 Solution

A skew-Hermitian matrix satisfies $A^{†}=-A$, where $†$ denotes the conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenSince $v^{†}v>0$, this gives $\boxed{\overline{\lambda }=-\lambda ,\text{ so }\lambda \in i\mathbb{R}}$.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. It preserves the Hermitian inner product and norm. Thus if $Uv=\mu v$,and therefore $\boxed{|\mu |=1}$. Zero is allowed among the purely imaginary eigenvalues in the first statement, but not among the unitary eigenvalues.