Although eta goes berserk on the unit circle, it really is 0 at e^(i pi rational) by virtue of infinitely many terms of the product formula vanishing identically. Just how emphatically 0 is clear from what you have to multiply by to keep it finite as you (nontangentially) approach the boundary. E.g., for rational = 2n/11, 0<n<11: 2 %pi 2 %i %pi n - ---------- ---------- 726 log(q) 11 limit %e sqrt(- log(q)) eta(%e q) q -> 1 %i %pi k(n) ---------- 2 %pi 132 = sqrt(-----) %e , 11 where k(1),k(2),... = -89, 28, -15, -14, 35, -24, 25, 26, 39, 100 (mod 264). There is probably a nifty formula for these if anyone wants to decrypt it. For more data, you should have no trouble guessing how to generalize to denominators besides 11. Mma 6.0 is surprisingly good at the numerics (but with the tau = log(q)/2/i/pi argument convention. tau = I/999 is plenty small enough.) Or I can crank out a few more k(n) sets if you tell me which you want. --rwg PS, this limit stuff applies also to (q;q)_oo = eta(q)/q^24.