From summing coefficients: z = 1 is a solution but z = 1 is not a solution.
Factors to: (z  1) (z^8 + z^7 + z^6 + 2z^5 + z^4 + 2z^3 + z^2 + z + 1)
From Vieta's equations,
the coefficient of z^8 is zero, so the sum of all 9 roots is zero.
the product of all 9 roots is 1.
For the product of factors to be 1, the 8 factors other than z=1, if considered in pairs:
(a) could be duplicate roots of (1,1), but summing coefficients of the 8th degree polynomial shows that there are no further roots of 1
(b) reciprocals. but if they were reciprocals, they would not sum to zero. (* exception below)
or
(c) complex conjugates with z=1
The only option is (c).
* exception: the problem with (b) is, 4 of the roots could be:
a, 1/a, a, 1/a then they could have sum=0 and product=1
So, this is not quite there yet.

Posted by Larry
on 20240819 10:14:11 