Can anyone help me figure out how the identity below was obtained?


$ \frac{1}{\sqrt{(e\_1-e\_3)(e\_2-e\_3)}} = R \prod \limits\_{n=1}^{\infty} \left(1 - \frac{1}{R^{4n}} \right)^{-4}\left(1 + \frac{1}{R^{2n}} \right)^{-4} = R\left(\sum\_{n=1}^{\infty} \frac{1}{R^{n(n-1)}}\right)^{-4}. \tag{1}$ 


(Komatu, "A coefficient problem for functions univalent in an annulus", [*Kodai Math. Sem. Rep.* **8**(2), 1956, p. 49-70.](https://doi.org/10.2996/kmj/1138843740), theorem 2 on page 57.)


$(e\_2-e\_3)= (\frac{\pi}{\omega\_1})^2 \theta\_2^4 \quad \textrm{and} \quad
 (e\_1-e\_3)= (\frac{\pi}{\omega\_1})^2 \theta\_3^4 \tag{2}$


<https://books.google.co.nz/books?id=zyxAb4ro-oMC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false> page no 178 & 179( Elliptic functions by Armitage & Eberlein)


$\theta\_2 = 2 q^\frac{1}{4}\prod \limits\_{n=1}^{\infty} (1-q^{2n}) (1+q^{2n})^2 
, \quad \theta\_3 = \prod \limits\_{n=1}^{\infty} (1-q^{2n}) (1+q^{2n-1})^2 \tag{3}$ and 


$\theta\_4 = \prod \limits\_{n=1}^{\infty} (1-q^{2n}) (1-q^{2n-1})^2 \tag{4}$
 page no 104 


As $(e\_2 -e\_3)(e\_1-e\_3)= (\frac{\pi}{\omega\_1})^4 q \prod \limits\_{n=1}^{\infty} (1-q^{2n})^8 (1+q^{2n})^8(1+q^{2n-1})^8 \tag{5}$


Where $ \sqrt{e\_2 -e\_3} \sqrt{e\_1-e\_3}= (\frac{\pi}{\omega})^2\theta\_2^2\theta\_3^2 \tag{6}$ by Abramowitz and Stegun


$\theta\_2^2 \theta\_3^2 = 4 q^\frac{1}{2} \prod \limits\_{n=1}^{\infty} (1-q^{2n})^4 (1+q^{n})^4 = \Big( \frac{\theta\_2(\sqrt{q })^2}{2} \Big)^2 = 4q^\frac{1}{2}\Big( \sum \limits\_{n=1}^{\infty}q^{n(n-1)/2}\Big)^4 \tag{7}$


$q= 1/R^2 \tag{8}$


$\sqrt{e\_2 -e\_3} \sqrt{e\_1-e\_3} \!=\! (\frac{\pi}{\omega\_1})^2 \! \frac{1}{R} \prod \limits\_{n=1}^{\infty} (1\!-\!R^{-4n})^4 (1\!+\!R^{-2n})^4 \!=\! (\frac{\pi}{\omega\_1})^2 \frac{1}{R}\Big(\sum \limits\_{n=1}^{\infty}R^{-n(n-1)} \Big)^4 \tag{9}$



Infinite sum and product associated with the Weierstrass elliptic function
From standard definitions of $e\_1,e\_2,e\_3$ and $\theta\_2$ in reference works such as Abramowitz and Stegun, (e.g., $\theta\_2(0,x)=2x^{1/4}\sum\_{n>0} x^{n(n-1)}$), it turns out that the following identity holds
$$(e\_1-e\_3)(e\_2-e\_3)=(\pi/(4\omega\_1))^4\theta\_2(0,\sqrt{q})^8$$ which does agree with the stated identity. Let $q:=1/R^2$, then if we define
$$Q\_0:=\prod\_{n>0}1-q^{2n},\;Q\_1:=\prod\_{n>0}1+q^{2n},
\;Q\_2:=\prod\_{n>0}1+q^{2n-1},\;Q\_3:=\prod\_{n>0}1-q^{2n-1},$$
then also $[e\_1,e\_2,e\_3]=(\pi/\omega\_1)^2Q\_0^4/12\;[Q\_2^8+Q\_3^8,\;16qQ\_1^8-Q\_3^8,\;-16qQ\_1^8-Q\_2^8].$


There are many standard identities between theta functions and infinite products such as $Q\_0,Q\_1,Q\_2,Q\_3.$ For example [DLMF equation 20.4.3](http://dlmf.nist.gov/20.4#E3). Also $1=Q\_1Q\_2Q\_3$, and $Q\_2^8=Q\_3^8+16qQ\_1^8.$
Note that $\theta\_2(0,q)\theta\_3(0,q)=\theta\_2(0,\sqrt{q})^2/2=2q^{1/4}Q\_0^2/Q\_3^2$.


Note that there are varying conventions. Some use $\ \omega = 2\omega\_1. \ $
For your equations you need to assume $\ \pi = \omega\_1 \ $ to avoid $\pi$ factors in equation (1) for example.


