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Let $S$ be the set of complex $N\times N$ matrices that are both unitary and hermitian.

I have observed the following fact. For any pair of matrices $A$, $B$ from $S$, the value of $\det(A+iB)$ is an imaginary number if $N \equiv 2 \text{ mod } 4$.

Any ideas how to prove this?

All I could do was compute $$\det(A+iB)^*=\det(A^\dagger-iB^\dagger)=\det(A^{-1}-iB^{-1})=\det(A-iB),$$ but this did not help me much.

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    $\begingroup$ Try putting $A=B=I$. $\endgroup$ – user647486 Apr 3 at 14:38
  • $\begingroup$ @user647486 the fact does not hold for all dimensions, sorry $\endgroup$ – thedude Apr 3 at 14:50
  • $\begingroup$ Take $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $B=I$. $\endgroup$ – user647486 Apr 3 at 15:38
  • $\begingroup$ Without computing a thing it was always clear that the property is obviously false, since an analytic function that returns values on a line must be constant. $\endgroup$ – user647486 Apr 3 at 17:37
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Consider$$[\det(A+iB)]^2=\det[(A+iB)^2]=\det[A^2+iAB+iBA-B^2]\tag1$$Now note that due to $A$ (similar for $B$) being Hermitian, $A^\dagger=A$. Due to unitarity, $A^\dagger=A^{-1}$. Combining, $$A=A^{-1}\implies A^2=\Bbb I$$ So $(1)$ becomes $$i^n\det(AB+BA)$$ If $n\equiv2\mod4$, then $i^n=-1$. So $$[\det(A+iB)]^2=-\det(AB+BA)\tag2$$ Thus, if we can show that $\det(AB+BA)\ge0$ for all $A,B\in S$, then your hypothesis holds. However, inspired by user647468's comment, we can take $B=\Bbb I\in S$, and $A_{ij}=(-1)^{i+1}\delta_{ij}$. Then $$\det(AB+BA)=2^n\prod_{i=1}^{n}(-1)^{i+1}=-2^n<0\tag3$$Hence the $RHS$ of $(2)$ can be positive, and so $\det(A+iB)$ is not necessarily imaginary.

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