= Solution
Choose the common <orthonormal eigenbasis> established in part (ii), and let $U$ be the <unitary matrix> of its columns. Then
$$
U^\dagger AU=\operatorname{diag}(a_1,\ldots,a_n),\qquad
U^\dagger BU=\operatorname{diag}(b_1,\ldots,b_n).
$$
Consequently
$$
\boxed{U^\dagger(A+iB)U=\operatorname{diag}(a_1+ib_1,\ldots,a_n+ib_n).}
$$
Thus $N$ is unitarily diagonalizable, even though it need not be Hermitian. As a consistency check, commutation also gives $NN^\dagger=N^\dagger N=A^2+B^2$, so $N$ is a <normal matrix>, as also follows from <unitary diagonalization of a normal matrix>.
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