Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - JEE MCQ

(a, b, c) z₁ = a + ib   and  z₂ = c + id.
ATQ  |zi|² = |z₂|² = 1
⇒​ a² + b² =1   and   c² + d² = 1. ....(1)

Also Re = 0 ⇒​ ac + bd=0

⇒ = α(say)  ....(2)

From (1) and (2), we get
b² α² + b² = c²α² + c² ⇒​ b² = c²;
Similarly   a²  = d²

∴      

and    

Also Re = ab + cd = (bα) b + c (-cα)
= α(b² -c² )=0

(a, d)   Let z₁ = a + ib, a > 0 and b ∈ R; z₂ = c+ id, d < 0, c∈R
then |z₁| = | z₂|
⇒ a₂ + b₂ = c₂+d₂
⇒ a₂- c₂ = d₂-b₂ ....(1)

Now ,  

=   

=   [Using (1)]

= purely imaginary number or zero in case a + c = b + d = 0.

(c) Let z₁ = r₁ (cos θ₁ +i sinθ₁) 

and z₂ = r₂ (cos θ₂ +i sinθ₂)

where  r₁ = |z₁|, r₂ =|z₂|, θ₁ = arg(z₁), θ₂= arg(z₂)

∴ z₁ + z₂ = r₁ (cos θ₁ + i sinθ₁ ) + r₂ (cos θ₂ +i sinθ₂)

= (r₁ cos θ₁+ r₂ cos θ₂) + i(r₁ sin θ₁+r₂ sin θ₂)

=  r₁² cos² θ₁ + r₂² cos² θ₂ + 2r₁r₂ cos θ₁ cosθ₂

+ r₁² sin² θ₁ + r₂² sin² θ₂ + 2r₁r₂ sin θ₁ sinθ₂

= r₁² + r₂² + 2r₁ r₂ cos ( θ₁ -θ₂)

and |z₁| + |z₂| = r₁+r₂

Since |z₁ + z₂| = | z₁ |+ |z₂| (given)

⇒ |z₁ + z₂|² = |z₁|² + | z₂ |²+ 2| z₁| |z₂|

⇒ r₁² + r₂²+2r₁ r₂ cos (θ₁ - θ ) = r12 + r₂²+2r₁r₂

⇒ cos (θ₁ -θ₂) = 1 ⇒ θ₁ -θ₂ =0

⇒ Arg (z₁) = arg (z₂)

(d) Let  z= 

Then by DeMoivre’s theorem, we have

Now,   

   =  [Using z⁷ = cos 2π + i sin 2π = 1]

(d) We have (1 + ω + ω²)⁷ = (- ω² -ω²)⁷

(-2)⁷ (ω²)⁷ = -128 ω¹⁴ = -128ω²

This forms a G.P.

Sum of G.P.= i(1 + i)

(d) Taking  – 3i common from C₂, we get

(∵ C₂ ≡ C₃)

⇒ x = 0,   y = 0

(a,c,d) Given that z = (1 – t) z₁+ tz₂ where 0 < t<1

⇒ z divides the join of z₁and z₂ internally in the ratio t : (1– t).

∴ z₁ , z and z₂ are collinear 

⇒ |z –z₁| + |z –z₂| = |z₁– z₂|

Also z = (1 – t) z₁+ tz₂

= t = purely real number

∴ arg = 0  ⇒ arg(z - z) = arg(z-z)

Also 

⇒

⇒ (z – z₁) (z₂ –z₁)

⇒  = 0

(c, d)

and wⁿ

∴ P contains all those points which lie on unit circle and have arguments

  and so on.

As z₁∈ P ∩ H₁ and z₂ ∈ P ∩ H₂, therefore z₁ and z₂ can have possible positions as shown in the figure.

∴ ∠z₁Oz₂ can be

(a, c, d) z =  = x + iy

⇒  x + iy 

⇒  

⇒   x² + y²

⇒   x² + y² - 

∴ Locus of z is a circle with centre  and radius

=  irrespective of ‘a’ +ve or –ve

Also for b = 0, a ≠ 0, we get, y = 0

∴ locus is x-axis

and for a = 0, b ≠ 0 we get x = 0

⇒ locus is y-axis.

∴ a, c, d are the correct options.