Formulate and prove an infinite product formula for \[ \sin^2(\sqrt z). \]
Solution
1. The function \(z\mapsto\sin^2(\sqrt z)\) is entire and has the power-series expression\[ \sin^2(\sqrt z)=\frac{1-\cos(2\sqrt z)}2. \]
Using the cosine series, \[ \cos(2\sqrt z) =\sum_{k=0}^{\infty}\frac{(-1)^k2^{2k}z^k}{(2k)!}, \] which is an entire power series in \(z\). Thus \((1-\cos(2\sqrt z))/2\) is entire. Since \(\sin^2 w=(1-\cos 2w)/2\), it equals \(\sin^2(\sqrt z)\) for either choice of the square root.
2. One has the infinite product\[ \boxed{ \sin^2(\sqrt z) =z\prod_{n=1}^{\infty} \left(1-\frac{z}{\pi^2n^2}\right)^2 }. \]
The sine product formula is \[ \frac{\sin u}{u} =\prod_{n=1}^{\infty} \left(1-\frac{u^2}{\pi^2n^2}\right). \] Squaring gives \[ \sin^2u =u^2\prod_{n=1}^{\infty} \left(1-\frac{u^2}{\pi^2n^2}\right)^2. \] Now put \(u^2=z\). Both sides depend only on \(u^2\), so the result is independent of the choice of square root and gives the displayed formula.
3. The product in 2 converges locally uniformly in \(z\).
For a compact set \(K\subset\mathbb C\), choose \(R\) with \(|z|\le R\) on \(K\). Then \[ \sum_{n=1}^{\infty} \sup_{z\in K}\left|\frac{z}{\pi^2n^2}\right| \le \frac{R}{\pi^2}\sum_{n=1}^{\infty}\frac1{n^2} <\infty. \] Hence \(\prod_n(1-z/(\pi^2n^2))\) converges uniformly on \(K\), and so does its square. This also directly exhibits the zero at \(z=0\) with multiplicity one and the zeros \(z=\pi^2n^2\) with multiplicity two.