\(\displaystyle \oint f(z)dz/(z^n-a) = \frac{2\pi i}{na^{((n-1)/n)}}\sum_{k=0}^{n-1}f(\zeta _{k})exp^{2\pi ik(n-1)/n}\)

where: \(\displaystyle \zeta_{k}=a^{1/n}exp^{2\pi ik/n}\) and k=0,1,2,...n-1

If true, this would be very interesting, at least to me, because it means a Fourier transform (or something very like one) is really just the sum of the residues of a function evaluated at the complex roots of \(\displaystyle a^{1/n}\)