= Solution
A <skew-Hermitian matrix> satisfies $A^\dagger=-A$, where $\dagger$ denotes the <conjugate transpose>. If $Av=\lambda v$ with $v\ne0$, then
$$
\overline\lambda\,v^\dagger v=v^\dagger A^\dagger v=-v^\dagger Av=-\lambda\,v^\dagger v.
$$
Since $v^\dagger v>0$, this gives $\boxed{\overline\lambda=-\lambda,\text{ so }\lambda\in i\mathbb R}$.
A <unitary matrix> satisfies $U^\dagger U=UU^\dagger=I$. It preserves the <Hermitian inner product> and <norm>. Thus if $Uv=\mu v$,
$$
\|v\|^2=\|Uv\|^2=|\mu|^2\|v\|^2,
$$
and therefore $\boxed{|\mu|=1}$. Zero is allowed among the purely imaginary <eigenvalues> in the first statement, but not among the unitary <eigenvalues>.
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