Check the below NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Complex Numbers and Quadratic Equations Class 11 Maths MCQs Questions with Answers to help students understand the concept very well. https://mcqquestions.guru/mcq-questions-for-class-11-maths-chapter-5/

## Complex Numbers and Quadratic Equations Class 11 MCQs Questions with Answers

**Complex Numbers Class 11 MCQ Question 1.**

The value of √(-16) is

(a) -4i

(b) 4i

(c) -2i

(d) 2i

## Answer

Answer: (b) 4i

Hint:

Given, √(-16) = √(16) × √(-1)

= 4i {since i = √(-1) }

**Complex Numbers Class 11 MCQ Questions And Answers Question 2.**

The value of √(-144) is

(a) 12i

(b) -12i

(c) ±12i

(d) None of these

## Answer

Answer: (a) 12i

Hint:

Given, √(-144) = √{(-1) × 144}

= √(-1) × √(144)

= i × 12 {Since √(-1) = i}

= 12i

So, √(-144) = 12i

**MCQ On Complex Numbers Class 11 Question 3.**

The value of √(-25) + 3√(-4) + 2√(-9) is

(a) 13i

(b) -13i

(c) 17i

(d) -17i

## Answer

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

**Class 11 Maths Chapter 5 MCQ With Answers Question 4.**

if z lies on |z| = 1, then 2/z lies on

(a) a circle

(b) an ellipse

(c) a straight line

(d) a parabola

## Answer

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.

**Class 11 Maths Chapter 5 MCQ Question 5.**

If ω is an imaginary cube root of unity, then (1 + ω – ω²)^{7} equals

(a) 128 ω

(b) -128 ω

(c) 128 ω²

(d) -128 ω²

## Answer

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 – ω^{2})^{7} = (-2ω^{2})^{7}

⇒ (1 + ω – ω^{2})^{7} = -128 ω^{14}

⇒ (1 + ω – ω^{2})^{7} = -128 ω^{12} × ω^{2}

⇒ (1 + ω – ω^{2})^{7} = -128 (ω^{3})^{4} ω^{2}

⇒ (1 + ω – ω^{2})^{7} = -128 ω^{2}

**MCQ Of Complex Numbers Class 11 Question 6.**

The least value of n for which {(1 + i)/(1 – i)}^{n} is real, is

(a) 1

(b) 2

(c) 3

(d) 4

## Answer

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}

= i^{n}

Now, in is real when n = 2 {since i2 = -1 }

So, the least value of n is 2

**MCQ Of Chapter 5 Maths Class 11 Question 7.**

Let z be a complex number such that |z| = 4 and arg(z) = 5π/6, then z =

(a) -2√3 + 2i

(b) 2√3 + 2i

(c) 2√3 – 2i

(d) -√3 + i

## Answer

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

**MCQ On Complex Numbers Class 11 Pdf Question 8.**

The value of i^{-999} is

(a) 1

(b) -1

(c) i

(d) -i

## Answer

Answer: (c) i

Hint:

Given, i^{-999}

= 1/i^{999}

= 1/(i^{996} × i³)

= 1/{(i^{4})^{249} × i^{3}}

= 1/{1^{249} × i^{3}} {since i^{4} = 1}

= 1/i^{3}

= i^{4}/i^{3} {since i^{4} = 1}

= i

So, i^{-999} = i

**MCQ Questions On Complex Numbers Class 11 Question 9.**

Let z_{1} and z_{2} be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z_{1} and z_{1} form an equilateral triangle. Then

(a) a² = b

(b) a² = 2b

(c) a² = 3b

(d) a² = 4b

## Answer

Answer: (c) a² = 3b

Hint:

Given, z_{1} and z_{2} be two roots of the equation z² + az + b = 0

Now, z_{1} + z_{2} = -a and z_{1} × z_{2} = b

Since z_{1} and z_{2} and z_{3} from an equilateral triangle.

