OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Exercise 9 (F)

Question 1. Find the square root of the following complex numbers.
(i) \( 3 + 4i \)
(ii) \( -8 + 6i \)
(iii) \( -40 - 42i \)
(iv) \( i \)
(v) \( \frac{2+3i}{5-4i} + \frac{2-3i}{5+4i} \)
Answer:
(i) To find the square root of \( 3 + 4i \):
Let \( \sqrt{3+4i} = x + iy \), where \( x, y \in \mathbb{R} \).
Square both sides:
\( 3 + 4i = (x + iy)^2 \)
\( \implies 3 + 4i = x^2 - y^2 + 2ixy \)
Comparing the real and imaginary parts:
\( x^2 - y^2 = 3 \) ... (1)
\( 2xy = 4 \) ... (2)
We also know that \( x^2 + y^2 = \sqrt{(x^2-y^2)^2 + (2xy)^2} \)
\( \implies x^2 + y^2 = \sqrt{3^2 + 4^2} \)
\( \implies x^2 + y^2 = \sqrt{9 + 16} \)
\( \implies x^2 + y^2 = \sqrt{25} \)
\( \implies x^2 + y^2 = 5 \) ... (3)
Adding equations (1) and (3):
\( (x^2 - y^2) + (x^2 + y^2) = 3 + 5 \)
\( \implies 2x^2 = 8 \)
\( \implies x^2 = 4 \)
\( \implies x = \pm 2 \)
Subtracting equation (1) from (3):
\( (x^2 + y^2) - (x^2 - y^2) = 5 - 3 \)
\( \implies 2y^2 = 2 \)
\( \implies y^2 = 1 \)
\( \implies y = \pm 1 \)
From equation (2), \( 2xy = 4 \), which means \( xy \) is positive. This tells us that \( x \) and \( y \) must have the same sign.
So, either \( x = 2, y = 1 \) or \( x = -2, y = -1 \).
Therefore, \( \sqrt{3+4i} = 2 + i \) or \( -(2 + i) \).

(ii) To find the square root of \( -8 + 6i \):
Let \( \sqrt{-8+6i} = x + iy \), where \( x, y \in \mathbb{R} \).
Square both sides:
\( -8 + 6i = (x + iy)^2 \)
\( \implies -8 + 6i = x^2 - y^2 + 2ixy \)
Comparing the real and imaginary parts:
\( x^2 - y^2 = -8 \) ... (1)
\( 2xy = 6 \) ... (2)
We also know that \( x^2 + y^2 = \sqrt{(x^2-y^2)^2 + (2xy)^2} \)
\( \implies x^2 + y^2 = \sqrt{(-8)^2 + 6^2} \)
\( \implies x^2 + y^2 = \sqrt{64 + 36} \)
\( \implies x^2 + y^2 = \sqrt{100} \)
\( \implies x^2 + y^2 = 10 \) ... (3)
Adding equations (1) and (3):
\( (x^2 - y^2) + (x^2 + y^2) = -8 + 10 \)
\( \implies 2x^2 = 2 \)
\( \implies x^2 = 1 \)
\( \implies x = \pm 1 \)
Subtracting equation (1) from (3):
\( (x^2 + y^2) - (x^2 - y^2) = 10 - (-8) \)
\( \implies 2y^2 = 18 \)
\( \implies y^2 = 9 \)
\( \implies y = \pm 3 \)
From equation (2), \( 2xy = 6 \), which means \( xy \) is positive. So, \( x \) and \( y \) must have the same sign.
Thus, either \( x = 1, y = 3 \) or \( x = -1, y = -3 \).
Therefore, \( \sqrt{-8+6i} = 1 + 3i \) or \( -(1 + 3i) \).

(iii) To find the square root of \( -40 - 42i \):
Let \( \sqrt{-40-42i} = x + iy \), where \( x, y \in \mathbb{R} \).
Square both sides:
\( -40 - 42i = (x + iy)^2 \)
\( \implies -40 - 42i = x^2 - y^2 + 2ixy \)
Comparing the real and imaginary parts:
\( x^2 - y^2 = -40 \) ... (1)
\( 2xy = -42 \) ... (2)
We also know that \( x^2 + y^2 = \sqrt{(x^2-y^2)^2 + (2xy)^2} \)
\( \implies x^2 + y^2 = \sqrt{(-40)^2 + (-42)^2} \)
\( \implies x^2 + y^2 = \sqrt{1600 + 1764} \)
\( \implies x^2 + y^2 = \sqrt{3364} \)
\( \implies x^2 + y^2 = 58 \) ... (3)
Adding equations (1) and (3):
\( (x^2 - y^2) + (x^2 + y^2) = -40 + 58 \)
\( \implies 2x^2 = 18 \)
\( \implies x^2 = 9 \)
\( \implies x = \pm 3 \)
Subtracting equation (1) from (3):
\( (x^2 + y^2) - (x^2 - y^2) = 58 - (-40) \)
\( \implies 2y^2 = 98 \)
\( \implies y^2 = 49 \)
\( \implies y = \pm 7 \)
From equation (2), \( 2xy = -42 \), which means \( xy \) is negative. So, \( x \) and \( y \) must have opposite signs.
Thus, either \( x = 3, y = -7 \) or \( x = -3, y = 7 \).
Therefore, \( \sqrt{-40-42i} = 3 - 7i \) or \( -3 + 7i \)
\( \implies \sqrt{-40-42i} = \pm (3 - 7i) \).

