 Hyperbolic-sine infinite product

This infinite product converges uniformly on compact subsets of the complex plane. Pair the factors at $iz$ and $-iz$ in the Weierstrass product for the reciprocal gamma function, and use the Gamma reflection formula together with $\Gamma (1+w)=w\Gamma (w)$. This gives $[\Gamma (1+iz)\Gamma (1-iz){]}^{-1}=\sinh (\pi z)/(\pi z)$ and the displayed product. The removable value at $z=0$ is one. Matching zeros alone would not exclude multiplication by a nonvanishing entire function; the normalized product fixes that ambiguity.