For $\chi$ a primitive Dirichlet character modulo $N$ satisfying $\chi(-1)=-1$, we have the (normalized) Eisenstein series of weight 1 \begin{equation}\tag{a} E_{1,\chi}(z) := \frac{N}{-4\pi i\,\tau(\overline{\chi})}\sum_{\substack{(n,m)\,\in\,\mathbb{Z}^2 \\ (n,m)\neq (0,0)}}\frac{\overline{\chi}(n)}{(n+mNz)}, \end{equation}
where $\tau(\overline{\chi})$ is the Gauss sum. We know that they are in $M_1(N,\chi)$ and have $q$-expansion
\begin{equation}\tag{b} E_{1,\chi}(z) = L(\chi,0)/2 + \sum_{n=1}^\infty \Big(\sum_{d|n}\chi(d)\Big)q^n, \quad \text{where }q=e^{2\pi i z}. \end{equation}
These series (and their generalization to other weights) are extremely important to number theory. My questions are:
$\textbf{1.}\;$ Let us suppose we have a primitive Dirichlet character with the "wrong" parity, $\chi(-1)=1$. In this case, $(a)$ simply adds up to $0$ (because the terms in the summation cancel out). However, can we still say something about the formal power series defined by $(b)$? I know it is not a modular form, but does it have any interesting properties? Is it of any relevance to number theory?
$\textbf{2.}\;$ Let us suppose our Dirichlet character $\chi$ satisfies $\chi(-1)=-1$, but is not primitive at level $N$ (say, it is of conductor $N'$). In this case, we can still define a series $E_{1,\chi}$ using $(a)$ at level $N$. But we could also define another series using $\chi^*$ (the primitive character attached to $\chi$), that is, we could consider the series $E_{1,\chi^*}$ defined by $(a)$ at level $N'$, and then regard it at level $N$ via the inclusion $M_1(N',\chi^*)\subset M_1(N,\chi)$. Notice $E_{1,\chi^*}$ and $E_{1,\chi}$ are related, but are not equal. For example, for each $t|(N/N')$ we have a different way to embed $E_{1,\chi^*}$ into level $N$, and we can write $E_{1,\chi}$ as a sum of these embeddings of $E_{1,\chi^*}$. So I wonder, is there a "correct" way to define the Eisenstein series of an imprimitive character? Is it $E_{1,\chi^*}$, $E_{1,\chi}$, or maybe something else? Whatever the definition is, do we still have the $q$-expansion given by $(b)$?