Roots of Unity and a Suspect Symmetry
A student is asked to find all complex z satisfying z⁴ = −16. They write z = 2cis(θ) and solve 4θ = π, giving θ = π/4, so z = 2cis(π/4) only. They then check by cubing... actually by raising to the fourth power and get 16cis(π) = −16, which is correct, so they conclude there is exactly one solution. A classmate claims there should be four and that the solutions form a square centred at the origin. The student insists the equation is degree 4 so 'at most four' but that the geometry forces one.
1. Resolve the disagreement by finding all four roots and describing their geometry.
Read the worked answer
Write −16 = 16cis(π + 2kπ). Then z = 2cis((π + 2kπ)/4) = 2cis(π/4 + kπ/2).
For k = 0,1,2,3: z = 2cis(π/4), 2cis(3π/2)... let us list carefully: π/4, 3π/4, 5π/4, 7π/4.
So the roots are 2cis(π/4), 2cis(3π/4), 2cis(5π/4), 2cis(7π/4), i.e. ±√2 ± √2 i.
These are four points on the circle |z| = 2 at arguments spaced π/2 apart, forming a square rotated 45° from the axes. The student found only the principal root and forgot the periodicity of the argument; the classmate is correct.
2. Without recomputing, explain how the sum of the four roots and the product of the four roots can be obtained from the polynomial z⁴ + 16 = 0, and verify against your roots.
Read the worked answer
For z⁴ + 16 = 0, the coefficient of z³ is 0, so the sum of roots is 0. The constant term is 16, and for a monic quartic the product of roots equals the constant term, so the product is 16.
Check: sum = 2cis(π/4) + 2cis(3π/4) + 2cis(5π/4) + 2cis(7π/4). The real parts are √2, −√2, −√2, √2, summing to 0; imaginary parts likewise cancel. Product: magnitudes multiply to 2⁴ = 16, arguments sum to π/4 + 3π/4 + 5π/4 + 7π/4 = 4π, giving cis(4π) = 1, so product = 16. Both agree with Vieta's formulas.