How to prove $L(s, \chi_0 \chi^*) = 0$ if and only if $L(s, \chi^*) = 0$?

Question

Let $\chi$ be a Dirichlet character modulo $q$ and suppose $\chi = \chi_0 \chi^*$ where $\chi_0$ is the principal character mod $q$ and $\chi^*$ is the primitive character inducing $\chi$.
I am wondering how can one prove that $L(s, \chi) = 0$ if and only if $L(s, \chi^*) = 0$? (I am only interested in $s$ in the critical strip) Any comments are appreciated. Thank you very much.

Answer 1

$$\begin{aligned}L(s,\chi_0 \chi^{\ast}) &= \prod_p \frac{1}{1-\chi(p)p^{-s}} \\ &= \prod_{p\mid q} \frac{1}{1-\chi(p)p^{-s}} \prod_{p\nmid q} \frac{1}{1-\chi(p)p^{-s}} \\ &= \prod_{p\mid q} \frac{1}{1-\chi(p)p^{-s}} \prod_{p \nmid q} \frac{1}{1-\chi^{\ast}(p)p^{-s}} \\ &= L(s,\chi^{\ast})\prod_{p\mid q} ({1-\chi^\ast(p)p^{-s}}) \end{aligned}$$

So the two $L$-functions differ by an elementary factor that is nonzero in the critical strip.