Chapter 1 Complex Numbers Miscellaneous Exercise 1 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1

(I) Select the correct answer from the given alternatives.

(II) Answer the following:

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

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

(iv) i(-4 – 3i)
Solution:

(v) (1 + 3i)²(3 + i)
Solution:
(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

(vi)
Solution:

(vii)
Solution:

(viii)
Solution:

(ix)
Solution:

(x)
Solution:

(ii) = 7 – i
Solution:
= 7 – i
x + iy = (7 – i)(2 + 3i)
x + iy = 14 + 21i – 2i – 3i²
x + iy = 14 + 19i – 3(-1)
x + iy = 17 + 19i
Equating real and imaginary parts, we get
∴ x = 17 and y = 19

(iii) (x + iy) (5 + 6i) = 2 + 3i
Solution:

(iv) 2x + i⁹y(2 + i) = x i⁷+ 10 i¹⁶
Solution:
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 + y i²= -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, we get
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) i¹³¹+ i⁴⁹
Solution:
i¹³¹+ i⁴⁹
= (i⁴)³². i³+ (i⁴)¹². i
= (1)³²(-i) + (1)¹². i
= -i + i
= 0

(ii) x⁴+ 9x³+ 35x²– x + 164, if x = -5 + 4i
Solution:

(ii) 15 – 8i
Solution:
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 =

When a = 4, b = = -1
When a = -4, b = = 1
∴ = ±(4 – i)

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

(iv) 18i
Solution:
Let √18i = 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 =


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
∴ √18i = ±(3 + 3i) = ±3(1 + i)

(v) 3 – 4i
Solution:
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 =

(vi) 6 + 8i
Solution:
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

(ii) 6 – i
Solution:

(iii)
Solution:

(iv)
Solution:

(v) 2i
Solution:

(vi) -3i
Solution:

(vii)
Solution:

(ii) z = -6 + √2 i
Solution:
z = -6 + √2 i
∴ a = -6, b = √2
i.e. a < 0, b > 0

(iii)
Solution:

(ii)
Solution:

(iii)
Solution: