 Solution

The shifted matrix $N=D-\lambda I$ is still a normal matrix, sinceand $N{N}^{†}$ has the same expansion except that its first term is $D{D}^{†}$, which equals $D^{†}D$. For the given eigenvector, $Ne=0$. Part (b), applied to $N$, therefore gives $N^{†}e=0$. Since $N^{†}=D^{†}-\overline{\lambda }I$,The corresponding eigenvalue of the Hermitian conjugate is the complex conjugate of the original eigenvalue.