τ = ζ3 = ( -1+√-3 )/2 :
k
= k( τ )
として
kk’
= 1
⇒
k
= ζ6 , ζ6^( -1 )
= ( 1+√-3 )/2 , ( 1-√-3 )/2
k’
= ζ6^( -1 ) , ζ6
= ( 1-√-3 )/2 , ( 1+√-3 )/2
次に
k’( √-3 = 2τ+1 ) = cos( π/12 )
= 1/k’( 2τ )
⇒
k’( 2τ )
= 1/cos( π/12 )
⇒
k( 2τ )
= √-1tan( π/12 )
更に
K’( k( √-3 = 2τ+1 ) )√-1/K( k( √-3 = 2τ+1 ) )
= √-3
⇒
K’( k( √-3 = 2τ+1 ) ) = K( cos( π/12 ) )
= √3K( k( √-3 = 2τ+1 ) )
= √3K( sin( π/12 ) )