Since not all a q are zero and the functions e H ζ (q) are linearly independent, it follows that
k is odd. Now the coefficient of α −k+1 becomes q e H ζ (q) (2b q − a q H ζ , d ) = 0.
Again by linear independence of the exponential functions, it follows that a q = 0 = b q for all q, which is a contradiction. 
A ′ d − B ′ d = X ω(α)e H ζ + ω(−α)e H ζ + H ζ ,d α .
Since both A ′ d , B ′ d have coefficients which are polynomial in α, this shows that the class ω(α) has property (b) of the preceding lemma. Suppose A, B both come from Euler **data** Q, P respectively, ie. 
A d = i * 0 Q v d and B d = i * 0 P v d . Suppose also that j * 0 (P d )| q = j * 0 (Q d )| q at α = λ/δ.T q X. (ii) deg α (A d − B d ) ≤ −2.
Then A = B.