ヤコビの虚数変換式
f ( 1 4 , τ 4 ) = − i ( τ / 4 ) e π i / 4 τ θ 3 ( 1 4 , τ 4 ) θ 3 ( 1 τ , − 4 τ ) = − i τ ∑ n = − ∞ ∞ e n 2 π i τ / 4 + n π i / 2 2 ∑ n = − ∞ ∞ e − 4 n 2 π i / τ + 2 n π i / τ e − π i / 4 τ = − i τ ∑ n = − ∞ ∞ i n e n 2 π i τ / 4 2 ∑ n = − ∞ ∞ e − ( 2 n − 1 / 2 ) 2 π i / τ = − i τ ∑ n = − ∞ ∞ i n e n 2 π i τ / 4 2 ( ∑ n = 1 ∞ e − ( 2 n − 1 / 2 ) 2 π i / τ + ∑ n = 1 ∞ e − ( − 2 n + 1 − 1 / 2 ) 2 π i / τ ) = − i τ ∑ n = − ∞ ∞ i n e n 2 π i τ / 4 2 ∑ n = 1 ∞ e − ( n − 1 / 2 ) 2 π i / τ {\displaystyle {\begin{aligned}f\left({\frac {1}{4}},{\frac {\tau }{4}}\right)&={\frac {{\sqrt {-i(\tau /4)}}e^{{\pi }i/4\tau }\vartheta _{3}\left({\frac {1}{4}},{\frac {\tau }{4}}\right)}{\vartheta _{3}\left({\frac {1}{\tau }},-{\frac {4}{\tau }}\right)}}={\frac {{\sqrt {-i\tau }}\sum _{n=-\infty }^{\infty }{e^{n^{2}{\pi }i\tau /4+n{\pi }i/2}}}{2\sum _{n=-\infty }^{\infty }{e^{{-4n^{2}\pi }i/\tau +2n{\pi }i/\tau }e^{-{\pi }i/4\tau }}}}\\&={\frac {{\sqrt {-i\tau }}\sum _{n=-\infty }^{\infty }{i^{n}e^{n^{2}{\pi }i{\tau }/4}}}{2\sum _{n=-\infty }^{\infty }{e^{-(2n-1/2)^{2}{\pi }i/\tau }}}}={\frac {{\sqrt {-i\tau }}\sum _{n=-\infty }^{\infty }{i^{n}e^{n^{2}{\pi }i{\tau }/4}}}{2\left(\sum _{n=1}^{\infty }{e^{-(2n-1/2)^{2}{\pi }i/\tau }}+\sum _{n=1}^{\infty }{e^{-(-2n+1-1/2)^{2}{\pi }i/\tau }}\right)}}\\&={\frac {{\sqrt {-i\tau }}\sum _{n=-\infty }^{\infty }{i^{n}e^{n^{2}{\pi }i{\tau }/4}}}{2\sum _{n=1}^{\infty }{e^{-(n-1/2)^{2}{\pi }i/\tau }}}}\\\end{aligned}}}
cn  ( i u , k ) = θ 4 ( 0 , τ ) θ 2 ( i u / 2 K , τ ) θ 2 ( 0 , τ ) θ 4 ( i u / 2 K , τ ) = θ 2 ( 0 , τ ′ ) θ 4 ( u / 2 K ′ , τ ′ ) θ 4 ( 0 , τ ′ ) θ 2 ( u / 2 K ′ , τ ′ ) = 1 cn  ( u , k ′ ) {\displaystyle \operatorname {cn} (iu,k)={\frac {\vartheta _{4}(0,\tau )\vartheta _{2}(iu/2K,\tau )}{\vartheta _{2}(0,\tau )\vartheta _{4}(iu/2K,\tau )}}={\frac {\vartheta _{2}(0,\tau ')\vartheta _{4}(u/2K',\tau ')}{\vartheta _{4}(0,\tau ')\vartheta _{2}(u/2K',\tau ')}}={\frac {1}{\operatorname {cn} (u,k')}}}
となり、楕円関数の虚数変数を得る。
先に示したように f ( v , τ ) {\displaystyle f(v,\tau )} は v {\displaystyle v} に依存しないので
f ( v , τ ) = f ( v , τ 4 ) = lim n → ∞ f ( v , τ 4 n ) = lim τ ′ → 0 f ( v , τ ′ ) = f ( v , 0 ) {\displaystyle f\left(v,\tau \right)=f\left(v,{\frac {\tau }{4}}\right)=\lim _{n\to \infty }f\left(v,{\frac {\tau }{4^{n}}}\right)=\lim _{\tau '\to 0}f\left(v,\tau '\right)=f(v,0)}
であり、 f ( v , τ ) {\displaystyle f(v,\tau )} は τ {\displaystyle \tau } にも依存しない定数である。その値は
f ( v , τ ) = f ( 0 , i ) = θ 3 ( 0 , i ) θ 3 ( 0 , i ) = 1 {\displaystyle f(v,\tau )=f(0,i)={\frac {\vartheta _{3}\left(0,i\right)}{\vartheta _{3}\left(0,i\right)}}=1}
である。
f ( v + 1 , τ ) = − i τ e π i v 2 / τ + 2 π i v / τ + π i / τ θ 3 ( v + 1 , τ ) θ 3 ( v τ + 1 τ , − 1 τ ) = − i τ e π i v 2 / τ + 2 π i v / τ + π i / τ θ 3 ( v , τ ) e π i / τ + 2 π i v / τ θ 3 ( v τ , − 1 τ ) = f ( v , τ ) {\displaystyle f(v+1,\tau )={\frac {{\sqrt {-i\tau }}e^{{\pi }iv^{2}/\tau +2{\pi }iv/\tau +{\pi }i/\tau }\vartheta _{3}\left(v+1,\tau \right)}{\vartheta _{3}\left({\frac {v}{\tau }}+{\frac {1}{\tau }},-{\frac {1}{\tau }}\right)}}={\frac {{\sqrt {-i\tau }}e^{{\pi }iv^{2}/\tau +2{\pi }iv/\tau +{\pi }i/\tau }\vartheta _{3}\left(v,\tau \right)}{e^{{\pi }i/\tau +2{\pi }iv/\tau }\vartheta _{3}\left({\frac {v}{\tau }},-{\frac {1}{\tau }}\right)}}=f(v,\tau )}