- Using Euler’s formula
\[ \sin \pi z = \frac {e ^ {i \pi z} - e ^ {- i \pi z}}{2 i}, \]
show that the complex zeros of sin πz are exactly at the integers, and that they are each of order 1.
Calculate the residue of 1/ sin πz at \(z = n \in \mathbb { Z }\)
Solution
By Euler’s formula, \[ \sin \pi z=0 \iff e^{i\pi z}=e^{-i\pi z} \iff e^{2\pi i z}=1. \] Write \(z=x+iy\). Then \[ e^{2\pi i z}=e^{2\pi i x}e^{-2\pi y}. \] If this equals \(1\), its modulus is \(1\), so \(e^{-2\pi y}=1\) and hence \(y=0\). Then \(e^{2\pi i x}=1\), so \(x\in\mathbb Z\). Thus the zeros are exactly the integers.
At \(z=n\in\mathbb Z\), \[ \frac{d}{dz}\sin\pi z=\pi\cos\pi z=\pi(-1)^n\ne0, \] so each zero is simple.
For a holomorphic function \(g\) with a simple zero at \(n\), the residue of \(1/g\) at \(n\) is \(1/g'(n)\). Therefore \[ \operatorname{Res}_{z=n}\frac1{\sin\pi z} =\frac1{\pi\cos\pi n} =\frac{(-1)^n}{\pi}. \]