Test: JEE Main 35 Year PYQs- Complex Numbers - JEE MCQ

(b) Let | z | = | ω | = r

(c) Given | z – 4 | < | z – 2 |
Let z = x + iy
⇒ | (x – 4) + iy) | < | (x – 2) + iy |
⇒ (x – 4)² + y² < (x – 2)² + y²
⇒ x² – 8x + 16 < x² – 4x + 4
⇒ 12 < 4x
⇒ x > 3
⇒ Re(z) > 3

(b) Let the circle be |z – z₀| = r. Then according to given conditions |z₀ – z₁| = r + a and |z₀ – z₂|= r + b.
Eliminating r, we get |z₀ – z₁| –|z₀ – z₂| = a – b.
∴ Locus of centre z₀ is |z – z₁| –|z – z₂| = a – b,
which represents a hyperbola.

Correct Answer is A.

z ² + az +b= 0 ;  z₁ + z₂ = - a & z₁ z₂=b 0, z₁,z₂ form an equilateral Δ 
∴ 0² + z₁² + z₂² = 0.z₁ + z₁.z₂+z₂ .0 (for an equilateral triangle, 
z₁² + z₂² + z₃² = z₁z₂+ z₂ z₃+ z₃z₁ )
⇒ z₁² + z₂²= z₁z²
⇒ ( z₁ + z₂ )²=3 z₁z²
∴a²= 3b

⇒

= ⇒ (i)^x = 1; 

arg zw = π ⇒ arg z + argw = π...(1)

= p+ iq ⇒ z = p³ + (iq)³ + 3 p (iq)( p+ iq)

⇒ x - iy = p³ - 3pq² + i(3p²q-q³)

∴ x = p³ - 3pq² ⇒ 

y = q³ - 3p²q ⇒ 

|z² - 1|=|z|²+1 ⇒|z² - 1|² = 

⇒  

⇒  

⇒  ⇒ 

⇒ z is purely imaginary

(x - 1)³ + 8=0  
⇒ (x - 1) = (-2) (1)^{1 /3}  
⇒ x – 1 = – 2 or -2ω or - 2ω²
or x = – 1 or 1 – 2ω or 1 – 2ω² .

|z₁ +z₂| = |z₁| + |z₂| ⇒ z₁ and z₂ are collinear and are to the same side of origin; hence arg z₁ – arg z₂ = 0.

As given w  ⇒ |w| = = 1

 ⇒ 

 ⇒ distance of z from origin and point  is

same hence z lies on bisector of the line joining points (0, 0) and (0, 1/3).
Hence z lies on a straight line.

= i × 0 – i                [∵ e^-2πi = 1] = – i

 z² +z + 1=0 ⇒ z = ω or ω²

So,

and

∴ The given sum = 1+1 + 4 + 1 + 1 + 4 = 12

z lies on or inside the circle with centre (–4, 0) and radius 3 units.

From the Argand diagram maximum value of | z + 1| is  6

Given S = {(x , y) : y = x + 1 and 0 < x < 2}
∵ x ≠ x  + 1 for any x ∈(0, 2) ⇒ (x, x) ∉ S
∴ S is not reflexive.
Hence S in not an equivalence relation.
Also T  ={x, y): x – y is an integer}
​∵ x – x = 0 is an integer " x ∈ R
∴ T is reflexive.
If  x – y is an integer then y – x is also an integer
∴T is symmetric If  x – y is an integer and y – z is an integer then (x – y) + (y– z) = x – z is also an integer.
∴ T is transitive Hence T is an equivalence relation.

Let z = x+ iy
|z -1| = |z +1| (x -1)² + y² = (x +1)²+y²
⇒ Re z =0 ⇒ x = 0
|z -1| = |z-i| (x-1)²+ y²= x² +(y -1)²
⇒ x =y
|z+1| = |z-i| (x+1)² + y² = x² +(y-1)²
Only (0, 0) will satisfy all conditions.
⇒ Number of complex number z  = 1

∵ Real part of roots is 1 Let roots are 1 + pi, 1 + q
∴ sum of roots = 1 + pi + 1 + qi = – a which is real
⇒  q = – p or root are 1 + pi and 1 – pi product of roots = 1 + p² = β∈ (1,)
p ≠ 0 as roots are distinct.

(1 + ω)⁷ = A + Bω;
​ (–ω²)⁷ = A + Bω – ω² = A + Bω;
1 + ω = A + Bω
⇒ A = 1, B = 1.

Now   is real ⇒  Im  = 0

⇒  

⇒​  Im [(x² – y² + 2ixy) (x – 1) – iy)] = 0

⇒  2xy (x – 1) – y (x² – y²) = 0

⇒ y(x² + y² – 2x) = 0 ⇒  y = 0;  x² + y² – 2x = 0

∴ z lies either on real axis or on a circle through origin.

Given | z | = 1, arg z = θ

As we know,  

∴   = arg (z) = θ

We know minimum value of 

Thus minimum value of 

Since, | Z |≥2 therefore 

⇒  

⇒  

⇒  

⇒ 

⇒ 

⇒  

⇒   

⇒ Point z₁ lies on circle of radius 2.

Rationalizing the given expression

For the given expression to be purely imaginary, real part of the above expression should be equal to zero.