Hadamard expansion of $\sin$

\(\sin(\pi z)\) has order 1, and its zero set is \(z_k = k\) for \(k\in \ZZ\). So one can write \[ \sin(\pi z) = ze^{az+b} \prod_{k\in \ZZ\smz} \qty{1 - {z\over k}}e^{z\over k} = ze^{az+b} \prod_{k\geq 1}\qty{1 - {z^2\over k^2}} .\] Determine \(e^b = \pi\) by considering \(\sin(\pi z)/z\) as \(z\to 0\), and use that \(\sin(\pi z)\) is odd and the product factor is even to conclude \(e^{az}\) is even and thus equal to 1. This yields \[ \sin(\pi z) = \pi z \prod_{k\geq 1}\qty{1- {z^2\over k^2}} .\]