Maharashtra Board Class 11 Maths Part 1 Chapter 3 Complex Numbers 3.3 Solutions

Question 1. If \( \omega \) is a complex cube root of unity, show that
(i) \( (2 - \omega)(2 - \omega^2) = 7 \)
(ii) \( (2 + \omega + \omega^2)^3 - (1 - 3\omega + \omega^2)^3 = 65 \)
(iii) \( \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} = \omega^2 \)
Answer:
Given that \( \omega \) is a complex cube root of unity.
\( \implies \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \)
\( \implies 1 + \omega = -\omega^2 \), \( 1 + \omega^2 = -\omega \), and \( \omega + \omega^2 = -1 \)

(i) To show: \( (2 - \omega)(2 - \omega^2) = 7 \)
L.H.S. = \( (2 - \omega)(2 - \omega^2) \)
\( = 4 - 2\omega^2 - 2\omega + \omega^3 \)
\( = 4 - 2(\omega^2 + \omega) + \omega^3 \)
Substituting \( \omega^2 + \omega = -1 \) and \( \omega^3 = 1 \):
\( = 4 - 2(-1) + 1 \)
\( = 4 + 2 + 1 \)
\( = 7 \)
\( = \) R.H.S.
Hence, \( (2 - \omega)(2 - \omega^2) = 7 \).

(ii) To show: \( (2 + \omega + \omega^2)^3 - (1 - 3\omega + \omega^2)^3 = 65 \)
L.H.S. = \( (2 + \omega + \omega^2)^3 - (1 - 3\omega + \omega^2)^3 \)
\( = (2 + (\omega + \omega^2))^3 - ((1 + \omega^2) - 3\omega)^3 \)
Substituting \( \omega + \omega^2 = -1 \) and \( 1 + \omega^2 = -\omega \):
\( = (2 - 1)^3 - (-\omega - 3\omega)^3 \)
\( = (1)^3 - (-4\omega)^3 \)
\( = 1 - (-64\omega^3) \)
Substituting \( \omega^3 = 1 \):
\( = 1 + 64(1) \)
\( = 65 \)
\( = \) R.H.S.
Hence, \( (2 + \omega + \omega^2)^3 - (1 - 3\omega + \omega^2)^3 = 65 \).

(iii) To show: \( \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} = \omega^2 \)
L.H.S. = \( \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} \)
We know that \( \omega^3 = 1 \) and \( \omega^4 = \omega \).
Let us rewrite the numerator \( a + b\omega + c\omega^2 \) by multiplying terms with powers of \( \omega \):
\( a + b\omega + c\omega^2 = a(1) + b(\omega) + c\omega^2 \)
\( = a\omega^3 + b\omega^4 + c\omega^2 \)
Taking \( \omega^2 \) common from the terms:
\( = \omega^2(a\omega + b\omega^2 + c) \)
\( = \omega^2(c + a\omega + b\omega^2) \)
Now, substitute this back into the L.H.S.:
L.H.S. = \( \frac{\omega^2(c + a\omega + b\omega^2)}{c + a\omega + b\omega^2} \)
Cancelling the common term \( (c + a\omega + b\omega^2) \) from numerator and denominator:
\( = \omega^2 \)
\( = \) R.H.S.
Hence, \( \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} = \omega^2 \).
In simple words: Since \( \omega \) is a cube root of unity, we use the standard properties \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \) to simplify complex expressions step-by-step into simple numbers.

🎯 Exam Tip: Always write down the basic properties \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \) at the beginning of your solution, as examiners look for these key relations when grading.

Question 1. Prove the following identities if \( \omega \) is a complex cube root of unity:
(i) \( 4 - 2\omega^2 - 2\omega + \omega^3 = 7 \)
(ii) \( (2 + \omega + \omega^2)^3 - (1 - 3\omega + \omega^2)^3 = 65 \)
(iii) \( \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} = \omega^2 \)
Answer:
(i)
\( \text{L.H.S.} = 4 - 2\omega^2 - 2\omega + \omega^3 \)
\( = 4 - 2(\omega^2 + \omega) + 1 \quad [ \because \omega^3 = 1 ] \)
\( = 4 - 2(-1) + 1 \quad [ \because 1 + \omega + \omega^2 = 0 \)
\( \implies \omega^2 + \omega = -1 ] \)
\( = 4 + 2 + 1 \)
\( = 7 \)
\( = \text{R.H.S.} \)

(ii)
\( \text{L.H.S.} = (2 + \omega + \omega^2)^3 - (1 - 3\omega + \omega^2)^3 \)
\( = [2 + (\omega + \omega^2)]^3 - [-3\omega + (1 + \omega^2)]^3 \)
\( = (2 - 1)^3 - (-3\omega - \omega)^3 \quad [ \because \omega + \omega^2 = -1 \text{ and } 1 + \omega^2 = -\omega ] \)
\( = 1^3 - (-4\omega)^3 \)
\( = 1 - (-64\omega^3) \)
\( = 1 + 64(1) \quad [ \times \omega^3 = 1 ] \)
\( = 65 \)
\( = \text{R.H.S.} \)

(iii)
\( \text{L.H.S.} = \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} \)
\( = \frac{a\omega^3 + b\omega^4 + c\omega^2}{c + a\omega + b\omega^2} \quad [ \because \omega^3 = 1 \text{ and } \omega^4 = \omega ] \)
\( = \frac{\omega^2(a\omega + b\omega^2 + c)}{c + a\omega + b\omega^2} \)
\( = \omega^2 \)
\( = \text{R.H.S.} \)
In simple words: We use the standard properties of the complex cube root of unity, namely \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \), to substitute and simplify the algebraic expressions step-by-step until the left-hand side matches the right-hand side.

🎯 Exam Tip: Clearly state the properties \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \) at the beginning of your proof to show the examiner which substitutions you are using.

Question 2. If \( \omega \) is a complex cube root of unity, find the value of:
(i) \( \omega + \frac{1}{\omega} \)
(ii) \( \omega^2 + \omega^3 + \omega^4 \)
(iii) \( (1 + \omega^2)^3 \)
(iv) \( (1 - \omega - \omega^2)^3 + (1 - \omega + \omega^2)^3 \)
(v) \( (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \)
Answer:
Since \( \omega \) is a complex cube root of unity, we have:
\( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \)

This also gives:
\( 1 + \omega^2 = -\omega \), \( 1 + \omega = -\omega^2 \), and \( \omega + \omega^2 = -1 \)

(i) \( \omega + \frac{1}{\omega} \)
\( = \frac{\omega^2 + 1}{\omega} \)
\( = \frac{-\omega}{\omega} \quad [ \because 1 + \omega^2 = -\omega ] \)
\( = -1 \)

(ii) \( \omega^2 + \omega^3 + \omega^4 \)
\( = \omega^2 + 1 + \omega \quad [ \because \omega^3 = 1 \text{ and } \omega^4 = \omega ] \)
\( = 1 + \omega + \omega^2 \)
\( = 0 \)

(iii) \( (1 + \omega^2)^3 \)
\( = (-\omega)^3 \quad [ \because 1 + \omega^2 = -\omega ] \)
\( = -\omega^3 \)
\( = -1 \quad [ \because \omega^3 = 1 ] \)

(iv) \( (1 - \omega - \omega^2)^3 + (1 - \omega + \omega^2)^3 \)
\( = [1 - (\omega + \omega^2)]^3 + [(1 + \omega^2) - \omega]^3 \)
\( = [1 - (-1)]^3 + [-\omega - \omega]^3 \quad [ \because \omega + \omega^2 = -1 \text{ and } 1 + \omega^2 = -\omega ] \)
\( = (1 + 1)^3 + (-2\omega)^3 \)
\( = 2^3 + (-8\omega^3) \)
\( = 8 - 8(1) \quad [ \times \omega^3 = 1 ] \)
\( = 8 - 8 \)
\( = 0 \)

(v) \( (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \)
Since \( \omega^4 = \omega^3 \cdot \omega = \omega \) and \( \omega^8 = (\omega^3)^2 \cdot \omega^2 = \omega^2 \), we get:
\( = (1 + \omega)(1 + \omega^2)(1 + \omega)(1 + \omega^2) \)
\( = [(1 + \omega)(1 + \omega^2)]^2 \)
\( = [ (-\omega^2)(-\omega) ]^2 \quad [ \because 1 + \omega = -\omega^2 \text{ and } 1 + \omega^2 = -\omega ] \)
\( = [ \omega^3 ]^2 \)
\( = [ 1 ]^2 \)
\( = 1 \)
In simple words: By substituting \( \omega^3 = 1 \) and using the relation \( 1 + \omega + \omega^2 = 0 \), we can reduce higher powers of \( \omega \) and simplify each expression to find its numerical value.

🎯 Exam Tip: For higher powers of \( \omega \), always divide the exponent by 3 and write it in terms of \( \omega^3 \) to simplify the expression quickly.

Question 2. Simplify the following expressions involving complex cube roots of unity:
(i) ...
(ii) \( \omega^2 + \omega^3 + \omega^4 \)
(iii) \( (1 + \omega^2)^3 \)
(iv) \( (1 - \omega - \omega^2)^3 + (1 - \omega + \omega^2)^3 \)
(v) \( (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \)
Answer:
(i) \( = \frac{-\omega}{\omega} = -1 \)

(ii) \( \omega^2 + \omega^3 + \omega^4 \)
\( = \omega^2 (1 + \omega + \omega^2) \)
\( = \omega^2 (0) \)
\( = 0 \)

(iii) \( (1 + \omega^2)^3 \)
\( = (-\omega)^3 \)
\( = -\omega^3 \)
\( = -1 \)

(iv) \( (1 - \omega - \omega^2)^3 + (1 - \omega + \omega^2)^3 \)
\( = [1 - (\omega + \omega^2)]^3 + [(1 + \omega^2) - \omega]^3 \)
\( = [1 - (-1)]^3 + (-\omega - \omega)^3 \)
\( = 2^3 + (-2\omega)^3 \)
\( = 8 - 8\omega^3 \)
\( = 8 - 8(1) \)
\( = 0 \)

(v) \( (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \)
\( = (1 + \omega)(1 + \omega^2)(1 + \omega)(1 + \omega^2) \) .....[\( \because \omega^3 = 1, \omega^4 = \omega, \omega^8 = \omega^2 \)]
\( = (-\omega^2)(-\omega)(-\omega^2)(-\omega) \)
\( = \omega^6 \)
\( = (\omega^3)^2 \)
\( = (1)^2 \)
\( = 1 \)
In simple words: We use the basic properties of the complex cube roots of unity, where the sum of the roots \( 1 + \omega + \omega^2 = 0 \) and the product \( \omega^3 = 1 \), to substitute and simplify each expression step-by-step.

🎯 Exam Tip: Always write down the standard identities \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \) in the margin of your answer sheet to avoid sign errors during substitution.

Question 3. If \( \alpha \) and \( \beta \) are the complex cube roots of unity, show that \( \alpha^2 + \beta^2 + \alpha\beta = 0 \).
Answer: Let the complex cube roots of unity be \( \alpha = \omega \) and \( \beta = \omega^2 \).
Substituting these values in the given expression:
L.H.S. \( = \alpha^2 + \beta^2 + \alpha\beta \)
\( = (\omega)^2 + (\omega^2)^2 + (\omega)(\omega^2) \)
\( = \omega^2 + \omega^4 + \omega^3 \)
Since \( \omega^3 = 1 \) and \( \omega^4 = \omega^3 \cdot \omega = \omega \), we can rewrite the expression as:
\( = \omega^2 + \omega + 1 \)
We know that the sum of the three cube roots of unity is zero, which gives \( 1 + \omega + \omega^2 = 0 \).
\( = 0 \)
\( = \) R.H.S.
Hence, \( \alpha^2 + \beta^2 + \alpha\beta = 0 \) is proved.
In simple words: Since the two complex cube roots of unity are \( \omega \) and \( \omega^2 \), we substitute them into the equation. Using the rule \( \omega^3 = 1 \), the terms simplify directly to \( 1 + \omega + \omega^2 \), which is always equal to zero.

🎯 Exam Tip: Clearly state your assumption \( \alpha = \omega \) and \( \beta = \omega^2 \) at the very beginning of your proof to make your steps logical and easy for the examiner to follow.

p>Question 3. Show that \( \alpha^2 + \beta^2 + \alpha\beta = 0 \), where \( \alpha \) and \( \beta \) are the complex cube roots of unity.
Answer: Let the complex cube roots of unity be:
\( \alpha = \frac{-1 + i\sqrt{3}}{2} \) and \( \beta = \frac{-1 - i\sqrt{3}}{2} \)
Multiplying the two roots:
\( \alpha\beta = \left(\frac{-1 + i\sqrt{3}}{2}\right)\left(\frac{-1 - i\sqrt{3}}{2}\right) \)
\( = \frac{(-1)^2 - (i\sqrt{3})^2}{4} \)
\( = \frac{1 - (-1)(3)}{4} \) ...[\( \because i^2 = -1 \)]
\( = \frac{1 + 3}{4} \)
\( \therefore \alpha\beta = 1 \)
Adding the two roots:
\( \alpha + \beta = \frac{-1 + i\sqrt{3}}{2} + \frac{-1 - i\sqrt{3}}{2} \)
\( = \frac{-1 + i\sqrt{3} - 1 - i\sqrt{3}}{2} \)
\( = \frac{-2}{2} \)
\( \dots \alpha + \beta = -1 \)
Now, substituting these values into the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = \alpha^2 + \beta^2 + \alpha\beta \)
\( = \alpha^2 + 2\alpha\beta + \beta^2 + \alpha\beta - 2\alpha\beta \) ...[Adding and subtracting \( 2\alpha\beta \)]
\( = (\alpha^2 + 2\alpha\beta + \beta^2) - \alpha\beta \)
\( = (\alpha + \beta)^2 - \alpha\beta \)
\( = (-1)^2 - 1 \)
\( = 1 - 1 \)
\( = 0 \)
\( = \text{R.H.S.} \)
Hence proved.
In simple words: We calculate the sum and product of the two complex cube roots of unity, which are -1 and 1. Substituting these into our simplified algebraic expression easily shows that the total value is 0.

🎯 Exam Tip: Always show the step-by-step calculation for the sum and product of roots to secure full marks before substituting them into the main equation.

Question 4. If \( x = a + b \), \( y = \alpha a + \beta b \) and \( z = a\beta + b\alpha \), where \( \alpha \) and \( \beta \) are the complex cube roots of unity, show that \( xyz = a^3 + b^3 \).
Answer: Given:
\( x = a + b \)
\( y = \alpha a + \beta b \)
\( z = a\beta + b\alpha \)
Since \( \alpha \) and \( \beta \) are the complex cube roots of unity, we have:
\( \alpha\beta = 1 \)
\( \alpha + \beta = -1 \)
\( \alpha^2 = \beta \) and \( \beta^2 = \alpha \)
First, let us find the product of \( y \) and \( z \):
\( yz = (\alpha a + \beta b)(a\beta + b\alpha) \)
\( = \alpha\beta a^2 + \alpha^2 ab + \beta^2 ab + \alpha\beta b^2 \)
Substituting \( \alpha\beta = 1 \), \( \alpha^2 = \beta \), and \( \beta^2 = \alpha \):
\( yz = (1)a^2 + \beta ab + \alpha ab + (1)b^2 \)
\( = a^2 + (\alpha + \beta)ab + b^2 \)
Substituting \( \alpha + \beta = -1 \):
\( yz = a^2 + (-1)ab + b^2 \)
\( = a^2 - ab + b^2 \)
Now, multiplying by \( x \):
\( xyz = x(yz) \)
\( = (a + b)(a^2 - ab + b^2) \)
Using the algebraic identity \( (a + b)(a^2 - ab + b^2) = a^3 + b^3 \):
\( xyz = a^3 + b^3 \)
\( \therefore \text{L.H.S.} = \text{R.H.S.} \)
Hence proved.
In simple words: We multiply the expressions for y and z first, using the special properties of complex cube roots to simplify it to \( a^2 - ab + b^2 \). Multiplying this result by x gives us the classic formula for the sum of two cubes.

🎯 Exam Tip: Clearly state the properties of the complex cube roots of unity (\( \alpha\beta = 1 \) and \( \alpha + \beta = -1 \)) at the beginning of your solution to make your proof easy to follow.

Question 4. If \(\alpha\) and \(\beta\) are the complex cube roots of unity, prove that \( (a + b)(\alpha a + \beta b)(\alpha b + \beta a) = a^3 + b^3 \).
Answer:
We know that the complex cube roots of unity are:
\(\alpha = \frac{-1 + i\sqrt{3}}{2}\) and \(\beta = \frac{-1 - i\sqrt{3}}{2}\)

First, let us find \(\alpha\beta\):
\(\alpha\beta = \left(\frac{-1 + i\sqrt{3}}{2}\right)\left(\frac{-1 - i\sqrt{3}}{2}\right)\)
\(\implies \alpha\beta = \frac{(-1)^2 - (i\sqrt{3})^2}{4}\)
\(\implies \alpha\beta = \frac{1 - (-1)(3)}{4} \quad \dots [\because i^2 = -1]\)
\(\implies \alpha\beta = \frac{1 + 3}{4}\)
\(\implies \alpha\beta = 1\)

Next, let us find \(\alpha + \beta\):
\(\alpha + \beta = \frac{-1 + i\sqrt{3}}{2} + \frac{-1 - i\sqrt{3}}{2}\)
\(\implies \alpha + \beta = \frac{-1 + i\sqrt{3} - 1 - i\sqrt{3}}{2}\)
\(\implies \alpha + \beta = \frac{-2}{2}\)
\(\implies \alpha + \beta = -1\)

Now, let us simplify the Left Hand Side (L.H.S.):
\(\text{L.H.S.} = (a + b)(\alpha a + \beta b)(\alpha b + \beta a)\)
\(\implies \text{L.H.S.} = (a + b)(\alpha\beta a^2 + \alpha^2 ab + \beta^2 ab + \alpha\beta b^2)\)
\(\implies \text{L.H.S.} = (a + b)[\alpha\beta(a^2) + (\alpha^2 + \beta^2)ab + \alpha\beta(b^2)]\)
\(\implies \text{L.H.S.} = (a + b)[1 \cdot (a^2) + (\alpha^2 + \beta^2)ab + 1 \cdot (b^2)] \quad \dots [\because \alpha\beta = 1]\)
\(\implies \text{L.H.S.} = (a + b)\{a^2 + [(\alpha + \beta)^2 - 2\alpha\beta]ab + b^2\}\)
\(\implies \text{L.H.S.} = (a + b)\{a^2 + [(-1)^2 - 2(1)]ab + b^2\} \quad \dots [\because \alpha + \beta = -1, \alpha\beta = 1]\)
\(\implies \text{L.H.S.} = (a + b)[a^2 + (1 - 2)ab + b^2]\)
\(\implies \text{L.H.S.} = (a + b)(a^2 - ab + b^2)\)
\(\implies \text{L.H.S.} = a^3 + b^3 = \text{R.H.S.}\)
Hence proved.
In simple words: By substituting the values of the complex cube roots \(\alpha\) and \(\beta\), we find their sum is \(-1\) and product is \(1\). Plugging these into the algebraic expansion simplifies the expression directly to \(a^3 + b^3\).

🎯 Exam Tip: Remember that the product of complex conjugate roots \(\alpha\beta\) is always \(1\), and their sum \(\alpha + \beta\) is \(-1\). Writing these relations down first secures step-marks.

Question 5. If \(\omega\) is a complex cube root of unity, then prove the following:
(i) \((\omega^2 + \omega - 1)^3 = -8\)
(ii) \((a + b) + (a\omega + b\omega^2) + (a\omega^2 + b\omega) = 0\)
Answer:
Since \(\omega\) is a complex cube root of unity, we have:
\(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\)
This also gives us:
\(1 + \omega = -\omega^2\), \(1 + \omega^2 = -\omega\), and \(\omega + \omega^2 = -1\)

(i) To prove: \((\omega^2 + \omega - 1)^3 = -8\)
\(\text{L.H.S.} = (\omega^2 + \omega - 1)^3\)
\(\implies \text{L.H.S.} = (-1 - 1)^3 \quad \dots [\because \omega^2 + \omega = -1]\)
\(\implies \text{L.H.S.} = (-2)^3\)
\(\implies \text{L.H.S.} = -8 = \text{R.H.S.}\)
Hence proved.

(ii) To prove: \((a + b) + (a\omega + b\omega^2) + (a\omega^2 + b\omega) = 0\)
\(\text{L.H.S.} = (a + b) + (a\omega + b\omega^2) + (a\omega^2 + b\omega)\)
\(\implies \text{L.H.S.} = a + b + a\omega + b\omega^2 + a\omega^2 + b\omega\)
\(\implies \text{L.H.S.} = a(1 + \omega + \omega^2) + b(1 + \omega + \omega^2)\)
\(\implies \text{L.H.S.} = a(0) + b(0) \quad \dots [\because 1 + \omega + \omega^2 = 0]\)
\(\implies \text{L.H.S.} = 0 = \text{R.H.S.}\)
Hence proved.
In simple words: Using the properties of the cube root of unity, we know that the sum of all three roots \(1 + \omega + \omega^2\) is \(0\), and \(\omega^2 + \omega\) equals \(-1\). Substituting these values makes both equations simplify to their right-hand sides very easily.

🎯 Exam Tip: Always look to group terms to form the identity \(1 + \omega + \omega^2 = 0\) or substitute \(\omega^2 + \omega = -1\) to simplify complex expressions quickly.

Question (ii) Prove that \( (a + b) + (a\omega + b\omega^2) + (a\omega^2 + b\omega) = 0 \)
Answer:
L.H.S. = \( (a + b) + (a\omega + b\omega^2) + (a\omega^2 + b\omega) \)
= \( (a + a\omega + a\omega^2) + (b + b\omega + b\omega^2) \)
= \( a(1 + \omega + \omega^2) + b(1 + \omega + \omega^2) \)
Since \( \omega \) is a complex cube root of unity, we use the standard identity \( 1 + \omega + \omega^2 = 0 \) to simplify the terms.
= \( a(0) + b(0) \)
= 0
= R.H.S.
In simple words: We group the 'a' terms and 'b' terms together. Since the sum of the cube roots of unity \( 1 + \omega + \omega^2 \) is equal to 0, both groups become 0, making the final sum 0.

🎯 Exam Tip: Always state the identity \( 1 + \omega + \omega^2 = 0 \) explicitly in your steps so the examiner knows exactly how you substituted zero.