= Solution
For a <polynomial> $f(x)=\sum_{j=0}^d a_jx^j$, the defining identity and uniqueness of the <adjoint operator> give
$$
f(T)^*=\sum_{j=0}^d\overline{a_j}(T^*)^j.
$$
Indeed, the <adjoint operator> of a sum is the sum of the <adjoint operators>, the adjoint of $aT$ is $\bar aT^*$, and the adjoint of a product $AB$ is $B^*A^*$. Since the <normal operator> $T$ commutes with $T^*$, each power of $T$ commutes with each power of $T^*$. Multiplying the two finite sums therefore gives
$$
\boxed{f(T)f(T)^*=f(T)^*f(T)}.
$$
Hence <polynomials of a normal operator are normal>, including constant <polynomials>.
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