NCERT Solutions For Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

NCERT Solutions Class 11 Maths Chapter 5 – Free PDF Download

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 4.

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are prepared by the expert teachers at BYJU’S. These NCERT Solutions of Maths help students in solving problems quickly, accurately and efficiently. Also, BYJU’S provides step-by-step solutions for all NCERT problems, thereby ensuring students understand them and clear their board exams with flying colours.
The chapter Complex Numbers and Quadratic Equations is categorised under the CBSE Syllabus for 2023-24 and includes different critical Mathematical theorems and formulae. The NCERT textbook has many practice problems to cover all these concepts, which would help students easily understand higher concepts in future. BYJU’S provides solutions for all these problems with proper explanations. These NCERT Solutions from BYJU’S help students who aim to clear their exams even with last-minute preparations. However, NCERT Solutions for Class 11 Maths are focused on mastering the concepts along with gaining broader knowledge.

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

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Access the exercises of Maths NCERT Class 11 Chapter 5

Exercise 5.1 Solutions 14 Questions
Exercise 5.2 Solutions 8 Questions
Exercise 5.3 Solutions 10 Questions
Miscellaneous Exercise on Chapter 5 Solutions 20 Questions; the summarisation of the topics discussed in Chapter 5 of the Class 11 NCERT curriculum is listed below.

Access NCERT Solutions for Class 11 Maths Chapter 5

Exercise 5.1 Page No: 103

Express each of the complex numbers given in Exercises 1 to 10 in the form a + ib.

1. (5i) (-3/5i)

Solution:

(5i) (-3/5i) = 5 x (-3/5) x i²

= -3 x -1 [i² = -1]

= 3

Hence,

(5i) (-3/5i) = 3 + i0

2. i⁹ + i¹⁹

Solution:

i⁹ + i¹⁹ = (i²)⁴. i + (i²)⁹. i

= (-1)⁴ . i + (-1)⁹ .i

= 1 x i + -1 x i

= i – i

= 0

Hence,

i⁹ + i¹⁹ = 0 + i0

3. i^-39

Solution:

i^-39 = 1/ i³⁹ = 1/ i^{4 x 9 + 3} = 1/ (1⁹ x i³) = 1/ i³ = 1/ (-i) [i⁴ = 1, i³ = -I and i² = -1]

Now, multiplying the numerator and denominator by i we get

i^-39 = 1 x i / (-i x i)

= i/ 1 = i

Hence,

i^-39 = 0 + i

4. 3(7 + i7) + i(7 + i7)

Solution:

3(7 + i7) + i(7 + i7) = 21 + i21 + i7 + i² 7

= 21 + i28 – 7 [i² = -1]

= 14 + i28

Hence,

3(7 + i7) + i(7 + i7) = 14 + i28

5. (1 – i) – (–1 + i6)

Solution:

(1 – i) – (–1 + i6) = 1 – i + 1 – i6

= 2 – i7

Hence,

(1 – i) – (–1 + i6) = 2 – i7

6.

Solution:

7.

Solution:

8. (1 – i)⁴

Solution:

(1 – i)⁴ = [(1 – i)²]²

= [1 + i² – 2i]²

= [1 – 1 – 2i]² [i² = -1]

= (-2i)²

= 4(-1)

= -4

Hence, (1 – i)⁴ = -4 + 0i

9. (1/3 + 3i)³

Solution:

Hence, (1/3 + 3i)³ = -242/27 – 26i

10. (-2 – 1/3i)³

Solution:

Hence,

(-2 – 1/3i)³ = -22/3 – 107/27i

Find the multiplicative inverse of each of the complex numbers given in Exercises 11 to 13.

11. 4 – 3i

Solution:

Let’s consider z = 4 – 3i

Then,

= 4 + 3i and

|z|² = 4² + (-3)² = 16 + 9 = 25

Thus, the multiplicative inverse of 4 – 3i is given by z^-1

12. √5 + 3i

Solution:

Let’s consider z = √5 + 3i

|z|² = (√5)² + 3² = 5 + 9 = 14

Thus, the multiplicative inverse of √5 + 3i is given by z^-1

13. – i

Solution:

Let’s consider z = –i

Thus, the multiplicative inverse of –i is given by z^-1

14. Express the following expression in the form of a + ib:

Solution:

Exercise 5.2 Page No: 108

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

1. z = – 1 – i √3

Solution:

2. z = -√3 + i

Solution:

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

3. 1 – i

Solution:

4. – 1 + i

Solution:

5. – 1 – i

Solution:

6. – 3

Solution:

7. 3 + i

Solution:

8. i

Solution:

Exercise 5.3 Page No: 109

Solve each of the following equations:

1. x² + 3 = 0

Solution:

