JEE Advanced Previous Year Questions (2018 - 2023): Complex Numbers | Mathematics (Maths) for JEE Main & Advanced PDF Download

2023

Q1: Let  . If A contains exactly one positive integer n, then the value of n is                    [JEE Advanced 2023 Paper 1]
Ans: 281

For positive integer

Q2: Let z be a complex number satisfying , where  denotes the complex conjugate of z. Let the imaginary part of z be nonzero.
Match each entry in List-I to the correct entries in List-II. 

The correct option is:
(a) (P)→(1)(Q)→(3)(R)→(5)(S)→(4)
(b) (P)→(2)(Q)→(1)(R)→(3)(S)→(5)
(c) (P)→(2)(Q)→(4)(R)→(5)(S)→(1)
(d) (P)→(2)(Q)→(3)(R)→(5)(S)→(4)               [JEE Advanced 2023 Paper 1]
Ans: (b)

Take conjugate both sides

Let

Put in (1)

Also

Now

|z + 1|² = 4 + 3 = 7
∴ (P)→(2)(Q)→(1)(R)→(3)(S)→(5)
∴ Option (b) is correct.

2022

Q1: Let z be a complex number with a non-zero imaginary part. If  is a real number, then the value of |z|² is _________.            [JEE Advanced 2022 Paper 1]
Ans: 0.49 to 0.51
For a complex number z = x + iy, it's conjugate . Now z is purely real when y = 0. 
When y = 0 then z = x + i × (0) = x and
∴   when z is purely real.
Now given,  is real

=

y = 0  not possible as given z is a complex number with non-zero imaginary part. 

Q2: Let  denote the complex conjugate of a complex number z and let . In the set of complex numbers, the number of distinct roots of the equation  is _________.     [JEE Advanced 2022 Paper 1]
Ans: 4
Let z = x + iy

Given,

Comparing both sides real part we get,

And comparing both sides imaginary part we get,

Adding equation (1) and (2) we get,

Case 1 : When x = 0 :
Put x =  0  at equation (1), we get

Case 2 : When y = −1/2 :
Put y = −1/2 in equation (1), we get

∴ Number of distinct  z = 4

Q3: Let denote the complex conjugate of a complex number z. If z is a non-zero complex number for which both real and imaginary parts of  are integers, then which of the following is/are possible value(s) of |z| ? 
(a)
(b)
(c)
(d)                   [JEE Advanced 2022 Paper 2]
Ans: (a)
Let, complex number  is a new complex number ω. 

Now, Let  where r = |z| and θ = argument

∴ Real part of

Imaginary part of

Given both Re(ω) and Im(ω) are integer. 
∴ Let Re(ω) = I₁
and Im(ω) = I₂

Now,

In option only positive sign is given so ignoring negative sign we get,

From option (A),

Comparing with (1), we get

Putting α = 45 in (1), we get

∴ Option (A) is correct. 
We can re-write

Comparing with option (B) we get,

Option (B) is incorrect.
Similarly option (C) and (D) also incorrect.

2021

Q1: Let θ₁, θ₂, ........, θ₁₀ = 2π. Define the complex numbers z₁ = e^iθ₁, z_k = z_k − 1e^iθ_k for k = 2, 3, ......., 10, where i = √−1. Consider the statements P and Q given below : 

Then,
(a) P is TRUE and Q is FALSE
(b) Q is TRUE and P is FALSE
(c) both P and Q are TRUE
(d) both P and Q are FALSE                           [JEE Advanced 2021 Paper 1]
Ans: (c)
Both P and Q are true.
∵ Length of direct distance ≤ length of arc
i.e. | z₂ − z₁ | = length of line AB ≤ length of arc AB. 

| z₃ − z₂ | = length of line BC ≤ length of arc BC.
∴ Sum of length of these 10 lines ≤ sum of length of arcs (i.e. 2π) (because θ₁ + θ₂ + θ₃ + .... + θ₁₀ = 2π (given)
∴ | z₂ − z₁ | + | z₃ − z₂ | + ..... + | z₁ − z₁₀ | ≤ 2π → P is true.
And | z_k² − z_k−1² | = | z_k − z_k − 1 | | z_k + z_k − 1 | 
As we know that,

≤ 4π → Q is true. 

