= Solution
Split $f=f_-+a_0+f_+$ into strictly negative and strictly positive Fourier parts, and let $A=T(f_-)$, $B=T(f_+)$. The Toeplitz multiplication rule is $T(g)T(h)=T(gh)$ when $g$ is coanalytic or $h$ is analytic. It follows directly because the discarded negative Fourier part cannot return to the Hardy space in those cases. Thus exponentiation within either subalgebra is exact, and
$$
T(e^f)T(e^{-f})=e^Ae^Be^{-A}e^{-B}.
$$
The constant terms cancel. If $S=T(z)$ is the unilateral shift, $A=\sum_{m>0}a_{-m}(S^*)^m$ and $B=\sum_{n>0}a_nS^n$. The <commutator> $[(S^*)^m,S^n]$ is finite <rank>, with <trace> $n$ when $m=n$ and zero otherwise. For equal powers it is the projection onto the first $n$ basis <vectors>; for unequal powers its <matrix> has no diagonal entries. Its <trace norm> is bounded by $2\min(m,n)$. Smoothness of $f$ makes the weighted double coefficient sum finite, so the <commutator> <series> converges in <trace norm> and
$$
\operatorname{Tr}[A,B]=\sum_{n>0}n a_na_{-n}.
$$
Applying part (c) proves the <Toeplitz exponential determinant identity>
$$
\boxed{\det(T(e^f)T(e^{-f}))=\exp\left(\sum_{n>0}na_na_{-n}\right).}
$$
In particular the product is an invertible identity-plus-trace-class operator, so the <determinant> in the formula is well-defined.
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