How to calculate a determinant of a 2x2 symmetry block matrix?

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I'd like to calculate the determinant of the matrix: $$ \begin{pmatrix} -A & B^\star \\ -B & A^\star \\ \end{pmatrix} $$ $A$, $B$ are $L\times L$ complex matrix.

I know that if $A$ and $B$ are real matrix, the determinant can be easily calculated: $$ \begin{pmatrix} -A & B \\ -B & A \\ \end{pmatrix} = \begin{pmatrix} A & B \\ B & A \\ \end{pmatrix} . \begin{pmatrix} -1 & 0 \\ 0 & 1 \\ \end{pmatrix} $$ and $$ \det \begin{pmatrix} A & B \\ B & A \\ \end{pmatrix} =\det(A-B)\det(A+B) $$

Is there a similar formula when $A$ and $B$ are complex matrix? Thank you for your help.

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Hint.

If $\mathbf {A^*}$ is invertible you can use the general result: $$ \det \begin{bmatrix} \mathbf A&\mathbf B\\ \mathbf C&\mathbf D \end{bmatrix}= \det (\mathbf D) \det(\mathbf A-\mathbf B\mathbf D^{-1}\mathbf C) $$ that is a consequence of the identity: $$ \begin{bmatrix} \mathbf A&\mathbf B\\ \mathbf C&\mathbf D \end{bmatrix} \begin{bmatrix} \mathbf I&\mathbf 0\\ \mathbf{-D^{-1}}\mathbf C&\mathbf D \end{bmatrix}= \begin{bmatrix} \mathbf A -\mathbf B\mathbf {D^{-1}}\mathbf C&\mathbf B\\ \mathbf 0&\mathbf D \end{bmatrix}= \begin{bmatrix} \mathbf I&\mathbf B\\ \mathbf 0&\mathbf D \end{bmatrix} \begin{bmatrix} \mathbf A -\mathbf B\mathbf {D^{-1}}\mathbf C&\mathbf 0\\ \mathbf 0&\mathbf D \end{bmatrix} $$