JEE Main · Mathematics ↓ Falling

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 3

Q18 jee_main_2026_28_january_evening Geometry of Complex Regions
Let A = z in C : |z - 2| ≤ 4 and B = z in C : |z - 2| + |z + 2| = 5. Then the max |z₁ - z₂| : z₁ in A and z₂ in B is
  • A. (15)/(2)
  • B. 8
  • C. (17)/(2)
  • D. 9

Solution

Core Logic

Analyze region A: |z - 2| ≤ 4 represents a solid disk with center (2, 0) and radius R=4. Its rightmost point on the real axis is (6, 0) and leftmost is (-2, 0).

Analyze locus B: |z - 2| + |z + 2| = 5 represents an ellipse with foci at (2, 0) and (-2, 0), and major axis length 2a = 5 ⇒ a = (5)/(2). The equation of the ellipse is (x²)/((5/2)²) + (y²)/(b²) = 1. Its leftmost vertex is at (-(5)/(2), 0).

Complex geometry loci map
Complex geometry loci map

Execution

To maximize |z₁ - z₂|, we need the maximum geometric distance between any point in the disk A and any point on the ellipse B. By visualizing the placement on the coordinate plane: The rightmost point of the circle A is z₁ = 6. The leftmost point of the ellipse B is z₂ = -(5)/(2). The distance between them is the maximum horizontal span since both shapes are symmetric around the real axis and their extremes occur on the real axis.

|z₁ - z₂| = 6 - (-(5)/(2)) = 6 + (5)/(2) = (17)/(2)
Pattern Recognition

Maximum distance problems between loci in the complex plane almost always resolve to finding the extreme diametric points along their shared axis of symmetry (usually the real axis).

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Conic Sections

Q75 jee_main_2025_02_april_evening Quadratic Equations
If the set of all a in R - 1, for which the roots of the equation (1 - a)x² + 2(a - 3)x + 9 = 0 are positive is (-∞, -α] [β, γ), then 2α + β + γ is equal to ____________.
Numerical Answer. Answer: 7 to 7

Solution

Related Formula
For a quadratic equation A x² + B x + C = 0 to have two positive real roots: 1. Real roots: D = B² - 4AC ≥ 0 2. Sum of roots: -(B)/(A) > 0 3. Product of roots: (C)/(A) > 0
Core Logic

We write down three systems of inequalities based on real and positive root conditions, find their intersection, and map the boundaries to solve for the parameters.

Step 1: Apply the discriminant condition (Real roots)

For real roots, the discriminant D ≥ 0:

D = [ 2(a - 3) ]² - 4(1 - a)(9) ≥ 0 4(a² - 6a + 9) - 36(1 - a) ≥ 0 (a² - 6a + 9) - 9(1 - a) ≥ 0 a² - 6a + 9 - 9 + 9 a ≥ 0 a² + 3a ≥ 0 a(a + 3) ≥ 0

Thus, the interval is:

a in (-∞, -3] [0, ∞) --- (1)
Step 2: Apply the sum of roots condition (Positive sum)

For positive roots, the sum of roots must be positive:

-(B)/(A) = (-2(a - 3))/(1 - a) = (2(a - 3))/(a - 1) > 0

Using the wavy curve method for (a-3)/(a-1) > 0:

a in (-∞, 1) (3, ∞) --- (2)
Step 3: Apply the product of roots condition (Positive product)

For positive roots, the product of roots must be positive:

(C)/(A) = (9)/(1 - a) > 0 1 - a > 0 a < 1

Thus, the interval is:

a in (-∞, 1) --- (3)
Step 4: Find the intersection of all conditions

Intersecting equations (1), (2), and (3):

  • First, intersect (2) and (3):
( (-∞, 1) (3, ∞) ) (-∞, 1) = (-∞, 1)
  • Next, intersect with (1):
( (-∞, -3] [0, ∞) ) (-∞, 1) = (-∞, -3] [0, 1)

Comparing this with (-∞, -α] [β, γ):

  • α = 3
  • β = 0
  • γ = 1
  • Now calculate the target sum:

2α + β + γ = 2(3) + 0 + 1 = 7
Pattern Recognition

Location of roots: When both roots are positive, checking sum and product signs along with D ≥ 0 is the standard and fastest set of inequalities, avoiding complex vertex projections.

Chapter Mix

Class 11 Chemistry: Practical Chemistry

Q jee_main_2025_02_april_morning Geometry of Complex Numbers
Let z be a complex number such that |z| = 1. If 2 + k²zk + z = kz, k in R, then the maximum distance of k + ik² from the circle |z - (1 + 2i)| = 1 is:
  • A. √(5) + 1
  • B. 2
  • C. 3
  • D. √(3) + 1

Solution

Related Formula

For a complex number lying on the unit circle:

|z| = 1 z z = 1 z = (1)/(z)

Maximum distance from a point P to a circle with center C and radius r is:

d = PC + r
Core Logic

Simplify the algebraic condition using z = 1/z to uniquely determine the value of the parameter k, then compute geometric distances.

