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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 5

Q55 jee_main_2025_04_april_evening Properties of Complex Numbers
Let the product of ω₁ = (8 + i) θ + (7 + 4 i) θ and ω₂ = (1 + 8 i) θ + (4 + 7 i) θ be α +iβ, i = √(-1). Let p and q be the maximum and the minimum values of α +β respectively.
  • A. 140
  • B. 130
  • C. 160
  • D. 150

Solution

Core Logic

Let's expand the terms by grouping real and imaginary parts explicitly:

ω₁ = (8 θ + 7 θ) + i( θ + 4 θ) ω₂ = ( θ + 4 θ) + i(8 θ + 7 θ)

Notice that if we let u = 8 θ + 7 θ and v = θ + 4 θ, then:

ω₁ = u + iv and ω₂ = v + iu
Step 1: Calculating the Product

Multiplying ω₁ and \omega_2:

ω₁ω₂ = (u + iv)(v + iu) = uv + iu² + iv² - uv = i(u² + v²)

Since the product is given as α + iβ:

α = 0

β = u² + v² = (8 θ + 7 θ)² + ( θ + 4 θ)²
Step 2: Simplifying the expression for alpha + beta

Expanding the terms for β:

β = (64 ²θ + 49 ²θ + 112 θ θ) + ( ²θ + 16 ²θ + 8 θ θ) α + β = 0 + β = 65 ²θ + 65 ²θ + 120 θ θ

Using the identity ²θ + ²θ = 1 and 2 θ θ = 2θ:

α + β = 65 + 60 2θ
Step 3: Max and Min Extrema Analysis

Since -1 ≤ 2θ ≤ 1:

Maximum value p = 65 + 60(1) = 125 Minimum value q = 65 + 60(-1) = 5

Sum of maximum and minimum bounds equals:

p + q = 125 + 5 = 130
Pattern Recognition

Observe the symmetric structure in complex variables: (u+iv) and (v+iu). Their product structurally completely cancels out the real component, saving you from a highly messy component expansion.

Chapter Mix

Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Trigonometric Functions

Q71 jee_main_2025_04_april_evening Cube Roots of Unity
If α is a root of the equation x² + x + 1 = 0 and Σk=1ⁿ(αk + 1αk)² = 20, then n is equal to
Numerical Answer. Answer: 11 to 11

Solution

Core Logic

The equation x² + x + 1 = 0 has complex roots which are the non-real cube roots of unity. Thus, we can set α = ω (where ω³ = 1 and 1 + ω + ω² = 0).

Let's analyze the general term block Tk = (ω^k + (1)/(ω^k))²:

Tk = (ω^k + ω-k)² = ω2k + ω-2k + 2 = ω2k + ω^k + 2

Because ω^k is periodic with period 3, let's examine the values of Tk for different values of k:

  • If k is a multiple of 3 (k=3m): ω2k = 1, ω^k = 1 Tk = 1 + 1 + 2 = 4.
  • If k is not a multiple of 3 (k=3m+1 or 3m+2): ω2k + ω^k = -1 Tk = -1 + 2 = 1.
Step 1: Evaluating periodic blocks

Every block of three consecutive terms (k = 1, 2, 3) contributes exactly:

Sum of a block = 1 + 1 + 4 = 6

We want the total summation to equal 20. Let's divide 20 by our block value 6:

20 = 3 × 6 + 2

This means the sum must consist of 3 full periodic blocks plus additional terms that add up to 2.

Step 2: Determining the final term count n

The number of terms in 3 full blocks is 3 × 3 = 9 terms, giving a sum of 18. To get the remaining value of 2, we look at the next terms:

  • Term 10 (k=10, not a multiple of 3) adds 1 Total = 18 + 1 = 19.
  • Term 11 (k=11, not a multiple of 3) adds 1 Total = 19 + 1 = 20.
  • Hence, the series terminates exactly at n = 11.

Pattern Recognition

Whenever complex roots of unity or cyclic properties show up inside series sums, group terms into blocks based on the underlying period length (3 here) to convert large sums into simple modular arithmetic arithmetic calculations.

Chapter Mix

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

Q73 jee_main_2025_04_april_morning Geometry of Complex Numbers
Let A = z in C : |z - 2 - i| = 3, B = z in C : Re(z - iz) = 2 and S = A B. Then Σz in S |z|² is equal to
Numerical Answer. Answer: 22 to 22

Solution

Related Formula

Magnitude squared representation:

|z|² = x² + y² for z = x + iy
Core Logic

Convert complex sets into Cartesian forms by setting z = x + iy: Set A: |(x-2) + i(y-1)| = 3 (x-2)² + (y-1)² = 9 (1) Set B: z - iz = (x+iy) - i(x+iy) = (x+y) + i(y-x). Re(z - iz) = 2 x + y = 2 y = 2 - x (2)

Step 1: Solve System Algebraically

Substitute (2) into (1):

(x - 2)² + (2 - x - 1)² = 9 (x - 2)² + (1 - x)² = 9 x² - 4x + 4 + 1 - 2x + x² = 9 2x² - 6x - 4 = 0 x² - 3x - 2 = 0

Roots are x1,2 = 3 ± √(17)2. Correspondingly, y = 2 - x y1,2 = 1 ∓ √(17)2.