Q2: For any complex number w = c + id, let arg⁡(ω)∈(−π, π], where i = √−1. Let α and β be real numbers such that for all complex numbers z = x + iy satisfying  , the ordered pair (x, y) lies on the circle x² + y² + 5x − 3y + 4 = 0, Then which of the following statements is (are) TRUE?
(a) α = −1
(b) αβ = 4
(c) αβ = −4
(d) β = 4               [JEE Advanced 2021 Paper 1]
Ans: (d)
Circle  x² + y² + 5x − 3y + 4 = 0 cuts the real axis (X-axis) at (−4, 0), (−1, 0).

 implies z is on arc and (− α, 0) and (− β, 0) subtend π/4 on z.
So, α = 1 and  β = 4
Hence, αβ = 1 × 4 = 4 and β = 4 

2020

Q1: For a complex number z, let Re(z) denote that real part of z. Let S be the set of all complex numbers z satisfying , where i = √−1. Then the minimum possible value of |z₁ − z₂|², where z₁, z₂∈S with Re(z₁) > 0 and Re(z₂) < 0 is _____     [JEE Advanced 2020 Paper 2]
Ans: 8
For a complex number z, it is given that,

So, either

Now, Case - I, if z²=0 and z = x + iy
So, x² - y² + 2ixy = 0
⇒ x² - y² = 0
and xy = 0
⇒ x = y  = 0
⇒ z = 0  which is not possible according to given conditions.
Case - II, if  and
z = x + iy
So,

⇒ xy = 1 is an equation of rectangular hyperbola and for minimum value of |z₁ − z₂|², the z₁ and z₂ must be vertices of the rectangular hyperbola.
Therefore,  z1 = 1 + i and z2 = -1 - i
∴ Minimum value of   |z₁ − z₂|²
= (1 + 1)² + (1 + 1)²
= 4 + 4
= 8

Q2: Let S be the set of all complex numbers z satisfying |z² + z + 1| = 1. Then which of the following statements is/are TRUE?
(a) |z + 1/2| ≤ 1/2 for all z ∈ S
(b) |z| ≤ 2 for all z ∈ S
(c) |z + 1/2| ≥ 1/2 for all z ∈ S
(d) The set S has exactly four elements           [JEE Advanced 2020 Paper 1]
Ans: (b) & (c)
It is given that the complex number satisfying

from Eqs. (i) and (ii), we get

2019

Q1: Let ω ≠ 1 be a cube root of unity. Then the maximum of the set  distinct non-zero integers} equals _____   [JEE Advanced 2019 Paper 1]
Ans: 3
Given, ω ≠ 1 be a cube root of unity, then

[as ω³ = 1) 

∵ a, b and c are distinct non-zero integers. For minimum value a= 1, b = 2 and c = 3 
∴

Q2: Let S be the set of all complex numbers z satisfying . If the complex number z₀ is such that  is the maximum of the set , then the principal argument of  is 
(a) π / 4
(b) 3π / 4
(c) - π / 2
(d) π / 2                            [JEE Advanced 2019 Paper 1]
Ans: (c)
The complex number z satisfying , which represents the region outside the circle (including the circumference) having centre (2, −1) and radius √5 units.

Now, for  is maximum.
When |z₀ − 1| is minimum. And for this it is required that z₀ ∈ S, such that z₀ is collinear with the points (2, −1) and (1, 0) and lies on the circumference of the circle |z − 2 + i| = √5. 
So let z₀ = x + iy, and from the figure 0 < x < 1 and y >0. 
So,

 is a positive real number, so  is purely negative imaginary number.

2018

Q1: Let s, t, r be non-zero complex numbers and L be the set of solutions  of the equation   where  = x − iy. Then, which of the following statement(s) is(are) TRUE?     [JEE Advanced 2018 Paper 2]
(a) If L has exactly one element, then |s| ≠ |t|
(b) If |s| = |t|, then L has infinitely many elements
(c) The number of elements in L ∩ {z:|z − 1 + i|=5} is at most 2
(d) If L has more than one element, then L has infinitely many elements             [JEE Advanced 2018 Paper 2]
Ans: (a), (c) & (d)
We have,

On taking conjugate,

On solving Eqs. (i) and (ii), we get

(a) For unique solutions of z
It is true
(b) If |s| = |t|, then  may or may not be zero. So, z may have no solutions.∴ L may be an empty set.
It is false.
(c) If elements of set L represents line, then this line and given circle intersect at maximum two point. Hence, it is true.
(d) In this case locus of z is a line, so L has infinite elements. Hence, it is true.