Given the quadratic equation,

x² + 3 = 0

On comparing it with ax² + bx + c = 0, we have

a = 1, b = 0, and c = 3

So, the discriminant of the given equation will be

D = b² – 4ac = 0² – 4 × 1 × 3 = –12

Hence, the required solutions are

2. 2x² + x + 1 = 0

Solution:

Given the quadratic equation,

2x² + x + 1 = 0

On comparing it with ax² + bx + c = 0, we have

a = 2, b = 1, and c = 1

So, the discriminant of the given equation will be

D = b² – 4ac = 1² – 4 × 2 × 1 = 1 – 8 = –7

Hence, the required solutions are

3. x² + 3x + 9 = 0

Solution:

Given the quadratic equation,

x² + 3x + 9 = 0

On comparing it with ax² + bx + c = 0, we have

a = 1, b = 3, and c = 9

So, the discriminant of the given equation will be

D = b² – 4ac = 3² – 4 × 1 × 9 = 9 – 36 = –27

Hence, the required solutions are

4. –x² + x – 2 = 0

Solution:

Given the quadratic equation,

–x² + x – 2 = 0

On comparing it with ax² + bx + c = 0, we have

a = –1, b = 1, and c = –2

So, the discriminant of the given equation will be

D = b² – 4ac = 1² – 4 × (–1) × (–2) = 1 – 8 = –7

Hence, the required solutions are

5. x² + 3x + 5 = 0

Solution:

Given the quadratic equation,

x² + 3x + 5 = 0

On comparing it with ax² + bx + c = 0, we have

a = 1, b = 3, and c = 5

So, the discriminant of the given equation will be

D = b² – 4ac = 3² – 4 × 1 × 5 =9 – 20 = –11

Hence, the required solutions are

6. x² – x + 2 = 0

Solution:

Given the quadratic equation,

x² – x + 2 = 0

On comparing it with ax² + bx + c = 0, we have

a = 1, b = –1, and c = 2

So, the discriminant of the given equation is

D = b² – 4ac = (–1)² – 4 × 1 × 2 = 1 – 8 = –7

Hence, the required solutions are

7. √2x² + x + √2 = 0

Solution:

Given the quadratic equation,

√2x² + x + √2 = 0

On comparing it with ax² + bx + c = 0, we have

a = √2, b = 1, and c = √2

So, the discriminant of the given equation is

D = b² – 4ac = (1)² – 4 × √2 × √2 = 1 – 8 = –7

Hence, the required solutions are

8. √3x² – √2x + 3√3 = 0

Solution:

Given the quadratic equation,

√3x² – √2x + 3√3 = 0

On comparing it with ax² + bx + c = 0, we have

a = √3, b = -√2, and c = 3√3

So, the discriminant of the given equation is

D = b² – 4ac = (-√2)² – 4 × √3 × 3√3 = 2 – 36 = –34

Hence, the required solutions are

9. x² + x + 1/√2 = 0

Solution:

Given the quadratic equation,

x² + x + 1/√2 = 0

It can be rewritten as,

√2x² + √2x + 1 = 0

On comparing it with ax² + bx + c = 0, we have

a = √2, b = √2, and c = 1

So, the discriminant of the given equation is

D = b² – 4ac = (√2)² – 4 × √2 × 1 = 2 – 4√2 = 2(1 – 2√2)

Hence, the required solutions are

10. x² + x/√2 + 1 = 0

Solution:

Given the quadratic equation,

x² + x/√2 + 1 = 0

It can be rewritten as,

√2x² + x + √2 = 0

On comparing it with ax² + bx + c = 0, we have

a = √2, b = 1, and c = √2

So, the discriminant of the given equation is

D = b² – 4ac = (1)² – 4 × √2 × √2 = 1 – 8 = -7

Hence, the required solutions are

Miscellaneous Exercise Page No: 112

1.

Solution:

2. For any two complex numbers z₁ and z₂, prove that

Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

Solution:

3. Reduce to the standard form.

Solution:

4.

Solution:

5. Convert the following into the polar form:

(i) , (ii) 

Solution:

Solve each of the equations in Exercises 6 to 9.

6. 3x² – 4x + 20/3 = 0

Solution:

Given the quadratic equation, 3x² – 4x + 20/3 = 0

It can be re-written as: 9x² – 12x + 20 = 0

On comparing it with ax² + bx + c = 0, we get

a = 9, b = –12, and c = 20

So, the discriminant of the given equation will be

D = b² – 4ac = (–12)² – 4 × 9 × 20 = 144 – 720 = –576

Hence, the required solutions are

7. x² – 2x + 3/2 = 0

Solution:

Given the quadratic equation, x² – 2x + 3/2 = 0

It can be re-written as 2x² – 4x + 3 = 0

On comparing it with ax² + bx + c = 0, we get

a = 2, b = –4, and c = 3

So, the discriminant of the given equation will be

D = b² – 4ac = (–4)² – 4 × 2 × 3 = 16 – 24 = –8

Hence, the required solutions are

8. 27x² – 10x + 1 = 0

Solution:

