Final answer:
To show that the n distinct n-th roots of a non-positive complex number z are s, sω, sω², ..., sωⁿ⁻¹, we use the fact that multiplying the n-th root s by each n-th root of unity produces all n distinct roots.
Step-by-step explanation:
The question asks us to show that the n distinct nth roots of a complex number z, where z ≤ 0, are s, sω, sω², sω³, ..., sωn-1. We know that if s is an nth root of z, then sⁿ = z.
We also know that the nth roots of unity are ω, ω², ..., ωn-1 where ω = ei(2π/n). Multiplying s by each of these will give us the distinct nth roots of z, because (sω)n = sⁿ·ωⁿ = z·1 = z, and similarly for other powers of ω up to ωn-1.