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

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) ω¹⁸
(ii) ω²¹
(iii) ω^-30
(iv) ω^-105
Solution:

Question 2.
If ω is the complex cube root of unity, show that
(i) (2 – ω)(2 – ω²) = 7
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
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) (1 + ω – ω²)⁶ = 64
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

(iii) (1 + ω)³ – (1 + ω²)³ = 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

(iv) (2 + ω + ω²)³ – (1 – 3ω + ω²)³ = 65
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

(v) (3 + 3ω + 5ω²)⁶ – (2 + 6ω + 2ω²)³ = 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

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

(vii) (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

(viii) (a – b)(a – bω)(a – bω²) = a³ – b³
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

(ix) (a + b)² + (aω + bω²)² + (aω²+ bω)² = 6ab
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

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.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
\(\omega+\frac{1}{\omega}=\frac{\omega^{2}+1}{\omega}=\frac{-\omega}{\omega}=-1\)

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

(iii) (1 + ω²)³
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
(1 + ω²)³
= (-ω)³
= -ω³
= -1

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

(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
(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 4.
If α and β are the complex cube roots of unity, show that
(i) α² + β² + αβ = 0
(ii) α⁴ + β⁴ + α^-1β^-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 = a³ + b³.
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
∴ x² + y² = 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 + y² = 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 + y² = (x – 4)² + y²
x² + 16x + 64 + y² = x² – 8x + 16 + y²
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
x²– 4x + 4 + y² – 4y + 4 = x² + 4x + 4 + y² + 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

x² + (y + 3)² = x² + (y – 6)²
y² + 6y + 9 = y² – 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: