Prove that the number: $z = \det(A+B) \det(\overline A-\overline B)$ is purely imaginary. Problem: Let $A_{n\times n}$ and $B_{n\times n}$ be complex unitary matrices, where n is an odd number. Prove that the number:
$$z=\det(A+B) \det(\overline A-\overline B)$$ is purely imaginary.
My idea:
We have that $AA^*=I$ and $BB^*=I$.
We also have that $\det A=\det A^*=(-1)^n \det A^*=-\det A$. Same for matrix $B$.
Am I free to say $A$ and $B$ are skew-symmetric matrices?
$$z=\det(A+B)\overline {\det(A-B)}$$
$$z=\det((A+B) {(A-B)^*)}$$
$$z=\det((A+B) {(A-B)^*)}$$
$$z=\det((A+B) {(A^*-B^*))}$$
$$z=\det((AA^*-AB^*+BA^*-BB^*)$$
$$z=\det((I-AB^*+BA^*-I)$$
$$z=\det((BA^*-AB^*)$$
This is where I am unsure of how to proceed. Am I free to say since matrices are unitary and skew-symmetric, their eigenvalues are purely imaginary thus $z$ must be as well? Is this statement correct?
Thank you all in advance.

As proceeding along the lines given in the hint by @Tsermo , we find that $$ z = det(BA^{*} -AB^{*}) =\Pi_i^{n} e_i $$ where $e_i$ are the eigenvalues of $(BA^{*}-AB^{*}) $ which are purely imaginary(as the matrix$(BA^{*}-AB^{*}) $ is skew symmetric of odd order).Since n is odd therefore the result follows.
By the way how do you say that A and B are skew symmetric? In fact $det (A) = \frac{1}{det(A^{*})}$
