If $X$ is a smooth finite type separated DM algebraic stack over $\mathbb C$ with coarse space $X^c$, then do we know whether the analytification of $X^c$ is the coarse space of the analytification of $X$? That is, do we have $$(X^c)^{an} = (X^{an})^c?$$

My stack X is actually even simpler, as its inertia stack I_X->X is finite etale. The coarse space $X^c$ of X is in this case the rigidification of X with respect to I_X, so that $X$ is an $I_X$-gerbe over $X^c$. I presume that the coarse space of $X^{an}$ should be the rigidification of $X^{an}$ with respect to $I_{X^{an}}$ in this case. Is that true?

"Definition." The coarse space of $X^{an}$ is a morphism of complex analytic stacks $X^{an} \to (X^{an})^c$ satisfying the usual universal properties, e.g., for a complex analytic space $V$ and morphism $X^{an}\to V$ there is a unique $(X^{an})^c\to V$ such that $X^{an}\to V$ factors via this morphism.

Idea: There is (by the universal property of $(X^{an})^c$) a morphism $(X^{an})^c\to (X^c)^{an}$. This morphism is finite of degree one, I think. Thus an application of "Zariski's Main Theorem" (in the analytic category) should conclude the proof. Does this work?