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.
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.