**Chapter 1 Complex Numbers Ex 1.1**

**Question 1.Simplify:(i) √-16 + 3√-25 + √-36 – √-625Solution:**

= 4i + 3(5i) + 6i – 25i

= 25i – 25i

= 0

**(ii) 4√-4 + 5√-9 – 3√-16Solution:**

**Question 2.Write the conjugates of the following complex numbers(i) 3 + iSolution:**

Conjugate of (3 + i) is (3 – i).

**(ii) 3 – iSolution:**

Conjugate of (3 – i) is (3 + i).

**(iii) √-5 – √7 iSolution:**

Conjugate of (√-5 – √7 i) is (√-5 + √7 i).

**(iv) -√-5Solution:**

-√-5 = -√5 × √-1 = -√5 i

Conjugate of (-√-5) is √5 i

**(v) 5iSolution:**Conjugate of (5i) is (-5i).

**(vi) √5 – iSolution:**

Conjugate of (√5 – i) is (√5 + i).

**(vii) √2 + √3 iSolution:**

Conjugate of (√2 + √3 i) is (√2 – √3 i)

**(viii) cos θ + i sin θSolution:**

Conjugate of (cos θ + i sin θ) is (cos θ – i sin θ)

**Question 3.Find a and b if(i) a + 2b + 2ai = 4 + 6iSolution:**a + 2b + 2ai = 4 + 6i

Equating real and imaginary parts, we get

a + 2b = 4 …..(i)

2a = 6 ……(ii)

∴ a = 3

Substituting, a = 3 in (i), we get

3 + 2b = 4

∴ b = 1/2

∴ a = 3 and b =

Check:

For a = 3 and b = 1/2

Consider, L.H.S. = a + 2b + 2ai

= 3 + 2(1/2) + 2(3)i

= 4 + 6i

= R.H.S.

**(ii) (a – b) + (a + b)i = a + 5iSolution:**

(a – b) + (a + b)i = a + 5i

Equating real and imaginary parts, we get

a – b = a ……(i)

a + b = 5 ……(ii)

From (i), b = 0

Substituting b = 0 in (ii), we get

a + 0 = 5

∴ a = 5

∴ a = 5 and b = 0

**(iii) (a + b) (2 + i) = b + 1 + (10 + 2a)iSolution:**

(a + b) (2 + i) = b + 1 + (10 + 2a)i

2(a + b) + (a + b)i = (b + 1) + (10 + 2a)i

Equating real and imaginary parts, we get

2(a + b) = b + 1

∴ 2a + b = 1 ……(i)

and a + b = 10 + 2a

-a + b = 10 …….(ii)

Subtracting equation (ii) from (i), we get

3a = -9

∴ a = -3

Substituting a = – 3 in (ii), we get

-(-3) + b = 10

∴ b = 7

∴ a = -3 and b = 7

**(iv) abi = 3a – b + 12iSolution:**abi = 3a – b + 12i

∴ 0 + abi = (3a – b) + 12i

Equating real and imaginary parts, we get

3a – b = 0

∴ 3a = b …..(i)

and ab = 12

∴ b = ……..(ii)

Substituting b = in (i), we get

3a =

3a

^{2}= 12

a

^{2}= 4

a = ±2

Solution:

**(vi) (a + ib) (1 + i) = 2 + iSolution:**(a + ib)(1 + i) = 2 + i

a + ai + bi + bi

^{2}= 2 + i

a + (a + b)i + b(-1) = 2 + i ……(∵ i

^{2}= -1)

(a – b) + (a + b)i = 2 + i

Equating real and imaginary parts, we get

a – b = 2 ……(i)

a + b = 1 …….(ii)

Adding equations (i) and (ii), we get

2a = 3

**Question 4.Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b:(i) (1 + 2i)(-2 + i)Solution:**

(1 + 2i)(-2 + i) = -2 + i – 4i + 2

= -2 – 3i + 2(-1) ……[∵ i

^{2}= -1]

∴ (1 + 2i)(-2 + i) = -4 – 3i

∴ a = -4 and b = -3

**(ii) (1 + i)(1 – i)-1Solution:**

Solution:

Solution:

Solution:

Solution:

**(vii) (1 + iSolution:**

Solution:

**(ix) (-√5 + 2√-4 ) + (1 – √-9 ) + (2 + 3i)(2 – 3i)Solution:**

(-√5 + 2√-4) + (1 – √-9) + (2 + 3i)(2 – 3i)

= (-√5 + 2√4.√-1) + (1 – √9.√-1) + 4 – 9

= [-√5 + 2(2)i] + (1 – 3i) + 4 – 9

= -√5 + 4i + 1 – 3i + 4 – 9(-1) ……[∵ = -1]

= (14 – √5) + i

∴ a = 14 – √5 and b = 1

**(x) (2 + 3i)(2 – 3i)Solution:**(2 + 3i)(2 – 3i)

= 4 – 9

= 4 – 9(-1) …[∵ = -1]

= 4 + 9

= 13

∴ (2 + 3i)(2 – 3i) = 13 + 0i

∴ a = 13 and b = 0

Solution:

**Question 5.Show that (-1 + √3i is a real number.Solution:**

**Question 6.Solution:**

**Question 7.Evaluate the following:Solution:**

**Question 8.Solution:**

= 1 – 1 + 1 – 1

= 0, which is a real number.

**Question 9.Find the value ofSolution:**

**Question 10.Solution:**

**Question 11.Solution:**

**Question 12.Solution:**

**Question 13.Is (1 + ) a real number? Justify your answer.Solution:**

**Question 14.Solution:**

**Question 15.Solution:**

**Question 16.Solution:**

**Question 17.Solution:**

**Question 18.**

Alternate Method:

Consider

**Question 19.**

a + 3i = (1 – i)(2 + ib)

= 2 + bi – 2i – bi

^{2}

= 2 + (b – 2)i – b(-1) ……[∵ i

^{2}= -1]

a + 3i = (2 + b) + (b – 2)i

Equating real and imaginary parts, we get

a = 2 + b and 3 = b – 2

a = 2 + b and b = 5

a = 2 + 5 = 7

5a – 7b = 5(7) – 7(5) = 35 – 35 = 0

**Question 20.Solution:**

**Question 21.Solution:**

∴ a + bi = 0 + i

Equating real and imaginary parts, we get

a = 0 and b = 1

**Question 22.Solution:**

**Question 23.Solution:**

**Question 24.Find the values of x and y which satisfy the following equations (x, y ∈ R)(i) (x + 2y) + (2x – 3y)i + 4i = 5Solution:**

(x + 2y) + (2x – 3y)i + 4i = 5

(x + 2y) + (2x – 3y)i = 5 – 4i

Equating real and imaginary parts, we get

x + 2y = 5 ……(i)

and 2x – 3y = -4 …..(ii)

Equation (i) x 2 – equation (ii) gives

7y = 14

∴ y = 2

Substituting y = 2 in (i), we get

x + 2(2) = 5

x + 4 = 5

∴ x = 1

∴ x = 1 and y = 2

Check:

For x = 1 and y = 2

Consider, L.H.S. = (x + 2y) + (2x – 3y)i + 4i

= (1 + 4) + (2 – 6)i + 4i

= 5 – 4i + 4i

= 5

= R.H.S.

**Solution:**(x + y) + (y – x – 2)i = 2i

(x + y) + (y – x – 2)i = 0 + 2i

Equating real and imaginary parts, we get

x + y = 0 and y – x – 2 = 2

∴ x + y = 0 …….(i)

and -x + y = 4 …….(ii)

Adding (i) and (ii), we get

2y = 4

∴ y = 2

Substituting y = 2 in (i), we get

x + 2 = 0

∴ x = -2

∴ x = -2 and y = 2

Solution:

(2x + 3y + 1) + (8 – 3x + 2y)i = 9 + 9i

Equating real and imaginary parts, we get

2x + 3y + 1 = 9 and 8 – 3x + 2y = 9

2x + 3y = 8 ……(i)

and 3x – 2y = – 1 ……(ii)

Equation (i) × 2 + equation (ii) × 3 gives

13x = 13

∴ x = 1

Substituting x = 1 in (i), we get

2(1) + 3y = 8

3y = 6

∴ y = 2

∴ x = 1 and y = 2

**(iv) If x(1 + 3i) + y(2 – i) – 5 + = 0, find x + ySolution:**

x(1 + 3i) + y(2 – i) – 5 + = 0

x + 3xi + 2y – yi – 5 – i = 0 ……[∵ = -i]

(x + 2y – 5) + (3x – y – 1)i = 0 + 0i

Equating real and imaginary parts, we get

x + 2y – 5 = 0 …..(i)

and 3x – y – 1 = 0 ……(ii)

Equation (i) + equation (ii) × 2 gives

7x – 7 = 0

7x = 1

∴ x = 1

Substituting x = 1 in (i), we get

1 + 2y – 5 = 0

2y = 4

y = 2

∴ x = 1 and y = 2

∴ x + y = 1 + 2 = 3

**(v) If x + 2i + 15y = 7x + i ^{3}(y + 4), find x + ySolution:**

Equating real and imaginary parts, we get

-6x – 15y = 0 and y + 6 = 0

-6x – 15y = 0 and y = -6

-6x – 15(-6) = 0

-6x + 90 = 0

∴ x = 15

∴ x + y = 15 – 6 = 9