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Cube roots and fourth roots of unity

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Cube roots and fourth roots of unity

Solve x3 = 1. One answer is x = 1.

Are there others? Yes, two more, and they are complex.

Let us find them.

1 · Cube roots of unity

Start with x3 = 1, so x3 − 1 = 0.

Factorise: (x − 1)(x2 + x + 1) = 0.

So x = 1 or x = (−1 ± √3 i)/2.

The two numbers with i are the imaginary cube roots of unity.

2 · Three properties of cube roots

Each imaginary root is the square of the other. Call one ω; the other is ω2.

The sum of all three roots is zero: 1 + ω + ω2 = 0.

The product of all three is one: 1 · ω · ω2 = ω3 = 1.

So ω = 1/ω2 and ω2 = 1/ω.

3 · Fourth roots of unity

From x4 = 1: (x2 − 1)(x2 + 1) = 0.

x2 = 1 gives x = ±1. x2 = −1 gives x = ±i.

The four roots are 1, −1, i, −i.

Their sum is 0. Their product is −1.

The two imaginary roots i and −i are conjugates.

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Notes, short questions and MCQs

Read the full notes: key terms, model answers and MCQs with answers

Cube roots and fourth roots of unity

Solve x3 = 1. One answer is x = 1.

Are there others? Yes, two more, and they are complex.

Let us find them.

1 · Cube roots of unity

Start with x3 = 1, so x3 − 1 = 0.

Factorise: (x − 1)(x2 + x + 1) = 0.

So x = 1 or x = (−1 ± √3 i)/2.

The two numbers with i are the imaginary cube roots of unity.

2 · Three properties of cube roots

Each imaginary root is the square of the other. Call one ω; the other is ω2.

The sum of all three roots is zero: 1 + ω + ω2 = 0.

The product of all three is one: 1 · ω · ω2 = ω3 = 1.

So ω = 1/ω2 and ω2 = 1/ω.

3 · Fourth roots of unity

From x4 = 1: (x2 − 1)(x2 + 1) = 0.

x2 = 1 gives x = ±1. x2 = −1 gives x = ±i.

The four roots are 1, −1, i, −i.

Their sum is 0. Their product is −1.

The two imaginary roots i and −i are conjugates.

Key terms

Cube roots of unity
The three numbers whose cube is 1: 1, ω, ω2.
ω (omega)
An imaginary cube root of unity. ω3 = 1 and 1 + ω + ω2 = 0.
Fourth roots of unity
The four numbers whose fourth power is 1: 1, −1, i, −i.

Short questions with model answers

  1. Q1. Prove x3 + y3 = (x + y)(x + ωy)(x + ω2y).

      Multiply the last two factors: x2 + (ω + ω2)xy + ω3y2. Use ω + ω2 = −1 and ω3 = 1. This is x2 − xy + y2. Then (x + y)(x2 − xy + y2) = x3 + y3. This is the book's Example 11.

    • Q2. Find the three cube roots of 8.

        8 = 23, so the roots are 2, 2ω and 2ω2. Check: (2ω)3 = 8ω3 = 8. This is like Exercise 1.4, question 1(i).

      Common mistakes

      • ✗ “The sum of the four fourth roots of unity is 1.”

        ✓ The sum is 0: 1 + (−1) + i + (−i) = 0.

      • ✗ “The product of the four fourth roots of unity is 1.”

        ✓ The book says −1: 1 × (−1) × i × (−i) = −1.

      MCQs

      1. 1. What is 1 + ω + ω2?

        1. (a) 0
        2. (b) 1
        3. (c) −1
        4. (d) 3
        Show answer

        (a) The sum of the three cube roots of unity is zero.

      2. 2. What is ω3?

        1. (a) 0
        2. (b) ω
        3. (c) 1
        4. (d) −1
        Show answer

        (c) ω is a cube root of 1, so ω3 = 1.

      3. 3. Which are the four fourth roots of unity?

        1. (a) 1, −1, i, −i
        2. (b) 1, 2, 3, 4
        3. (c) 1, ω, ω2, ω3
        4. (d) 1, −1 only
        Show answer

        (a) From (x2 − 1)(x2 + 1) = 0.

      4. 4. What is the product of the four fourth roots of unity?

        1. (a) 1
        2. (b) 0
        3. (c) i
        4. (d) −1
        Show answer

        (d) i × (−i) = 1. Then 1 × (−1) × 1 = −1.

      5. 5. Which equation gives the cube roots of unity?

        1. (a) x2 = 1
        2. (b) x3 = 1
        3. (c) x3 = 3
        4. (d) 3x = 1
        Show answer

        (b) Cube roots of 1 satisfy x3 = 1.

      Quick revision

      • Cube roots of 1: 1, ω, ω2. Sum 0. Product 1.
      • Fourth roots of 1: 1, −1, i, −i. Sum 0. Product −1.

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