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MCQ Class 11 Mathematics Complex Numbers with Answers - Set - 2
Question 1:
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
Question 2 :
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
Correct Answer – (C)
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 3 :
if x + 1/x = 1 find the value of x2000 + 1/x2000 is
(a) 0
(b) 1
(c) -1
(d) None of these
Correct Answer – (C)
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 4 :
If {(1 + i)/(1 – i)}n = 1 then the least value of n is
(a) 1
(b) 2
(c) 3
(d) 4
Correct Answer – (D)
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
Question 5 :
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
Correct Answer – (A)
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
MCQ Class 11 Mathematics Complex Numbers with Answer
Question 6 :
(1 – w + w²)×(1 – w² + w4)×(1 – w4 + w8) × …………… to 2n factors is equal to
(a) 2n
(b) 22n
(c) 23n
(d) 24n
Correct Answer – (B)
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 7 :
The value of √(-144) is
(a) 12i
(b) -12i
(c) ±12i
(d) None of these
Correct Answer – (A)
Given, √(-144) = √{(-1)×144}
= √(-1) × √(144)
= i × 12 {Since √(-1) = i}
= 12i
So, √(-144) = 12i
Question 8 :
If arg (z) < 0, then arg (-z) – arg (z) =
(a) π
(b) -π
(c) -π/2
(d) π/2
Correct Answer – (A)
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 9 :
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
Correct Answer – (B)
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
Question 10 :
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
Correct Answer – (D)
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.
- 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
