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 ListI to the correct entries in ListII.
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.
Q1: Let z be a complex number with a nonzero 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 nonzero 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 nonzero 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) = I_{1}
and Im(ω) = I_{2}
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 rewrite
Comparing with option (B) we get,
Option (B) is incorrect.
Similarly option (C) and (D) also incorrect.
Q1: Let θ_{1}, θ_{2}, ........, θ_{10} = 2π. Define the complex numbers z_{1} = e^{iθ1}, z_{k} = z_{k − 1}e^{iθk }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.  z_{2} − z_{1}  = length of line AB ≤ length of arc AB.
 z_{3} − z_{2}  = length of line BC ≤ length of arc BC.
∴ Sum of length of these 10 lines ≤ sum of length of arcs (i.e. 2π) (because θ_{1} + θ_{2} + θ_{3} + .... + θ_{10} = 2π (given)
∴  z_{2} − z_{1}  +  z_{3} − z_{2}  + ..... +  z_{1} − z_{10}  ≤ 2π → P is true.
And  z_{k}^{2} − z_{k}_{−1}^{2}  =  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 x^{2} + y^{2} + 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 x^{2} + y^{2} + 5x − 3y + 4 = 0 cuts the real axis (Xaxis) 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
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 z_{1} − z_{2}^{2}, where z_{1}, z_{2}∈S with Re(z_{1}) > 0 and Re(z_{2}) < 0 is _____ [JEE Advanced 2020 Paper 2]
Ans: 8
For a complex number z, it is given that,
So, either
Now, Case  I, if z^{2}=0 and z = x + iy
So, x^{2 } y^{2} + 2ixy = 0
⇒ x^{2 } y^{2} = 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 z_{1} − z_{2}^{2}, the z_{1} and z_{2} must be vertices of the rectangular hyperbola.
Therefore, z1 = 1 + i and z2 = 1  i
∴ Minimum value of z_{1} − z_{2}^{2}
= (1 + 1)^{2} + (1 + 1)^{2}
= 4 + 4
= 8
Q2: Let S be the set of all complex numbers z satisfying z^{2} + 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
Q1: Let ω ≠ 1 be a cube root of unity. Then the maximum of the set distinct nonzero integers} equals _____ [JEE Advanced 2019 Paper 1]
Ans: 3
Given, ω ≠ 1 be a cube root of unity, then
[as ω^{3} = 1)
∵ a, b and c are distinct nonzero 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 z_{0} 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 z_{0} − 1 is minimum. And for this it is required that z_{0} ∈ S, such that z_{0} is collinear with the points (2, −1) and (1, 0) and lies on the circumference of the circle z − 2 + i = √5.
So let z_{0} = x + iy, and from the figure 0 < x < 1 and y >0.
So,
is a positive real number, so is purely negative imaginary number.
Q1: Let s, t, r be nonzero 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.
209 videos443 docs143 tests

1. What is the importance of studying complex numbers in JEE Advanced? 
2. How are complex numbers represented in the complex plane? 
3. How can complex numbers be expressed in polar form? 
4. How are complex numbers added and multiplied? 
5. How can complex numbers be used to solve equations? 
209 videos443 docs143 tests


Explore Courses for JEE exam
