Find the residue at \(\omega_n \da e^{\pi i \over n}\) of \[ f(z) = {1\over z^n + 1} .\]
Solution
Check that \(\dd{}{z} z^n+1 = nz^{n-1}\neq 0\) for \(z\neq 0\), so this has no repeated roots since \(z=0\) is not a root. Thus all of the poles are simple, and the residue of \(1/q\) at a simple zero \(z_0\) of \(q\) is \(1/q'(z_0)\): \[ \Res_{z=\omega_n} {1\over z^n + 1} &= {1 \over nz^{n-1}}\evalfrom_{z=\omega_n} \\ &= {1\over n\omega_n^{n-1}} \\ &= {\omega_n^{1-n}\over n} \\ &= -{\omega_n \over n} ,\] which follows from expanding \[ \omega_n^{1-n} = e^{i\pi (1-n) \over n} = e^{i\pi\over n}e^{-i\pi n \over n} = e^{i\pi \over n}\cdot (-1) = -\omega_n .\].