Complex Numbers and Quadratic Equations Class 11 MCQs Questions with Answers
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) }
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
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
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.
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 – ω2)7 = (-2ω2)7
⇒ (1 + ω – ω2)7 = -128 ω14
⇒ (1 + ω – ω2)7 = -128 ω12 × Ï‰2
⇒ (1 + ω – ω2)7 = -128 (ω3)4 ω2
⇒ (1 + ω – ω2)7 = -128 ω2
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
= in
Now, in is real when n = 2 {since i2 = -1 }
So, the least value of n is 2
Question 7.
Let z be a complex number such that |z| = 4 and arg(z) = 5Ï€/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 5Ï€/6 + i × sin 5Ï€/6)
⇒ z = 4(-√3/2 + i/2)
⇒ z = -2√3 + 2i
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/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
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
Answer
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
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
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.
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
Question 13.
Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary
(a) θ = nÏ€ ± Ï€/2 where n is an integer
(b) θ = nÏ€ ± Ï€/3 where n is an integer
(c) θ = nÏ€ ± Ï€/4 where n is an integer
(d) None of these
Answer
Answer: (b) θ = nÏ€ ± Ï€/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 = nÏ€ ± Ï€/3 where n is an integer
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
⇒ in = 1
Now, in is 1 when n = 4
So, the least value of n is 4
Question 15.
If arg (z) < 0, then arg (-z) – arg (z) =
(a) π
(b) -Ï€
(c) -Ï€/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) = Ï€ {since sin Ï€ + i cos Ï€ = -1, So arg(-1) = Ï€}
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
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
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
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 + w4)×(1 – w4 + w8) × …………… to 2n factors is equal to
(a) 2n
(b) 22n
(c) 23n
(d) 24n
Answer
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
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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