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

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 the students in solving the 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 exams with flying colours.
The Chapter Complex Numbers and Quadratic Equations includes different critical mathematical theorems and formulae. The NCERT textbook has many practice problems to cover all these concepts, which would help students to easily understand higher concepts in future. BYJU’S provides solutions for all these problems by explaining all the steps with proper explanations. These NCERT Solutions of BYJU’S help students who aim to clear their exams even with the last-minute preparations. However, NCERT Solutions For Class 11 Maths are focused on mastering the concepts along with gaining broader knowledge.

## Download PDF of NCERT Solutions for Class 11 Maths Chapter 5- Complex Numbers and Quadratic Equations

### Access Answers of Maths NCERT Class 11 Chapter 5 – Complex Numbers and Quadratic Equations                         Exercise:
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 summarization of the topics discussed in Chapter 5 of 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 number given in the 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/ i4 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 the Exercises 11 to 13.

11. 4 – 3i

Solution:

Let’s consider z = 4 – 3i

Then,

= 4 + 3and

|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:

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:

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:

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:

x2 + – 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:

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:

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:

√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:

√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:

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:

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 zRe z2 – Im z1 Im z2

Solution: 3. Reduce to the standard form Solution: 4. Solution:  5. Convert the following in the polar form:

(i) , (ii) Solution:  Solve each of the equation in Exercises 6 to 9.

6. 3x2 – 4x + 20/3 = 0

Solution:

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

On comparing it with ax2 + bx = 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 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 = 2x

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).

 Also Access NCERT Exemplar for Class 11 Maths Chapter 5 CBSE Notes for Class 11 Maths Chapter 5

#### 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 in practicing 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 include 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 the 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 this exercise, 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, i4k = 1, i4k + 1 = i, i4k + 2 = – 1, i4k + 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.