CBSE Class 11 Maths – Chapter 5 Complex Numbers and Quadratic Equations- Study Materials

Answer: (b) 4i
Hint:
Given, √(-16) = √(16) × √(-1)
= 4i {since i = √(-1) }

Answer: (a) 12i
Hint:
Given, √(-144) = √{(-1) × 144}
= √(-1) × √(144)
= i × 12 {Since √(-1) = i}
= 12i
So, √(-144) = 12i

Answer: (c) 17i
Hint:
Given, √(-25) + 3√(-4) + 2√(-9)
= √{(-1) × (25)} + 3√{(-1) × 4} + 2√{(-1) × 9}
= √(-1) × √(25) + 3{√(-1) × √4} + 2{√(-1) × √9}
= 5i + 3×2i + 2×3i {since √(-1) = i}
= 5i + 6i + 6i
= 17i
So, √(-25) + 3√(-4) + 2√(-9) = 17i

Answer: (a) a circle
Hint:
Let w = 2/z
Now, |w| = |2/z|
=> |w| = 2/|z|
=> |w| = 2
This shows that w lies on a circle with center at the origin and radius 2 units.

Answer: (d) -128 ω²
Hint:
Given ω is an imaginary cube root of unity.
So 1 + ω + ω² = 0 and ω³ = 1
Now, (1 + ω – ω²)⁷ = (-ω² – ω²)⁷
⇒ (1 + ω – ω²)⁷ = (-2ω²)⁷
⇒ (1 + ω – ω²)⁷ = -128 ω¹⁴
⇒ (1 + ω – ω²)⁷ = -128 ω¹² × ω²
⇒ (1 + ω – ω²)⁷ = -128 (ω³)⁴ ω²
⇒ (1 + ω – ω²)⁷ = -128 ω²

Answer: (b) 2
Hint:
Given, {(1 + i)/(1 – i)}ⁿ
= [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]ⁿ
= [{(1 + i)²}/{(1 – i²)}]ⁿ
= [(1 + i² + 2i)/{1 – (-1)}]ⁿ
= [(1 – 1 + 2i)/{1 + 1}]ⁿ
= [2i/2]ⁿ
= iⁿ
Now, in is real when n = 2 {since i2 = -1 }
So, the least value of n is 2

Answer: (a) -2√3 + 2i
Hint:
Let z = r(cos θ + i × sin θ)
Then r = 4 and θ = 5π/6
So, z = 4(cos 5π/6 + i × sin 5π/6)
⇒ z = 4(-√3/2 + i/2)
⇒ z = -2√3 + 2i

Answer: (c) i
Hint:
Given, i^-999
= 1/i⁹⁹⁹
= 1/(i⁹⁹⁶ × i³)
= 1/{(i⁴)²⁴⁹ × i³}
= 1/{1²⁴⁹ × i³} {since i⁴ = 1}
= 1/i³
= i⁴/i³ {since i⁴ = 1}
= i
So, i^-999 = i

Answer: (c) a² = 3b
Hint:
Given, z₁ and z₂ be two roots of the equation z² + az + b = 0
Now, z₁ + z₂ = -a and z₁ × z₂ = b
Since z₁ and z₂ and z₃ from an equilateral triangle.
⇒ z₁² + z₂² + z₃² = z₁ × z₂ + z₂ × z₃ + z₁ × z₃
⇒ z₁²+ z₂² = z₁ × z₂ {since z₃ = 0}
⇒ (z₁ + z₂)² – 2z₁ × z₂ = z₁ × z₂
⇒ (z₁ + z₂)² = 2z₁ × z₂ + z₁ × z₂
⇒ (z₁ + z₂)² = 3z₁ × z₂
⇒ (-a)² = 3b
⇒ a² = 3b

Answer: (d) no value of x
Hint:
Given complex number = sin x + i cos 2x
Conjugate of this number = sin x – i cos 2x
Now, sin x + i cos 2x = sin x – i cos 2x
⇒ sin x = cos x and sin 2x = cos 2x {comparing real and imaginary part}
⇒ tan x = 1 and tan 2x = 1
Now both of them are not possible for the same value of x.
So, there exist no value of x

Answer: (d) a hyperbola
Hint:
Let z = x + iy
Now, z² = (x + iy)²
⇒ z² = x² – y² + 2xy
Given, Im(z²) = k
⇒ 2xy = k
⇒ xy = k/2 which is a hyperbola.

