= Solution
The <adjoint operator> $\alpha^*$ is defined by
$$
\langle\alpha v,w\rangle=\langle v,\alpha^*w\rangle
\qquad(v,w\in V).
$$
Since $V$ is a nonzero finite-dimensional complex vector space, $\alpha$ has an <eigenvector> $v$ with eigenvalue $\lambda$. Normality gives
$$
\|(\alpha-\lambda I)x\|^2
=\|(\alpha^*-\overline\lambda I)x\|^2
$$
for every $x$, by expanding both sides and using $\alpha\alpha^*=\alpha^*\alpha$. Taking $x=v$ yields
$$
\boxed{\alpha v=\lambda v,\qquad
\alpha^*v=\overline\lambda v}.
$$
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