NCERT Solutions For Class 11 Chapter – 5 Exercise – 5.1

NCERT Solutions For Class 11 Math Chapter – 5 Exercise – 5.1

 

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/ 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 the 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: