Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Ex 1.2 Questions and Answers.

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

Question 1.

Find the square root of the following complex numbers:

(i) -8 – 6i

Solution:

Let \(\sqrt{-8-6 i}\) = a + bi, where a, b ∈ R.

Squaring on both sides, we get

-8 – 6i = (a + bi)^{2}

-8 – 6i = a^{2} + b^{2}i^{2} + 2abi

-8 – 6i = (a^{2} – b^{2}) + 2abi …..[∵ i^{2} = -1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = -8 and 2ab = -6

(ii) 7 + 24i

Solution:

Let \(\sqrt{7+24 i}\) = a + bi, where a, b ∈ R.

Squaring on both sides, we get

7 + 24i = (a + bi)^{2}

7 + 24i = a^{2} + b^{2}i^{2} + 2abi

7 + 24i = (a^{2} – b^{2}) + 2abi …..[∵ i^{2} = -1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 7 and 2ab = 24

(iii) 1 + 4√3 i

Solution:

Let \(\sqrt{1+4 \sqrt{3} i}\) = a + bi, where a, b ∈ R.

Squaring on both sides, we get

1 + 4√3 i = (a + bi)^{2}

1 + 4√3i = a^{2} + b^{2}i^{2} + 2abi

1 + 4√3i = (a^{2} – b^{2}) + 2abi …..[∵ i^{2} = -1]

Equating real arid imaginary parts, we get

a^{2} – b^{2} = 1 and 2ab = 4√3

(iv) 3 + 2√10 i

Solution:

Let \(\sqrt{3+2 \sqrt{10}} i\) = a + bi, where a, b ∈ R.

Squaring on both sides, we get

3 + 2√10 i = a^{2} + b^{2}i^{2} + 2abi

3 + 2√10 i = (a^{2} – b^{2}) + 2abi ……[∵ i^{2} = -1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 3 and 2ab = 2√10

(v) 2(1 – √3 i)

Solution:

Let \(\sqrt{2(1-\sqrt{3} i)}\) = a + bi, where a, b ∈ R.

Squaring on both sides, we get

2(1 – √3 i) = a^{2} + b^{2}i^{2} + 2abi

2 – 2√3 i = (a^{2} – b^{2}) + 2abi ….[∵ i^{2} = -1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 2 and 2ab = -2√3

a^{2} – b^{2} = 2 and b = \(-\frac{\sqrt{3}}{a}\)

Question 2.

Solve the following quadratic equations:

(i) 8x^{2} + 2x + 1 = 0

Solution:

Given equation is 8x^{2} + 2x + 1 = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 8, b = 2, c = 1

Discriminant = b^{2} – 4ac

= (2)^{2} – 4 × 8 × 1

= 4 – 32

= -28 < 0

So, the given equation has complex roots.

These roots are given by

(ii) 2x^{2} – √3 x + 1 = 0

Solution:

Given equation is 2x^{2} – √3 x + 1 = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 2, b = -√3, c = 1

Discriminant = b^{2} – 4ac

= (-√3)^{2} – 4 × 2 × 1

= 3 – 8

= -5 < 0

So, the given equation has complex roots.

These roots are given by

(iii) 3x^{2} – 7x + 5 = 0

Solution:

Given equation is 3x^{2} – 7x + 5 = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 3, b = -7, c = 5

Discriminant = b^{2} – 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 \(\frac{7+\sqrt{11} \mathrm{i}}{6}\) and \(\frac{7-\sqrt{11} \mathrm{i}}{6}\)

(iv) x^{2} – 4x + 13 = 0

Solution:

Given equation is x^{2} – 4x + 13 = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 1, b = -4, c = 13

Discriminant = b^{2} – 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.

Question 3.

Solve the following quadratic equations:

(i) x^{2} + 3ix + 10 = 0

Solution:

Given equation is x^{2} + 3ix + 10 = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 1, b = 3i, c = 10

Discriminant = b^{2} – 4ac

= (3i)^{2} – 4 × 1 × 10

= 9i^{2} – 40

= -9 – 40 ……[∵ i^{2} = -1]

= -49 < 0

So, the given equation has complex roots.

These roots are given by

∴ x = 2i or x = -5i

∴ The roots of the given equation are 2i and -5i.

(ii) 2x^{2} + 3ix + 2 = 0

Solution:

Given equation is 2x^{2} + 3ix + 2 = 0

Comparing with ax + bx + c = 0, we get

a = 2, b = 3i, c = 2

Discriminant = b^{2} – 4ac

= (3i)^{2} – 4 × 2 × 2

= 9i^{2} – 16

= -9 – 16 …..[∵ i^{2} = -1]

= -25 < 0

So, the given equation has complex roots.

These roots are given by

∴ The roots of the given equation are \(\frac{1}{2}\)i and -2i.

(iii) x^{2} + 4ix – 4 = 0

Solution:

Given equation is x^{2} + 4ix – 4 = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 1, b = 4i, c = -4

Discriminant = b^{2} – 4ac

= (4i)^{2} – 4 × 1 × (-4)

= 16i^{2} + 16

= -16 + 16 …..[∵ i^{2} = -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) ix^{2} – 4x – 4i = 0

Solution:

ix^{2} – 4x – 4i = 0

Multiplying throughout by i, we get

i2x^{2} – 4ix – 4i^{2} = 0

-x^{2} – 4ix + 4 = 0 …[∵ i^{2} = -1]

x^{2} + 4ix – 4 = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 1, b = 4i, c = -4

Discriminant = b^{2} – 4ac

= (4i)^{2} – 4 × 1 × (-4)

= 16i^{2} + 16

= -16 + 16

= 0

So, the given equation has equal roots.

These roots are given by

∴ The root of the given equation is -2i

Question 4.

Solve the following quadratic equations:

(i) x^{2} – (2 + i) x – (1 – 7i) = 0

Solution:

Given equation is x^{2} – (2 + i)x – (1 – 7i) = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 1, b = -(2 + i), c = -(1 – 7i)

Discriminant = b^{2} – 4ac

= [-(2 + i)]^{2} – 4 × 1 × -(1 – 7i)

= 4 + 4i + i^{2} + 4 – 28i

= 4 + 4i – 1 + 4 – 28i …..[∵ i^{2} = – 1]

= 7 – 24i

So, the given equation has complex roots.

These roots are given by

Let \(\sqrt{7-24 i}\) = a + bi, where a, b ∈ R

Squaring on both sides, we get

7 – 24i = a^{2} + i^{2}b^{2} + 2abi

7 – 24i = a^{2} – b^{2} + 2abi

Equating real and imaginary parts, we get

a^{2} – b^{2} = 7 and 2ab = -24

(ii) x^{2} – (3√2 + 2i) x + 6√2 i = 0

Solution:

Given equation is x^{2} – (3√2 + 2i) x + 6√2 i = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 1, b = -(3√2 + 2i), c = 6√2i

Discriminant = b^{2} – 4ac

= [-(3√2 + 2i)]^{2} – 4 × 1 × 6√2 i

= 18 + 12√2i + 4i^{2} – 24√2 i

= 18 – 12√2 i – 4 ……[∵ i^{2} = -1]

= 14 – 12√2 i

So, the given equation has complex roots.

These roots are given by

(iii) x^{2} – (5 – i) x + (18 + i) = 0

Solution:

Given equation is x^{2} – (5 – i)x + (18 + i) = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 1, b = -(5 – i), c = 18 + i

Discriminant = b^{2} – 4ac

= [-(5 – i)]^{2} – 4 × 1 × (18 + i)

= 25 – 10i + i^{2} – 72 – 4i

= 25 – 10i – 1 – 72 – 4i ……[∵ i^{2} = -1]

= -48 – 14i

So, the given equation has complex roots.

These roots are given by

(iv) (2 + i) x^{2} – (5 – i) x + 2(1 – i) = 0

Solution:

Given equation is (2 + i) x^{2} – (5 – i) x + 2(1 – i) = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 2 + i, b = -(5 – i), c = 2(1 – i)

Discriminant = b^{2} – 4ac

= [-(5 – i)]^{2} – 4 × (2 + i) × 2(1 – i)

= 25 – 10i + i^{2} – 8(2 + i) (1 – i)

= 25 – 10i + i^{2} – 8(2 – 2i + i – i^{2})

= 25 – 10i – 1 – 8(2 – i + 1) …..[∵ i^{2} = -1]

= 25 – 10i – 1 – 16 + 8i – 8

= -2i

So, the given equation has complex roots.

These roots are given by

Let \(\sqrt{-2 i}\) = a + bi, where a, b ∈ R

Squaring on both sides, we get

-2i = a^{2} + b^{2}i^{2} + 2abi

-2i = a^{2} – b^{2} + 2abi

Equating real and imaginary parts, we get

a^{2} – b^{2} = 0 and 2ab = -2

a^{2} – b^{2} = 0 and b = \(-\frac{1}{a}\)

Question 5.

Find the value of

(i) x^{3} – x^{2} + x + 46, if x = 2 + 3i

Solution:

x = 2 + 3i

x – 2 = 3i

(x – 2)^{2} = 9i^{2}

x^{2} – 4x + 4 = 9(-1) …..[∵ i^{2} = -1]

x^{2} – 4x + 13 = 0 …..(i)

Dividend = Divisor × Quotient + Remainder

∴ x^{3} – x^{2} + x + 46 = (x^{2} – 4x + 13) (x + 3) + 7

= 0(x + 3) + 7 …..[from(i)]

= 7

Alternate Method:

x = 2 + 3i

α = 2 + 3i, \(\bar{\alpha}\) = 2 – 3i

α\(\bar{\alpha}\) = (2 + 3i)(2 – 3i)

= 4 – 6i + 6i – 9i^{2}

= 4 – 9(-1)

= 4 + 9

= 13

α + \(\bar{\alpha}\) = 2 + 3i + 2 – 3i = 4

∴ Standard form of quadratic equation,

x^{2} – (Sum of roots) x + Product of roots = 0

x^{2} – 4x + 13 = 0

Dividend = Divisor × Quotient + Remainder

∴ x^{3} – x^{2} + x + 46 = (x^{2} – 4x + 13).(x + 3) + 7

= 0(x + 3) + 7 …..[From (i)]

= 7

(ii) 2x^{3} – 11x^{2} + 44x + 27, if x = \(\frac{25}{3-4 i}\)

Solution:

Dividend = Divisor × Quotient + Remainder

2x^{3} – 11x^{2} + 44x + 27 = (x^{2} – 6x + 25)(2x + 1) + 2

= 0.(2x + 1) + 2 …..[From (i)]

= 0 + 2

= 2

(iii) x^{3} + x^{2} – x + 22, if x = \(\frac{5}{1-2 i}\)

Solution:

Dividend = Divisor × Quotient + Remainder

x^{3} + x^{2} – x + 22 = (x^{2} – 2x + 5)(x + 3) + 7

= 0.(x + 3) + 7 …..[From (i)]

= 0 + 7

= 7

(iv) x^{4} + 9x^{3} + 35x^{2} – x + 4, if x = -5 + √-4

Solution:

x = -5 + √-4

x + 5 = √-4

x + 5 = √4 √-1

x + 5 = 2i

(x + 5)^{2} = 4i^{2}

x^{2} + 10x + 25 = 4(-1) ….[∵ i^{2} = -1]

x^{2} + 10x + 29 = 0 …..(i)

Dividend = Divisor × Quotient + Remainder

x^{4} + 9x^{3} + 35x^{2} – x + 4 = (x^{2} + 10x + 29) (x^{2} – x + 16) – 132x – 460

= 0.(x^{2} – x + 16) – 132x – 460 …..[From (i)]

= -132 (-5 + 2i) – 460

= 660 – 264i – 460

= 200 – 264i

(v) 2x^{4} + 5x^{3} + 7x^{2} – x + 41, if x = -2 – √3i

Solution:

Dividend = Divisor × Quotient + Remainder

2x^{4} + 5x^{3} + 7x^{2} – x + 41 = (x^{2} + 4x + 7) (2x^{2} – 3x + 5) + 6

= 0(2x^{2} – 3x + 5) + 6 ……[From (i)]

= 0 + 6

= 6