CBSE Class 11 Mathematics Chapter 5 Complex Numbers Multiple Choice Questions with Answers. MCQ Class 11 Mathematics Complex Numbers with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 11 Mathematics Complex Numbers MCQs with Answers to know their preparation level.
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MCQ Class 11 Mathematics Complex Numbers with Answers - Set - 4
Question 1:
The value of (1 – i)² is
(a) i
(b) -i
(c) 2i
(d) -2i
Correct Answer – (D)
Given, (1 – i)² = 1 + i² – 2i
= 1 + (-1) – 2i
= 1 – 1 – 2i
= -2i
Question 2 :
The polar form of -1 + i is
(a) √2(cos π/2 + i × sin π/2)
(b) √2(cos π/4 + i × sin π/4)
(c) √2(cos 3π/2 + i × sin 3π/2)
(d) √2(cos 3π/4 + i × sin 3π/4)
Correct Answer – (D)
The polar form of a com plex number = r(cos θ + i × sin θ)
Given, complex number = -1 + i
Let x + iy = -1 + i
Now, x = -1, y = 1
Now, r = √{(-1)² + 1²} = √(1 + 1) = √2
and tan θ = y/x
⇒ tan θ = 1/(-1)
⇒ tan θ = -1
⇒ θ = 3π/4
Now, polar form is √2(cos 3π/4 + i × sin 3π/4)
Question 3 :
The value of i9 + i10 + i11 + i12 is
(a) i
(b) 2i
(c) 0
(d) 1
Correct Answer – (C)
Given, i9 + i10 + i11 + i12
= i9 (1 + i + i2 + i3 )
= i9 (1 + i – 1 – i ) {since i2 = (-1) and i4 = 1}
= i9 × 0
= 0
Question 4 :
The value of {-√(-1)}4n+3, n ∈ N is
(a) i
(b) -i
(c) 1
(d) -1
Correct Answer – (A)
Given, {-√(-1)}4n+3
= {-i}4n+3 {since √(-1) = i}
= {-i}4n × {-i}³
= {(-i)4}ⁿ ×(-i³) {since i4 = 1}
= 1ⁿ × (-i × i²)
= -i × (-1) {since i² = -1}
= i
Question 5 :
The modulus of 5 + 4i is
(a) 41
(b) -41
(c) √41
(d) -√41
Correct Answer – (C)
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
MCQ Class 11 Mathematics Complex Numbers with Answer
Question 6 :
For all complex numbers z1, z2 satisfying |z1| = 12 and |z2 – 3 – 4i| = 5, the minimum value of |z1 – z2| is
(a) 0
(b) 2
(c) 7
(d) 17
Correct Answer – (B)
Given For all complex numbers z1, z2 satisfying |z1| = 12 and |z2 – 3 – 4i| = 5
Now, mod(z1) = 12 represents a circle centred at 0 and radius 12
mod(z2 – 3 – 4i) = 5 represents a circle centred at (3, 4) and radius 5
This circle passes through the origin. Distance of diametrically opposite end is 10
So, the minimum value (z1 – z2) = 2
Question 7 :
If a = cos α + i sin α and b = cos β + i sin β , then the value of 1/2(ab + 1/ ab) is
(a) sin (α + β)
(b) cos (α + β)
(c) sin (α – β)
(d) cos (α – β)
Correct Answer – (B)
Given a = cos α + i sin α and b = cos β + i sin β
Now, 1/a = 1/(cos α + i sin α)
⇒ 1/a = {1 × (cos α – i sin α)/{(cos α + i sin α) × (cos α + i sin α)}
⇒ 1/a = (cos α – i sin α)/(cos² α + i sin² α)
⇒ 1/a = (cos α – i sin α)
Again, 1/b = 1/(cos β + i sin β)
⇒ 1/b = {1 × (cos β – i sin β)/{(cos β + i sin β) × (cos β + i sin β)}
⇒ 1/b = (cos β – i sin β)/(cos² β + i sin² β)
⇒ 1/b = (cos β – i sin β)
Now, ab = (cos α + i sin α) × (cos β + i sin β)
⇒ ab = cos α × cos β + i cos α × sin β + i sin α × cos β – sin α × sin β
Again, 1/ab = (cos α – i sin α) × (cos β – i sin β)
⇒ 1/ab = cos α × cos β – i cos α × sin β – i sin α × cos β – sin α × sin β
Now, ab + 1/ab = cos α × cos β + i cos α × sin β + i sin α × cos β – sin α × sin β + cos α × cos β – i cos α × sin β – i sin α × cos β – sin α × sin β
⇒ ab + 1/ab = 2(cos α × cos β – sin α × sin β)
⇒ 1/2(ab + 1/ ab) = 2(cos α × cos β – sin α × sin β)/2
⇒ 1/2(ab + 1/ ab) = cos α × cos β – sin α × sin β
⇒ 1/2(ab + 1/ ab) = cos(α + β)
Question 8 :
If ω is cube root of unity (ω ≠ 1) , then the least value of n where n is a positive integer such that (1 + ω²)ⁿ = (1 + ω4)ⁿ is
(a) 2
(b) 3
(c) 5
(d) 6
Correct Answer – (B)
Given ω is an imaginary cube root of unity.
So 1 + ω + ω² = 0 and ω³ = 1
Now, (1 + ω²)ⁿ = (1 + ω4)ⁿ
⇒ (-1)ⁿ ×(ω)ⁿ = (1 + ω × ω³)ⁿ
⇒ (-1)ⁿ × (ω)ⁿ = (1 + ω)ⁿ
⇒ (-1)ⁿ × (ω)ⁿ = (-ω²)ⁿ
⇒ (-1)ⁿ × (ω)ⁿ = (-1)ⁿ × ω²ⁿ
⇒ ωⁿ = ω²ⁿ
Since ω³ = 1, So the least value of n is 3
Question 9 :
The modulus of 1 + i√3 is
(a) 1
(b) 2
(c) 3
(d) None of these
Correct Answer – (B)
Let Z = 1 + i√3
Now modulus of Z is calculated as
|Z| = √{1² + (√3)²}
⇒ |Z| = √(1 + 3)
⇒ |Z| = √4
⇒ |Z| = 2
So, the modulus of 1 + i√3 is 2
Question 10 :
The real part of the complex number √9 + √(-16) is
(a) 3
(b) -3
(c) 4
(d) -4
Correct Answer – (A)
Given, √9 + √(-16) = √9 + √(16) × √(-1)
= 3 + 4i {since i = √(-1)}
So, the real part of the complex number is 3
- NCERT Solutions Class 11 Mathematics Complex Numbers with Answers : Exercise 5.1
- NCERT Solutions Class 11 Mathematics Complex Numbers with Answers : Exercise 5.2
- NCERT Solutions Class 11 Mathematics Complex Numbers with Answers : Exercise 5.3
- NCERT Solutions Class 11 Mathematics Complex Numbers with Answers : Exercise 5 Misc
- NCERT Solutions Class 11 Mathematics Textbook download
