1. ## Infinite Product

i am having some difficult to expand e^z - 1 in an infinite product. I was using the identity e^z-1 = 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^z-1) = 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

2. 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

$\displaystyle \chi$ $\displaystyle \sigma$

3. thank you so much