(iv) To find the square root of \( i \):
Let \( \sqrt{i} = x + iy \), where \( x, y \in \mathbb{R} \). We can write \( i \) as \( 0 + 1i \).
Square both sides:
\( i = (x + iy)^2 \)
\( \implies 0 + 1i = x^2 - y^2 + 2ixy \)
Comparing the real and imaginary parts:
\( x^2 - y^2 = 0 \) ... (1)
\( 2xy = 1 \) ... (2)
We also know that \( x^2 + y^2 = \sqrt{(x^2-y^2)^2 + (2xy)^2} \)
\( \implies x^2 + y^2 = \sqrt{0^2 + 1^2} \)
\( \implies x^2 + y^2 = \sqrt{1} \)
\( \implies x^2 + y^2 = 1 \) ... (3)
Adding equations (1) and (3):
\( (x^2 - y^2) + (x^2 + y^2) = 0 + 1 \)
\( \implies 2x^2 = 1 \)
\( \implies x^2 = \frac{1}{2} \)
\( \implies x = \pm \frac{1}{\sqrt{2}} \)
Subtracting equation (1) from (3):
\( (x^2 + y^2) - (x^2 - y^2) = 1 - 0 \)
\( \implies 2y^2 = 1 \)
\( \implies y^2 = \frac{1}{2} \)
\( \implies y = \pm \frac{1}{\sqrt{2}} \)
From equation (2), \( 2xy = 1 \), which means \( xy \) is positive. So, \( x \) and \( y \) must have the same sign.
If \( x = \frac{1}{\sqrt{2}} \), then \( y = \frac{1}{\sqrt{2}} \).
If \( x = -\frac{1}{\sqrt{2}} \), then \( y = -\frac{1}{\sqrt{2}} \).
Therefore, \( \sqrt{i} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \) or \( -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \).
This can also be written as \( \sqrt{i} = \pm \frac{1}{\sqrt{2}}(1+i) \).

(v) To find the square root of \( z = \frac{2+3i}{5-4i} + \frac{2-3i}{5+4i} \):
First, simplify the complex number \( z \):
\[ z = \frac{(2+3i)(5+4i) + (2-3i)(5-4i)}{(5-4i)(5+4i)} \] \[ z = \frac{(10 + 8i + 15i + 12i^2) + (10 - 8i - 15i + 12i^2)}{5^2 - (4i)^2} \] Since \( i^2 = -1 \):
\[ z = \frac{(10 + 23i - 12) + (10 - 23i - 12)}{25 - (-16)} \] \[ z = \frac{(-2 + 23i) + (-2 - 23i)}{25 + 16} \] \[ z = \frac{-2 + 23i - 2 - 23i}{41} \] \[ z = \frac{-4}{41} \] Now we need to find the square root of \( z = \frac{-4}{41} \). Since this is a real negative number, its square roots will be purely imaginary.
\[ \sqrt{z} = \sqrt{\frac{-4}{41}} \] \[ \implies \sqrt{z} = \sqrt{-1 \cdot \frac{4}{41}} \] \[ \implies \sqrt{z} = \pm i \sqrt{\frac{4}{41}} \] \[ \implies \sqrt{z} = \pm \frac{2}{\sqrt{41}}i \]In simple words: To find the square root of a complex number, we assume it's \( x + iy \), then square both sides and compare the real and imaginary parts to solve for \( x \) and \( y \). The sign of \( xy \) helps us determine if \( x \) and \( y \) have the same or opposite signs. For part (v), we first simplify the complex fraction into a single number, then find its square root.

🎯 Exam Tip: Remember to always consider both positive and negative values for \( x \) and \( y \) when taking square roots. The sign of \( xy \) is crucial for pairing the correct \( x \) and \( y \) values. Always simplify complex fractions before finding their square root.

Question 2. If \( \omega \) is a cube root of unity, then
(i) \( \omega + \omega^2 = ... \)
(ii) \( 1 + \omega = ... \)
(iii) \( 1 + \omega^3 = ... \)
Answer:
We know that if \( \omega \) is a cube root of unity, then \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).
Also, the complex cube roots of unity are \( \omega = \frac{-1+\sqrt{3}i}{2} \) and \( \omega^2 = \frac{-1-\sqrt{3}i}{2} \).

(i) \( \omega + \omega^2 \)
From \( 1 + \omega + \omega^2 = 0 \), we can write \( \omega + \omega^2 = -1 \).
Alternatively, using the expressions:
\[ \omega + \omega^2 = \frac{-1+\sqrt{3}i}{2} + \frac{-1-\sqrt{3}i}{2} \] \[ \implies \omega + \omega^2 = \frac{-1+\sqrt{3}i - 1 - \sqrt{3}i}{2} \] \[ \implies \omega + \omega^2 = \frac{-2}{2} \] \[ \implies \omega + \omega^2 = -1 \]
(ii) \( 1 + \omega \)
From \( 1 + \omega + \omega^2 = 0 \), we can write \( 1 + \omega = -\omega^2 \).
Alternatively, using the expressions:
\[ 1 + \omega = 1 + \frac{-1+\sqrt{3}i}{2} \] \[ \implies 1 + \omega = \frac{2 + (-1+\sqrt{3}i)}{2} \] \[ \implies 1 + \omega = \frac{1+\sqrt{3}i}{2} \] We know that \( \omega^2 = \frac{-1-\sqrt{3}i}{2} \). The result \( \frac{1+\sqrt{3}i}{2} \) is \( -\omega^2 \).
So, \( 1 + \omega = -\omega^2 \).

(iii) \( 1 + \omega^3 \)
We know that \( \omega^3 = 1 \).
So, \( 1 + \omega^3 = 1 + 1 = 2 \).
In simple words: For cube roots of unity, there are two main rules: the sum of all three roots (1, ω, ω²) is zero, and \( \omega \) raised to the power of 3 (or any multiple of 3) is one. We use these rules to find the values.

🎯 Exam Tip: Always remember the two fundamental properties of cube roots of unity: \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \). These identities simplify almost all problems related to cube roots of unity.

Question 3. If 1, \( \omega \), \( \omega^2 \) are of unity, prove that
(i) \( (1 + \omega^2)^4 = \omega \)
(ii) \( (1 + \omega - \omega^2)^3 = -8 \)
(iii) \( (1 - \omega)(1 - \omega^2) = 3 \)
(iv) \( \frac{1}{1+\omega}+\frac{1}{1+\omega^2} = 1 \)
Answer:
Given that 1, \( \omega \), \( \omega^2 \) are the cube roots of unity. Therefore, we know that \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).

(i) Prove \( (1 + \omega^2)^4 = \omega \):
From \( 1 + \omega + \omega^2 = 0 \), we have \( 1 + \omega^2 = -\omega \).
L.H.S. \( = (1 + \omega^2)^4 \)
\( = (-\omega)^4 \)
\( = \omega^4 \)
We can write \( \omega^4 = \omega^3 \cdot \omega \).
Since \( \omega^3 = 1 \):
\( = 1 \cdot \omega \)
\( = \omega \)
L.H.S. \( = \) R.H.S. Hence proved.

(ii) Prove \( (1 + \omega - \omega^2)^3 = -8 \):
From \( 1 + \omega + \omega^2 = 0 \), we have \( 1 + \omega = -\omega^2 \).
L.H.S. \( = (1 + \omega - \omega^2)^3 \)
\( = (-\omega^2 - \omega^2)^3 \)
\( = (-2\omega^2)^3 \)
\( = (-2)^3 (\omega^2)^3 \)
\( = -8 \omega^6 \)
We can write \( \omega^6 = (\omega^3)^2 \).
Since \( \omega^3 = 1 \):
\( = -8 (1)^2 \)
\( = -8 \)
L.H.S. \( = \) R.H.S. Hence proved.

(iii) Prove \( (1 - \omega)(1 - \omega^2) = 3 \):
L.H.S. \( = (1 - \omega)(1 - \omega^2) \)
\( = 1 - \omega^2 - \omega + \omega^3 \)
\( = 1 - (\omega + \omega^2) + \omega^3 \)
From \( 1 + \omega + \omega^2 = 0 \), we have \( \omega + \omega^2 = -1 \). And \( \omega^3 = 1 \).
\( = 1 - (-1) + 1 \)
\( = 1 + 1 + 1 \)
\( = 3 \)
L.H.S. \( = \) R.H.S. Hence proved.

(iv) Prove \( \frac{1}{1+\omega}+\frac{1}{1+\omega^2} = 1 \):
From \( 1 + \omega + \omega^2 = 0 \), we have \( 1 + \omega = -\omega^2 \) and \( 1 + \omega^2 = -\omega \).
L.H.S. \( = \frac{1}{1+\omega}+\frac{1}{1+\omega^2} \)
\( = \frac{1}{-\omega^2} + \frac{1}{-\omega} \)
\( = -\frac{1}{\omega^2} - \frac{1}{\omega} \)
To combine these, find a common denominator, which is \( \omega^2 \):
\( = \frac{-\omega - \omega^2}{\omega^2} \)
\( = \frac{-(\omega + \omega^2)}{\omega^2} \)
Since \( \omega + \omega^2 = -1 \):
\( = \frac{-(-1)}{\omega^2} \)
\( = \frac{1}{\omega^2} \)
We can multiply the numerator and denominator by \( \omega \):
\( = \frac{1 \cdot \omega}{\omega^2 \cdot \omega} \)
\( = \frac{\omega}{\omega^3} \)
Since \( \omega^3 = 1 \):
\( = \frac{\omega}{1} \)
\( = \omega \)
Hold on, the target result is 1. Let's re-examine the step where we combined \( \frac{1}{-\omega^2} + \frac{1}{-\omega} \).
Alternatively, from \( 1+\omega+\omega^2 = 0 \), we have \( \omega^2 = -(1+\omega) \) and \( \omega = -(1+\omega^2) \). Let's try cross-multiplication:
L.H.S. \( = \frac{1}{1+\omega}+\frac{1}{1+\omega^2} \)
\( = \frac{(1+\omega^2) + (1+\omega)}{(1+\omega)(1+\omega^2)} \)
\( = \frac{1+\omega^2+1+\omega}{1+\omega^2+\omega+\omega^3} \)
\( = \frac{2 + (\omega+\omega^2)}{1 + (\omega+\omega^2) + \omega^3} \)
Substitute \( \omega+\omega^2 = -1 \) and \( \omega^3 = 1 \):
\( = \frac{2 + (-1)}{1 + (-1) + 1} \)
\( = \frac{1}{1} \)
\( = 1 \)
L.H.S. \( = \) R.H.S. Hence proved.
In simple words: When working with cube roots of unity, the key is to use the two main properties: \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \). By replacing terms using these rules, we can simplify complex expressions and show they are equal to the given values.

🎯 Exam Tip: For proofs involving cube roots of unity, always start by listing the properties \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \). Use these properties to simplify expressions, replacing \( 1+\omega \) with \( -\omega^2 \), \( 1+\omega^2 \) with \( -\omega \), or powers of \( \omega \) with their equivalent simpler forms.

Question 4.
(i) \( (1 - \omega - \omega^2)^6 = 64 \)
(ii) \( (1 + \omega - \omega^2)(1 - \omega + \omega^2) = 4 \)
Answer:
Given that \( \omega \) is a cube root of unity, we know \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).

(i) Prove \( (1 - \omega - \omega^2)^6 = 64 \):
L.H.S. \( = (1 - \omega - \omega^2)^6 \)
\( = [1 - (\omega + \omega^2)]^6 \)
From \( 1 + \omega + \omega^2 = 0 \), we know \( \omega + \omega^2 = -1 \).
\( = [1 - (-1)]^6 \)
\( = [1 + 1]^6 \)
\( = 2^6 \)
\( = 64 \)
L.H.S. \( = \) R.H.S. Hence proved.

(ii) Prove \( (1 + \omega - \omega^2)(1 - \omega + \omega^2) = 4 \):
From \( 1 + \omega + \omega^2 = 0 \), we have \( 1 + \omega = -\omega^2 \) and \( 1 + \omega^2 = -\omega \).
L.H.S. \( = (1 + \omega - \omega^2)(1 - \omega + \omega^2) \)
Substitute the identities:
\( = (-\omega^2 - \omega^2)(-\omega - \omega) \)
\( = (-2\omega^2)(-2\omega) \)
\( = (-2) \cdot (-2) \cdot \omega^2 \cdot \omega \)
\( = 4 \omega^3 \)
Since \( \omega^3 = 1 \):
\( = 4 \cdot 1 \)
\( = 4 \)
L.H.S. \( = \) R.H.S. Hence proved.
In simple words: These proofs use the basic relationships of cube roots of unity. By grouping terms and substituting \( \omega + \omega^2 = -1 \) or \( 1 + \omega = -\omega^2 \), expressions simplify quickly. Remember that \( \omega^3 \) always equals 1, which helps reduce higher powers.

🎯 Exam Tip: Always look for ways to substitute \( 1 + \omega + \omega^2 = 0 \) or \( \omega^3 = 1 \) to simplify terms. Pay attention to signs, especially when distributing negative terms like \( -( \omega + \omega^2 ) \).

