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Complex Numbers and Quadratic Equations Class 11 MCQs Questions 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-π
(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]²
= [i19 + 1/i25]²
= [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
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