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Complex Numbers and Quadratic Equations

Meghalaya Board · Class 11 · Mathematics

Flashcards for Complex Numbers and Quadratic Equations — Meghalaya Board Class 11 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.

101 questions20 flashcards5 concepts

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20 Flashcards
Card 1Introduction to Complex Numbers

Solve: x² + 1 = 0

Answer

Step 1: Rearrange to standard form → x² = -1 Step 2: Take square root of both sides → x = ±√(-1) Step 3: Since √(-1) = i → x = ±i Answer: x = i or x = -i Note: This shows why we need complex numbers -

Card 2Addition of Complex Numbers

Add the complex numbers: (3 + 4i) + (2 - 7i)

Answer

Step 1: Group real parts and imaginary parts separately Real parts: 3 + 2 = 5 Imaginary parts: 4i + (-7i) = -3i Step 2: Combine results Answer: 5 - 3i Rule: (a + bi) + (c + di) = (a + c) + (b + d)i

Card 3Multiplication of Complex Numbers

Multiply: (2 + 3i)(4 - 5i)

Answer

Step 1: Use FOIL method = 2(4) + 2(-5i) + 3i(4) + 3i(-5i) = 8 - 10i + 12i - 15i² Step 2: Remember i² = -1 = 8 - 10i + 12i - 15(-1) = 8 - 10i + 12i + 15 Step 3: Combine like terms = (8 + 15) + (-10 + 1

Card 4Conjugate and Modulus

Find the conjugate and modulus of z = 5 - 12i

Answer

Conjugate z̄: Step 1: Change sign of imaginary part z̄ = 5 - (-12i) = 5 + 12i Modulus |z|: Step 2: Use formula |z| = √(a² + b²) |z| = √(5² + (-12)²) = √(25 + 144) = √169 = 13 Answers: z̄ = 5 + 12i,

Card 5Powers of i

Calculate i⁴⁷

Answer

Step 1: Use the pattern of powers of i i¹ = i, i² = -1, i³ = -i, i⁴ = 1 Pattern repeats every 4 powers Step 2: Divide exponent by 4 47 ÷ 4 = 11 remainder 3 So i⁴⁷ = i³ Step 3: From the pattern i³ =

Card 6Division of Complex Numbers

Divide: (6 + 8i) ÷ (3 - 4i)

Answer

Step 1: Multiply numerator and denominator by conjugate of denominator (6 + 8i)/(3 - 4i) × (3 + 4i)/(3 + 4i) Step 2: Multiply numerator (6 + 8i)(3 + 4i) = 18 + 24i + 24i + 32i² = 18 + 48i - 32 = -14

Card 7Square Roots of Negative Numbers

Find √(-16)

Answer

Step 1: Factor out the negative sign √(-16) = √(-1 × 16) = √(-1) × √(16) Step 2: Use definition √(-1) = i = i × √(16) = i × 4 Step 3: Simplify = 4i Answer: √(-16) = 4i General rule: √(-a) = i√(a)

Card 8Quadratic Equations with Complex Solutions

Solve: x² - 4x + 5 = 0 using quadratic formula

Answer

Step 1: Identify coefficients a = 1, b = -4, c = 5 Step 2: Calculate discriminant D = b² - 4ac = (-4)² - 4(1)(5) = 16 - 20 = -4 Step 3: Since D < 0, solutions are complex x = [-b ± √D]/2a = [4 ± √(-

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What are the important topics in Complex Numbers and Quadratic Equations for Meghalaya Board Class 11 Mathematics?
Complex Numbers and Quadratic Equations covers several key topics that are frequently asked in Meghalaya Board Class 11 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Complex Numbers and Quadratic Equations — Meghalaya Board Class 11 Mathematics?
Understand the core concepts first, then work through the 101 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
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There are 20 flashcards for Complex Numbers and Quadratic Equations covering key definitions, formulas, and concepts. Use them daily for 10–15 minutes for best results.

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