Test: Complex Numbers & Quadratic Equations - JEE MCQ

∣z−5i∣/∣z+5i| = 1
 ∣z−5i∣=∣z+5i| (Using definition ∣z−z1∣=∣z−z2∣ given Perpendicular bisector of z1 and z2.)
⇒ Perpendicular bisector of points (0,5) and (0,−5). which lies on x-axis.

 Let, z = x + iy
Putting in the given inequality,
|(x−4)−iy| < |(x−2)+iy|
⇒ [(x−4)² + y²]11/2 < [(x−2)² + y²]^1/2
squaring both sides,
⇒ (x−4)² + y² < (x−2)² + y²
⇒ −8x+16 < −4x+4
⇒ 4

x = (√3+i)/2 
x³ = 1/8(√3+i)³
Apply formula (a+b)³ = a³ + b³ + 3a²b + 3ab²
= (3√3 + i³ + 3*3*i + 3*√3*i²)/8 
= (3√3 - i + 9i - 3√3)/8 
= 8i/8
= i

(√3 + i)¹⁰ = a + ib
Z = √3 + i = rcosθ + i rsinθ
⇒ √3 = rcosθ   i = rsinθ
⇒ (√3)² + (1)² = r²cos²θ + r²sin²θ
⇒ 4 = r²
⇒ r = 2
tan = 1/√3    
⇒ tan π/6
Therefore, Z = √3 + i = 2(cos π/6 + i sin π/6)
(Z)¹⁰ = √3 + i = (2cos π/6 + 2i sin π/6)¹⁰
= 2¹⁰ (cos π/6 + i sin π/6)¹⁰
2¹⁰ (cos 10π/6 + i sin 10π/6)
= 2¹⁰ (cos(2π - π/3) + i sin(2π - π/3))
= 2¹⁰ (cos π/3 - i sin π/3)
= 2¹⁰ (1/2 - i√3/2)
2⁹(1 - i√3)
a = 2⁹ = 512
b = - 2⁹(√3) = -512√3

Let u be multiplicative inverse
zu = 1
u = 1/z
u = 1/(x+iy)
Rationalise it 
[1/(x+iy)]*[(x-iy)/(x-iy)]
= (x-iy)/(x²+y²)
u = x/(x²+y²) +i(-y)/(x²+y²)

=

Let, z = x + iy

Putting in the given inequality,

∣(x−4) − iy∣ < ∣(x−2) + iy∣

squaring both sides,

⇒ (x−4)² + y² < (x−2)² + y²
⇒ −8x + 16 < −4x + 4
⇒ 4x > 12
⇒ x > 3
⇒ Re(z) > 3

AC = z₃ = z₁ eiπ
= z₁ (cosπ + i sinπ)
= z₃ = z₁(-1 + i(0))
= z₃ = -z₁
AC = z₁ - z₃
BC = z₂ - z₄
(z₁ - z₃)/(z₂ - z₄) = k
(z₁ - z₃) = eⁱπ/2(z₂ - z4)
(z₁ - z₃) k(cosπ/2 + sinπ/2) (z₂ - z₄)
z₁ - z₃ = ki(z₂ - z₄)
z₁ - z₃ = ik(z₂ - z₄)
 

Since w is the cube root of unity.
(1+w)⁹ =A+Bw
⇒(−w²)⁹ =A+Bw
⇒−(w³)⁶ = A+Bw
⇒−1=A+Bw
∴ A= -1 & B=0

This is Equation of a Circle.

√-8 * √-18 * √-4
= 2√2(√-1) × 3√2(√-1) × 2(√-1)
= 2√2(i) × 3√2(i) × 2(i)
= (2√2 * 3√2 * 2) *(i * i * i)
= (24)(-i2 * i)
= 24(-i)
= -24i