Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 3 Complex Numbers Ex 3.3 Questions and Answers.
Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3
Solution & Step-by-Step Answer:
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. =
= ……[∵ ω3= 1, ω4= ω]
=
= ω2
= R.H.S.
Solution & Step-by-Step Answer:
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 + = = = -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
Solution & Step-by-Step Answer:
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.
Solution & Step-by-Step Answer:
x = a + b, y = αa + βb, z = aβ + bα α and β are the complex cube roots of unity.
Solution & Step-by-Step Answer:
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.