Show that the complex zeros of \(f(z) \da \sin(\pi z)\) are exactly \(\ZZ\), and each is order 1. Calculate the residue of \(1/\sin(\pi x)\) at \(z=n\in \ZZ\).
Solution
Goal: Show that the complex zeros of \(f(z) = \sin(\pi z)\) are exactly \(\ZZ\), each of order 1, and compute the residue of \(1/\sin(\pi z)\) at \(z = n \in \ZZ\).
1. The zeros of \(\sin(\pi z)\) in \(\CC\) are exactly the integers.
\(\sin(\pi z) = (e^{i\pi z} - e^{-i\pi z})/2i\), so \(\sin(\pi z) = 0\) iff \(e^{2i\pi z} = 1\) iff \(2i\pi z = 2\pi i k\) for some \(k \in \ZZ\) iff \(z = k \in \ZZ\).
2. Each zero \(z = n\) is simple.
\(\ddd{z}\sin(\pi z) = \pi \cos(\pi z)\), and \(\cos(\pi n) = (-1)^n \neq 0\) for \(n \in \ZZ\).
3. \(1/\sin(\pi z)\) has a simple pole at each \(z = n \in \ZZ\).
The denominator vanishes to order exactly 1 at each integer (<1>2) and the numerator is the nonzero constant 1.
4. The residue of \(1/\sin(\pi z)\) at \(z = n\) is \((-1)^n/\pi\).
For a simple pole, \(\Res_{z = n} \frac{1}{\sin(\pi z)} = \lim_{z \to n} (z - n)\frac{1}{\sin(\pi z)} = \frac{1}{\sin'(\pi n)} = \frac{1}{\pi \cos(\pi n)} = \frac{1}{\pi (-1)^n} = \frac{(-1)^n}{\pi}\).
5. Q.E.D.
<1>1–<1>2 identify the zeros and their orders; <1>3–<1>4 give the residues.