NCERT Solutions Class 11th Maths Chapter – 5 Complex Numbers and Quadratic Equations Exercise 5.1
| Textbook | NCERT |
| class | Class – 11th |
| Subject | Mathematics |
| Chapter | Chapter – 5 |
| Chapter Name | Complex Numbers and Quadratic Equations |
| grade | Class 11th Maths solution |
| Medium | English |
| Source | last doubt |
NCERT Solutions Class 11th Maths Chapter – 5 Complex Numbers and Quadratic Equations Exercise 5.1
?Chapter – 5?
✍Complex Numbers and Quadratic Equations✍
?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/ 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 + 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:
