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 2020-04-13, 20:29 #9 Nick     Dec 2012 The Netherlands 2×33×31 Posts OK, I think I understand the confusion. Let's use your example of q=7 with $$\zeta=e^{2\pi i/7}$$. Then we have $\zeta^0=1,\zeta^1=\zeta,\zeta^2,\zeta^3,\zeta^4,\zeta^5,\zeta^6$ all distinct and $$\zeta^7=1$$ again. We also have the equation $\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1=0$ so it follows that $-1=\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta$ Thus you can use $$\zeta^0$$ to $$\zeta^5$$ inclusive OR $$\zeta^1$$ to $$\zeta^6$$ inclusive and still equate coefficients (they are linearly independent). I hope this helps!