= Solution
For any complex $\lambda$, the shifted <linear operator> $T=\alpha-\lambda I$ is normal: expansion of $T^*T-TT^*$ leaves just $\alpha^*\alpha-\alpha\alpha^*=0$. Apply part (i) to this shifted <normal operator>:
$$
\ker(\alpha-\lambda I)=\ker(\alpha^*-\bar\lambda I).
$$
Hence a nonzero vector is an <eigenvector> of $\alpha$ with <eigenvalue> $\lambda$ if and only if it is an <eigenvector> of $\alpha^*$ with <eigenvalue> $\bar\lambda$. \b[The eigenspaces coincide, with conjugate eigenvalues.]
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