Use the following formula to show that the complex zeros of \(\sin(\pi z)\) are exactly the integers, and they are each of order 1: \[ \sin \pi z=\frac{e^{i \pi z}-e^{-i \pi z}}{2 i} .\]
Calculate the residue of \({1\over \sin(\pi z)}\) at \(z=n\in \ZZ\).
Solution
From \[ \sin(\pi z)=\frac{e^{i\pi z}-e^{-i\pi z}}{2i}, \] we have \(\sin(\pi z)=0\) exactly when \[ e^{2\pi i z}=1. \] Writing \(z=x+iy\), the modulus of the left side is \(e^{-2\pi y}\), so \(y=0\); then \(e^{2\pi i x}=1\), hence \(x\in\mathbb Z\). Thus the zeros are exactly the integers.
Moreover \[ \frac{d}{dz}\sin(\pi z)=\pi\cos(\pi z), \] and at \(z=n\) this equals \(\pi(-1)^n\ne0\), so every zero is simple. Therefore \(1/\sin(\pi z)\) has a simple pole at \(n\) with residue \[ \boxed{ \operatorname{Res}_{z=n}\frac1{\sin(\pi z)} =\frac1{\pi\cos(\pi n)} =\frac{(-1)^n}{\pi}.} \]