“…)...)c c/2 n instead of exp(πc(n−k + )i/2) as above and putting |D| = 1. Thus(20) Ψ(z 1 , ..., z n ) = −Γ((n/2)−1)(P * (z 1 , ..., z n )−cl0) 1−(n/2) [(...(c )...)c c/2 n ] * /(4π n/2 ) for 3 ≤ n, while (21) Ψ(z 1 , z 2 ) = 4 −1 [c * Ln(P * (z 1 , z 2 ) − cl0) for n = 2, since c * j = c −1 j for |c j | = 1, y j q j = y j (c c/q 2 )...)dc c/2 n q n ] = dq 1 ...dq n [(...(c )...)c c/2 n | = 1. 36.…”