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(1 + i)(1 + 2i) = 1 × (1 +2i) + i(1 + 2i)
= 1 + 2i + i + 2i2
= 1 + 3i - 2
= -1 + 3i
∴ (1 + i)(1 + 2i) = -1 + 3i
We have,
We have (x + iy)(2 - 3i) = 4 + i
⟹ x(2 - 3i) + iy ((2 - 3i) = 4 + i
⟹ 2x - 3xi + 2yi + 3y = 4 + i
⟹ 2x + 3y + i (-3x + 2y) = 4 + i
Equating the real and imaginary parts we get
2x + 3y = 4 ......(i)
- 3x + 2y = 1 ......(ii)
Multiplying (i) by 3 and (ii) by 2 and adding
6x - 6x - 9y + 4y = 12 + 2
⟹ 13y = 14
Substituting the value of y in (i), we get
Hence
(3x - 2iy) (2 + i)2 = 10(1 + i)
⟹ (3x – 2iy)(22 + i2 + 2 × 2 × i) = 10 + 10i
⟹ (3x – 2iy)(4 – 1 + 4i) = 10 + 10i
⟹ 3x(3 + 4i) – 2iy(3 + 4i) = 10 + 10i
⟹ 9x + 12xi – 6yi + 8y = 10 + 10i
⟹ 9x + 8y+i(12x – 6y) = 10 + 10i Equating the real andim aginary parts we get
9x + 8y = 10 ...........(i)
12x - 6y = 10 ...........(ii)
Multiplying (i) by 6 and (ii) by 8 and adding
54x + 96x + 48y – 48y = 60 + 80
⟹ 150x = 140
Substituting value of x in (i) we get
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Chapter 13: Complex Numbers – Exercise 13.4...
Chapter 13: Complex Numbers – Exercise 13.3...
Chapter 13: Complex Numbers – Exercise 13.1...