= Solution
Reading the coefficients of the <linear map> gives
$$
\boxed{A=\begin{pmatrix}e^{i\theta}&1\\1&e^{-i\phi}\end{pmatrix},\qquad
\det A=e^{i(\theta-\phi)}-1
=2i\sin\!\left(\frac{\theta-\phi}2\right)e^{i(\theta-\phi)/2}.}
$$
Under the inverse identification, the four standard real basis vectors map respectively to $(1,0)$, $(i,0)$, $(0,1)$ and $(0,i)$ in $\mathbb C^2$. Applying $A$ and taking real and imaginary parts therefore gives the columns
$$
\begin{aligned}
\mathcal B e_1&=(\cos\theta,\sin\theta,1,0)^T,\\
\mathcal B e_2&=(-\sin\theta,\cos\theta,0,1)^T,\\
\mathcal B e_3&=(1,0,\cos\phi,-\sin\phi)^T,\\
\mathcal B e_4&=(0,1,\sin\phi,\cos\phi)^T.
\end{aligned}
$$
Thus \b[the real <matrix> is]
$$
\boxed{B=\begin{pmatrix}
\cos\theta&-\sin\theta&1&0\\
\sin\theta&\cos\theta&0&1\\
1&0&\cos\phi&\sin\phi\\
0&1&-\sin\phi&\cos\phi
\end{pmatrix}.}
$$
For verification, write its blocks as $\begin{pmatrix}R_\theta&I\\I&R_{-\phi}\end{pmatrix}$, where $R_\psi$ is the planar rotation <matrix>. Block elimination gives $\det B=\det(R_{\theta-\phi}-I)=2-2\cos(\theta-\phi)$. Therefore \b[the determinant identity is]
$$
\boxed{\det B=4\sin^2\!\left(\frac{\theta-\phi}2\right)=|\det A|^2.}
$$
This is an instance of the <realification determinant identity>. The identification is real-linear; it is not an identification of complex vector-space dimensions.
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