Chapter 1 Complex Numbers Ex 1.2 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2

(ii) 7 + 24i
Solution:
Let = a + bi, where a, b ∈ R.
Squaring on both sides, we get
7 + 24i = (a + bi)²
7 + 24i = a²+ b²i²+ 2abi
7 + 24i = (a²– b²) + 2abi …..[∵ i²= -1]
Equating real and imaginary parts, we get
a²– b²= 7 and 2ab = 24

(iii) 1 + 4√3 i
Solution:
Let = a + bi, where a, b ∈ R.
Squaring on both sides, we get
1 + 4√3 i = (a + bi)²
1 + 4√3i = a²+ b²i²+ 2abi
1 + 4√3i = (a²– b²) + 2abi …..[∵ i²= -1]
Equating real arid imaginary parts, we get
a²– b²= 1 and 2ab = 4√3

(iv) 3 + 2√10 i
Solution:
Let = a + bi, where a, b ∈ R.
Squaring on both sides, we get
3 + 2√10 i = a²+ b²i²+ 2abi
3 + 2√10 i = (a²– b²) + 2abi ……[∵ i²= -1]
Equating real and imaginary parts, we get
a²– b²= 3 and 2ab = 2√10

(v) 2(1 – √3 i)
Solution:
Let = a + bi, where a, b ∈ R.
Squaring on both sides, we get
2(1 – √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 =

(ii) 2x²– √3 x + 1 = 0
Solution:
Given equation is 2x²– √3 x + 1 = 0
Comparing with ax²+ bx + c = 0, we get
a = 2, b = -√3, c = 1
Discriminant = b²– 4ac
= (-√3)²– 4 × 2 × 1
= 3 – 8
= -5 < 0
So, the given equation has complex roots.
These roots are given by

(iii) 3x²– 7x + 5 = 0
Solution:
Given equation is 3x²– 7x + 5 = 0
Comparing with ax²+ bx + c = 0, we get
a = 3, b = -7, c = 5
Discriminant = b²– 4ac
= (-7)²– 4 × 3 × 5
= 49 – 60
= -11 < 0
So, the given equation has complex roots.
These roots are given by

The roots of the given equation are and

(iv) x²– 4x + 13 = 0
Solution:
Given equation is x²– 4x + 13 = 0
Comparing with ax²+ bx + c = 0, we get
a = 1, b = -4, c = 13
Discriminant = b²– 4ac
= (-4)²– 4 × 1 × 13
= 16 – 52
= -36 < 0
So, the given equation has complex roots.
These roots are given by

∴ The roots of the given equation are 2 + 3i and 2 – 3i.

(ii) 2x²+ 3ix + 2 = 0
Solution:
Given equation is 2x²+ 3ix + 2 = 0
Comparing with ax + bx + c = 0, we get
a = 2, b = 3i, c = 2
Discriminant = b²– 4ac
= (3i)²– 4 × 2 × 2
= 9i²– 16
= -9 – 16 …..[∵ i²= -1]
= -25 < 0
So, the given equation has complex roots.
These roots are given by

∴ The roots of the given equation are i and -2i.

(iii) x²+ 4ix – 4 = 0
Solution:
Given equation is x²+ 4ix – 4 = 0
Comparing with ax²+ bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b²– 4ac
= (4i)²– 4 × 1 × (-4)
= 16i²+ 16
= -16 + 16 …..[∵ i²= -1]
= 0
So, the given equation has equal roots.
These roots are given by

∴ x = -2i
∴ The root of the given equation is -2i.

(iv) ix²– 4x – 4i = 0
Solution:
ix²– 4x – 4i = 0
Multiplying throughout by i, we get
i2x²– 4ix – 4i²= 0
-x²– 4ix + 4 = 0 …[∵ i²= -1]
x²+ 4ix – 4 = 0
Comparing with ax²+ bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b²– 4ac
= (4i)²– 4 × 1 × (-4)
= 16i²+ 16
= -16 + 16
= 0
So, the given equation has equal roots.
These roots are given by

∴ The root of the given equation is -2i

(ii) x²– (3√2 + 2i) x + 6√2 i = 0
Solution:
Given equation is x²– (3√2 + 2i) x + 6√2 i = 0
Comparing with ax²+ bx + c = 0, we get
a = 1, b = -(3√2 + 2i), c = 6√2i
Discriminant = b²– 4ac
= [-(3√2 + 2i)]²– 4 × 1 × 6√2 i
= 18 + 12√2i + 4i²– 24√2 i
= 18 – 12√2 i – 4 ……[∵ i²= -1]
= 14 – 12√2 i
So, the given equation has complex roots.
These roots are given by

(iii) x²– (5 – i) x + (18 + i) = 0
Solution:
Given equation is x²– (5 – i)x + (18 + i) = 0
Comparing with ax²+ bx + c = 0, we get
a = 1, b = -(5 – i), c = 18 + i
Discriminant = b²– 4ac
= [-(5 – i)]²– 4 × 1 × (18 + i)
= 25 – 10i + i²– 72 – 4i
= 25 – 10i – 1 – 72 – 4i ……[∵ i²= -1]
= -48 – 14i
So, the given equation has complex roots.
These roots are given by

(iv) (2 + i) x²– (5 – i) x + 2(1 – i) = 0
Solution:
Given equation is (2 + i) x²– (5 – i) x + 2(1 – i) = 0
Comparing with ax²+ bx + c = 0, we get
a = 2 + i, b = -(5 – i), c = 2(1 – i)
Discriminant = b²– 4ac
= [-(5 – i)]²– 4 × (2 + i) × 2(1 – i)
= 25 – 10i + i²– 8(2 + i) (1 – i)
= 25 – 10i + i²– 8(2 – 2i + i – i²)
= 25 – 10i – 1 – 8(2 – i + 1) …..[∵ i²= -1]
= 25 – 10i – 1 – 16 + 8i – 8
= -2i
So, the given equation has complex roots.
These roots are given by

Let = a + bi, where a, b ∈ R
Squaring on both sides, we get
-2i = a²+ b²i²+ 2abi
-2i = a²– b²+ 2abi
Equating real and imaginary parts, we get
a²– b²= 0 and 2ab = -2
a²– b²= 0 and b =

Alternate Method:
x = 2 + 3i
α = 2 + 3i, = 2 – 3i
α = (2 + 3i)(2 – 3i)
= 4 – 6i + 6i – 9i²
= 4 – 9(-1)
= 4 + 9
= 13
α + = 2 + 3i + 2 – 3i = 4
∴ Standard form of quadratic equation,
x²– (Sum of roots) x + Product of roots = 0
x²– 4x + 13 = 0

Dividend = Divisor × Quotient + Remainder
∴ x³– x²+ x + 46 = (x²– 4x + 13).(x + 3) + 7
= 0(x + 3) + 7 …..[From (i)]
= 7

(ii) 2x³– 11x²+ 44x + 27, if x =
Solution:


Dividend = Divisor × Quotient + Remainder
2x³– 11x²+ 44x + 27 = (x²– 6x + 25)(2x + 1) + 2
= 0.(2x + 1) + 2 …..[From (i)]
= 0 + 2
= 2

(iii) x³+ x²– x + 22, if x =
Solution:

Dividend = Divisor × Quotient + Remainder
x³+ x²– x + 22 = (x²– 2x + 5)(x + 3) + 7
= 0.(x + 3) + 7 …..[From (i)]
= 0 + 7
= 7

(iv) x⁴+ 9x³+ 35x²– x + 4, if x = -5 + √-4
Solution:
x = -5 + √-4
x + 5 = √-4
x + 5 = √4 √-1
x + 5 = 2i
(x + 5)²= 4i²
x²+ 10x + 25 = 4(-1) ….[∵ i²= -1]
x²+ 10x + 29 = 0 …..(i)

Dividend = Divisor × Quotient + Remainder
x⁴+ 9x³+ 35x²– x + 4 = (x²+ 10x + 29) (x²– x + 16) – 132x – 460
= 0.(x²– x + 16) – 132x – 460 …..[From (i)]
= -132 (-5 + 2i) – 460
= 660 – 264i – 460
= 200 – 264i

(v) 2x⁴+ 5x³+ 7x²– x + 41, if x = -2 – √3i
Solution:

Dividend = Divisor × Quotient + Remainder
2x⁴+ 5x³+ 7x²– x + 41 = (x²+ 4x + 7) (2x²– 3x + 5) + 6
= 0(2x²– 3x + 5) + 6 ……[From (i)]
= 0 + 6
= 6