Show that the complex zeros of \(f(z) \da \sin(\pi z)\) are exactly \(\ZZ\), and each is order 1. Calculate the residue of \(1/\sin(\pi z)\) at \(z=n\in \ZZ\).
Solution
Write \[ f(z) = \sin(\pi z) = \inverseof{(2i)} (e^{i\pi z} - e^{-i\pi z}) = 0 \iff e^{i 2\pi z} = 1 = e^{i 2k\pi} \iff 2\pi z = 2k\pi \iff z=k\in \ZZ .\] To see that these zeros are order one, write \[ \sin(\pi z) &= \sin(\pi(z-k) + k\pi) \\ &= \pm \sin(\pi(z-k)) \\ &= \pm\qty{ \pi(z-k) - {\pi^3\over 3!}(z-k)^3 + \cdots } \\ &= (z-k)^1 \cdot \pm \qty{ \pi - {\pi^3\over 3!}(z-k)^2 + \cdots } \da (z-k)g(z) \\ \] where \(g(k) = \pm \pi \neq 0\), making \(z=k\) an order 1 zero.
For the residues: \[ \Res_{z=k} \csc(\pi z) = \lim_{z\to k}{z-k\over \sin(\pi z)} = {1\over \pi\cos(\pi k)} = {(-1)^k\over \pi} .\]