= Solution
Put $D=A-iB$, so $D^\dagger=A+iB$. Because the <Hermitian matrices> $A,B$ <commute>,
$$
D^\dagger D=DD^\dagger=A^2+B^2.
$$
For every nonzero <vector> $v$,
$$
v^\dagger(A^2+B^2)v=\|Av\|^2+\|Bv\|^2>0,
$$
since $A$ is an <invertible matrix>. Thus $D$ and $D^\dagger$ are <invertible matrices>, justifying the inverse in the definition of $U$. The <conjugate transpose> of $U=D^\dagger D^{-1}$ is $(D^\dagger)^{-1}D$. Since $D$ and $D^\dagger$ <commute>, their inverses do too, and
$$
\boxed{U^\dagger U=(D^\dagger)^{-1}DD^\dagger D^{-1}=I.}
$$
Hence $U$ is a <unitary matrix>.
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