Question 5. Prove that \( (3 + 5\omega + 3\omega^2)^6 = (3 + 5\omega^2 + 3\omega)^6 = 64 \)
Answer:
Given that \( \omega \) is a cube root of unity, we know \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).

Part 1: Prove \( (3 + 5\omega + 3\omega^2)^6 = 64 \)
Consider the expression \( (3 + 5\omega + 3\omega^2)^6 \).
We can rearrange the terms and factor:
\( = [3(1 + \omega^2) + 5\omega]^6 \)
From \( 1 + \omega + \omega^2 = 0 \), we know \( 1 + \omega^2 = -\omega \).
\( = [3(-\omega) + 5\omega]^6 \)
\( = [-3\omega + 5\omega]^6 \)
\( = [2\omega]^6 \)
\( = 2^6 \omega^6 \)
\( = 64 (\omega^3)^2 \)
Since \( \omega^3 = 1 \):
\( = 64 (1)^2 \)
\( = 64 \)

Part 2: Prove \( (3 + 5\omega^2 + 3\omega)^6 = 64 \)
Consider the expression \( (3 + 5\omega^2 + 3\omega)^6 \).
We can rearrange the terms and factor:
\( = [3(1 + \omega) + 5\omega^2]^6 \)
From \( 1 + \omega + \omega^2 = 0 \), we know \( 1 + \omega = -\omega^2 \).
\( = [3(-\omega^2) + 5\omega^2]^6 \)
\( = [-3\omega^2 + 5\omega^2]^6 \)
\( = [2\omega^2]^6 \)
\( = 2^6 (\omega^2)^6 \)
\( = 64 \omega^{12} \)
We can write \( \omega^{12} = (\omega^3)^4 \).
Since \( \omega^3 = 1 \):
\( = 64 (1)^4 \)
\( = 64 \)
Both expressions simplify to 64. Hence proved.
In simple words: This problem asks us to show that two different complex expressions, when raised to the power of 6, both equal 64. We solve it by using the basic properties of cube roots of unity to simplify the expressions inside the brackets first, then calculate the power. Grouping similar terms helps make the substitutions clearer.

🎯 Exam Tip: When faced with polynomial expressions involving cube roots of unity, group terms to apply the \( 1 + \omega + \omega^2 = 0 \) identity effectively. This simplifies the base of the power, making the final calculation much easier.

Question 6. Prove that \( \omega^{28} + \omega^{29} + 1 = 0 \)
Answer:
Given that \( \omega \) is a cube root of unity, we know that \( \omega^3 = 1 \).
L.H.S. \( = \omega^{28} + \omega^{29} + 1 \)
To simplify powers of \( \omega \), we divide the exponent by 3 and use the remainder. For example, \( \omega^n = \omega^{3k+r} = (\omega^3)^k \cdot \omega^r = 1^k \cdot \omega^r = \omega^r \).
For \( \omega^{28} \): \( 28 = 3 \times 9 + 1 \), so \( \omega^{28} = \omega^1 = \omega \).
For \( \omega^{29} \): \( 29 = 3 \times 9 + 2 \), so \( \omega^{29} = \omega^2 \).
Substitute these simplified forms back into the expression:
\( = \omega + \omega^2 + 1 \)
We know that for cube roots of unity, \( 1 + \omega + \omega^2 = 0 \).
\( = 0 \)
L.H.S. \( = \) R.H.S. Hence proved.
In simple words: To simplify high powers of \( \omega \), we divide the exponent by 3 and keep only the remainder as the new power. For example, \( \omega^{28} \) becomes \( \omega^1 \). After simplifying all terms, we use the rule that \( 1 + \omega + \omega^2 \) is always zero.

🎯 Exam Tip: For large powers of \( \omega \), always reduce the exponent modulo 3. That is, divide the exponent by 3 and use the remainder as the new exponent of \( \omega \). This makes calculations much faster and simpler.

Question 7. If 1, \( \omega \), \( \omega^2 \) are the cube roots of unity, prove that \( \omega^n + \omega^{2n} = 2 \) or \( -1 \) according as \( n \) is a multiple of 3 or any other integer.
Answer:
We are given that 1, \( \omega \), \( \omega^2 \) are the cube roots of unity. This means \( \omega = \frac{-1+\sqrt{3}i}{2} \) and \( \omega^2 = \frac{-1-\sqrt{3}i}{2} \). We also know \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).
We need to prove that \( \omega^n + \omega^{2n} \) is either 2 or -1, depending on whether \( n \) is a multiple of 3.

Case-I: When \( n \) is a multiple of 3.
Let \( n = 3k \) for some integer \( k \).
Then \( \omega^n + \omega^{2n} = \omega^{3k} + \omega^{2(3k)} \)
\( = \omega^{3k} + \omega^{6k} \)
\( = (\omega^3)^k + (\omega^3)^{2k} \)
Since \( \omega^3 = 1 \):
\( = (1)^k + (1)^{2k} \)
\( = 1 + 1 \)
\( = 2 \)
So, if \( n \) is a multiple of 3, \( \omega^n + \omega^{2n} = 2 \).

Case-II: When \( n \) is not a multiple of 3.
If \( n \) is not a multiple of 3, then \( n \) can be of the form \( 3k+1 \) or \( 3k+2 \) for some integer \( k \).

Subcase-II (a): When \( n = 3k+1 \).
Then \( \omega^n + \omega^{2n} = \omega^{3k+1} + \omega^{2(3k+1)} \)
\( = \omega^{3k+1} + \omega^{6k+2} \)
\( = \omega^{3k} \cdot \omega^1 + \omega^{6k} \cdot \omega^2 \)
\( = (\omega^3)^k \cdot \omega + (\omega^3)^{2k} \cdot \omega^2 \)
Since \( \omega^3 = 1 \):
\( = (1)^k \cdot \omega + (1)^{2k} \cdot \omega^2 \)
\( = \omega + \omega^2 \)
From \( 1 + \omega + \omega^2 = 0 \), we know \( \omega + \omega^2 = -1 \).
So, if \( n = 3k+1 \), \( \omega^n + \omega^{2n} = -1 \).

Subcase-II (b): When \( n = 3k+2 \).
Then \( \omega^n + \omega^{2n} = \omega^{3k+2} + \omega^{2(3k+2)} \)
\( = \omega^{3k+2} + \omega^{6k+4} \)
\( = \omega^{3k} \cdot \omega^2 + \omega^{6k} \cdot \omega^4 \)
\( = (\omega^3)^k \cdot \omega^2 + (\omega^3)^{2k} \cdot \omega^3 \cdot \omega \)
Since \( \omega^3 = 1 \):
\( = (1)^k \cdot \omega^2 + (1)^{2k} \cdot 1 \cdot \omega \)
\( = \omega^2 + \omega \)
From \( 1 + \omega + \omega^2 = 0 \), we know \( \omega + \omega^2 = -1 \).
So, if \( n = 3k+2 \), \( \omega^n + \omega^{2n} = -1 \).

Conclusion: If \( n \) is a multiple of 3, \( \omega^n + \omega^{2n} = 2 \). If \( n \) is any other integer (not a multiple of 3), \( \omega^n + \omega^{2n} = -1 \). Hence proved.
In simple words: We are testing how the sum \( \omega^n + \omega^{2n} \) behaves depending on what kind of number \( n \) is. If \( n \) can be divided by 3, the sum is 2. If \( n \) cannot be divided by 3, the sum is -1. This is because powers of \( \omega \) repeat every three steps.

🎯 Exam Tip: This question tests your understanding of the periodicity of powers of \( \omega \). Always classify \( n \) into \( 3k \), \( 3k+1 \), or \( 3k+2 \) to show the different outcomes based on the properties \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

Prove the Following:

Question 8. \( (1 - \omega + \omega^2)(1 + \omega - \omega^2)(1 - \omega - \omega^2) = 8. \)
Answer:
Given that \( \omega \) is a cube root of unity, we know \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).
We need to prove \( (1 - \omega + \omega^2)(1 + \omega - \omega^2)(1 - \omega - \omega^2) = 8 \).
L.H.S. \( = (1 - \omega + \omega^2)(1 + \omega - \omega^2)(1 - \omega - \omega^2) \)
Use the identity \( 1 + \omega + \omega^2 = 0 \):
\( 1 + \omega^2 = -\omega \)
\( 1 + \omega = -\omega^2 \)
\( \omega + \omega^2 = -1 \)

Substitute these into the factors:
First factor: \( 1 - \omega + \omega^2 = (1 + \omega^2) - \omega = (-\omega) - \omega = -2\omega \).
Second factor: \( 1 + \omega - \omega^2 = (1 + \omega) - \omega^2 = (-\omega^2) - \omega^2 = -2\omega^2 \).
Third factor: \( 1 - \omega - \omega^2 = 1 - (\omega + \omega^2) = 1 - (-1) = 1 + 1 = 2 \).

Now multiply these three simplified factors:
L.H.S. \( = (-2\omega)(-2\omega^2)(2) \)
\( = (-2)(-2)(2) \cdot \omega \cdot \omega^2 \)
\( = 8 \omega^3 \)
Since \( \omega^3 = 1 \):
\( = 8 \cdot 1 \)
\( = 8 \)
L.H.S. \( = \) R.H.S. Hence proved.
In simple words: This problem involves simplifying three terms multiplied together. We use the rule that \( 1 + \omega + \omega^2 \) is zero to change parts of each term, making them much simpler. Once simplified, we multiply them, remembering that \( \omega^3 \) is 1.

🎯 Exam Tip: When dealing with products of multiple factors involving cube roots of unity, simplify each factor individually using \( 1 + \omega + \omega^2 = 0 \) before multiplying. This prevents calculation errors and makes the solution straightforward.

Question 9.
(i) \( (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) ... \) to \( 2n \) factors \( = 1 \)
(ii) \( (1 - \omega + \omega^2)(1 - \omega^2 + \omega^4)(1 - \omega^4 + \omega^8)... \) to \( 2n \) factors \( = 2^{2n} \)
Answer:
Given that \( \omega \) is a cube root of unity, we know \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).

(i) Prove \( (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) ... \) to \( 2n \) factors \( = 1 \):
Let the given expression be P.
P \( = (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) ... \) to \( 2n \) factors
First, simplify the terms using \( \omega^3 = 1 \):
\( \omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega \)
\( \omega^8 = \omega^6 \cdot \omega^2 = (\omega^3)^2 \cdot \omega^2 = 1^2 \cdot \omega^2 = \omega^2 \)
So, the factors repeat a pattern of \( (1+\omega) \) and \( (1+\omega^2) \).
The expression becomes:
P \( = (1 + \omega)(1 + \omega^2)(1 + \omega)(1 + \omega^2) ... \) to \( 2n \) factors.
Since there are \( 2n \) factors, there will be \( n \) pairs of \( (1+\omega) \) and \( (1+\omega^2) \).
P \( = [(1 + \omega)(1 + \omega^2)]^n \)
Now, expand the inner product:
\( (1 + \omega)(1 + \omega^2) = 1 + \omega^2 + \omega + \omega^3 \)
\( = 1 + (\omega + \omega^2) + \omega^3 \)
Using \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \):
\( = 1 + (-1) + 1 \)
\( = 1 \)
Substitute this back into the expression for P:
P \( = (1)^n \)
P \( = 1 \)
Hence proved.

(ii) Prove \( (1 - \omega + \omega^2)(1 - \omega^2 + \omega^4)(1 - \omega^4 + \omega^8)... \) to \( 2n \) factors \( = 2^{2n} \):
Let the given expression be Q.
Q \( = (1 - \omega + \omega^2)(1 - \omega^2 + \omega^4)(1 - \omega^4 + \omega^8)... \) to \( 2n \) factors.
Simplify the powers of \( \omega \):
\( \omega^4 = \omega \)
\( \omega^8 = \omega^2 \)
The factors repeat a pattern:
First factor: \( 1 - \omega + \omega^2 \)
Using \( 1 + \omega + \omega^2 = 0 \), we have \( 1 + \omega^2 = -\omega \).
So, \( 1 - \omega + \omega^2 = (1 + \omega^2) - \omega = -\omega - \omega = -2\omega \).

Second factor: \( 1 - \omega^2 + \omega^4 = 1 - \omega^2 + \omega \)
Using \( 1 + \omega + \omega^2 = 0 \), we have \( 1 + \omega = -\omega^2 \).
So, \( 1 - \omega^2 + \omega = (1 + \omega) - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2 \).

Since the powers of \( \omega \) repeat \( \omega, \omega^2, \omega, \omega^2, ... \), the terms in the product will alternate between \( (-2\omega) \) and \( (-2\omega^2) \).
The expression Q has \( 2n \) factors, so there are \( n \) factors of \( (-2\omega) \) and \( n \) factors of \( (-2\omega^2) \).
Q \( = (-2\omega)^n (-2\omega^2)^n \)
\( = (-2)^n \omega^n \cdot (-2)^n (\omega^2)^n \)
\( = (-2)^n (-2)^n \cdot \omega^n \cdot \omega^{2n} \)
\( = ((-2) \cdot (-2))^n \cdot \omega^{n+2n} \)
\( = (4)^n \cdot \omega^{3n} \)
\( = 4^n \cdot (\omega^3)^n \)
Since \( \omega^3 = 1 \):
\( = 4^n \cdot (1)^n \)
\( = 4^n \)
We need to show it is \( 2^{2n} \). Note that \( 4^n = (2^2)^n = 2^{2n} \).
Q \( = 2^{2n} \)
Hence proved.
In simple words: For part (i), we see that the terms in the product repeat. Each pair of terms multiplies to 1, so if there are \( 2n \) terms, there are \( n \) such pairs, giving a total product of 1. For part (ii), the terms also repeat, simplifying to \( -2\omega \) and \( -2\omega^2 \). When we multiply these \( 2n \) terms, we get \( 4^n \), which is the same as \( 2^{2n} \).

🎯 Exam Tip: When a product has many factors involving powers of \( \omega \), first simplify the powers \( \omega^4, \omega^8, \omega^{12}, \) etc., to \( \omega, \omega^2, 1, \) using \( \omega^3=1 \). Then, simplify the binomial factors using \( 1+\omega+\omega^2=0 \) to identify repeating patterns or pairs that simplify easily.

Question 10. Prove that \( \frac{a+b\omega+c\omega^2}{b+c\omega+a\omega^2} = \omega \)
Answer:
Given that \( \omega \) is a cube root of unity, we know \( \omega^3 = 1 \).
We need to prove \( \frac{a+b\omega+c\omega^2}{b+c\omega+a\omega^2} = \omega \).
L.H.S. \( = \frac{a+b\omega+c\omega^2}{b+c\omega+a\omega^2} \)
We want to transform the numerator to be \( \omega \) times the denominator. Let's try multiplying the numerator by \( \omega \) in a way that uses \( \omega^3=1 \).
We can rewrite \( a \) as \( a\omega^3 \) since \( \omega^3 = 1 \).
L.H.S. \( = \frac{a\omega^3+b\omega+c\omega^2}{b+c\omega+a\omega^2} \)
Now, factor out \( \omega \) from the numerator:
\( = \frac{\omega(a\omega^2+b+c\omega)}{b+c\omega+a\omega^2} \)
Rearranging the terms in the parenthesis in the numerator, we get:
\( = \frac{\omega(b+c\omega+a\omega^2)}{b+c\omega+a\omega^2} \)
Since the term \( (b+c\omega+a\omega^2) \) is present in both numerator and denominator, and assuming it's not zero, we can cancel it out.
\( = \omega \)
L.H.S. \( = \) R.H.S. Hence proved.
In simple words: To prove this, we make the top part of the fraction look like the bottom part, multiplied by \( \omega \). We do this by changing 'a' into '\( a\omega^3 \)' (because \( \omega^3 \) is 1), and then factoring out \( \omega \) from the numerator. This makes the remaining part of the numerator the same as the denominator, allowing us to cancel it out.

🎯 Exam Tip: For rational expressions involving cube roots of unity, the trick is often to make the numerator a multiple of the denominator (or vice-versa) by strategically using the identity \( \omega^3 = 1 \). This might involve multiplying a term by \( \omega^3 \) to change its power and allow for factoring.

Question 11. Prove that \( \frac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2}+\frac{a+b\omega+c\omega^2}{b+c\omega+a\omega^2} = -1 \)
Answer:
Given that \( \omega \) is a cube root of unity, we know \( \omega^3 = 1 \).
We need to prove \( \frac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2}+\frac{a+b\omega+c\omega^2}{b+c\omega+a\omega^2} = -1 \).
Let's simplify each fraction separately.

First fraction: \( \frac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2} \)
To make the numerator similar to the denominator, we can multiply the denominator by \( \omega \):
\( \omega(c+a\omega+b\omega^2) = c\omega+a\omega^2+b\omega^3 \)
Since \( \omega^3 = 1 \):
\( = c\omega+a\omega^2+b \)
This is the same as the numerator \( a+b\omega+c\omega^2 \) but with terms rearranged. So, if we factor \( \frac{1}{\omega} \) from the denominator, it will match the numerator.
Alternatively, multiply numerator and denominator by \( \omega \).
\( = \frac{\omega(a+b\omega+c\omega^2)}{\omega(c+a\omega+b\omega^2)} \)
\( = \frac{a\omega+b\omega^2+c\omega^3}{c\omega+a\omega^2+b\omega^3} \)
Since \( \omega^3 = 1 \):
\( = \frac{a\omega+b\omega^2+c}{c\omega+a\omega^2+b} \)
This now matches the denominator of the original first fraction. So, \( \frac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2} \) can be seen as \( \frac{1}{\omega} \).
Let's prove this more directly by manipulating the denominator.
Denominator \( = c+a\omega+b\omega^2 \)
Multiply by \( \omega^2 \): \( \omega^2(c+a\omega+b\omega^2) = c\omega^2+a\omega^3+b\omega^4 \)
\( = c\omega^2+a+b\omega \)
This is exactly the numerator. So, \( \frac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2} = \frac{1}{\omega^2} \).

Second fraction: \( \frac{a+b\omega+c\omega^2}{b+c\omega+a\omega^2} \)
This is the same as Question 10. We found it simplifies to \( \omega \).
Let's reconfirm:
Denominator \( = b+c\omega+a\omega^2 \)
Multiply by \( \omega \): \( \omega(b+c\omega+a\omega^2) = b\omega+c\omega^2+a\omega^3 \)
\( = b\omega+c\omega^2+a \)
This is exactly the numerator. So, \( \frac{a+b\omega+c\omega^2}{b+c\omega+a\omega^2} = \frac{1}{\omega} \).

Now add the two simplified fractions:
L.H.S. \( = \frac{1}{\omega^2} + \frac{1}{\omega} \)
Find a common denominator, which is \( \omega^2 \):
\( = \frac{1 + \omega}{\omega^2} \)
From \( 1 + \omega + \omega^2 = 0 \), we know \( 1 + \omega = -\omega^2 \).
\( = \frac{-\omega^2}{\omega^2} \)
\( = -1 \)
L.H.S. \( = \) R.H.S. Hence proved.
In simple words: This problem asks us to add two fractions involving cube roots of unity. We simplify each fraction first by finding a way to make its numerator match its denominator by using the rule \( \omega^3 = 1 \). After simplifying, the fractions become \( \frac{1}{\omega^2} \) and \( \frac{1}{\omega} \). Adding these together and using the rule \( 1 + \omega + \omega^2 = 0 \) gives us -1.

🎯 Exam Tip: When dealing with sums of rational expressions like this, focus on simplifying each term individually. The key is to strategically multiply the numerator and/or denominator by powers of \( \omega \) to make them identical, leveraging \( \omega^3=1 \), so that the fraction simplifies to a power of \( \omega \).

Question 12. If \( \omega \) is a cube root of unity and \( n \) is a positive integer which is not a multiple of 3, then show that \( (1 + \omega^n + \omega^{2n}) = 0 \).
Answer:
Given that \( \omega \) is a cube root of unity, we know \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).
We are given that \( n \) is a positive integer but not a multiple of 3. This means \( n \) can be of the form \( 3k+1 \) or \( 3k+2 \) for some integer \( k \ge 0 \).

Case-I: When \( n = 3k+1 \).
We need to find \( 1 + \omega^n + \omega^{2n} \). Substitute \( n = 3k+1 \):
\( = 1 + \omega^{3k+1} + \omega^{2(3k+1)} \)
\( = 1 + \omega^{3k} \cdot \omega^1 + \omega^{6k+2} \)
\( = 1 + (\omega^3)^k \cdot \omega + (\omega^3)^{2k} \cdot \omega^2 \)
Since \( \omega^3 = 1 \):
\( = 1 + (1)^k \cdot \omega + (1)^{2k} \cdot \omega^2 \)
\( = 1 + \omega + \omega^2 \)
Since \( 1 + \omega + \omega^2 = 0 \):
\( = 0 \)

Case-II: When \( n = 3k+2 \).
Substitute \( n = 3k+2 \) into the expression \( 1 + \omega^n + \omega^{2n} \):
\( = 1 + \omega^{3k+2} + \omega^{2(3k+2)} \)
\( = 1 + \omega^{3k} \cdot \omega^2 + \omega^{6k+4} \)
\( = 1 + (\omega^3)^k \cdot \omega^2 + (\omega^3)^{2k} \cdot \omega^3 \cdot \omega^1 \)
Since \( \omega^3 = 1 \):
\( = 1 + (1)^k \cdot \omega^2 + (1)^{2k} \cdot 1 \cdot \omega \)
\( = 1 + \omega^2 + \omega \)
Since \( 1 + \omega + \omega^2 = 0 \):
\( = 0 \)

In both cases, when \( n \) is not a multiple of 3, the expression \( 1 + \omega^n + \omega^{2n} \) equals 0. Hence proved.
In simple words: This problem shows that if \( n \) is any number that isn't a multiple of 3, then the sum \( 1 + \omega^n + \omega^{2n} \) will always be zero. This is because the powers of \( \omega \) repeat in a cycle of three (\( \omega^1, \omega^2, \omega^3=1 \)). So, \( \omega^n \) will either be \( \omega \) or \( \omega^2 \), and \( \omega^{2n} \) will be the other, ensuring that we always end up with \( 1 + \omega + \omega^2 \), which is zero.

🎯 Exam Tip: This is a fundamental property. The expression \( 1 + \omega^n + \omega^{2n} \) evaluates to 3 if \( n \) is a multiple of 3, and 0 if \( n \) is not a multiple of 3. This pattern is essential for solving many problems with higher powers of \( \omega \).

Question 13. Show that \( (x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z) = x^2 + y^2 + z^2 - yz - zx - xy \).
Answer:
Given that \( \omega \) is a cube root of unity, we know \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).
We need to prove \( (x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z) = x^2 + y^2 + z^2 - yz - zx - xy \).
L.H.S. \( = (x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z) \)
Multiply the two factors using the distributive property:
\( = x(x + \omega^2 y + \omega z) + \omega y(x + \omega^2 y + \omega z) + \omega^2 z(x + \omega^2 y + \omega z) \)
\( = x^2 + x\omega^2 y + x\omega z + \omega y x + \omega^3 y^2 + \omega^2 y z + \omega^2 z x + \omega^4 z y + \omega^3 z^2 \)
Group terms and simplify powers of \( \omega \): \( \omega^3 = 1 \), \( \omega^4 = \omega \).
\( = x^2 + x\omega^2 y + x\omega z + \omega xy + y^2 + \omega^2 y z + \omega^2 x z + \omega y z + z^2 \)
Now, group terms by \( xy, yz, zx \) and factor out common parts:
\( = x^2 + y^2 + z^2 + (x\omega^2 y + \omega xy) + (x\omega z + \omega^2 x z) + (\omega^2 y z + \omega y z) \)
\( = x^2 + y^2 + z^2 + xy(\omega^2 + \omega) + xz(\omega + \omega^2) + yz(\omega^2 + \omega) \)
Using the identity \( \omega + \omega^2 = -1 \):
\( = x^2 + y^2 + z^2 + xy(-1) + xz(-1) + yz(-1) \)
\( = x^2 + y^2 + z^2 - xy - xz - yz \)
This is the R.H.S. Hence proved.
In simple words: This problem asks us to multiply two complex expressions and show the result. We multiply each term carefully, using the distributive property. Then, we simplify any powers of \( \omega \) (like \( \omega^3 \) or \( \omega^4 \)) and group similar terms. Finally, we use the rule that \( \omega + \omega^2 \) is -1 to get the final answer. This is an important identity in complex numbers.

