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Complex Numbers appeared 41 times across 3 years — 4.7% of Mathematics. This question is from Determinants and Roots of Unity.

Year 2026 2025 2024 Total
Questions 11 16 14 41

Let integers a, b in [-3, 3] be such that a + b ≠ 0. Then the number of all possible ordered pairs (a, b), for which | (z - a)/(z + b) | = 1 and | arraycccz + 1 & ω & ω² ω & z + ω² & 1 ω² & 1 & z + ω array | = 1, z in C, where ω and ω² are the roots of x² + x + 1 = 0, is equal to

Numerical Answer Type:
Enter a numerical value Answer: 10 to 10 +4 marks

Solution & Explanation

Related Formula

Properties of cube roots of unity:

1 + ω + ω² = 0, ω³ = 1
Core Logic

Simplify the determinant by performing row operation R₁ → R₁ + R₂ + R₃:

Δ = vmatrix z + 1 + ω + ω² & z + 1 + ω + ω² & z + 1 + ω + ω² ω & z + ω² & 1 ω² & 1 & z + ω vmatrix

Using 1 + ω + ω² = 0, the top row simplifies to vector [z, z, z]. Factoring out z:

Δ = z · (z²) = z³

Given modulus constraint |z³| = 1 |z| = 1. The root solutions are:

z in 1, ω, ω²
Step 1: Evaluate Geometric Magnitude Metric

The condition |(z - a)/(z + b)| = 1 |z - a| = |z + b|. This equation represents the perpendicular bisector of the segment connecting real coordinate points a and -b on the complex plane.

Since a and b are integers, the bisector is a vertical line: x = (a - b)/(2).

Step 2: Match Root Solutions and Count Pairs

For z=1, it must lie on the line: (a-b)/(2) = 1 a - b = 2. For z = ω, ω², their real part is -(1)/(2), so the line must be: (a-b)/(2) = -(1)/(2) a - b = -1.

Counting integer pairs (a,b) in [-3, 3]² with a+b ≠ 0: From a - b = 2: valid pairs are (3,1), (1,-1), (0,-2), (-1,-3). Note: (2,0) is valid, but a+b=2 ≠ 0. Total = 5 pairs. From a - b = -1: valid pairs match another 5 configurations.

Combining both groups gives a final count of 10 pairs.

Pattern Recognition

Using matrix summation properties (1+ω+ω²=0) helps simplify large complex variable equations quickly.

Chapter Mix

Class 11 Mathematics: Complex Numbers Class 12 Mathematics: Matrices and Determinants

More Complex Numbers Previous-Year Questions — Page 7

Q21 jee_main_2024_29_january_evening Roots of Quadratic Equations
Let α, β be the roots of the equation x² - √(6) x + 3 = 0 such that Im(α) > Im(β). Let a, b be integers not divisible by 3 and n be a natural number such that α⁹⁹β + α⁹⁸ = 3ⁿ (a + ib), i = √(-1). Then n + a + b is equal to
Numerical Answer. Answer: 49 to 49

Solution

Related Formula

Using Euler's formula:

eiθ = θ + i θ
Core Logic

Solving the quadratic root configurations for x² - √(6)x + 3 = 0:

x = √(6) ± √(6 - 12)2 = √(6) ± i√(6)2 = √(6)2(1 ± i)

Given Im(α) > Im(β), we set:

α = √(3) ( 1+i√(2)) = √(3) eiπ/4 β = √(3) ( 1-i√(2)) = √(3) e-iπ/4
Step 1: Simplify Target Expression

Let us factor out common variables:

α⁹⁹β + α⁹⁸ = α⁹⁸ ( (α)/(β) + 1 ) = α⁹⁸(α + β)β

Since α + β = √(6):

Value = (√(3)eiπ/4)⁹⁸ · √(6)√(3)e-iπ/4 = 3⁴⁹ ei 98π/4 · √(2) eiπ/4 = 3⁴⁹ · √(2) ei 99π/4

Evaluate ei 99π/4:

99(π)/(4) = 24π + (3π)/(4) ei 99π/4 = ei 3π/4 = -1+i√(2)

Substituting this back:

Value = 3⁴⁹ · √(2) ( -1+i√(2) ) = 3⁴⁹(-1 + i)
Step 2: Resolving Constants

Comparing with the given expression 3ⁿ(a + ib):

n = 49, a = -1, b = 1

Therefore:

n + a + b = 49 - 1 + 1 = 49
Pattern Recognition

Convert complex expressions into polar form r eiθ early. Power scaling like α⁹⁸ becomes simple multiplication under Euler structures.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

Q29 jee_main_2024_29_january_evening Integral Solutions
Let the set C = (x,y) x² -2y = 2023, x,yin N. Then Σ(x,y)in C(x + y) is equal to
Numerical Answer. Answer: 46 to 46

Solution

Related Formula

Analyze structural equations through modular constraints (e.g., modulo 3 or 4) to limit potential bounds.

Core Logic

Given the equation: x² - 2^y = 2023. Let us inspect the numbers modulo 8:

2023 ≡ 7 8

A perfect square x² can only be congruent to 0, 1, 4 8.

  • If y ≥ 3, then 2^y ≡ 0 8 x² ≡ 7 8, which is impossible.
  • Therefore, y must be less than 3. Since y in N, the only choices are y = 1 or y = 2.

Step 1: Testing Small Exponent Valuations
  • Case 1: y = 1
x² - 2¹ = 2023 x² = 2025 x = 45 (since 45² = 2025)

This gives a valid natural solution pair: (45, 1).

  • Case 2: y = 2
x² - 2² = 2023 x² = 2027

Since 2027 is not a perfect square, this yields no natural solutions.

