(Sorry for my poor english..) Let $F(z)=\sum_{n=1}^{\infty} a(n)q^n\in S_{2k}(\Gamma_0(N),\chi_0))$ be a newform with trivial character $\chi_0$. For $\text{Re}(s)>>0$, we can define \begin{equation} L(F,s):=\sum_{n=1}^{\infty} \frac{a(n)}{n^s} \end{equation} and by analytic continuation, we can define modular $L$-function. And many mathematicians also research the quadratic twists of modular L-function. That is, for a fundamental discriminant $D$, let $\chi_{D}$ be a character of $\mathbb{Q}(\sqrt{D})$, then for $\text{Re}(s)>>0$, \begin{equation} L(F\otimes \chi_{D}, s)=\sum_{n=1}^{\infty} \frac{a(n)\chi_{D}(n)}{n^s} \end{equation}. (And by analytic continuation, we can define $L(F\otimes \chi_{D},s).$)
I think many mathematicians are interested in $L(F\otimes \chi_{D},k)$ with $(D,N)=1$. For example, Lemma 9.2 in Ono's book "The web of modularity : Arithmetic of the coefficients of Modular forms and $q$-series", \begin{equation} \Lambda(F\otimes \chi_{D},s)=\epsilon\cdot \chi_{D}(-N)\Lambda(F\otimes \chi_{D},2k-s) \end{equation} where \begin{equation} \Lambda(F,s):=(2\pi)^{-s}\Gamma(s)N^{s/2}L(F,s) \end{equation} and there is an $\epsilon\in \{\pm 1\}$ such that \begin{equation} \Lambda(F,s)=\epsilon \Lambda(F,2k-s). \end{equation} My question is..
Is it true for when $(D,N)\neq 1$? And why are many mathematicians interested in $(D,N)=1$?