
Infinite Product
i am having some difficult to expand e^z  1 in an infinite product. I was using the identity e^z1 = 2ie^(z/2)sin(z/2i) but i couldn't really figure it out.
also, similar question im trying to prove an identity
1/(e^z1) = 1/z  1/2 + 2z ∑(n=1 to infinite) 1/(z^2+4π^2n^2) by taking logarithmic derivatives in e^z  1
I do not know what to do after... plz help me (Crying)

Your approach is quite correct and You can obtain the desired result applying the following theorem...
Weierstrass factorization theorem  Wikipedia, the free encyclopedia
Starting from the identity...
$\displaystyle e^{z}1 = 2i\cdot e^{\frac{z}{2}}\cdot \sin \frac{z}{2i}$ (1)
... and remembering that...
$\displaystyle \sin p= p\cdot \prod_{n=1}^{\infty} \{1\frac{p^{2}}{(n\pi)^{2}}\}$ (2)
... setting in (2) $\displaystyle p=\frac{z}{2i}$ and inserting it in (1) You obtain...
$\displaystyle e^{z}1 = e^{\frac{z}{2}}\cdot z\cdot \prod_{n=1}^{\infty} \{1+\frac{z^{2}}{(2n\pi)^{2}}\}$ (3)
Using (3) You should have no difficulties to resolve the second part of the problem...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
