$ is a zero of $L(s, \overline{\chi})$, and so $1-\overline{\rho}$ is a zero of $L(s, \chi)$ (since $\overline{L(s, \chi)}=L(\overline{s}, \overline{\chi})$). But, in general, $\rho$ and $1-\overline{\rho}$ are not symmetrically placed with respect to the critical line. Am I wrong?

Answer 1

Write $\rho = \beta + i \gamma$. Then $1- \overline{\rho} = (1-\beta) + i \gamma$ is indeed the reflection of $\rho$ with respect to the critical line $\mathrm{Re}(s) = 1/2$.

Notice that if $\chi$ is a real character, then $L(\rho, \chi) = 0$ iff $1 - \rho$ is also a zero of $L(s, \chi)$ iff $\overline{\rho}$ is a zero of $L(s, \chi)$.