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Class 11 Maths NCERT Solutions

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Complex Numbers and Quadratic Equations Class 11 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 4, Complex Numbers and Quadratic Equations — 14 questions from Ex 4.1, each worked through step by step in the CBSE marking pattern. Algebra of complex numbers, modulus and argument, the polar form, and the quadratic formula over the complex plane.

Class:11Subject:MathsChapter:4
4 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 11 Maths Chapter 4?

Chapter 4 carries 1 exercise question, numbered Ex 4.1. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

Complex numbers extend the real number line to the complex plane: a complex number is z = a + ib with i² = −1. This chapter practises the four operations, powers of i, the multiplicative inverse, modulus, conjugate and polar form, and then uses complex numbers to solve quadratic equations with negative discriminants.

Board pattern

The shortcut i⁴ = 1 collapses every power of i to a remainder mod 4. The multiplicative inverse of a + ib is (a − ib)/(a² + b²) — always rationalise the denominator. For quadratic equations, compute D = b² − 4ac first; if D < 0 write √−D as (√|D|)i and finish with the quadratic formula.
02

Exercise 4.1 — Operations on Complex Numbers

14Exercise questions

Step-by-step solution

  1. 1(5i)(−3/5 i) = 5 × (−3/5) × i × i = −3i².
  2. 2i² = −1, so −3i² = 3.
  3. 3In a + ib form: 3 + 0i.

Final answer

3 + 0i.

Step-by-step solution

  1. 1Use i⁴ = 1: i⁹ = i⁴·² · i = i, and i¹⁹ = i⁴·⁴ · i³ = i³ = −i.
  2. 2i⁹ + i¹⁹ = i + (−i) = 0.
  3. 3In a + ib form: 0 + 0i.

Final answer

0 + 0i.

Step-by-step solution

  1. 139 = 4·9 + 3, so i³⁹ = i³ = −i.
  2. 2i⁻³⁹ = 1/(−i). Multiply numerator and denominator by i: 1/(−i) × i/i = i/(−i²) = i.
  3. 3In a + ib form: 0 + i.

Final answer

0 + i.

Step-by-step solution

  1. 1Expand: 3(7 + 7i) = 21 + 21i, and i(7 + 7i) = 7i + 7i².
  2. 2Sum = 21 + 21i + 7i + 7i² = 21 + 28i + 7(−1).
  3. 3= 14 + 28i.

Final answer

14 + 28i.

Step-by-step solution

  1. 1Subtract term by term: (1 − i) + (1 − 6i).
  2. 2= 1 + 1 − i − 6i = 2 − 7i.

Final answer

2 − 7i.

Step-by-step solution

  1. 1Separate real and imaginary parts: (1/5 − 4) + i(2/5 − 5/2).
  2. 2Real: 1/5 − 4 = (1 − 20)/5 = −19/5.
  3. 3Imaginary: 2/5 − 5/2 = (4 − 25)/10 = −21/10.
  4. 4Form: −19/5 + (−21/10)i.

Final answer

−19/5 − 21/10 i.

Step-by-step solution

  1. 1Add the first two brackets: (1/3 + 4) + i(7/3 + 1/3) = 13/3 + 8/3 i.
  2. 2Subtract (−4/3 + i): 13/3 + 8/3 i + 4/3 − i.
  3. 3= (13/3 + 4/3) + i(8/3 − 1) = 17/3 + 5/3 i.

Final answer

17/3 + 5/3 i.

Step-by-step solution

  1. 1(1 − i)² = 1 − 2i + i² = −2i.
  2. 2(1 − i)⁴ = (−2i)² = 4i² = −4.
  3. 3Form: −4 + 0i.

Final answer

−4 + 0i.

Step-by-step solution

  1. 1Expand (a + b)³ with a = 1/3, b = 3i.
  2. 2a³ = 1/27; 3a²b = 3(1/9)(3i) = i; 3ab² = 3(1/3)(9i²) = 9i² = −9; b³ = 27i³ = −27i.
  3. 3Sum: 1/27 − 9 + i − 27i = (1 − 243)/27 − 26i.
  4. 4= −242/27 − 26i.

Final answer

−242/27 − 26i.

Step-by-step solution

  1. 1Expand (a + b)³ with a = −2, b = −i/3.
  2. 2a³ = −8; 3a²b = 3(4)(−i/3) = −4i; 3ab² = 3(−2)(i²/9) = −6/9 (−1) = 2/3; b³ = −i³/27 = i/27.
  3. 3Sum: −8 + 2/3 − 4i + i/27 = −22/3 + i(−4 + 1/27) = −22/3 − 107/27 i.

Final answer

−22/3 − 107/27 i.

Step-by-step solution

  1. 1Inverse = 1/(4 − 3i). Rationalise: multiply by (4 + 3i)/(4 + 3i).
  2. 21/(4 − 3i) × (4 + 3i)/(4 + 3i) = (4 + 3i)/(16 + 9).
  3. 3= (4 + 3i)/25 = 4/25 + 3/25 i.
  4. 4Powers-of-i check: modulus² of 4−3i is 25 ⟶ inverse modulus 1/5. Correct.

Final answer

4/25 + 3/25 i.

Step-by-step solution

  1. 1Inverse = 1/(√5 + 3i). Rationalise with (√5 − 3i).
  2. 2= (√5 − 3i)/(5 + 9) = (√5 − 3i)/14.
  3. 3= √5/14 − 3/14 i.

Final answer

√5/14 − 3/14 i.

Step-by-step solution

  1. 1Inverse = 1/(−i). Multiply numerator and denominator by i: 1/(−i) × i/i = i/(−i²) = i/1 = i.
  2. 2Check: (−i)(i) = −i² = 1. ✓

Final answer

i.

Step-by-step solution

  1. 1Denominator: (√3 + √2 i) − (√3 − √2 i) = 2√2 i.
  2. 2Numerator: (3 + i√5)(3 − i√5) = 3² − (i√5)² = 9 + 5 = 14.
  3. 3Expression = 14/(2√2 i) = 7/(√2 i).
  4. 4Rationalise: 7/(√2 i) × i/i = 7i/(√2 i²) = −7i/√2.
  5. 5Form: 0 + (−7/√2)i.

Final answer

0 − (7/√2)i.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Modulus

Conjugate product

Division

Polar form

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • To divide, multiply numerator and denominator by the conjugate of the denominator — that is what turns the denominator into a real number.
  • The discriminant decides the nature of the roots: negative gives a conjugate pair a ± bi, never a repeated root.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Maths Chapter 4 (Complex Numbers and Quadratic Equations)?

There are 1 exercise question in this chapter, numbered Ex 4.1. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Complex Numbers and Quadratic Equations Class 11 Maths?

The formulas this chapter's questions actually turn on are: Modulus, Conjugate product, Division, Polar form. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Complex Numbers and Quadratic Equations important for JEE Main?

Moderate — mostly mechanical manipulation worth secure marks, and the modulus-argument and quadratic questions appear in boards and in the early JEE Main paper.

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Previous: Trigonometric FunctionsNext: Linear InequalitiesComplex Numbers and Quadratic Equations revision notesAll Class 11 Maths NCERT Solutions

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