Complex Numbers — Previous Year Questions (Class 11 Mathematics)
Complex numbers extend real numbers to solve equations with no real solutions. Master operations, moduli, and arguments.
TL;DR: Complex numbers extend real numbers to solve equations with no real solutions. Master operations, moduli, and arguments.
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Complex numbers extend real numbers to solve equations with no real solutions. Master operations, moduli, and arguments.
Complex Numbers — Previous Year Questions with Solutions
Q (2023, 2 marks): Simplify (3 + 4i)/(2 - i) and express in the form a + bi.
Answer: (3 + 4i)/(2 - i)
Multiply numerator and denominator by the conjugate of denominator (2 + i):
= [(3 + 4i)(2 + i)]/[(2 - i)(2 + i)]
Numerator: (3 + 4i)(2 + i) = 3(2) + 3(i) + 4i(2) + 4i(i)
= 6 + 3i + 8i + 4i²
= 6 + 11i + 4(-1)
= 6 + 11i - 4
= 2 + 11i
Denominator: (2 - i)(2 + i) = 2² - i² = 4 - (-1) = 4 + 1 = 5
Result: (2 + 11i)/5 = 2/5 + 11i/5
Final Answer: 2/5 + 11i/5
Q (2022, 2 marks): If z = 1 + i, find z² and z³.
Answer: z = 1 + i
z² = (1 + i)² = 1² + 2(1)(i) + i² = 1 + 2i + (-1) = 2i
z³ = z² × z = 2i × (1 + i) = 2i + 2i² = 2i + 2(-1) = 2i - 2 = -2 + 2i
Final Answer: z² = 2i, z³ = -2 + 2i
Q (2023, 3 marks): Find the modulus and argument of the complex number z = -1 + i√3.
Answer: z = -1 + i√3
Modulus: |z| = √[(-1)² + (√3)²] = √[1 + 3] = √4 = 2
Argument: z lies in second quadrant (real part negative, imaginary part positive)
tan(θ) = (√3)/(-1) = -√3
In standard position, arg(z) = π - π/3 = 2π/3 (or 120°)
Alternatively, arg(z) = π - arctan(√3/1) = π - π/3 = 2π/3
Final Answer: |z| = 2, arg(z) = 2π/3
Q (2021, 3 marks): Solve the equation z² - 2z + 2 = 0.
Answer: Using quadratic formula: z = [-b ± √(b² - 4ac)]/(2a)
with a = 1, b = -2, c = 2
z = [2 ± √(4 - 8)]/2
z = [2 ± √(-4)]/2
z = [2 ± 2i]/2
z = 1 ± i
Solutions: z₁ = 1 + i, z₂ = 1 - i
Final Answer: z = 1 + i or 1 - i
Q (2022, 3 marks): If z₁ = 3 + 4i and z₂ = 1 - 2i, find z₁ + z₂, z₁ - z₂, and z₁ × z₂.
Answer: z₁ = 3 + 4i, z₂ = 1 - 2i
z₁ + z₂ = (3 + 4i) + (1 - 2i) = 4 + 2i
z₁ - z₂ = (3 + 4i) - (1 - 2i) = 3 + 4i - 1 + 2i = 2 + 6i
z₁ × z₂ = (3 + 4i)(1 - 2i)
= 3(1) + 3(-2i) + 4i(1) + 4i(-2i)
= 3 - 6i + 4i - 8i²
= 3 - 2i - 8(-1)
= 3 - 2i + 8
= 11 - 2i
Final Answer: z₁ + z₂ = 4 + 2i; z₁ - z₂ = 2 + 6i; z₁ × z₂ = 11 - 2i
Q (2023, 3 marks): Prove that the product of a complex number and its conjugate equals the square of its modulus: z × z̄ = |z|².
Answer: Let z = a + bi, where a, b are real.
Conjugate: z̄ = a - bi
Product: z × z̄ = (a + bi)(a - bi)
= a² - abi + abi - b²i²
= a² - b²i²
= a² - b²(-1)
= a² + b²
Modulus: |z| = √(a² + b²)
|z|² = (√(a² + b²))² = a² + b²
Therefore: z × z̄ = |z|²
Proof complete.
Final Answer: Proved that z × z̄ = |z|² = a² + b²
Frequently Asked Questions
What is the geometric interpretation of complex numbers?
Complex numbers can be represented as points in the complex plane (Argand diagram): the real part is the x-coordinate and the imaginary part is the y-coordinate. Complex number z = a + bi corresponds to point (a, b). Addition corresponds to vector addition; multiplication involves rotation and scaling. The modulus |z| is the distance from origin; the argument is the angle from positive real axis.
What is Euler's formula and why is it important?
Euler's formula: e^(iθ) = cos(θ) + i·sin(θ). This connects exponential and trigonometric functions. Complex number z = r(cos(θ) + i·sin(θ)) can be written as z = re^(iθ), making calculations with complex numbers easier, especially powers and roots. It's fundamental in electrical engineering, signal processing, and quantum mechanics.
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