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.

## 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

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

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

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

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

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

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

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

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

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

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

MCQ Questions On Complex Numbers Class 11 Question 9.
Let z1 and z2 be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z1 and z1 form an equilateral triangle. Then
(a) a² = b
(b) a² = 2b
(c) a² = 3b
(d) a² = 4b

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

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: (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

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: (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: (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

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

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

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 x2000 + 1/x2000 is
(a) 0
(b) 1
(c) -1
(d) None of these

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

Question 17.
The value of √(-144) is
(a) 12i
(b) -12i
(c) ±12i
(d) None of these

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: (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

Factor complex polynomials calculator is generally used to solve the complex equation so that it is easier to work with simpler terms.

Question 19.
(1 – w + w²)×(1 – w² + w4)×(1 – w4 + w8) × …………… to 2n factors is equal to
(a) 2n
(b) 22n
(c) 23n
(d) 24n

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

Question 20.
The modulus of 5 + 4i is
(a) 41
(b) -41
(c) √41
(d) -√41