Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 / Analysis / 5 / a / 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 $k$,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.