Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Ex 1.4 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4

Question 1.

Find the value of

(i) ω^{18}

(ii) ω^{21}

(iii) ω^{-30}

(iv) ω^{-105}

Solution:

Question 2.

If ω is the complex cube root of unity, show that

(i) (2 – ω)(2 – ω^{2}) = 7

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

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) (1 + ω – ω^{2})^{6} = 64

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

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

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

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

(v) (3 + 3ω + 5ω^{2})^{6} – (2 + 6ω + 2ω^{2})^{3} = 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

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

(vii) (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

(viii) (a – b)(a – bω)(a – bω^{2}) = a^{3} – b^{3}

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

(ix) (a + b)^{2} + (aω + bω^{2})^{2} + (aω^{2}+ bω)^{2} = 6ab

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

Question 3.

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

(i) ω + \(\frac{1}{\omega}\)

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

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

(ii) ω^{2} + ω^{3} + ω^{4}

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

ω^{2} + ω^{3} + ω^{4}

= ω^{2}(1 + ω + ω^{2})

= ω^{2}(0)

= 0

(iii) (1 + ω^{2})^{3}

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

(1 + ω^{2})^{3}

= (-ω)^{3}

= -ω^{3}

= -1

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

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

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

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

= [1 – (-1)]^{2} + (-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})

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

(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 4.

If α and β are the complex cube roots of unity, show that

(i) α^{2} + β^{2} + αβ = 0

(ii) α^{4} + β^{4} + α^{-1}β^{-1} = 0

Solution:

α and β are the complex cube roots of unity.

Question 5.

If x = a + b, y = αa + βb and z = aβ + bα, where α and β are 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.

∴ α = \(\frac{-1+i \sqrt{3}}{2}\) and β = \(\frac{-1-i \sqrt{3}}{2}\)

Question 6.

Find the equation in cartesian coordinates of the locus of z if

(i) |z| = 10

Solution:

Let z = x + iy

|z| = 10

|x + iy| = 10

\(\sqrt{x^{2}+y^{2}}\) = 10

∴ x^{2} + y^{2} = 100

(ii) |z – 3| = 2

Solution:

Let z = x + iy

|z – 3| = 2

|x + iy – 3| = 2

|(x – 3) + iy| = 2

\(\sqrt{(x-3)^{2}+y^{2}}\) = 2

∴ (x – 3)^{2} + y^{2} = 4

(iii) |z – 5 + 6i| = 5

Solution:

Let z = x + iy

|z – 5 + 6i| = 5

|x + iy – 5 + 6i| = 5

|(x – 5) + i(y + 6)| = 5

\(\sqrt{(x-5)^{2}+(y+6)^{2}}\) = 5

∴ (x – 5)^{2} + (y + 6)^{2} = 25

(iv) |z + 8| = |z – 4|

Solution:

Let z = x + iy

|z + 8| = |z – 4|

|x + iy + 8| = |x + iy – 4|

|(x + 8) + iy | = |(x – 4) + iy|

\(\sqrt{(x+8)^{2}+y^{2}}=\sqrt{(x-4)^{2}+y^{2}}\)

(x + 8)^{2} + y^{2} = (x – 4)^{2} + y^{2}

x^{2} + 16x + 64 + y^{2} = x^{2} – 8x + 16 + y^{2}

16x + 64 = -8x + 16

24x + 48 = 0

∴ x + 2 = 0

(v) |z – 2 – 2i | = |z + 2 + 2i|

Solution:

Let z = x + iy

|z – 2 – 2i| = |z + 2 + 2i|

|x + iy – 2 – 2i | = |x + iy + 2 + 2i |

|(x – 2) + i(y – 2)| = |(x + 2) + i(y + 2)|

\(\sqrt{(x-2)^{2}+(y-2)^{2}}=\sqrt{(x+2)^{2}+(y+2)^{2}}\)

(x – 2)^{2} + (y – 2)^{2} = (x + 2)^{2} + (y + 2)^{2}

x^{2}– 4x + 4 + y^{2} – 4y + 4 = x^{2} + 4x + 4 + y^{2} + 4y + 4

-4x – 4y = 4x + 4y

8x + 8y = 0

x + y = 0

y = -x

(vi) \(\frac{|z+3 i|}{|z-6 i|}=1\)

Solution:

Let z = x + iy

x^{2} + (y + 3)^{2} = x^{2} + (y – 6)^{2}

y^{2} + 6y + 9 = y^{2} – 12y + 36

18y – 27 = 0

2y – 3 = 0

Question 7.

Use De Moivre’s theorem and simplify the following:

(i) \(\frac{(\cos 2 \theta+i \sin 2 \theta)^{7}}{(\cos 4 \theta+i \sin 4 \theta)^{3}}\)

Solution:

(ii) \(\frac{\cos 5 \theta+i \sin 5 \theta}{(\cos 3 \theta-i \sin 3 \theta)^{2}}\)

Solution:

(iii) \(\frac{\left(\cos \frac{7 \pi}{13}+i \sin \frac{7 \pi}{13}\right)^{4}}{\left(\cos \frac{4 \pi}{13}-i \sin \frac{4 \pi}{13}\right)^{6}}\)

Solution:

Question 8.

Express the following in the form a + ib, a, b ∈ R, using De Moivre’s theorem.

(i) (1 – i)^{5}

Solution:

(ii) (1 + i)^{6}

Solution:

(iii) (1 – √3 i)^{4}

Solution:

(iv) (-2√3 – 2i)^{5}

Solution: