Choosing the correct branch when solving a complex equation
A student solving z³ = 8 − 8√3 i writes the right-hand side in polar form and finds one solution, then stops, claiming 'a cube root is unique, so there is only one answer.' The teacher points out that a nonzero complex number has exactly three distinct cube roots, spaced evenly around a circle in the Argand plane, and that the student's single answer is only one of them. The student's working shows modulus |8 − 8√3 i| = √(64 + 192) = √256 = 16 and argument θ = −π/3 (since the point lies in the fourth quadrant with tan θ = −√3). The student then writes z = 16^(1/3) cis(−π/9) and declares the problem finished.
1. Find all three cube roots of 8 − 8√3 i in exact polar form, and explain why there are exactly three.
Read the worked answer
Write 8 − 8√3 i = 16 cis(−π/3 + 2kπ) for integer k, since the modulus is 16 and the argument is −π/3.
Then z = 16^(1/3) cis((−π/3 + 2kπ)/3) = 2 cis(−π/9 + 2kπ/3) for k = 0, 1, 2 (2³ = 8, so 16^(1/3) = 2·2^(1/3)? No - check: 2³ = 8, so 16^(1/3) is not 2. Correct value: 16^(1/3) = 2·2^(1/3) = 2^(4/3) ≈ 2.520. Let me restate carefully: 16 = 2⁴, so 16^(1/3) = 2^(4/3) ≈ 2.520.)
So the three roots are z_k = 2^(4/3) cis(−π/9 + 2kπ/3), k = 0, 1, 2:
- z₀ = 2^(4/3) cis(−π/9)
- z₁ = 2^(4/3) cis(−π/9 + 2π/3) = 2^(4/3) cis(5π/9)
- z₂ = 2^(4/3) cis(−π/9 + 4π/3) = 2^(4/3) cis(11π/9)
There are exactly three because the argument −π/3 + 2kπ produces three distinct arguments modulo 2π once divided by 3; adding k = 3 returns the same root as k = 0. The fundamental theorem of algebra guarantees z³ − (8 − 8√3 i) = 0 has exactly three complex roots counted with multiplicity, and they are distinct here.
2. The student's claimed answer was z = 16^(1/3) cis(−π/9). Identify the error in that expression and verify the correct modulus numerically.
Read the worked answer
The argument −π/9 is correct for one root: (−π/3)/3 = −π/9. The error is not the argument but the modulus: the student wrote 16^(1/3) but likely intended 2. The correct modulus is 16^(1/3) = 2^(4/3) ≈ 2.520, not 2. If the student had written 2 cis(−π/9), that would be wrong because 2³ = 8 ≠ 16.
Numerical check: 2.520³ ≈ 16.00, confirming the modulus. Also verify the argument: the cube of cis(−π/9) has argument 3(−π/9) = −π/3, matching the target.
So the corrected single root is z₀ = 2^(4/3) cis(−π/9) ≈ 2.520(cos(−20°) + i sin(−20°)) ≈ 2.368 − 0.862i. Cubing this numerically gives approximately 8 − 13.86i, i.e. 8 − 8√3 i, confirming correctness. The student's main conceptual error was stopping at one root; the secondary error was the modulus slip.