Does $\prod_{n=1}^\infty (1-\frac{z^2}{(2n\pi i)^2})=\frac{e^z-1}{z}$?

Question

I was thinking if the next statement is true: Let $f: \mathbb{C} \rightarrow \mathbb{C}$, $f(0)=1$, $f(\pm a_n)=0$, for all $n \in [1, \infty]$. If this is true, so $f(z)=\prod_{n=1}^\infty (1-\frac{z^2}{a_n^2})$. Does the product $\prod_{n=1}^\infty (1+\frac{z^2}{4n^2\pi^2})=\frac{e^z-1}{z}$? If not, for what functions the statement isn't true?

Answer 1

Hint

Let $z=2 i x$ and consider now $$\prod_{n=1}^\infty \left(1-\frac{x^2}{\pi ^2 n^2}\right)$$ which looks to be very close to a simple infinite product (ggogle for that).

Even without Google, think about a function which cancels for any $x=\pm n \pi$.

Does the bell start ringing ?