Seven roots and conjugate synthesis
Find the complete set, then use symmetry without losing an exceptional exponent
- All roots For a nonzero target, seven root directions have spacing 2π/7.
- Conjugate factor A pair Rcis θ,Rcis(−θ) gives x²−2Rcos θ x+R²
- Power sum For seventh roots Rωʲ, the sum of mth powers is 0 unless 7 divides m, when it is 7Rᵐ
Worked example
Find the sum of cubes of the six nonreal roots of z⁷=128.
- \(\displaystyle \sum_{j=0}^6(2\omega^j)^3=0\)7 does not divide 3, so the full geometric sum vanishes
- \(\displaystyle \sum_{j=1}^6(2\omega^j)^3=-8\)Remove the real root cubed, 2³=8.
Common mistakes
Difficulty levels
Questions about this topic
Are symbolic root expressions allowed?+
The roots are given by exact radius and rational angle-coefficient fields. A different argument convention can represent the same roots, but these fields explicitly request [0,2π)
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Specialist Mathematics, chapter 3: Complex numbers