This is not a research level question. But due to some reason I can't ask this question on Math Stack Exchange. So, I am asking this question here.


By definition we know that we can measure the multiplicity of a root of a function $f(z)$ as follow.


If the a root of some function $f(z)$ is $\alpha\_i$, then the multiplicity of that root is $a\_i$ for which $\lim \limits\_{x \to \alpha\_i} |\frac{f(z)}{(z-\alpha\_i)^{a\_i}}|$ has a finite value. 


So, following this we can see that all the roots ($(2n+1)i\pi$, 
) of the polynomial $e^z+1$ has multiplicity $1$. 


So, $$e^z+1= 2\prod\_{n=-\infty}^{\infty} \left(1-\frac{z}{(2n+1)i\pi}\right)$$.....(1)


Which further can be written as 


$$e^z+1= 2\prod\_{n=0}^{\infty} \left(1+\frac{z^2}{((2n+1)i\pi)^2}\right)$$ 


But this only consists of $z^{2k}$ not the odd powers.


If we directly check this from (1) we see the coefficients of $z^3$, we see they are $0$ which can be directly concluded from (1). 


But actually $e^z+1=2+z+\frac{z^2}{2!}+....$. So, what am I doing wrong?



A question on multiplicity of complex polynomial
Your product on the right does not converge for any $z \neq 0.$ Never fear, [the Weierstrass factorization theorem](https://www.wikiwand.com/en/Weierstrass_factorization_theorem) is your friend.