Step 2: Evaluate Sum of Square Magnitudes

Since S consists of the two intersection points z₁, z₂:

Σz in S |z|² = (x₁² + y₁²) + (x₂² + y₂²) = (x₁² + x₂²) + (y₁² + y₂²)

Using identities from quadratic equation x² - 3x - 2 = 0 (x₁+x₂ = 3, x₁x₂ = -2): x₁² + x₂² = (3)² - 2(-2) = 13. Since y = 2-x, y² = 4 - 4x + x² y₁² + y₂² = 8 - 4(3) + 13 = 9.

Σz in S |z|² = 13 + 9 = 22
Pattern Recognition

Avoid explicitly using radical root approximations. Summing symmetric expressions directly through standard Vieta coefficient sum shortcuts preserves clean fractions.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

Q61 jee_main_2025_07_april_evening Locus of a Complex Number
If the locus of z in C, such that R e ( z - 12 z + i) + R e ( z - 12 z - i) = 2, is a circle of radius r and center (a, b) then (15ab)/(r²) is equal to:
  • A. 24
  • B. 12
  • C. 18
  • D. 16

Solution

Related Formula

For a complex number w, Re(w) = Re( w). Hence:

Re(w) + Re( w) = 2Re(w)
Core Logic

Notice that z - 12 z - i is the exact complex conjugate of (z - 1)/(2z + i). Thus, the given equation simplifies directly via complex identities to:

2Re((z - 1)/(2z + i)) = 2 Re((z - 1)/(2z + i)) = 1
Step 1: Substitute z = x + iy

Let z = x + iy:

((x - 1) + iy)/(2x + i(2y + 1))

To find the real part, multiply numerator and denominator by the conjugate of the denominator:

Re[ (((x - 1) + iy)(2x - i(2y + 1)))/(4x² + (2y + 1)²) ] = 1 (2x(x - 1) + y(2y + 1))/(4x² + (2y + 1)²) = 1
Step 2: Expand and Arrange Circle Equation

Expanding the expression:

2x² - 2x + 2y² + y = 4x² + 4y² + 4y + 1 2x² + 2y² + 2x + 3y + 1 = 0

Dividing full equation by 2:

x² + y² + x + (3)/(2)y + (1)/(2) = 0
Step 3: Extract Center and Radius
Center (a, b) = (-(1)/(2), -(3)/(4)) r² = g² + f² - c = ((1)/(2))² + ((3)/(4))² - (1)/(2) = (1)/(4) + (9)/(16) - (1)/(2) = (5)/(16)

Evaluating (15ab)/(r²):

(15 · (-(1)/(2)) · (-(3)/(4)))/((5)/(16)) = ((45)/(8))/((5)/(16)) = 18
Pattern Recognition

Recognizing that Re(w) + Re( w) = 2Re(w) avoids complex algebraic division on the second fractional expression completely.

Chapter Mix

Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Circles

Q56 jee_main_2025_24_jan_morning Algebraic Properties of Complex Roots
If α and β are the roots of the equation 2z² - 3z - 2i = 0 , where i = √(-1) , then 16 · Re( α¹⁹ + β¹⁹ + α¹¹ + β¹¹α¹⁵ + β¹⁵) · Im( α¹⁹ + β¹⁹ + α¹¹ + β¹¹α¹⁵ + β¹⁵) is equal to :
  • A. 398
  • B. 312
  • C. 409
  • D. 441

Solution

Related Formula

Since α and β are roots of 2z² - 3z - 2i = 0, they satisfy the quadratic equation directly, meaning:

2α² - 3α - 2i = 0 2(α - (i)/(α)) = 3 α - (i)/(α) = (3)/(2)

Similarly for β:

β - (i)/(β) = (3)/(2)
Core Logic

Square the baseline relation to transition to higher exponential powers:

(α - (i)/(α))² = ((3)/(2))² α² - (1)/(α²) - 2i = (9)/(4) α² - (1)/(α²) = (9)/(4) + 2i

Squaring once more to isolate the fourth powers:

(α² - (1)/(α²))² = ((9)/(4) + 2i)² α⁴ + (1)/(α⁴) - 2 = (81)/(16) - 4 + 9i α⁴ + (1)/(α⁴) = (49)/(16) + 9i
Step 1: Simplify the Target Expression Fraction

Rearrange the given complex algebraic fraction by factoring out powers:

α¹⁹ + α¹¹ + β¹⁹ + β¹¹α¹⁵ + β¹⁵ = α¹⁵(α⁴ + (1)/(α⁴)) + β¹⁵(β⁴ + (1)/(β⁴))α¹⁵ + β¹⁵

Since both α and β satisfy the exact same symmetric relational identity:

α⁴ + (1)/(α⁴) = β⁴ + (1)/(β⁴) = (49)/(16) + 9i

Substitute this uniform value back into the algebraic expression:

= (α¹⁵ + β¹⁵)((49)/(16) + 9i)α¹⁵ + β¹⁵ = (49)/(16) + 9i
Step 2: Extract Real and Imaginary Components

From our simplified expression:

Re = (49)/(16) Im = 9

Now, substitute these into the evaluation formula:

Result = 16 · ((49)/(16)) · 9 = 49 · 9 = 441
Pattern Recognition

Symmetric rational polynomials in roots α, β that can be split into identical numeric multipliers for αⁿ and βⁿ allow direct cancellation of the polynomial bases without evaluating the individual roots explicitly.

Chapter Mix

Class 11 Mathematics: Complex Numbers

More Complex Numbers Questions — jee_main_2025_29_jan_evening

Practice all Complex Numbers previous-year questions →

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