= Solution
The shifted <matrix> $N=D-\lambda I$ is still a <normal matrix>, since
$$
N^\dagger N=D^\dagger D-\lambda D^\dagger-\overline\lambda D+|\lambda|^2I,
$$
and $NN^\dagger$ has the same expansion except that its first term is $DD^\dagger$, which equals $D^\dagger D$. For the given <eigenvector>, $Ne=0$. Part (b), applied to $N$, therefore gives $N^\dagger e=0$. Since $N^\dagger=D^\dagger-\overline\lambda I$,
$$
\boxed{D^\dagger e=\overline\lambda e.}
$$
The corresponding <eigenvalue> of the <Hermitian conjugate> is the <complex conjugate> of the original <eigenvalue>.
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