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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., theorem 2 on page 57.)

$(e_2 -e_3)= (\frac{\pi}{\omega_1})^2 \theta_2^4 \quad \textrm{and} \quad (e_1-e_3)= (\frac{\pi}{\omega_1})^2 \theta_4^4 \tag{2}$

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

$\theta_4 = \prod \limits_{n=1}^{\infty} (1-q^{2n}) (1-q^{2n-1}) \tag{4}$ page no 104

As $(e_2 -e_3)(e_1-e_3)= (\frac{\pi}{\omega_1})^4 (16 q) \prod \limits_{n=1}^{\infty} (1-q^{2n})^8 (1+q^{2n})^8(1-q^{2n-1})^8 \tag{5}$

Where $ \sqrt{e_2 -e_3} \sqrt{e_1-e_3}= (\frac{\pi}{2\omega})^2\theta_2^2\theta_3^2 \tag{6}$ by Abramowitz and Stegun

$\theta_2^2 \theta_3^2 = 4 q^\frac{1}{2} \prod \limits_{n=1}^{\infty} (1-q^{2n})^4 (1+q^{n})^4 = \Big( \frac{\theta_2(\sqrt{q })^2}{2} \Big)^2 = 4q^\frac{1}{2} \sum \limits_{n=1}^{\infty}q^{2n(n-1)} \tag{7}$

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

$\sqrt{e_2 -e_3} \sqrt{e_1-e_3}= \frac{1}{R}\prod \limits_{n=1}^{\infty} (1-R^{-4n})^4 (1+R^{-2n})^4 = \frac{1}{R}\Big(\sum \limits_{n=1}^{\infty}R^{-n(n-1)} \Big)^4 \tag{9}$

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., 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

$\theta_2^2 \theta_3^2 = 4 q^\frac{1}{2} \prod \limits_{n=1}^{\infty} (1-q^{2n})^4 (1+q^{n})^4 = \Big( \frac{\theta_2(\sqrt{q })^2}{2} \Big)^2 = 4q^\frac{1}{2}\Big( \sum \limits_{n=1}^{\infty}q^{n(n-1)/2}\Big)^4 \tag{7}$

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

$\sqrt{e_2 -e_3} \sqrt{e_1-e_3} \!=\! (\frac{\pi}{\omega_1})^2 \! \frac{1}{R} \prod \limits_{n=1}^{\infty} (1\!-\!R^{-4n})^4 (1\!+\!R^{-2n})^4 \!=\! (\frac{\pi}{\omega_1})^2 \frac{1}{R}\Big(\sum \limits_{n=1}^{\infty}R^{-n(n-1)} \Big)^4 \tag{9}$

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

$$\frac{1}{\sqrt{(e_1-e_3)(e_2-e_3)}} = R \prod \limits_{n=1}^{\infty} \left(1 - \frac{1}{R^{4n}} \right)^{-4}\left(1 + \frac{1}{R^{2n}} \right)^{-4} = R\left(\sum_{n=1}^{\infty} \frac{1}{R^{n(n-1)}}\right)^{-4}.$$

(Komatu, "A coefficient problem for functions univalent in an annulus", Kodai Math. Sem. Rep. 8(2), 1956, p. 49-70., theorem 2 on page 57.)

$(e_2 -e_3)= (\frac{\pi}{\omega_1})^2 \theta_2^4 $ and $(e_1-e_3)= (\frac{\pi}{\omega_1})^2 \theta_4^4 $

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)

$\theta_2 = 2 q^\frac{1}{4}\prod \limits_{n=1}^{\infty} (1-q^{2n}) (1+q^{2n})^2$ , $\theta_3 = \prod \limits_{n=1}^{\infty} (1-q^{2n}) (1+q^{2n-1})^2$ and

$\theta_4 = \prod \limits_{n=1}^{\infty} (1-q^{2n}) (1-q^{2n-1})$ page no 104

As $(e_2 -e_3)(e_1-e_3)= (\frac{\pi}{\omega_1})^4 (16 q \prod \limits_{n=1}^{\infty} (1-q^{2n})^8 (1+q^{2n})^8(1-q^{2n-1})^8$

$\sqrt{e_2 -e_3} \sqrt{e_1-e_3}= (\frac{\pi}{2\omega})^2\theta_2^2\theta_3^2$ by Abaramowitz 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 = ( \frac{\theta_2(\sqrt{q })^2}{2})^2 = 4q^\frac{1}{2} \sum \limits_{n=1}^{\infty}q^{2n(n-1)} $

$q= 1/R^2$

$\sqrt{e_2 -e_3} \sqrt{e_1-e_3}= \frac{1}{R}\prod \limits_{n=1}^{\infty} (1-R^{-4n})^4 (1+R^{-2n})^4 = \frac{1}{R}(\sum \limits_{n=1}^{\infty}R^{-n(n-1)} )^4$

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., theorem 2 on page 57.)

