Wouter wrote: << Let q[n] represent the n-th root of -1, or exp(2 i pi/n) given that Sum(j=1 to n, q[n]^j) equals zero by symmetry, it is evident that (q[1]+q[2]+..+q[n])^k also equals zero for any k. . . .

...

Did you intend to omit the 2: q[n] := exp(i pi/n) (so that q[n]^n = -1) ? or perhaps q[n] represents the n-th root of +1 ? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele