Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3

Question 1.
If ω is a complex cube root of unity, show that
(i) (2 – ω)(2 – ω²) = 7
(ii) (2 + ω + ω²)³ – (1 – 3ω + ω²)³ = 65
(iii) \(\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^{2}\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^{2}}\) = ω²
Solution:
ω is the complex cube root of unity.
∴ ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
(i) L.H.S. = (2 – ω)(2 – ω²)
= 4 – 2ω² – 2ω + ω³
= 4 – 2(ω² + ω) + 1
= 4 – 2(-1) + 1
= 4 + 2 + 1
= 7
= R.H.S.

(ii) L.H.S. = (2 + ω + ω²)³ – (1 – 3ω + ω²)³
= [2 + (ω + ω²)]³ – [-3ω + (1 + ω²)]³
= (2 – 1)³ – (-3ω – ω)³
= 13 – (-4ω)³
= 1 + 64ω³
= 1 + 64(1)
= 65
= R.H.S.

(iii) L.H.S. =\(\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^{2}\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^{2}}\)
= \(\frac{a \omega^{3}+b \omega^{4}+c \omega^{2}}{c+a \omega+b \omega^{2}}\) ……[∵ ω³ = 1, ω⁴ = ω]
= \(\frac{\omega^{2}\left(c+a \omega+b \omega^{2}\right)}{c+a \omega+b \omega^{2}}\)
= ω²
= R.H.S.

Question 2.
If ω is a complex cube root of unity, find the value of
(i) ω + \(\frac{1}{\omega}\)
(ii) ω² + ω³ + ω⁴
(iii) (1 + ω²)³
(iv) (1 – ω – ω²)³ + (1 – ω + ω²)³
(v) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)
Solution:
ω is the complex cube root of unity.
∴ ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
(i) ω + \(\frac{1}{\omega}\)
= \(\frac{\omega^{2}+1}{\omega}\)
= \(\frac{-\omega}{\omega}\)
= -1

(ii) ω² + ω³ + ω⁴
= ω² (1 + ω + ω²)
= ω² (0)
= 0

(iii) (1 + ω²)³
= (-ω)³
= -ω³
= -1

(iv) (1 – ω – ω²)³ + (1 – ω + ω²)³
= [1 – (ω + ω²)]³ + [(1 + ω²) – ω]³
= [1 – (-1)]³ + (-ω – ω)³
= 2³ + (-2ω)³
= 8 – 8ω³
= 8 – 8(1)
= 0

(v) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)
= (1 + ω)(1 + ω²)(1 + ω)(1 + ω²) …..[∵ ω³ = 1, ω⁴ = ω]
= (-ω²)(-ω)(-ω²)(-ω)
= ω⁶
= (ω³)²
= (1)²
= 1

Question 3.
If α and β are the complex cube roots of unity, show that α² + β² + αβ = 0.
Solution:
α and β are the complex cube roots of unity.

∴ α – β = -1
L.H.S. = α² + β² + αβ
= α² + 2αβ + β² + αβ – 2αβ ……[Adding and subtracting 2αβ]
= (α² + 2αβ + β²) – αβ
= (α + β)² – αβ
= (-1)² – 1
= 1 – 1
= 0
= R.H.S.

Question 4.
If x = a + b, y = αa + βb and z = aβ + bα, where α and β are the complex cube roots of unity, show that xyz = a³ + b³.
Solution:
x = a + b, y = αa + βb, z = aβ + bα
α and β are the complex cube roots of unity.

Question 5.
If ω is a complex cube root of unity, then prove the following:
(i) (ω² + ω – 1)³ = -8
(ii) (a + b) + (aω + bω²) + (aω² + bω) = 0
Solution:
ω is the complex cube root of unity.
∴ ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
(i) L.H.S. = (ω² + ω – 1)³
= (-1 – 1)³
= (-2)³
= -8
= R.H.S.

(ii) L.H.S. = (a + b) + (aω + bω²) + (aω² + bω)
= (a + aω + aω²) + (b + bω + bω²)
= a(1 + ω + ω²) + b(1 + ω + ω²)
= a(0) + b(0)
= 0
= R.H.S.