 Concurrence in the magic basis

For a normalized pure two-qubit state written in a magic basis, concurrence is $\mathcal{C}=|{\sum }_j{b}_j^2|$, while $\det {\rho }_A={\mathcal{C}}^2/4$. This bilinear sum uses no complex conjugation inside its square; replacing it by ${\sum }_j|b_j{|}^2$ would incorrectly give maximal concurrence for every state.