= Solution
For any $0<r<1$, a <Laurent series> converges uniformly on the circle $|z|=r$, so termwise <contour integration> is valid. Multiplication by $z^{-k-1}$ and integration gives, for every <integer> $k$,
$$
\boxed{a_k=\frac1{2\pi i}\int_{|z|=r}\frac{f(z)}{z^{k+1}}\,dz.}
$$
Indeed, the integral of $z^{j-k-1}$ is $2\pi i$ for $j=k$ and zero otherwise. These coefficients depend only on $f$, so any two Laurent expansions must agree coefficient by coefficient. This proves uniqueness, including all terms of the <principal part>.
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