﻿ Class 11 NCERT Math Solution
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Exercise - 5.1

Question-1 :- Express the complex number in the form of a + bi:
(5i)(-3i/5)

Solution :-
```  We have, (5i)(-3i/5)
= -15/5 x i2
= -3 x (-1)             [i² = -1]
= 3
```

Question-2 :-  Express the complex number in the form of a + bi:
i9 + i19

Solution :-
```  We have, i9 + i19
= (i²)4.i + (i²)9.i
= i[(-1)4 + (-1)9]        [i² = -1]
= i[1 - 1]
= 0
```

Question-3 :- Express the complex number in the form of a + bi:
i-39

Solution :-
```  We have, i-39
= 1/i39
= 1/(i²)19.i
= 1/(-1)19.i     [i² = -1]
= 1/(-i)
= 1/(-i) x (-i)/(-i)
= -i/i2
= -i/(-1)
= i
```

Question-4 :-  Express the complex number in the form of a + bi:
3(7 + i7) + i (7 + i7)

Solution :-
```  We have, 3(7 + i7) + i (7 + i7)
= 21 + 21i + 7i + 7i2
= 21 + 28i + 7 x (-1)           [i² = -1]
= 21 - 7 + 28i
= 14 + 28i
```

Question-5 :-  Express the complex number in the form of a + bi:
(1 – i) – ( –1 + i6)

Solution :-
```  We have, (1 – i) – ( –1 + i6)
= 1 - i + 1 - 6i
= 2 - 7i
```

Question-6 :-  Express the complex number in the form of a + bi:

Solution :-
```  We have, (1/5 + 2i/5) - (4 + 5i/2)
= 1/5 + 2i/5 - 4 - 5i/2
= (1/5 - 4) + (2i/5 - 5i/2)
= (1 - 20)/5 + (4i - 25i)/10
= -19/5 + (-21i)/10
= -19/5 - 21i/10
```

Question-7 :-  Express the complex number in the form of a + bi:

Solution :-
```  We have, [(1/3 + 7i/3) + (4 + i/3)] - (-4/3 + i)
= 1/3 + 7i/3 + 4 + i/3 + 4/3 - i
= (1/3 + 4/3 + 4) + (7i/3 + i/3 - i)
= (1 + 4 + 12)/3 + (7i + i - 3i)/3
= 17/3 + 5i/3
```

Question-8 :-  Express the complex number in the form of a + bi:
(1 - i)4

Solution :-
```  We have, (1 - i)4
= (1 - i)² x (1 - i)²
By using Property, (z₁ + z₂)2 = z₁2 + 2 z₁.z₂ + z₂2
= [1² + i² + 2i] x [1² + i² + 2i]           [i² = -1]
= [1 - 1 + 2i] x [1 - 1 + 2i]
= 2i x 2i
= 4i²
= 4 x (-1)                                  [i² = -1]
= -4
```

Question-9 :-  Express the complex number in the form of a + bi:
(1/3 + 3i)³

Solution :-
```  We have, (1/3 + 3i)3
By using Property, (z₁ + z₂)3 = z₁3 + z₂3 + 3 z₁2.z₂ + 3 z₁.z₂2
= (1/3)³ + (3i)³ + 3 x (1/3)² x 3i + 3 x 1/3 x (3i)²
= 1/27 + 27i³ + 9i x 1/9 + 9i²
= 1/27 + 27 x (-i) + i + 9 x (-1)            [i² = -1; i³ = -i]
= 1/27 - 27i + i - 9
= (1/27 - 9) - 26i
= -242/27 - 26i
```

Question-10 :- Express the complex number in the form of a + bi:
(-2 - i/3)³

Solution :-
```  We have, (-2 - i/3)³
By using Property, (z₁ + z₂)3 = z₁3 + z₂3 + 3 z₁2.z₂ + 3 z₁.z₂2
= (-2)³ + (-i/3)³ + 3 x (-2)² x (-i/3) + 3 x (-2) x (-i/3)²
= -8 + (-i)³/27 - 4i - 6 x i²/9
= -8 + i/27 - 4i + 2/3          [i² = -1; (-i)³ = i]
= (-8 + 2/3) + (i/27 - 4i)
= -22/3 - 107i/27
```

Question-11 :-  Find the multiplicative inverse of the complex numbers:
4 - 3i

Solution :-
```  Let z = 4 – 3i
Conjugate z = 4 + 3i
|z|2 = 42 + (-3)2 = 16 + 9 = 25
Therefore, the multiplicative inverse of 4 - 3i is given by z-1 = z/|z|2
z-1 = (4 + 3i)/25 = 4/25 + 3i/25
```

Question-12 :- Find the multiplicative inverse of the complex numbers:
√5 + 3i

Solution :-
```  Let z = √5 + 3i
Conjugate z = √5 - 3i
|z|2 = √52 + 32 = 5 + 9 = 14
Therefore, the multiplicative inverse of √5 + 3i is given by z-1 = z/|z|2
z-1 = (√5 - 3i)/14 = √5/14 - 3i/14
```

Question-13 :-  Find the multiplicative inverse of the complex numbers:
-i

Solution :-
```  Let z = 0 – 1.i
Conjugate z = 0 + 1.i
|z|2 = 0 + (-1)2 = 0 + 1 = 1
Therefore, the multiplicative inverse of -i is given by z-1 = z/|z|2
z-1 = (0 + i)/1 = i
```

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

Solution :-
```
```
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