Cube and Fourth Roots of Unity: PECTAA Class 11 Maths, Punjab Board

What is ∛1? Most students say 1 and stop. The book shows there are three answers, and two of them are complex. Learn the one trick, ω³ = 1, and the whole exercise becomes short.

Section 1.4 of Unit 1 in the PECTAA Mathematics 11 book is about the roots of unity. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Cube roots of unity. Let x be a cube root of 1. Then x³ = 1, so x³ − 1 = 0. Factorise:

(x − 1)(x² + x + 1) = 0

• Either x − 1 = 0, so x = 1.
• Or x² + x + 1 = 0. By the quadratic formula, x = (−1 ± √(1 − 4))/2 = (−1 ± √3 i)/2.

So the three cube roots of unity are 1, (−1 + √3 i)/2 and (−1 − √3 i)/2. The book calls the last two the imaginary cube roots of unity, since they contain i.

Properties of cube roots of unity (section 1.4.1). Write ω (omega) for one imaginary root.

1. Each complex cube root of unity is the square of the other. If ω = (−1 + √3 i)/2, then ω² = (−1 − √3 i)/2.
2. The sum of all three cube roots is zero: 1 + ω + ω² = 0.
3. The product of all three is one: 1 · ω · ω² = ω³ = 1. So each imaginary root is the reciprocal of the other.

Fourth roots of unity (section 1.4.2). Let x⁴ = 1. Then x⁴ − 1 = 0, so (x² − 1)(x² + 1) = 0.

• x² − 1 = 0 gives x = ±1.
• x² + 1 = 0 gives x = ±i.

The four fourth roots of unity are 1, −1, i and −i.

Properties of fourth roots of unity (section 1.4.3).

1. The sum of all four is zero: 1 + (−1) + i + (−i) = 0.
2. The real roots 1 and −1 are additive inverses of each other.
3. The imaginary roots i and −i are conjugates of each other.
4. The product of all four is −1: 1 × (−1) × i × (−i) = −1.

Key points

• Two facts do the work. ω³ = 1 and 1 + ω + ω² = 0. From the second, ω + ω² = −1.
• Reduce powers by 3. Since ω³ = 1, write the power as 3k + r. Then ωⁿ = ωʳ. For example ω⁷ = ω, ω⁸ = ω² and ω⁹ = 1.
• Cube roots and fourth roots differ. Cube roots sum to 0 and multiply to 1. Fourth roots sum to 0 and multiply to −1.

Worked examples

Example 1 (book Example 11). Prove that x³ + y³ = (x + y)(x + ωy)(x + ω²y).

1. Right side = (x + y)[x² + (ω + ω²)xy + ω³y²].
2. Use ω³ = 1 and ω + ω² = −1.
3. Right side = (x + y)(x² − xy + y²) = x³ + y³. Proved.

Example 2 (in the style of Exercise 1.4, question 1). Find the three cube roots of 8.

1. x³ = 8 = 2³, so x = 2 times a cube root of unity.
2. The roots are 2, 2ω and 2ω².
3. In full: 2, −1 + √3 i and −1 − √3 i.

Example 3 (in the style of Exercise 1.4, question 2). Find the four fourth roots of 16 and show that their sum is zero.

1. x⁴ = 16 = 2⁴, so x = 2 times a fourth root of unity.
2. The roots are 2, −2, 2i and −2i.
3. Sum: 2 + (−2) + 2i + (−2i) = 0.

Example 4 (in the style of Exercise 1.4, question 4). Evaluate ω⁷ + (ω²)⁷.

1. ω⁷ = ω⁶ · ω = ω.
2. (ω²)⁷ = ω¹⁴ = ω¹² · ω² = ω².
3. Sum = ω + ω² = −1.

Example 5 (in the style of Exercise 1.4, question 7). Evaluate the sum of ω^(2k) for k = 0 to 5.

1. The terms are 1, ω², ω⁴, ω⁶, ω⁸, ω¹⁰.
2. Reduce: 1, ω², ω, 1, ω², ω.
3. Group in threes: (1 + ω + ω²) + (1 + ω + ω²) = 0 + 0 = 0.

Common mistakes in the exam

1. Stopping at x = 1. A cube root question has three roots.
2. Writing ω² = −ω. That is false. The facts are ω² = 1/ω and ω + ω² = −1.
3. Getting the sign of the product wrong. Cube roots multiply to 1. Fourth roots multiply to −1.
4. Not reducing powers. Use ω³ = 1 first. Do not expand big powers.
5. Forgetting the factor when the number is not 1. The cube roots of 8 are 2, 2ω and 2ω², not just ω, ω² and 1.
6. Mixing up ω and i. ω is a cube root of 1. i is a fourth root of 1.

Exam technique

• Start with x³ − 1 = (x − 1)(x² + x + 1) so the marker sees the derivation.
• Write "ω³ = 1 and 1 + ω + ω² = 0" at the top of the page before you use them.
• In proofs, start from one side and finish at the other, as the book does in Example 11.
• Check your board's notice for the sections in your paper.

Practice questions

1. Find the three cube roots of −8. Answer: −2, −2ω and −2ω².
2. If ω is an imaginary cube root of unity, find 1 + ω + ω². Answer: 0.
3. Evaluate ω¹⁰ + ω²⁰. Answer: ω¹⁰ = ω and ω²⁰ = ω², so the sum is ω + ω² = −1.
4. Find the product of the four fourth roots of 81. Answer: The roots are ±3 and ±3i. Product = 3 × (−3) × 3i × (−3i) = (−9)(−9i²) = (−9)(9) = −81.
5. Show that 1 + ωⁿ + ω²ⁿ = 3 when n is a multiple of 3. Answer: If n = 3k, ωⁿ = (ω³)ᵏ = 1 and ω²ⁿ = 1. So the sum is 1 + 1 + 1 = 3.

Ready to practise this? Continue in 1st Year Mathematics.