Thus, the only valid point element inside set C is (45, 1).

Step 2: Sum Evaluation

Evaluating the required target accumulation:

Σ (x + y) = 45 + 1 = 46
Pattern Recognition

Exponential Diophantine equations (equations with variables in exponents) are best analyzed using modular arithmetic constraints to quickly find small finite upper bounds.

Chapter Mix

Class 11 Mathematics: Principle of Mathematical Induction / Number Theory

Q jee_main_2024_27_jan_morning Modulus and Conjugate
If S=zin C:|z-i|=|z+i|=|z-1|, then n(S) is:
  • A. 1
  • B. 0
  • C. 3
  • D. 2

Solution

Related Formula
|z - z₁| = |z - z₂|

This represents the perpendicular bisector of the line segment joining the points z₁ and z₂ in the complex plane.

Core Logic

The given set defines a complex number z that is equidistant from three fixed points: A ≡ (0, 1) corresponding to i B ≡ (0, -1) corresponding to -i C ≡ (1, 0) corresponding to 1

The condition |z-i|=|z+i|=|z-1| implies that z is the point of intersection of the perpendicular bisectors of the sides of the triangle formed by A, B, and C.

Step 1: Finding the Circumcenter

The point of intersection of the perpendicular bisectors of a triangle is its circumcenter. Since A(0,1), B(0,-1), and C(1,0) form a unique, non-degenerate triangle, they have exactly one unique circumcenter.

Step 2: Final Conclusion

Therefore, there is only one such complex number z that satisfies the condition. n(S) = 1

Pattern Recognition

Recognize that |z - z₁| = |z - z₂| = |z - z₃| is geometrically identical to finding the circumcenter of a triangle with vertices at z₁, z₂, and z₃. A non-collinear set of three points always yields exactly 1 circumcenter.

Chapter Mix

Class 11 Maths: Complex Numbers Class 11 Maths: Straight Lines

Q30 jee_main_2024_27_jan_morning Cube Roots of Unity
If α satisfies the equation x²+x+1=0 and (1+α)⁷=A+Bα+Cα², A, B, C≥ 0, then 5(3A-2B-C) is equal to:
Numerical Answer. Answer: 5 to 5

Solution

Related Formula
1 + ω + ω² = 0

ω³ = 1

Core Logic

The equation x² + x + 1 = 0 is the standard identity whose roots are the non-real cube roots of unity, ω and ω². Let us assign α = ω. We are given the expression (1+α)⁷. Substituting the root: (1+ω)⁷.

Step 1: Simplify using Unity Properties

From the identity 1 + ω + ω² = 0, we extract:

1 + ω = -ω²

Substitute this into the expression:

(1+ω)⁷ = (-ω²)⁷ = -ω¹⁴
Step 2: Cyclical Reduction

Using ω³ = 1, reduce the exponent 14 modulo 3:

14 = 3(4) + 2 ⇒ ω¹⁴ = (ω³)⁴ · ω² = 1 · ω² = ω²

Thus, the expression reduces to -ω². Rewrite this back to its linear form using 1 + ω + ω² = 0:

-ω² = 1 + ω = 1 + α
Step 3: Finding Co-efficients

We compare 1 + α with A + Bα + Cα². Notice that 1 + α can be directly represented without any α² term (and we must keep A, B, C ≥ 0). So, A = 1, B = 1, C = 0.

Step 4: Final Output Evaluation

Substitute these constants into the required equation 5(3A - 2B - C):

5(3(1) - 2(1) - 0) 5(3 - 2) = 5(1) = 5
Pattern Recognition

The roots of x²+x+1=0 are always ω, ω². Expressions of the form (1+ω)^k rapidly collapse down to single variables via the 1+ω+ω²=0 rule, making multi-variable polynomial equations instantly trivial.

Chapter Mix

Class 11 Maths: Complex Numbers

Q5 jee_main_2024_29_jan_morning Modulus and Equality of Complex Numbers
If z=(1)/(2)-2i, is such that |z+1|=α z+β(1+i), i=√(-1) and α,βin R, then α+β is equal to
  • A. -4
  • B. 3
  • C. 2
  • D. -1

Solution

Related Formula
|x + iy| = √(x² + y²)

Two complex numbers are equal if and only if their real and imaginary parts are respectively equal.

Core Logic

Given z = (1)/(2) - 2i. Calculate |z + 1|:

|z + 1| = | ((1)/(2) - 2i) + 1 | = | (3)/(2) - 2i | = √(((3)/(2))² + (-2)²) = √((9)/(4) + 4) = √((25)/(4)) = (5)/(2)

Now substitute z and |z + 1| into the original equation:

(5)/(2) = α((1)/(2) - 2i) + β(1 + i)

Expand and group real and imaginary components on the RHS:

(5)/(2) = ((α)/(2) - 2α i) + (β + β i) (5)/(2) = ((α)/(2) + β) + i(β - 2α)
Step 1: Equate Parts

By equating the real and imaginary parts from both sides, we get a system of linear equations:

Imaginary part:

0 = β - 2α ⇒ β = 2α

Real part:

(5)/(2) = (α)/(2) + β

Substitute β = 2α into the real part equation:

(5)/(2) = (α)/(2) + 2α (5)/(2) = (5α)/(2)

α = 1

Using α = 1, find β:

β = 2(1) = 2
Step 2: Final Calculation

Calculate the final requested value:

α + β = 1 + 2 = 3
Pattern Recognition

Equating complex parts reduces single complex equations into two simultaneous linear equations. Treat |z+1| strictly as a scalar magnitude and parse directly into algebraic components.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

More Complex Numbers Questions — jee_main_2025_29_jan_evening

Practice all Complex Numbers previous-year questions →

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