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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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)

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

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