= Solution
To avoid confusing the diagonalizing <matrix> with the exponential, write $A=WDW^\dagger$ from part (b). The convergent <matrix exponential> series gives
$$
E=e^A=W e^D W^\dagger=W\operatorname{diag}(e^{\lambda_1},\ldots,e^{\lambda_n})W^\dagger.
$$
Each $\lambda_j$ is purely imaginary, so $|e^{\lambda_j}|=1$. The diagonal exponential is therefore a <unitary matrix>, and
$$
\boxed{E^\dagger E=W\operatorname{diag}(|e^{\lambda_j}|^2)W^\dagger=I.}
$$
Hence the matrix called $U=\exp(A)$ in this part is unitary. The series identity $(e^A)^\dagger=e^{A^\dagger}=e^{-A}$ gives the same conclusion directly and also covers repeated <eigenvalues>.
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