The question is kind of self contained, but let me develop a bit further. 


Assume K is a CM field of degree $2g$, that is, a quadratic imaginary extension of a totally real field. A CM type of $K$ is a set $\Phi$ consisting of $g$ complex embeddings of $K$ such that $\mathrm{Hom}(K, \mathbb{C})=\Phi \cup \overline{\Phi}$. 


Given $(K, \Phi)$, there always exists a $g$-dimensional abelian variety $A$ such that $\mathrm{End}(A) \otimes \mathbb{Q}=K$ and that $K$ acts on $H^0(A, \Omega^1\_A)$ through $\Phi$. One easily constructs as a complex torus, starting from the embedding $K \subset \mathbb{C}^g$ given by $\Phi$. 


The following question seems much more subtle:


*Is there always such an abelian variety of the form $A=\mathrm{Jac}(C)$ for a smooth projective curve $C$?*



can all CM types be realized by Jacobians?
``Given $(K,\Phi)$ , there always exists a $g$ -dimensional abelian variety $A$ such that $End(A)\otimes Q=K$ and that $K$ acts on $H^0(A,\Omega^1)$ through $\Phi$. One easily constructs as a complex torus, starting from the embedding $K\subset C^g$ given by $\Phi$."


Actually, this is not always the case. For example, if $K$ is a quartic CM-field containing an imaginary quadratic subfield (i.e., $K$ is a compositum of two imaginary quadratic fields) then it is not isomorphic to the endomorphism algebra of any abelian surface (or a complex torus), see Sect. 5 of arXiv:1312.0377 [math.NT].


However, if $[K;Q] \le 6$ and there exists a simple complex $g$-dimensional CM-abelian variety $B$ (with $g\le 3$) of CM-type $(K,\Phi)$ then there exists a jacobian $J$ isogenous to $B$. (It's true, because the Siegel upper half-plane $H\_g$ is the $Sp(2g,R)$-orbit, $Sp(2g,Q)$ is everywhere dense in $Sp(2g,R)$ in the classical topology, and the ``Torelli locus" is open in $H\_g$ if $g\le 3$ and therefore meets every (dense) $Sp(2g,Q)$-orbit; compare with Remark 3 at the end of Sect. 2 in arXiv:0912.4325 [math.NT]). This implies $End(J)\otimes Q=K$ and the CM-type of $J$ is $\Phi$.


For big $g$ the situation seems to be murky; however, as far as I understand, it is expected that not every $K$ is isomorphic to the endomorphism algebra of a $g$-dimensional jacobian.


