Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 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
Maharashtra Board Solution
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 & Step-by-Step Answer:
(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
Question 2
Maharashtra Board Solution
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) (iii) (iv) (v) (vi) (2 + 3i)(2 – 3i) (vii)
Solution & Step-by-Step Answer:
Question 3
Maharashtra Board Solution
Show that (-1 + √3i)3 is a real number.
Solution & Step-by-Step Answer:
(-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
Maharashtra Board Solution
Evaluate the following: (i) i35 (ii) i888 (iii) i93 (iv) i116 (v) i403 (vi) (vii) i30 + i40 + i50 + i60
Solution & Step-by-Step Answer:
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) (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
Question 5
Maharashtra Board Solution
Show that 1 + i10 + i20 + i30 is a real number.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
Find the value of (i) i49 + i68 + i89 + i110 (ii) i + i2 + i3 + i4
Solution & Step-by-Step Answer:
(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
Maharashtra Board Solution
Find the value of 1 + i2 + i4 + i6 + i8 + …… + i20.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
Find the values of x and y which satisfy the following equations (x, y ∈ R): (i) (x + 2y) + (2x – 3y)i + 4i = 5 (ii)
Solution & Step-by-Step Answer:
(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)
(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
Question 9
Maharashtra Board Solution
Find the value of: (i) x3 – x2 + x + 46, if x = 2 + 3i (ii) 2x3 – 11x2 + 44x + 27, if x =
Solution & Step-by-Step Answer:
(i) x = 2 + 3i ∴ x – 2 = 3i ∴ (x – 2)2 = 9i2 ∴ x2 – 4x + 4 = 9(-1) …..[∵ i2 = -1] ∴ x2 – 4x + 13 = 0 ……(i) ∴ x3 – x2 + x + 46 = (x2 – 4x + 13)(x + 3) + 7 = 0(x + 3) + 7 ……[From (i)] = 7
(ii) x =
∴ x = 3 + 4i
∴ x – 3 = 4i
∴ (x – 3)2= 16i2
∴ x2– 6x + 9 = 16(-1) …….[∵ i2= -1]
∴ x2– 6x + 25 = 0 …….(i)
∴ 2x3– 11x2+ 44x + 27
= (x2– 6x + 25) (2x + 1) + 2
= 0. (2x + 1) + 2 ……[From (i)]
= 0 + 2
= 2