⇒ z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} × z_{2} + z_{2} × z_{3} + z_{1} × z_{3}

⇒ z_{1}^{2}+ z_{2}^{2} = z_{1} × z_{2} {since z_{3} = 0}

⇒ (z_{1} + z_{2})² – 2z_{1} × z_{2} = z_{1} × z_{2}

⇒ (z_{1} + z_{2})² = 2z_{1} × z_{2} + z_{1} × z_{2}

⇒ (z_{1} + z_{2})² = 3z_{1} × z_{2}

⇒ (-a)² = 3b

⇒ a² = 3b

**Complex Numbers MCQs With Solution Question 10.**

The complex numbers sin x + i cos 2x are conjugate to each other for

(a) x = nπ

(b) x = 0

(c) x = (n + 1/2) π

(d) no value of x

## Answer

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

**Class 11 Complex Numbers MCQ Question 11.**

The curve represented by Im(z²) = k, where k is a non-zero real number, is

(a) a pair of striaght line

(b) an ellipse

(c) a parabola

(d) a hyperbola

## Answer

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.

**Class 11 Maths Complex Numbers MCQ Question 12.**

The value of x and y if (3y – 2) + i(7 – 2x) = 0

(a) x = 7/2, y = 2/3

(b) x = 2/7, y = 2/3

(c) x = 7/2, y = 3/2

(d) x = 2/7, y = 3/2

## Answer

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

**Complex Numbers MCQs With Solution Pdf Question 13.**

Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary

(a) θ = nπ ± π/2 where n is an integer

(b) θ = nπ ± π/3 where n is an integer

(c) θ = nπ ± π/4 where n is an integer

(d) None of these

## Answer

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

**Class 11 Maths Chapter 5 MCQ Questions Question 14.**

If {(1 + i)/(1 – i)}^{n} = 1 then the least value of n is

(a) 1

(b) 2

(c) 3

(d) 4

## Answer

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

⇒ i^{n} = 1

Now, i^{n} is 1 when n = 4

So, the least value of n is 4

**Class 11 Maths MCQ Chapter 5 Question 15.**

If arg (z) < 0, then arg (-z) – arg (z) =

(a) π

(b) -π

(c) -π/2

(d) π/2

## Answer

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}

Question 16.

if x + 1/x = 1 find the value of x^{2000} + 1/x^{2000} is

(a) 0

(b) 1

(c) -1

(d) None of these

## Answer

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

x^{2000} + 1/x^{2000} = (-w)^{2000} + 1/(-w)^{2000}

= w^{2000} + 1/w^{2000}

= w^{2000} + 1/w^{2000}

= {(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, x^{2000} + 1/x^{2000} = -1

Question 17.

The value of √(-144) is

(a) 12i

(b) -12i

(c) ±12i

(d) None of these

## Answer

Answer: (a) 12i

Hint:

Given, √(-144) = √{(-1)×144}

= √(-1) × √(144)

= i × 12 {Since √(-1) = i}

= 12i

So, √(-144) = 12i

Question 18.

If the cube roots of unity are 1, ω, ω², then the roots of the equation (x – 1)³ + 8 = 0 are

(a) -1, -1 + 2ω, – 1 – 2ω²

(b) – 1, -1, – 1

(c) – 1, 1 – 2ω, 1 – 2ω²

(d) – 1, 1 + 2ω, 1 + 2ω²

## Answer

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

Question 19.

(1 – w + w²)×(1 – w² + w^{4})×(1 – w^{4} + w^{8}) × …………… to 2n factors is equal to

(a) 2^{n}

(b) 2^{2n}

(c) 2^{3n}

(d) 2^{4n}

## Answer

Answer: (b) 2^{2n}

Hint:

Given, (1 – w + w²)×(1 – w² + w^{4})×(1 – w^{4} + w^{8}) × …………… to 2n factors

= (1 – w + w^{2})×(1 – w^{2} + w )×(1 – w + w^{2}) × …………… to 2n factors

{Since w^{4} = w, w^{8} = w^{2}}

= (-2w) × (-2w²) × (-2w) × (-2w²)× …………… to 2n factors

= (2² w³)×(2² w³)×(2² w³) …………… to 2n factors

= (2²)^{n} {since w³ = 1}

= 2^{2n}

Question 20.

The modulus of 5 + 4i is

(a) 41

(b) -41

(c) √41

(d) -√41

## Answer

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

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