Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2

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 = a2 + b2i2 + 2abi
-8 – 6i = (a2 – b2) + 2abi …..[∵ i2 = -1]
Equating real and imaginary parts, we get
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q1 (i)

(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 = a2 + b2i2 + 2abi
7 + 24i = (a2 – b2) + 2abi …..[∵ i2 = -1]
Equating real and imaginary parts, we get
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q1 (ii)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q1 (ii).1

(iii) 1 + 4√3i
Solution:
Let \(\sqrt{1+4 \sqrt{3} i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
1 + 4√3i = (a + bi)2
1 + 4√3i = a2 + b2i2 + 2abi
1 +4√3i = (a2 – b2) + 2abi ……[∵ i2 = -1]
Equating real and imaginary parts, we get
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q1 (iii)

(iv) 3 + 2√10i
Solution:
Let \(\sqrt{3+2 \sqrt{10}} i\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
3 + 2√10i = (a + bi)2
3 + 2√10i = a2 + b2i2 + 2abi
3 + 2√10i = (a2 – b2) + 2abi …..[∵ i2 = -1]
Equating real and imaginary parts, we get
a2 – b2 = 3 and 2ab = 2√10
a2 – b2 = 3 and b = \(\frac{\sqrt{10}}{a}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q1 (iv)

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2

(v) 2(1 – √3i)
Solution:
Let \(\sqrt{2(1-\sqrt{3} i)}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
2(1 – √3i) = (a + bi)2
2(1 – √3i) = a2 + b2i2 + 2abi
2 – 2√3i = (a2 – b2) + 2abi …..[∵ i2 = -1]
Equating real and imaginary parts, we get
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q1 (v)

Question 2.
Solve the following quadratic equations.
(i) 8x2 + 2x + 1 = 0
Solution:
Given equation is 8x2 + 2x + 1 = 0
Comparing with ax2 + bx + c = 0, we get
a = 8, b = 2, c = 1
Discriminant = b2 – 4ac
= (2)2 – 4 × 8 × 1
= 4 – 32
= -28 < 0
So, the given equation has complex roots.
These roots are given by
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q2 (i)
∴ the roots of the given equation are \(\frac{-1+\sqrt{7} \mathrm{i}}{8}\) and \(\frac{-1-\sqrt{7} \mathrm{i}}{8}\)

(ii) 2x2 – √3x + 1 = 0
Solution:
Given equation is 2x2 – √3x + 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
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q2 (ii)
∴ the roots of the given equation are \(\frac{\sqrt{3}+\sqrt{5} i}{4}\) and \(\frac{\sqrt{3}-\sqrt{5} i}{4}\)

(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
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q2 (iii)
∴ the roots of the given equation are \(\frac{7+\sqrt{11} i}{6}\) and \(\frac{7-\sqrt{11} i}{6}\)

(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
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q2 (iv)
∴ the roots of the given equation are 2 + 3i and 2 – 3i.

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2

Question 3.
Solve the following quadratic equations.
(i) x2 + 3ix + 10 = 0
Solution:
Given equation is x2 + 3ix + 10 = 0
Comparing with ax2 + bx + c = 0, we get
a = 1, b = 3i, c = 10
Discriminant = b2 – 4ac
= (3i)2 – 4 × 1 × 10
= 9i2 – 40
= -9 – 40 …..[∵ i2 = -1]
= -49
So, the given equation has complex roots.
These roots are given by
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q3 (i)
∴ x = 2i or x = -5i
∴ the roots of the given equation are 2i and -5i.
Check:
If x = 2i and x = -5i satisfy the given equation, then our answer is correct.
L.H.S. = x2 + 3ix + 10
= (2i)2 + 3i(2i) + 10i
= 4i2 + 6i2 + 10
= 10i2 + 10
= -10 + 10 ……[∵ i2 = -1]
= 0
= R.H.S.
L.H.S. = x2 + 3ix + 10
= (-5i)2 + 3i(-5i) + 10
= 25i2 – 15i2 + 10
= 10i2 + 10
= -10 + 10 …..[∵ i2 = -1]
= 0
= R.H.S.
Thus, our answer is correct.

(ii) 2x2 + 3ix + 2 = 0
Solution:
Given equation is 2x2 + 3ix + 2 = 0
Comparing with ax2 + bx + c = 0, we get
a = 2, b = 3i, c = 2
Discriminant = b2 – 4ac
= (3i)2 – 4 × 2 × 2
= 9i2 – 16
= -9 – 16
= -25 < 0
So, the given equation has complex roots.
These roots are given by
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q3 (ii)
∴ the roots of the given equation are \(\frac{1}{2}\)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
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q3 (iii)
∴ the roots of the given equation are -2i and -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 …..[∵ i2 = -1]
= 0
So, the given equation has equal roots.
These roots are given by
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q3 (iv)
∴ the roots of the given equation are -2i and -2i.

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2

Question 4.
Solve the following quadratic equations.
(i) x2 – (2 + i) x – (1 – 7i) = 0
Solution:
Given equation is x2 – (2 + i)x – (1 – 7i) = 0
Comparing with ax2 + bx + c = 0, we get
a = 1, b = -(2 + i), c = -(1 – 7i)
Discriminant = b2 – 4ac
= [-(2 + i)]2 – 4 × 1 × -(1 – 7i)
= 4 + 4i + i2 + 4 – 28i
= 4 + 4i – 1 + 4 – 28i …….[∵ i2 = -1]
= 7 – 24i
So, the given equation has complex roots.
These roots are given by
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (i)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (i).1
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (i).2

(ii) x2 – (3√2 + 2i) x + 6√2i = 0
Solution:
Given equation is x2 – (3√2 + 2i) x + 6√2i = 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√2i
= 18 + 12√2i + 4i2 – 24√2i
= 18 – 12√2i – 4 …..[∵ i2 = -1]
= 14 – 12√2i
So, the given equation has complex roots.
These roots are given by
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (ii)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (ii).1
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (ii).2

(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
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (iii)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (iii).1

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2

(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
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (iv)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (iv).1
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2 Q4 (iv).2

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Question 1.
Write the conjugates of the following complex numbers:
(i) 3 + i
(ii) 3 – i
(iii) -√5 – √7i
(iv) -√-5
(v) 5i
(vi) √5 – i
(vii) √2 + √3i
Solution:
(i) Conjugate of (3 + i) is (3 – i)
(ii) Conjugate of (3 – i) is (3 + i)
(iii) Conjugate of (-√5 – √7i) is (-√5 + √7i)
(iv) -√-5 = -√5 × √-1 = -√5i
Conjugate of -√-5 is √5i
(v) Conjugate of 5i is -5i
(vi) Conjugate of √5 – i is √5 + i
(vii) Conjugate of √2 + √3i is √2 – √3i

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Question 2.
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)
(ii) \(\frac{\mathrm{i}(4+3 \mathrm{i})}{(1-\mathrm{i})}\)
(iii) \(\frac{(2+i)}{(3-i)(1+2 i)}\)
(iv) \(\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}\)
(v) \(\frac{2+\sqrt{-3}}{4+\sqrt{-3}}\)
(vi) (2 + 3i)(2 – 3i)
(vii) \(\frac{4 i^{8}-3 i^{9}+3}{3 i^{11}-4 i^{10}-2}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2.1
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2.2
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2.3
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2.4
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2.5

Question 3.
Show that (-1 + √3i)3 is a real number.
Solution:
(-1 + √3i)3
= (-1)3 + 3(-1)2 (√3i) + 3(-1)(√3i)2 +(√3i)3 [∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3]
= -1 + 3√3i – 3(3i2) + 3√3 i3
= -1 + 3√3i – 3(-3) – 3√3i [∵ i2 = -1, i3 = -1]
= -1 + 9
= 8, which is a real number.

Question 4.
Evaluate the following:
(i) i35
(ii) i888
(iii) i93
(iv) i116
(v) i403
(vi) \(\frac{1}{i^{58}}\)
(vii) i30 + i40 + i50 + i60
Solution:
We know that, i2 = -1, i3 = -i, i4 = 1
(i) i35 = (i4)8 (i2) i = (1)8 (-1) i = -i
(ii) i888 = (i4)222 = (1)222 = 1
(iii) i93 = (i4)23 . i = (1)23 . i = i
(iv) i116 = (i4)29 = (1)29 = 1
(v) i403 = (i4)100 (i2) i = (1)100 (-1) i = -i
(vi) \(\frac{1}{i^{88}}=\frac{1}{\left(i^{4}\right)^{14} \cdot i^{2}}=\frac{1}{(1)^{14}(-1)}=-1\)
(vii) i30 + i40 + i50 + i60
= (i4)7 i2 + (i4)10 + (i4)12 i2 + (i4)15
= (1)7 (-1) + (1)10 + (1)12 (-1) + (1)15
= -1 + 1 – 1 + 1
= 0

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Question 5.
Show that 1 + i10 + i20 + i30 is a real number.
Solution:
1 + i10 + i20 + i30
= 1 + (i4)2 . i2 + (i4)5 + (i4)7 . i2
= 1 + (1)2 (-1) + (1)5 + (1)7 (-1) [∵ i4 = 1, i2 = -1]
= 1 – 1 + 1 – 1
= 0, which is a real number.

Question 6.
Find the value of
(i) i49 + i68 + i89 + i110
(ii) i + i2 + i3 + i4
Solution:
(i) i49 + i68 + i89 + i110
= (i4)12 . i + (i4)17 + (i4)22 . i + (i4)27 . i2
= (1)12 . i + (1)17 + (1)22 . i + (1)27(-1) ……[∵ i4 = 1, i2 = -1]
= i + 1 + i – 1
= 2i

(ii) i + i2 + i3 + i4
= i + i2 + i2 . i + i4
= i – 1 – i + 1 [∵ i2 = -1, i4 = 1]
= 0

Question 7.
Find the value of 1 + i2 + i4 + i6 + i8 + …… + i20.
Solution:
1 + i2 + i4 + i6 + i8 + ….. + i20
= 1 + (i2 + i4) + (i6 + i8) + (i10 + i12) + (i14 + i16) + (i18 + i20)
= 1 + [i2 + (i2)2] + [(i2)3 + (i2)4] + [(i2)5 + (i2)6] + [(i2)7 + (i2)8] + [(i2)9 + (i2)10]
= 1 + [-1 + (- 1)2] + [(-1)3 + (-1)4] + [(-1)5 + (-1)6] + [(-1)7 + (-1)8] + [(-1)9 + (-1)10] [∵ i2 = -1]
= 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1)
= 1 + 0 + 0 + 0 + 0 + 0
= 1

Question 8.
Find the values of x and y which satisfy the following equations (x, y ∈ R):
(i) (x + 2y) + (2x – 3y)i + 4i = 5
(ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\)
Solution:
(i) (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) × 2 – equation (ii) gives
7y = 14
∴ y = 2
Putting y- 2 in (i), we get
x + 2(2) = 5
∴ x + 4 = 5
∴ x = 1
∴ x = 1 and y = 2
Check:
If x = 1 and y = 2 satisfy the given condition, then our answer is correct.
L.H.S. = (x + 2y) + (2x – 3y)i + 4i
= (1 + 4) + (2 – 6)i + 4i
= 5 – 4i + 4i
= 5
= R.H.S.
Thus, our answer is correct.

(ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q8
(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
Putting y = 2 in (i), we get
x + 2 = 0
∴ x = -2
∴ x = -2 and y = 2

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Question 9.
Find the value of:
(i) x3 – x2 + x + 46, if x = 2 + 3i
(ii) 2x3 – 11x2 + 44x + 27, if x = \(\frac{25}{3-4 i}\)
Solution:
(i) x = 2 + 3i
∴ x – 2 = 3i
∴ (x – 2)2 = 9i2
∴ x2 – 4x + 4 = 9(-1) …..[∵ i2 = -1]
∴ x2 – 4x + 13 = 0 ……(i)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q9
∴ x3 – x2 + x + 46 = (x2 – 4x + 13)(x + 3) + 7
= 0(x + 3) + 7 ……[From (i)]
= 7

(ii) x = \(\frac{25}{3-4 i}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q9.1
∴ x = 3 + 4i
∴ x – 3 = 4i
∴ (x – 3)2 = 16i2
∴ x2 – 6x + 9 = 16(-1) …….[∵ i2 = -1]
∴ x2 – 6x + 25 = 0 …….(i)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q9.2
∴ 2x3 – 11x2 + 44x + 27
= (x2 – 6x + 25) (2x + 1) + 2
= 0 . (2x + 1) + 2 ……[From (i)]
= 0 + 2
= 2