(In reply to the roots of the imaginary
by Dej Mar)
Hi, Dej Mar
IMHO, It was not z6 + z4 + z3 + 2z2 +1 = 0 to be solved:
it was either z6 + z4 + z3 + z2 + z + 1 = 0 (i)
or z6 +z5 + z4 + z3 + z2 + z+ 1 = 0 (ii)
(ii) is simple and interesting
(i) more difficult and less interesting
Both have positive and negative imaginary parts, in (ii) you can get the product and the angle without listing the roots.
So - either solve both corrected version or wait for KS's reply.
Still - may be I'm wrong, but it was a funny way to write 2z2....