MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers

Students are advised to solve the Complex Numbers and Quadratic Equations Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Complex Numbers and Quadratic Equations Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Complex Numbers and Quadratic Equations Class 11 with answers provided with detailed solutions by looking below.

Question 1.
Let z1 and z2 be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z1 and z2 form an equilateral triangle. Then
(a) a² = b
(b) a² = 2b
(c) a² = 3b
(d) a² = 4b

Answer

Answer: (c) a² = 3b
Given, z1 and z1 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 2.
The value of ii is
(a) 0
(b) e
(c) 2e-π/2
(d) e-π/2

Answer

Answer: (d) e-π/2
Let A = ii
⇒ log A = i log i
⇒ log A = i log(0 + i)
⇒ log A = i [log 1 + i tan-1 ∞]
⇒ log A = i [0 + i π/2]
⇒ log A = -π/2
⇒ A = e-π/2


Question 3.
The value of √(-25) + 3√(-4) + 2√(-9) is
(a) 13 i
(b) -13 i
(c) 17 i
(d) -17 i

Answer

Answer: (c) 17 i
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
= 17 i
So, √(-25) + 3√(-4) + 2√(-9) = 17 i


Question 4.
If the cube roots of unity are 1, ω and ω², then the value of (1 + ω / ω²)³ is
(a) 1
(b) -1
(c) ω
(d) ω²

Answer

Answer: (b) -1
Given, the cube roots of unity are 1, ω and ω²
So, 1 + ω + ω² = 0
and ω³ = 1
Now, {(1 + ω)/ ω²}³ = {-ω²/ ω²}³ = {-1}³ = -1


Question 5.
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
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 6.
The value of [i19 + (1/i)25]² is
(a) -1
(b) -2
(c) -3
(d) -4

Answer

Answer: (d) -4
Given, [i19 + (1/i)25
= [i19 + 1/i25
= [i16 × i³ + 1/(i24 × i)]²
= [1 × i³ + 1/(1 × i)]² {since i4 = 1}
= [i³ + 1/i]²
= [i² × i + 1/i]²
= [(-1) × i + 1/i]² {since i² = -1}
= [-i + 1/i]²
= [-i + i4 /i]²
= [-i + i³]²
= [-i + i² × i]²
= [-i + (-1) × i]²
= [-i – i]²
= [-2i]²
= 4i²
= 4 × (-1)
= -4
So, [i19 + (1/i)25]² = -4


Question 7.
If z and w be two complex numbers such that |z| ≤ 1, |w| ≤ 1 and |z + iw| = |z – iw| = 2, then z equals {w is congugate of w}
(a) 1 or i
(b) i or – i
(c) 1 or – 1
(d) i or – 1

Answer

Answer: (c) 1 or – 1
Given |z + iw| = |z – iw| = 2 {w is congugate of w}
⇒ |z – (-iw)| = |z – (iw)| = 2
⇒ |z – (-iw)| = |z – (-iw)|
So, z lies on the perpendicular bisector of the line joining -iw and -iw.
Since, -iw is the mirror in the x-axis, the locus of z is the x-axis.
Let z = x + iy and y = 0
⇒ |z| < 1 and x² + 0² < 0
⇒ -1 ≤ x ≤ 1
So, z may take value 1 or -1


Question 8.
The value of {-√(-1)}4n+3, n ∈ N is
(a) i
(b) -i
(c) 1
(d) -1

Answer

Answer: (a) i
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 9.
Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is real
(a) π
(b) nπ
(c) nπ/2
(d) 2nπ

Answer

Answer: (b) nπ
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 real if sin θ = 0
⇒ sin θ = sin nπ
⇒ θ = nπ


Question 10.
If i = √(-1) then 4 + 5(-1/2 + i√3/2)334 + 3(-1/2 + i√3/2)365 is equals to
(a) 1 – i√3
(b) -1 + i√3
(c) i√3
(d) -i√3

Answer

Answer: (c) i√3
Given, 4 + 5(-1/2 + i√3/2)334 + 3(-1/2 + i√3/2)365
= 4 + 5w334 + 3w365 {since w = -1/2 + i√3/2}
= 4 + 5w + 3w² {since w³ = 1}
= 4 + 5(-1/2 + i√3/2) + 3(-1/2 – i√3/2) {since w² = (-1/2 – i√3/2)}
= i√3


Question 11.
The real part of the complex number √9 + √(-16) is
(a) 3
(b) -3
(c) 4
(d) -4

Answer

Answer: (a) 3
Given, √9 + √(-16) = √9 + √(16) × √(-1)
= 3 + 4i {since i = √(-1)}
So, the real part of the complex number is 3


Question 12.
The modulus of 5 + 4i is
(a) 41
(b) -41
(c) √41
(d) -√41

Answer

Answer: (c) √41
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 13.
The modulus of 1 + i√3 is
(a) 1
(b) 2
(c) 3
(d) None of these

Answer

Answer: (b) 2
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 14.
The value of {-√(-1)}4n+3, n ∈ N is
(a) i
(b) -i
(c) 1
(d) -1

Answer

Answer: (a) i
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 15.
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

Answer

Answer: (b) 3
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 16.
The value of i9 + i10 + i11 + i12 is
(a) i
(b) 2i
(c) 0
(d) 1

Answer

Answer: (c) 0
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 17.
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)

Answer

Answer: (b) cos (α + β)
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 18.
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)

Answer

Answer: (d) √2(cos 3π/4 + i × sin 3π/4)
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 19.
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

Answer

Answer: (b) 2
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 20.
The value of (1 – i)² is
(a) i
(b) -i
(c) 2i
(d) -2i

Answer

Answer: (d) -2i
Given, (1 – i)² = 1 + i² – 2i
= 1 + (-1) – 2i
= 1 – 1 – 2i
= -2i


 


0 Comments

Leave a Reply

Avatar placeholder

Your email address will not be published. Required fields are marked *