= Solution
The specified Bell-state <phases> give
$$
a_1=\frac{b_2+ib_3}{\sqrt2},\quad
a_4=\frac{b_2-ib_3}{\sqrt2},\quad
a_2=\frac{ib_1+b_4}{\sqrt2},\quad
a_3=\frac{ib_1-b_4}{\sqrt2}.
$$
Therefore
$$
a_1a_4-a_2a_3
=\frac12(b_2^2+b_3^2+b_1^2+b_4^2),
\qquad
\boxed{\det\rho_A=\frac14\left|\sum_{j=1}^4b_j^2\right|^2.}
$$
The sum contains complex squares, not squared moduli. Define $q=\sum_jb_j^2$ and $\mathcal C=|q|$. Normalization gives $\sum_j|b_j|^2=1$, so the <triangle inequality> ensures $0\leq\mathcal C\leq1$.
The <characteristic polynomial> is $\lambda^2-\lambda+\mathcal C^2/4$. Hence
$$
\boxed{\lambda_\pm=\frac12\left(1\pm\sqrt{1-\mathcal C^2}\right),\qquad
S_A=-\lambda_+\ln\lambda_+-\lambda_-\ln\lambda_-.}
$$
Here logarithms are natural and $0\ln0$ means its limiting value zero. Using logarithms to base two gives entropy in bits instead. <Product states> have $\mathcal C=0$, <spectrum> $(1,0)$ and entropy zero; <maximally entangled states> have $\mathcal C=1$, <spectrum> $(1/2,1/2)$ and entropy $\ln2$.
This phased <Bell basis> is a <magic basis>, and $\mathcal C$ is the pure-state <concurrence>. The paper's ordering is a permutation of the usual magic-basis ordering, which leaves $|\sum_jb_j^2|$ unchanged.<Entanglement of a Pair of Quantum Bits by Scott Hill and William K. Wootters>[Eqs.(3)–(6), magic basis, concurrence and binary entropy]{ref}
\Image[/past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2016/paper-1-concurrence-entropy.png]
{title=Original two-qubit calculation of the reduced eigenvalues and entropy as functions of concurrence}
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