﻿ Class 11 NCERT Math Solution
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TOPICS
Unit-5(Examples)

Example-1 :-  If 4x + i(3x – y) = 3 + i (– 6), where x and y are real numbers, then find the values of x and y.

Solution :-
```  We have, 4x + i (3x – y) = 3 + i (–6) ... (1)
Equating the real and the imaginary parts of (1), we get 4x = 3, 3x – y = – 6,
Now,  4x = 3; x = 3/4 and
3 X 3/4 - y = -6; 9/4 - y = -6 ; y = 9/4 + 6; y = 33/4.
```

Example-2 :-  Express the following in the form of a + bi:
(i) (-5i)(i/8)   (ii) (-i)(2i)(-i/8)3

Solution :-
```(i) (-5i)(i/8)
= (-5/8)(i2)
= (-5/8) x (-1)
= 5/8

(ii) (-i)(2i)(-i/8)3
= (-2)(i2)(-i/8)3
= (-2)(-1)[(-i)3/512]
= 2 x i/512
= i/256.
```

Example-3 :-  Express (5 – 3i)3 in the form a + ib.

Solution :-
```  By using Identity: (z₁ - z₂)3 = z₁3 - z₂3 - 3 z₁2.z₂ + 3 z₁.z₂2
We have, (5 – 3i)3
= 53 – 3 × 52 × (3i) + 3 × 5 (3i)2 – (3i)3
= 125 –  225i – 135 + 27i
= – 10 – 198i.
```

Example-4 :-  Express (√-3 + √-2)(2√3 - i) in the form of a + ib.

Solution :-
```  We have, (-√3 + √-2)(2√3 - i)
= (-√3 + √2i)(2√3 - i)
= -√3 x 2√3 + √3i + √2i x 2√3 - √2i2
= -3 x 2 + √3i + 2√6i - √2 x (-1)
= -6 + (√3 + 2√6)i + √2
= (-6 + √2) + √3(1 + 2√2)i
```

Example-5 :-  Find the multiplicative inverse of 2 – 3i.

Solution :-
```  Let z = 2 – 3i
Conjugate z = 2 + 3i
|z|2 = 22 + (-3)2 = 4 + 9 = 13
Therefore, the multiplicative inverse of 2 - 3i is given by z-1 = z/|z|2
z-1 = (2 + 3i)/13 = 2/13 + 3i/13
```

Example-6 :- Express the following in the form a + ib
(i) (5 + √2i)/(1 - √2i)    (ii) i-35

Solution :-
```(i)  We have, (5 + √2i)/(1 - √2i)
= (5 + √2i)/(1 - √2i) x (1 + √2i)/(1 + √2i)
= (5 + √2i) x (1 + √2i)/[12 - (√2i)2]
= (5 + 5√2i + √2i + 2i2)/(1 - 2i2)
= (5 + 6√2i - 2 )/(1 + 2)
= (3 + 6√2i)/3
= 3(1 + 2√2i)/3
= 1 + 2√2i
```
```(ii)  We have, i-35
= 1/i35
= 1/(i2)17.i
= 1/(-1)17.i
= -1/i
= -1/i x i/i
= -i/i2
= -i/(-1)
= i
```

Example-7 :-  Represent the complex number z = 1 + i√3 in the polar form.

Solution :-
```  We have, z = 1 + i√3
Now, 1 = r cos θ, √3 = r sin θ
By squaring and adding, we get
r2 cos2 θ + r2 sin2 θ = 12 + (√3)2
r2 (cos2 θ + sin2 θ) = 1 + 3
r2 (cos2 θ + sin2 θ) = 4
r x 1 = √4
r = 2

Therefore, 1 = r cos θ and √3 = r sin θ
cos θ = 1/2 and sin θ = √3/2, which gives θ = π/3
Therefore, required polar form is z = r[cosθ + i sinθ] = 2[cos(π/3) + i sin(π/3)]
```

Example-8 :-  Convert the complex number -16/(1 + i√3) into polar form.

Solution :-
```  We have, z = -16/(1 + i√3)
z = -16/(1 + i√3) x (1 - i√3)/(1 - i√3)
z = -16(1 - i√3)/[12 - (i√3)2]
z = (-16 + i 16√3)/(1 + 3)
z = (-16 + i 16√3)/4
z = -4 + i 4√3

Now, -4 = r cos θ, 4√3 = r sin θ
By squaring and adding, we get
r2 cos2 θ + r2 sin2 θ = (-4)2 + (4√3)2
r2 (cos2 θ + sin2 θ) = 16 + 48
r2 (cos2 θ + sin2 θ) = 64
r x 1 = √64
r = 8

Therefore, -4 = r cos θ and 4√3 = r sin θ
cos θ = -4/8 = -1/2 and
sin θ = 4√3/8 = √3/2,
which gives θ = π - π/3 = 2π/3
Therefore, required polar form is z = r[cosθ + i sinθ] = 8[cos(2π/3) + i sin(2π/3)]
```

Example-9 :-  Solve x2 + 2 = 0.

Solution :-
```  We have, x2 + 2 = 0
x2 = -2
x = ±√-2 = ±√2i
```

Example-10 :-  Solve x2 + x + 1= 0.

Solution :-
```  We have, x2 + x + 1= 0
Now, a = 1; b = 1; c = 1
b2 - 4ac = 12 - 4 x 1 x 1 = 1 - 4 = -3
Therefore, the solutions are given by x = (-1 ± √-3)/2 = (-1 ± √3i)/2
```

Example-11 :-  Solve √5x2 + x + √5 = 0.

Solution :-
```  We have, √5x2 + x + √5 = 0
Now, a = √5; b = 1; c = √5
b2 - 4ac = 12 - 4 x √5 x √5 = 1 - 20 = -19
Therefore, the solutions are given by x = (-1 ± √-19)/2√5 = (-1 ± √19i)/2√5
```
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