デデキントのイータ関数
であるが、 τ {\displaystyle \tau } が純虚数であれば両辺ともに実数であるから、
η ( − 1 τ ) = − i τ η ( τ ) {\displaystyle {\begin{aligned}\eta \left(-{\frac {1}{\tau }}\right)&={\sqrt {-i\tau }}\eta (\tau )\\\end{aligned}}}
である。また、
η ( τ + 1 ) = e π i ( τ + 1 ) / 12 ∏ m = 1 ∞ ( 1 − e 2 π i ( τ + 1 ) m ) = e π i / 12 e π i τ / 12 ∏ m = 1 ∞ ( 1 − e 2 π i τ m ) = e π i / 12 η ( τ ) {\displaystyle {\begin{aligned}\eta \left(\tau +1\right)&=e^{\pi {i}(\tau +1)/12}\prod _{m=1}^{\infty }(1-e^{2\pi {i}(\tau +1)m})\\&=e^{\pi {i}/12}e^{\pi {i}\tau /12}\prod _{m=1}^{\infty }(1-e^{2\pi {i}\tau {m}})\\&=e^{\pi {i}/12}\eta \left(\tau \right)\\\end{aligned}}}
テータ関数の虚数変換式により
η 3 ( − 1 τ ) = 1 2 θ 2 ( 0 , − 1 τ ) θ 3 ( 0 , − 1 τ ) θ 4 ( 0 , − 1 τ ) = 1 2 − i τ θ 4 ( 0 , τ ) − i τ θ 3 ( 0 , τ ) − i τ θ 2 ( 0 , τ ) = i τ 3 η 3 ( τ ) {\displaystyle {\begin{aligned}\eta ^{3}\left(-{\frac {1}{\tau }}\right)&={\frac {1}{2}}\vartheta _{2}\left(0,-{\frac {1}{\tau }}\right)\vartheta _{3}\left(0,-{\frac {1}{\tau }}\right)\vartheta _{4}\left(0,-{\frac {1}{\tau }}\right)\\&={\frac {1}{2}}{\sqrt {-i\tau }}\vartheta _{4}(0,\tau ){\sqrt {-i\tau }}\vartheta _{3}(0,\tau ){\sqrt {-i\tau }}\vartheta _{2}(0,\tau )\\&={\sqrt {i\tau ^{3}}}\eta ^{3}(\tau )\\\end{aligned}}}