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 – ω^{2}) = 7

(ii) (2 + ω + ω^{2})^{3} – (1 – 3ω + ω^{2})^{3} = 65

(iii) \(\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^{2}\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^{2}}\) = ω^{2}

Solution:

ω is the complex cube root of unity.

∴ ω^{3} = 1 and 1 + ω + ω^{2} = 0

Also, 1 + ω^{2} = -ω, 1 + ω = -ω^{2} and ω + ω^{2} = -1

(i) L.H.S. = (2 – ω)(2 – ω^{2})

= 4 – 2ω^{2} – 2ω + ω^{3}

= 4 – 2(ω^{2} + ω) + 1

= 4 – 2(-1) + 1

= 4 + 2 + 1

= 7

= R.H.S.

(ii) L.H.S. = (2 + ω + ω^{2})^{3} – (1 – 3ω + ω^{2})^{3}

= [2 + (ω + ω^{2})]^{3} – [-3ω + (1 + ω^{2})]^{3}

= (2 – 1)^{3} – (-3ω – ω)^{3}

= 13 – (-4ω)^{3}

= 1 + 64ω^{3}

= 1 + 64(1)

= 65

= R.H.S.

(iii) L.H.S. =\(\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^{2}\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^{2}}\)

= \(\frac{a \omega^{3}+b \omega^{4}+c \omega^{2}}{c+a \omega+b \omega^{2}}\) ……[∵ ω^{3} = 1, ω^{4} = ω]

= \(\frac{\omega^{2}\left(c+a \omega+b \omega^{2}\right)}{c+a \omega+b \omega^{2}}\)

= ω^{2}

= R.H.S.

Question 2.

If ω is a complex cube root of unity, find the value of

(i) ω + \(\frac{1}{\omega}\)

(ii) ω^{2} + ω^{3} + ω^{4}

(iii) (1 + ω^{2})^{3}

(iv) (1 – ω – ω^{2})^{3} + (1 – ω + ω^{2})^{3}

(v) (1 + ω)(1 + ω^{2})(1 + ω^{4})(1 + ω^{8})

Solution:

ω is the complex cube root of unity.

∴ ω^{3} = 1 and 1 + ω + ω^{2} = 0

Also, 1 + ω^{2} = -ω, 1 + ω = -ω^{2} and ω + ω^{2} = -1

(i) ω + \(\frac{1}{\omega}\)

= \(\frac{\omega^{2}+1}{\omega}\)

= \(\frac{-\omega}{\omega}\)

= -1

(ii) ω^{2} + ω^{3} + ω^{4}

= ω^{2} (1 + ω + ω^{2})

= ω^{2} (0)

= 0

(iii) (1 + ω^{2})^{3}

= (-ω)^{3}

= -ω^{3}

= -1

(iv) (1 – ω – ω^{2})^{3} + (1 – ω + ω^{2})^{3}

= [1 – (ω + ω^{2})]^{3} + [(1 + ω^{2}) – ω]^{3}

= [1 – (-1)]^{3} + (-ω – ω)^{3}

= 2^{3} + (-2ω)^{3}

= 8 – 8ω^{3}

= 8 – 8(1)

= 0

(v) (1 + ω)(1 + ω^{2})(1 + ω^{4})(1 + ω^{8})

= (1 + ω)(1 + ω^{2})(1 + ω)(1 + ω^{2}) …..[∵ ω^{3} = 1, ω^{4} = ω]

= (-ω^{2})(-ω)(-ω^{2})(-ω)

= ω^{6}

= (ω^{3})^{2}

= (1)^{2}

= 1

Question 3.

If α and β are the complex cube roots of unity, show that α^{2} + β^{2} + αβ = 0.

Solution:

α and β are the complex cube roots of unity.

∴ α – β = -1

L.H.S. = α^{2} + β^{2} + αβ

= α^{2} + 2αβ + β^{2} + αβ – 2αβ ……[Adding and subtracting 2αβ]

= (α^{2} + 2αβ + β^{2}) – 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 = a^{3} + b^{3}.

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) (ω^{2} + ω – 1)^{3} = -8

(ii) (a + b) + (aω + bω^{2}) + (aω^{2} + bω) = 0

Solution:

ω is the complex cube root of unity.

∴ ω^{3} = 1 and 1 + ω + ω^{2} = 0

Also, 1 + ω^{2} = -ω, 1 + ω = -ω^{2} and ω + ω^{2} = -1

(i) L.H.S. = (ω^{2} + ω – 1)^{3}

= (-1 – 1)^{3}

= (-2)^{3}

= -8

= R.H.S.

(ii) L.H.S. = (a + b) + (aω + bω^{2}) + (aω^{2} + bω)

= (a + aω + aω^{2}) + (b + bω + bω^{2})

= a(1 + ω + ω^{2}) + b(1 + ω + ω^{2})

= a(0) + b(0)

= 0

= R.H.S.