Answer: (a) x = 7/2, y = 2/3
Hint:
Given, (3y – 2) + i(7 – 2x) = 0
Compare real and imaginary part, we get
3y – 2 = 0
⇒ y = 2/3
and 7 – 2x = 0
⇒ x = 7/2
So, the value of x = 7/2 and y = 2/3

Answer: (b) θ = nπ ± π/3 where n is an integer
Hint:
Given,
(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 + 2i × sin θ)×(1 – 2i × sin θ)}/(1 – 4i² × sin² θ)
(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 – 4sin² θ) + 8i × sin θ}/(1 + 4sin² θ) …………. 1
Now, equation 1 is imaginary if
3 – 4sin² θ = 0
⇒ 4sin² θ = 3
⇒ sin² θ = 3/4
⇒ sin θ = ±√3/2
⇒ θ = nπ ± π/3 where n is an integer

Answer: (d) 4
Hint:
Given, {(1 + i)/(1 – i)}ⁿ = 1
⇒ [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]ⁿ = 1
⇒ [{(1 + i)²}/{(1 – i²)}]ⁿ = 1
⇒ [(1 + i² + 2i)/{1 – (-1)}]ⁿ = 1
⇒ [(1 – 1 + 2i)/{1 + 1}]ⁿ = 1
⇒ [2i/2]ⁿ = 1
⇒ iⁿ = 1
Now, iⁿ is 1 when n = 4
So, the least value of n is 4

Answer: (a) π
Hint:
Given, arg (z) < 0
Now, arg (-z) – arg (z) = arg(-z/z)
⇒ arg (-z) – arg (z) = arg(-1)
⇒ arg (-z) – arg (z) = π {since sin π + i cos π = -1, So arg(-1) = π}

Answer: (c) -1
Hint:
Given x + 1/x = 1
⇒ (x² + 1) = x
⇒ x² – x + 1 = 0
⇒ x = {-(-1) ± √(1² – 4 × 1 × 1)}/(2 × 1)
⇒ x = {1 ± √(1 – 4)}/2
⇒ x = {1 ± √(-3)}/2
⇒ x = {1 ± √(-1)×√3}/2
⇒ x = {1 ± i√3}/2 {since i = √(-1)}
⇒ x = -w, -w²
Now, put x = -w, we get
x²⁰⁰⁰ + 1/x²⁰⁰⁰ = (-w)²⁰⁰⁰ + 1/(-w)²⁰⁰⁰
= w²⁰⁰⁰ + 1/w²⁰⁰⁰
= w²⁰⁰⁰ + 1/w²⁰⁰⁰
= {(w³)⁶⁶⁶ × w²} + 1/{(w³)⁶⁶⁶ × w²}
= w² + 1/w² {since w³ = 1}
= w² + w³ /w²
= w² + w
= -1 {since 1 + w + w² = 0}
So, x²⁰⁰⁰ + 1/x²⁰⁰⁰ = -1

Answer: (a) 12i
Hint:
Given, √(-144) = √{(-1)×144}
= √(-1) × √(144)
= i × 12 {Since √(-1) = i}
= 12i
So, √(-144) = 12i

Answer: (c) – 1, 1 – 2ω, 1 – 2ω²
Hint:
Note that since 1, ω, and ω² are the cube roots of unity (the three cube roots of 1), they are the three solutions to x³ = 1 (note: ω and ω² are the two complex solutions to this)
If we let u = x – 1, then the equation becomes
u³ + 8 = (u + 2)(u² – 2u + 4) = 0.
So, the solutions occur when u = -2 (giving -2 = x – 1 ⇒ x = -1), or when:
u² – 2u + 4 = 0,
which has roots, by the Quadratic Formula, to be u = 1 ± i√3
So, x – 1 = 1 ± i√3
⇒ x = 2 ± i√3
Now, x³ = 1 when x³ – 1 = (x – 1)(x² + x + 1) = 0, giving x = 1 and
x² + x + 1 = 0
⇒ x = (-1 ± i√3)/2
If we let ω = (-1 – i√3)/2 and ω₂ = (-1 + i√3)/2
then 1 – 2ω and 1 – 2ω² yield the two complex solutions to (x – 1)³ + 8 = 0
So, the roots of (x – 1)³ + 8 are -1, 1 – 2ω, and 1 – 2ω²

Answer: (b) 2²ⁿ
Hint:
Given, (1 – w + w²)×(1 – w² + w⁴)×(1 – w⁴ + w⁸) × …………… to 2n factors
= (1 – w + w²)×(1 – w² + w )×(1 – w + w²) × …………… to 2n factors
{Since w⁴ = w, w⁸ = w²}
= (-2w) × (-2w²) × (-2w) × (-2w²)× …………… to 2n factors
= (2² w³)×(2² w³)×(2² w³) …………… to 2n factors
= (2²)ⁿ {since w³ = 1}
= 2²ⁿ

Answer: (c) √41
Hint:
Let Z = 5 + 4i
Now modulus of Z is calculated as
|Z| = √(5² + 4²)
⇒ |Z| = √(25 + 16)
⇒ |Z| = √41
So, the modulus of 5 + 4i is √41