Chapter 3 Complex Numbers Miscellaneous Exercise 3 Solutions

Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 3 Complex Numbers Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3

(ii) (2i³)²
= 4i⁶
= 4(i²)³
= 4(-1)³…..[∵ i²= -1]
= -4
= -4 + 0i

(iii) (2 + 3i)(1 – 4i) = 2 – 8i + 3i – 12i²
= 2 – 5i – 12(-1) ……[∵ i²= -1]
= 14 – 5i

(iv) i(-4 – 3i)
= (-4i – 3i²)
= [-4i – 3(-1)] ……[∵ i²= -1]
= (3 – 4i)
= – 10i

(v) (1 + 3i)²(3 + i)
= (1 + 6i + 9i²) (3 + i)
= (1 + 6i – 9)(3 + i) ……[∵ i²= -1]
= (-8 + 6i)(3 + i)
= -24 – 8i + 18i + 6i²
= -24 + 10i + 6(-1)
= -24 + 10i – 6
= -30 + 10i

(ii) (1 – 3i) x + (2 + 5i) y = 7 + i
∴ (x + 2y) + (-3x + 5y)i = 7 + i
Equating real and imaginary parts, we get
x + 2y = 7 ……..(i)
and -3x + 5y = 1 ……..(ii)
Equation (i) × 3 + equation (ii) gives
11y = 22
∴ y = 2
Putting y = 2 in (i), we get
x + 2(2) = 7
∴ x = 3
∴ x = 3 and y = 2

(iii) = 7 – i
∴ x + iy = (7 – i)(2 + 3i)
∴ x + iy = 14 + 21i – 2i – 3i²
∴ x + iy = 14 + 19i – 3(-1) …..[∵ i²= -1]
∴ x + iy = 17 + 19i
Equating real and imaginary parts, we get
x = 17 and y = 19

(iv) (x + iy)(5 + 6i) = 2 + 3i

Equating real and imaginary parts, we get
x = and y =

(v) 2x + i⁹y (2 + i) = x i⁷+ 10 i¹⁶
∴ 2x + (i⁴)². i. y (2 + i) = x (i²)³. i + 10. (i⁴)⁴
∴ 2x + (1)². iy (2 + i) = x (-1)³. i + 10 (1)⁴……[∵ i²= -1, i⁴= 1]
∴ 2x + 2yi + yi²= -xi + 10
∴ 2x + 2yi – y + xi = 10
∴ (2x – y) + (x + 2y)i = 10 + 0.i
Equating real and imaginary parts, we get
2x – y = 10 ……(i)
and x + 2y = 0 ……..(ii)
Equation (i) × 2 + equation (ii) gives
5x = 20
∴ x = 4
Putting x = 4 in (i), we get
2(4) – y = 10
∴ y = 8 – 10
∴ y = -2
∴ x = 4 and y = -2

(ii) x =

x³– 5x²+ 4x + 8
= (x²– 6x + 10)(x + 1) – 2
= 0. (x + 1) – 2 ……[From (i)]
= 0 – 2
∴ x³– 5x²+ 4x + 8 = -2

(iii) x = 1 – 4i
∴ x – 1 = -4i
∴ (x – 1)²= 16i²
∴ x²– 2x + 1 = -16 ……[∵ i²= -1]
∴ x²– 2x + 17 = 0 ……(i)

∴ x³– 3x²+ 19x – 20
= (x²– 2x + 17) (x – 1) – 3
= 0. (x – 1) – 3 ….[From (i)]
= 0 – 3
= -3
∴ x³– 3x²+ 19x – 20 = -3

(ii) Let = a + bi, where a, b ∈ R
Squaring on both sides, we get
15 – 8i = a²+ b²i²+ 2abi
∴ 15 – 8i = (a²– b²) + 2abi ……[∵ i²= -1]
Equating real and imaginary parts, we get
a²– b²= 15 and 2ab = -8
∴ a²– b²= 15 and b =
∴ a²– () = 15
∴ a²– = 15
∴ a⁴– 16 = 15a²
∴ a⁴– 15a²– 16 = 0
∴ (a²– 16)(a²+ 1) = 0
∴ a²= 16 or a²= -1
But a ∈ R, a²≠ -1
∴ a²= 16
∴ a = ±4
When a = 4, b = = -1
When a = -4, b = = 1
= ±(4 – i)

(iii) Let = a + bi, where a, b ∈ R.
Squaring on both sides, we get
2 – 2√3i = a²+ b²i²+ 2abi
∴ 2 – 2√3i = a²– b²+ 2abi …..[∵ i²= -1]
Equating real and imaginary parts, we get
a²– b²= 2 and 2ab = 2√3
∴ a²– b²= 2 and b =
∴ a²– = 2
∴ a²– = 2
∴ a⁴– 3 = 2a²
∴ a⁴– 2a²– 3 = 0
∴ (a²– 3)(a²+ 1) = 0
∴ a²= 3 or a²= -1
But a ∈ R, a²≠ -1
∴ a²= 3
∴ a = ±√3
When a = √3, b = = 1
When a = -√3, b = = -1
∴ = ±(√3 + i)

(iv) Let = a + bi, where a, b ∈ R
Squaring on both sides, we get
18i = a²+ b²i²+ 2abi
∴ 0 + 18i = a²– b²+ 2abi …..[∵ i²= -1]
Equating real and imaginary parts, we get
a²– b²= 0 and 2ab = 18
∴ a²– b²= 0 and b =
∴ a²– = 0
∴ a²– = 0
∴ a⁴– 81 = 0
∴ (a²– 9) (a²+ 9) = 0
∴ a²= 9 or a²= -9
But a ∈ R, a²≠ -9
∴ a²= 9
∴ a = ±3
When a = 3, b = = 3
When a = 3, b = = -3
∴ = ±3(1 + i)

(v) Let = a + bi, where a, b ∈ R
Squaring on both sides, we get
3 – 4i = a²+ b²i²+ 2abi
∴ 3 – 4i = a²– b²+ 2abi ……[∵ i²= -1]
Equating real and imaginary parts, we get
a²– b²= 3 and 2ab = -4
∴ a²– b²= 3 and b =
∴ a²– = 3
∴ a²– = 3
∴ a⁴– 4 = 3a²
∴ a⁴– 3a²– 4 = 0
∴ (a²– 4)(a²+ 1) = 0
∴ a²= 4 or a²= -1
But, a ∈ R, a²≠ -1
∴ a²= 4
∴ a = ±2
When a = 2, b = = -1
When a = -2, b = = 1
∴ = ±(2 – i)

(vi) Let = a + bi, where a, b ∈ R
Squaring on both sides, we get
6 + 8i = a²+ b²i²+ 2abi
∴ 6 + 8i = a²– b²+ 2abi ……[∵ i²= -1]
Equating real and imaginary parts, we get
a²– b²= 6 and 2ab = 8
∴ a²– b²= 6 and b =
∴ a²– = 6
∴ a²– = 6
∴ a⁴– 16 = 6a²
∴ a⁴– 6a²– 16 = 0
∴ (a²– 8)(a²+ 2) = 0
∴ a²= 8 or a²= -2
But a ∈ R, a²≠ -2
∴ a²= 8
∴ a = ±2√2
When a = 2√2, b = = √2
When a = -2√2, b = = -√2
∴ = ±(2√2 + √2i) = ±√2(2 + i)