Practice Questions with Answers - Multiple Choice Questions

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

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

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

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

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

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

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

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

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: (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

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

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