Given the quadratic equation, 27x² – 10x + 1 = 0

On comparing it with ax² + bx + c = 0, we get

a = 27, b = –10, and c = 1

So, the discriminant of the given equation will be

D = b² – 4ac = (–10)² – 4 × 27 × 1 = 100 – 108 = –8

Hence, the required solutions are

9. 21x² – 28x + 10 = 0

Solution:

Given the quadratic equation, 21x² – 28x + 10 = 0

On comparing it with ax² + bx + c = 0, we have

a = 21, b = –28, and c = 10

So, the discriminant of the given equation will be

D = b² – 4ac = (–28)² – 4 × 21 × 10 = 784 – 840 = –56

Hence, the required solutions are

10. If z₁ = 2 – i, z₂ = 1 + i, find

Solution:

Given, z₁ = 2 – i, z₂ = 1 + i

11.

Solution:

12. Let z₁ = 2 – i, z₂ = -2 + i. Find

(i) , (ii) 

Solution:

13. Find the modulus and argument of the complex number.

Solution:

14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.

Solution:

Let’s assume z = (x – iy) (3 + 5i)

And,

(3x + 5y) – i(5x – 3y) = -6 -24i

On equating real and imaginary parts, we have

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

Performing (i) x 3 + (ii) x 5, we get

(9x + 15y) + (25x – 15y) = -18 + 120

34x = 102

x = 102/34 = 3

Putting the value of x in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

y = -3

Therefore, the values of x and y are 3 and –3, respectively.

15. Find the modulus of

Solution:

16. If (x + iy)³ = u + iv, then show that

Solution:

17. If α and β are different complex numbers with |β| = 1, then find

Solution:

18. Find the number of non-zero integral solutions of the equation |1 – i|^x = 2^x.

Solution:

Therefore, 0 is the only integral solution of the given equation.

Hence, the number of non-zero integral solutions of the given equation is 0.

19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that

(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²

Solution:

20. If, then find the least positive integral value of m.

Solution:

Thus, the least positive integer is 1.

Therefore, the least positive integral value of m is 4 (= 4 × 1).

NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations

Chapter 5 of Class 11 Complex Numbers and Quadratic Equations has 3 exercises and a miscellaneous exercise to help the students practise the required number of problems to understand all the concepts. The topics and sub-topics discussed in the PDF of NCERT Solutions for Class 11 of this chapter include
5.1 Introduction
We know that some of the quadratic equations have no real solutions. That means the solution of such equations includes complex numbers. Here, we have found the solution of a quadratic equation ax² + bx + c = 0 where D = b² – 4ac < 0.
5.2 Complex Numbers
Definition of complex numbers, examples and explanations about the real and imaginary parts of complex numbers have been discussed in this section. Class 11 Maths NCERT Supplementary Exercise Solutions PDF helps the students to understand the questions in detail.
5.3 Algebra of Complex Numbers
5.3.1 Addition of two complex numbers

5.3.2 Difference of two complex numbers

5.3.3 Multiplication of two complex numbers

5.3.4 Division of two complex number

5.3.5 Power of i

5.3.6 The square roots of a negative real number

5.3.7 Identities

After studying these exercises, students are able to understand the basic BODMAS operations on complex numbers, along with their properties, power of i, square root of a negative real number and identities of complex numbers.
5.4 The Modulus and the Conjugate of a Complex Number
The detailed explanation provides the modulus and conjugate of a complex number with solved examples.
5.5 Argand Plane and Polar Representation
5.5.1 Polar representation of a complex number

In this section, it has been explained how to write the ordered pairs for the given complex numbers, the definition of a Complex plane or Argand plane and the polar representation of the ordered pairs in terms of complex numbers.

1. A number of the form a + ib, where a and b are real numbers, is called a complex number, “a” is called the real part, and “b” is called the imaginary part of the complex number
2. Let z₁ = a + ib and z₂ = c + id. Then
   1. z₁ + z₂ = (a + c) + i (b + d)
   2. z₁z₂ = (ac – bd) + i (ad + bc)
3. For any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there exists a complex number, denoted by 1/z or z–1, called the multiplicative inverse of z
4. For any integer k, i^4k = 1, i^{4k + 1} = i, i^{4k + 2} = – 1, i^{4k + 3} = – i
5. The polar form of the complex number z = x + iy is r (cosθ + i sinθ)
6. A polynomial equation of n degree has n roots.

Disclaimer – 

Dropped Topics – 

5.5.1 Polar Representation of a Complex Number
5.6 Quadratic Equation
Example 11 and Exercise 5.3
Examples 13, 15, 16
Ques. 5–8, 9 and 13 (Miscellaneous Exercise)
Last three points in the Summary
5.7 Square-root of a Complex Number