 # NCERT Solutions for Class 11 Maths Chapter 5- Complex Numbers and Quadratic Equations Exercise 5.1

The answers to the questions given in the first exercise of Chapter 5, NCERT Class 11 Textbook are provided here. Chapter 5 Complex Numbers and Quadratic Equations of Class 11 Maths is categorised under the CBSE Syllabus for 2022-23. Exercise 5.1 of NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations is based on the following topics:

1. Complex Numbers
2. Algebra of Complex Numbers
1. Addition of two complex numbers
2. Difference of two complex numbers
3. Multiplication of two complex numbers
4. Division of two complex numbers
5. Power of i
6. The square roots of a negative real number
7. Identities
3. The Modulus and the Conjugate of a Complex Number

For students to score high marks in the finals, practising with NCERT Solutions is mandatory. These NCERT Solutions of Class 11 Maths are helpful for students to score well in the board examination, as it works for them as a reference tool to do the revision.

### Download the PDF of NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Exercise 5.1     ### Access other exercise solutions of Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations

Exercise 5.2 Solutions 8 Questions

Exercise 5.3 Solutions 10 Questions

Miscellaneous Exercise on Chapter 5 Solutions 20 Questions

#### Access Solutions for Class 11 Maths Chapter 5.1 Exercise

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/ 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 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: #### 1 Comment

1. VIKAS Chauhan

This app help me very much
Thank you byjus