Question 1 – Find the three cube roots

In this question, we have five different numbers: 8, –8, –27, 64, and –125. For each number, we have to find its three cube roots.

Cube roots means we want all numbers that when multiplied by themselves three times give the original number.

Every real number has one real cube root and two complex cube roots. The complex cube roots are conjugates of each other.

To find the cube roots, we first write the number in polar form. Then we use the cube root formula. We add 360° or 2π to the angle to get three different roots.

For positive numbers like 8 and 64, the real cube root is positive. For negative numbers like –8 and –27, the real cube root is negative.

The two complex roots will be the real cube root multiplied by ω and ω², where ω is the cube root of unity.

Question 2 – Find the four fourth roots

In this question, we have three numbers: 16, 81, and 625. For each number, we have to find its four fourth roots.

Fourth roots means numbers that when multiplied by themselves four times give the original number.

Every positive real number has two real fourth roots and two complex fourth roots. The complex roots are conjugates.

To find the fourth roots, we write the number in polar form. Then we use the fourth root formula. We add 360° or 2π to the angle and divide by 4. This gives us four different roots.

The four roots are equally spaced on a circle. They are 90° apart. If you add all four roots, their sum is always zero. We have to show this for each case.

So for 16, the four fourth roots will be 2, –2, 2i, and –2i. Their sum is 2 – 2 + 2i – 2i = 0. Similarly for 81, the roots will be 3, –3, 3i, –3i. Sum is zero. For 625, the roots will be 5, –5, 5i, –5i. Sum is zero.

Question 3 – Cube roots of unity property

In this question, we are given that 1, ω, and ω² are the cube roots of unity. We know that ω³ = 1 and 1 + ω + ω² = 0.

We have to show that 1 + ωⁿ + ω²ⁿ = 3 when n is a multiple of 3.

When n is a multiple of 3, we can write n = 3k. Then ωⁿ = ω³ᵏ = (ω³)ᵏ = 1ᵏ = 1. Similarly, ω²ⁿ = (ωⁿ)² = 1² = 1.

So 1 + ωⁿ + ω²ⁿ = 1 + 1 + 1 = 3. That is all. This property is very useful in many problems.

If n is not a multiple of 3, the sum becomes zero, but that is not asked here.

Question 4 – Evaluate expressions

This question has two parts. Both expressions have numbers that look like cube roots of unity.

First, we notice that (–1 + √–3)/2 is actually ω, the complex cube root of unity. Because ω = (–1 + i√3)/2. Similarly, (–1 – √–3)/2 is ω².

So in part (i), we have ω⁷ + (ω²)⁷.

Since ω³ = 1, we can reduce the powers. ω⁷ = ω⁶ × ω = (ω³)² × ω = 1² × ω = ω. Similarly, (ω²)⁷ = ω¹⁴ = ω¹² × ω² = (ω³)⁴ × ω² = 1⁴ × ω² = ω².

So ω⁷ + ω¹⁴ = ω + ω². And we know that 1 + ω + ω² = 0, so ω + ω² = –1. Therefore the answer is –1.

For part (ii), it is the same number without dividing by 2. So we have (–1 + √–3) = 2ω and (–1 – √–3) = 2ω². Then we raise them to the 5th power and add. We will use the same reduction method.

Question 5 – Product of many factors

This question looks complicated but it is actually a pattern. We have to multiply many factors of the form (1 – ω + ω²), (1 – ω² + ω⁴), and so on.

First, we notice that 1 – ω + ω² can be simplified. Since 1 + ω + ω² = 0, we can write 1 + ω² = –ω. So 1 – ω + ω² = (1 + ω²) – ω = (–ω) – ω = –2ω.

Similarly, we can simplify each factor. Each factor reduces to something like 2 or –2 times some power of ω.

When we multiply 2n factors together, all the ω powers will cancel out because their sum becomes a multiple of 3. We will be left with 2 to the power 2n, meaning 2²ⁿ.

This is a proof question. We have to show the pattern step by step.

Question 6 – Prove that expression equals –1

In this question, we have to prove that ((i + √3)/2)⁸ + ((i – √3)/2)⁸ = –1.

First, we notice that (i + √3)/2 is actually another form related to cube roots of unity. If we multiply ω by i, we get something like this.

Alternatively, we can write (i + √3)/2 as i times something. There is a relation: ω² = (–1 – i√3)/2. This expression is similar but with i and √3 swapped.

We can also notice that (i + √3)/2 squared gives something simple. Then we raise to the 8th power means we square multiple times.

The second term is the conjugate of the first term. When we add them, the imaginary parts cancel and the real parts add. We will get –1 as the answer.

This is a standard problem. Once we find the pattern, the calculation becomes very short.

Question 7 – Evaluate sum of ω to powers

In this question, we have to find the sum of ω⁰ + ω² + ω⁴ + ω⁶ + ω⁸ + ω¹⁰. That is k from 0 to 5 of ω²ᵏ.

First, we write out the terms: ω⁰ = 1, ω², ω⁴, ω⁶, ω⁸, ω¹⁰.

Since ω³ = 1, we can reduce each power:

• ω⁶ = (ω³)² = 1² = 1

• ω⁸ = ω⁶ × ω² = 1 × ω² = ω²

• ω¹⁰ = ω⁹ × ω = (ω³)³ × ω = 1³ × ω = ω

So the sum becomes 1 + ω² + ω + 1 + ω² + ω.

Group the terms: 1 + 1 = 2. ω + ω = 2ω. ω² + ω² = 2ω².

So the sum is 2(1 + ω + ω²). And we know that 1 + ω + ω² = 0.

Therefore the sum is 2 × 0 = 0.

Question 8 – Prove a fraction equals ω

In this question, we have to prove that (a + bω² + cω) divided by (aω² + bω + c) equals ω.

We are given that ω is a cube root of unity. We know that ω³ = 1 and 1 + ω + ω² = 0.

We also know that ω² = 1/ω and ω = 1/ω².

In the numerator, we have a + bω² + cω. In the denominator, we have aω² + bω + c.

We can multiply numerator and denominator by something to simplify. Or we can factor ω from the numerator or denominator.

If we multiply the numerator by ω, we get aω + bω³ + cω² = aω + b(1) + cω² = b + cω² + aω. This is not exactly the denominator.

Instead, we notice that if we divide numerator and denominator by ω, we get the desired result. This is a standard proof. We will show that numerator = ω × denominator.

Question 9 – Prove another fraction equals ω

This question is similar to question 8 but with higher powers. We have to prove that (aω¹² + bω¹⁷ + cω¹⁹) divided by (aω¹⁴ + bω²² + cω³⁰) equals ω.

First, we reduce all the powers of ω using ω³ = 1.

For ω¹²: 12 is a multiple of 3, so ω¹² = 1.
For ω¹⁷: 17 ÷ 3 gives remainder 2, so ω¹⁷ = ω².
For ω¹⁹: 19 ÷ 3 gives remainder 1, so ω¹⁹ = ω.

So the numerator becomes a(1) + b(ω²) + c(ω) = a + bω² + cω.

Now the denominator:
ω¹⁴: 14 ÷ 3 gives remainder 2, so ω¹⁴ = ω².
ω²²: 22 ÷ 3 gives remainder 1, so ω²² = ω.
ω³⁰: 30 is a multiple of 3, so ω³⁰ = 1.

So the denominator becomes a(ω²) + b(ω) + c(1) = aω² + bω + c.

Now this is exactly the same as question 8. So the fraction equals ω.

This question shows how reducing powers of ω makes complicated expressions very simple.