🎯 Exam Tip: When multiplying complex expressions like these, carry out the multiplication systematically. Be careful with powers of \( \omega \) (using \( \omega^3=1 \)) and remember to substitute \( \omega+\omega^2=-1 \) to simplify sums of \( \omega \) terms.

Question 14. Show that \( x^3 + y^3 = (x + y)(\omega x + \omega^2 y)(\omega^2 x + \omega y) \).
Answer:
Given that \( \omega \) is a cube root of unity, we know \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).
We need to prove \( x^3 + y^3 = (x + y)(\omega x + \omega^2 y)(\omega^2 x + \omega y) \).
Let's start with the R.H.S. and expand it.
R.H.S. \( = (x + y)(\omega x + \omega^2 y)(\omega^2 x + \omega y) \)
First, multiply the last two factors:
\( (\omega x + \omega^2 y)(\omega^2 x + \omega y) = \omega x (\omega^2 x + \omega y) + \omega^2 y (\omega^2 x + \omega y) \)
\( = \omega^3 x^2 + \omega^2 xy + \omega^4 yx + \omega^3 y^2 \)
Simplify powers of \( \omega \): \( \omega^3 = 1 \) and \( \omega^4 = \omega \).
\( = 1 \cdot x^2 + \omega^2 xy + \omega xy + 1 \cdot y^2 \)
\( = x^2 + y^2 + xy(\omega^2 + \omega) \)
Using the identity \( \omega^2 + \omega = -1 \):
\( = x^2 + y^2 + xy(-1) \)
\( = x^2 + y^2 - xy \)
Now, substitute this back into the full R.H.S. expression:
R.H.S. \( = (x + y)(x^2 - xy + y^2) \)
This is the standard algebraic identity for the sum of two cubes, \( a^3 + b^3 \).
\( = x^3 + y^3 \)
This is the L.H.S. Hence proved.
In simple words: This problem shows an important identity for \( x^3 + y^3 \) using complex numbers. We start by multiplying the three factors on the right side. The key is to simplify the powers of \( \omega \) (like \( \omega^3 \) and \( \omega^4 \)) and use the property that \( \omega + \omega^2 \) equals -1. After these steps, the expression simplifies to the familiar \( (x+y)(x^2-xy+y^2) \), which is \( x^3+y^3 \).

🎯 Exam Tip: This is a factorization of \( x^3+y^3 \) using cube roots of unity. Recognizing that \( (\omega x + \omega^2 y)(\omega^2 x + \omega y) \) simplifies to \( x^2 - xy + y^2 \) is the crucial step. Remember to use \( \omega^3 = 1 \) and \( \omega+\omega^2 = -1 \) diligently.

Question 15. If 1, \( \omega \), \( \omega^2 \) are cube roots of unity, prove that 1, \( \omega \), \( \omega^2 \) are vertices of an equilateral triangle.
Answer:
Given that 1, \( \omega \), \( \omega^2 \) are the cube roots of unity. These can be represented as complex numbers in the complex plane.
Let A = 1, B = \( \omega \), C = \( \omega^2 \).
Their values are: \( 1 \), \( \omega = \frac{-1+\sqrt{3}i}{2} \), and \( \omega^2 = \frac{-1-\sqrt{3}i}{2} \).
For the points to form an equilateral triangle, the distance between any two points must be equal.

Distance AB \( = |1 - \omega| \)
\( = \left|1 - \left(\frac{-1+\sqrt{3}i}{2}\right)\right| \)
\( = \left|\frac{2 - (-1+\sqrt{3}i)}{2}\right| \)
\( = \left|\frac{2 + 1 - \sqrt{3}i}{2}\right| \)
\( = \left|\frac{3 - \sqrt{3}i}{2}\right| \)
\( = \frac{1}{2}|3 - \sqrt{3}i| \)
\( = \frac{1}{2}\sqrt{3^2 + (-\sqrt{3})^2} \)
\( = \frac{1}{2}\sqrt{9 + 3} \)
\( = \frac{1}{2}\sqrt{12} \)
\( = \frac{1}{2} (2\sqrt{3}) \)
\( = \sqrt{3} \)

Distance BC \( = |\omega - \omega^2| \)
\( = \left|\left(\frac{-1+\sqrt{3}i}{2}\right) - \left(\frac{-1-\sqrt{3}i}{2}\right)\right| \)
\( = \left|\frac{-1+\sqrt{3}i + 1 + \sqrt{3}i}{2}\right| \)
\( = \left|\frac{2\sqrt{3}i}{2}\right| \)
\( = |\sqrt{3}i| \)
\( = \sqrt{0^2 + (\sqrt{3})^2} \)
\( = \sqrt{3} \)

Distance CA \( = |\omega^2 - 1| \)
\( = \left|\left(\frac{-1-\sqrt{3}i}{2}\right) - 1\right| \)
\( = \left|\frac{-1-\sqrt{3}i - 2}{2}\right| \)
\( = \left|\frac{-3-\sqrt{3}i}{2}\right| \)
\( = \frac{1}{2}|-3-\sqrt{3}i| \)
\( = \frac{1}{2}\sqrt{(-3)^2 + (-\sqrt{3})^2} \)
\( = \frac{1}{2}\sqrt{9 + 3} \)
\( = \frac{1}{2}\sqrt{12} \)
\( = \frac{1}{2} (2\sqrt{3}) \)
\( = \sqrt{3} \)

Since \( |AB| = |BC| = |CA| = \sqrt{3} \), the three points (1, \( \omega \), \( \omega^2 \)) form an equilateral triangle.

In simple words: We can represent the cube roots of unity (1, \( \omega \), and \( \omega^2 \)) as points on a graph called the complex plane. To show they form an equilateral triangle, we need to prove that the distance between any two of these points is the same. By calculating the distance using the formula \( |z_1 - z_2| \) for each pair, we found all distances are \( \sqrt{3} \), which confirms they form an equilateral triangle.

🎯 Exam Tip: To prove points form an equilateral triangle in the complex plane, calculate the distances between all pairs of points using the modulus formula \( |z_1 - z_2| \). If all three distances are equal, the triangle is equilateral. Visualizing the points can help confirm the result.