JEE Main · Mathematics ↓ Falling

Complex Numbers appeared 41 times across 3 years — 4.7% of Mathematics. This question is from Properties of Complex Numbers.

Year 2026 2025 2024 Total
Questions 11 16 14 41

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.

Solution & Explanation

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

Reference Study Guides

More Complex Numbers Previous-Year Questions — Page 9

Q26 jee_main_2024_31_jan_morning Properties of Modulus and Argument
If α denotes the number of solutions of |1 - i|^x = 2^x and β = ((|z|)/( (z))), where z = (π)/(4) (1 + i)⁴ ( 1 - √(π) i√(π) + i + √(π) - i1 + √(π) i), i = √(-1), then the distance of the point (α, β) from the line 4x - 3y = 7 is
Numerical Answer. Answer: 3 to 3

Solution

Core Logic
|1 - i|^x = 2^x (√(2))^x = 2^x 2x/2 = 2^x

This implies (x)/(2) = x x = 0. There is exactly 1 solution, so α = 1.

Step 1: Simplify complex number z
(1+i)⁴ = ((1+i)²)² = (1 + i² + 2i)² = (2i)² = -4

Thus, z = -π ( (1-√(π)i)(√(π)-i)π + 1 + (√(π)-i)(1-√(π)i)1 + π )

Step 2: Simplify Bracket

Let's expand the terms directly: z = (π)/(4)(-4) [ √(π) - π i - i - √(π)π + 1 + √(π) - i - π i - √(π)1 + π ]

= -π [ (-i(π+1))/(π+1) + (-i(π+1))/(π+1) ] = -π [ -i - i ] = 2π i
Step 3: Find beta

For z = 2π i: |z| = 2π and (z) = (π)/(2).

β = (|z|)/( (z)) = (2π)/(π/2) = 4
Step 4: Distance from Line

Distance of point (α, β) = (1, 4) from the line 4x - 3y - 7 = 0:

D = |4(1) - 3(4) - 7|√(4² + (-3)²) = (|4 - 12 - 7|)/(5) = (|-15|)/(5) = 3
Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines

More Complex Numbers Questions — jee_main_2025_04_april_evening

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