$(e_2 -e_3)= (\frac{\pi}{\omega_1})^2 \theta_2^4 \quad \textrm{and} \quad (e_1-e_3)= (\frac{\pi}{\omega_1})^2 \theta_4^4 \tag{2}$

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

$\theta_4 = \prod \limits_{n=1}^{\infty} (1-q^{2n}) (1-q^{2n-1}) \tag{4}$ page no 104

As $(e_2 -e_3)(e_1-e_3)= (\frac{\pi}{\omega_1})^4 (16 q) \prod \limits_{n=1}^{\infty} (1-q^{2n})^8 (1+q^{2n})^8(1-q^{2n-1})^8 \tag{5}$

Where $ \sqrt{e_2 -e_3} \sqrt{e_1-e_3}= (\frac{\pi}{2\omega})^2\theta_2^2\theta_3^2 \tag{6}$ by Abramowitz and Stegun

$\theta_2^2 \theta_3^2 = 4 q^\frac{1}{2} \prod \limits_{n=1}^{\infty} (1-q^{2n})^4 (1+q^{n})^4 = \Big( \frac{\theta_2(\sqrt{q })^2}{2} \Big)^2 = 4q^\frac{1}{2} \sum \limits_{n=1}^{\infty}q^{2n(n-1)} \tag{7}$

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

$\sqrt{e_2 -e_3} \sqrt{e_1-e_3}= \frac{1}{R}\prod \limits_{n=1}^{\infty} (1-R^{-4n})^4 (1+R^{-2n})^4 = \frac{1}{R}\Big(\sum \limits_{n=1}^{\infty}R^{-n(n-1)} \Big)^4 \tag{9}$

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

$$\frac{1}{\sqrt{(e_1-e_3)(e_2-e_3)}} = R \prod \limits_{n=1}^{\infty} \left(1 - \frac{1}{R^{4n}} \right)^{-4}\left(1 + \frac{1}{R^{2n}} \right)^{-4} = R\left(\sum_{n=1}^{\infty} \frac{1}{R^{n(n-1)}}\right)^{-4}.$$

(Komatu, "A coefficient problem for functions univalent in an annulus", Kodai Math. Sem. Rep. 8(2), 1956, p. 49-70., theorem 2 on page 57.)

$(e_2 -e_3)= (\frac{\pi}{\omega_1})^2 \theta_2^4 $ and $(e_1-e_3)= (\frac{\pi}{\omega_1})^2 \theta_4^4 $

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)

$\theta_2 = 2 q^\frac{1}{4}\prod \limits_{n=1}^{\infty} (1-q^{2n}) (1+q^{2n})^2$ , $\theta_3 = \prod \limits_{n=1}^{\infty} (1-q^{2n}) (1+q^{2n-1})^2$ and

$\theta_4 = \prod \limits_{n=1}^{\infty} (1-q^{2n}) (1-q^{2n-1})$ page no 104

As $(e_2 -e_3)(e_1-e_3)= (\frac{\pi}{\omega_1})^4 (16 q \prod \limits_{n=1}^{\infty} (1-q^{2n})^8 (1+q^{2n})^8(1-q^{2n-1})^8$

$\sqrt{e_2 -e_3} \sqrt{e_1-e_3}= (\frac{\pi}{2\omega})^2\theta_2^2\theta_3^2$ by Abaramowitz 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 = ( \frac{\theta_2(\sqrt{q })^2}{2})^2 = 4q^\frac{1}{2} \sum \limits_{n=1}^{\infty}q^{2n(n-1)} $

$q= 1/R^2$

$\sqrt{e_2 -e_3} \sqrt{e_1-e_3}= (\frac{\pi}{\omega})^2 \frac{1}{R}\prod \limits_{n=1}^{\infty} (1-R^{-4n})^4 (1+R^{-2n})^4 = (\frac{\pi}{\omega})^2 \frac{1}{R}\sum \limits_{n=1}^{\infty}R^{-4n(n-1)} $

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

$$\frac{1}{\sqrt{(e_1-e_3)(e_2-e_3)}} = R \prod \limits_{n=1}^{\infty} \left(1 - \frac{1}{R^{4n}} \right)^{-4}\left(1 + \frac{1}{R^{2n}} \right)^{-4} = R\left(\sum_{n=1}^{\infty} \frac{1}{R^{n(n-1)}}\right)^{-4}.$$

(Komatu, "A coefficient problem for functions univalent in an annulus", Kodai Math. Sem. Rep. 8(2), 1956, p. 49-70., theorem 2 on page 57.)

$(e_2 -e_3)= (\frac{\pi}{\omega_1})^2 \theta_2^4 $ and $(e_1-e_3)= (\frac{\pi}{\omega_1})^2 \theta_4^4 $

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)

$\theta_2 = 2 q^\frac{1}{4}\prod \limits_{n=1}^{\infty} (1-q^{2n}) (1+q^{2n})^2$ , $\theta_3 = \prod \limits_{n=1}^{\infty} (1-q^{2n}) (1+q^{2n-1})^2$ and

$\theta_4 = \prod \limits_{n=1}^{\infty} (1-q^{2n}) (1-q^{2n-1})$ page no 104

As $(e_2 -e_3)(e_1-e_3)= (\frac{\pi}{\omega_1})^4 (16 q \prod \limits_{n=1}^{\infty} (1-q^{2n})^8 (1+q^{2n})^8(1-q^{2n-1})^8$

$\sqrt{e_2 -e_3} \sqrt{e_1-e_3}= (\frac{\pi}{2\omega})^2\theta_2^2\theta_3^2$ by Abaramowitz 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 = ( \frac{\theta_2(\sqrt{q })^2}{2})^2 = 4q^\frac{1}{2} \sum \limits_{n=1}^{\infty}q^{2n(n-1)} $

$q= 1/R^2$

$\sqrt{e_2 -e_3} \sqrt{e_1-e_3}= \frac{1}{R}\prod \limits_{n=1}^{\infty} (1-R^{-4n})^4 (1+R^{-2n})^4 = \frac{1}{R}(\sum \limits_{n=1}^{\infty}R^{-n(n-1)} )^4$