Step 1: Solve for k

Cross-multiply the given expression:

2 + k²z = kz(k + z) = k²z + kz z

Since z z = |z|² = 1:

2 + k²z = k²z + k(1) k = 2
Step 2: Locate Point and Circle Parameters

Substitute k=2 into the target point expression P = k + ik²:

P = 2 + 4i ≡ (2,4)

The circle equation is |z - (1 + 2i)| = 1, which represents a circle centered at C = (1, 2) with radius r = 1.

Step 3: Compute Geometric Distances

Find the Euclidean distance between P(2,4) and center C(1,2):

PC = √((2-1)² + (4-2)²) = √(1 + 4) = √(5)

The maximum distance from the point to the circle boundary is:

d = PC + r = √(5) + 1
Pattern Recognition

Notice how k²z cancels perfectly on both sides during expansion due to the unique property of uni-modular complex numbers (z z=1), rendering the calculation of k trivial.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

Q jee_main_2025_02_april_morning Theory of Equations
Let Pₙ = αⁿ + βⁿ, n in N. If P₁₀ = 123, P₉ = 76, P₈ = 47 and P₁ = 1, then the quadratic equation having roots (1)/(α) and (1)/(β) is:
  • A. x² - x + 1 = 0
  • B. x² + x - 1 = 0
  • C. x² - x - 1 = 0
  • D. x² + x + 1 = 0

Solution

Related Formula

Newton's Sums for the roots of a quadratic equation ax² + bx + c = 0:

a Pₙ + b Pₙ₋₁ + c Pₙ₋₂ = 0
Core Logic

Observe the recurrence relation from the given numerical values of Pₙ to construct the base quadratic equation satisfied by α and β, then invert the roots.

Step 1: Identify the Linear Recurrence Relation

Compare the provided sequence values:

P₈ + P₉ = 47 + 76 = 123 = P₁₀

This fits the general sequence relation:

Pₙ = Pₙ₋₁ + Pₙ₋₂ Pₙ - Pₙ₋₁ - Pₙ₋₂ = 0
Step 2: Construct the Base Quadratic Equation

The characteristic equation corresponding to this recurrence relation is:

x² - x - 1 = 0

Thus, α and \(\beta\) are roots of x² - x - 1 = 0, giving sum α+β = 1 and product αβ = -1 (which matches P₁ = α+β = 1).

Step 3: Construct Equation with Reciprocal Roots

To find the equation with roots (1)/(α) and (1)/(β), apply the transformations:

Sum of new roots = (1)/(α) + (1)/(β) = (α + β)/(αβ) = (1)/(-1) = -1 Product of new roots = (1)/(αβ) = (1)/(-1) = -1

The new quadratic equation is:

x² - (Sum)x + (Product) = 0 x² - (-1)x + (-1) = 0 x² + x - 1 = 0
Pattern Recognition

The recurrence pattern Pₙ = Pₙ₋₁ + Pₙ₋₂ is the Fibonacci sequence recurrence line. Its roots generate the Golden Ratio layout from x²-x-1=0. Inverting roots swaps the coefficients of x² and the constant term, yielding x²+x-1=0 instantly.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations Class 11 Mathematics: Sequences and Series

Q65 jee_main_2025_03_april_evening Geometry of Complex Numbers
If z₁, z₂, z₃ in C are the vertices of an equilateral triangle, whose centroid is z₀, then Σk=1³ (zk - z₀)² is equal to
  • A. 0
  • B. 1
  • C. i
  • D. -i

Solution

Related Formula

For any equilateral triangle with vertices z₁, z₂, z₃:

z₁² + z₂² + z₃² = z₁ z₂ + z₂ z₃ + z₃ z₁ --- (1)

Centroid equation:

z₀ = (z₁ + z₂ + z₃)/(3) z₁ + z₂ + z₃ = 3z₀
Core Logic

Let's expand the target sum:

Σk=1³ (zk - z₀)² = (z₁ - z₀)² + (z₂ - z₀)² + (z₃ - z₀)²
Step 1: Expansion and Algebraic Grouping

Expanding each quadratic term:

= (z₁² + z₂² + z₃²) - 2z₀(z₁ + z₂ + z₃) + 3z₀²

Substitute z₁ + z₂ + z₃ = 3z₀:

= (z₁² + z₂² + z₃²) - 2z₀(3z₀) + 3z₀² = z₁² + z₂² + z₃² - 3z₀²
Step 2: Resolving using Equilateral Condition

Substitute 3z₀² = 3 ((z₁+z₂+z₃)/(3))² = ((z₁+z₂+z₃)²)/(3):

= (z₁² + z₂² + z₃²) - (z₁² + z₂² + z₃² + 2(z₁z₂ + z₂z₃ + z₃z₁))/(3) = (2(z₁² + z₂² + z₃²) - 2(z₁z₂ + z₂z₃ + z₃z₁))/(3) = (2)/(3)(z₁² + z₂² + z₃² - (z₁ z₂ + z₂ z₃ + z₃ z₁))

Using condition (1) for equilateral triangles, the terms inside the parentheses equal 0. Thus: = 0

Pattern Recognition

This is a standard invariant of equilateral triangles. Any translation to center of mass coordinates leaves the shape invariant, making the sum of squares of coordinate vectors relative to the centroid equal to zero.

Chapter Mix

Class 11 Mathematics: Complex Numbers

More Complex Numbers Questions — jee_main_2025_29_jan_evening

Practice all Complex Numbers previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)