11th Commerce Maths 1 Chapter 3 Exercise 3.1 Answers Maharashtra Board

Complex Numbers Class 11 Commerce Maths 1 Chapter 3 Exercise 3.1 Answers Maharashtra Board

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

Std 11 Maths 1 Exercise 3.1 Solutions Commerce Maths

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

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:

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 = a ³ + 3a ² b + 3ab ² + b ³ ]
= -1 + 3√3i – 3(3i ² ) + 3√3 i ³
= -1 + 3√3i – 3(-3) – 3√3i [∵ i ² = -1, i ³ = -1]
= -1 + 9
= 8, which is a real number.

Question 4.
Evaluate the following:
(i) i ³⁵
(ii) i ⁸⁸⁸
(iii) i ⁹³
(iv) i ¹¹⁶
(v) i ⁴⁰³
(vi) \(\frac{1}{i^{58}}\)
(vii) i ³⁰ + i ⁴⁰ + i ⁵⁰ + i ⁶⁰
Solution:
We know that, i ² = -1, i ³ = -i, i ⁴ = 1
(i) i ³⁵ = (i ⁴ ) ⁸ (i ² ) i = (1) ⁸ (-1) i = -i
(ii) i ⁸⁸⁸ = (i ⁴ ) ²²² = (1) ²²² = 1
(iii) i ⁹³ = (i ⁴ ) ²³ . i = (1) ²³ . i = i
(iv) i ¹¹⁶ = (i ⁴ ) ²⁹ = (1) ²⁹ = 1
(v) i ⁴⁰³ = (i ⁴ ) ¹⁰⁰ (i ² ) 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) i ³⁰ + i ⁴⁰ + i ⁵⁰ + i ⁶⁰
= (i ⁴ ) ⁷ i ² + (i ⁴ ) ¹⁰ + (i ⁴ ) ¹² i ² + (i ⁴ ) ¹⁵
= (1) ⁷ (-1) + (1) ¹⁰ + (1) ¹² (-1) + (1) ¹⁵
= -1 + 1 – 1 + 1
= 0

Question 5.
Show that 1 + i ¹⁰ + i ²⁰ + i ³⁰ is a real number.
Solution:
1 + i ¹⁰ + i ²⁰ + i ³⁰
= 1 + (i ⁴ ) ² . i ² + (i ⁴ ) ⁵ + (i ⁴ ) ⁷ . i ²
= 1 + (1) ² (-1) + (1) ⁵ + (1) ⁷ (-1) [∵ i ⁴ = 1, i ² = -1]
= 1 – 1 + 1 – 1
= 0, which is a real number.

Question 6.
Find the value of
(i) i ⁴⁹ + i ⁶⁸ + i ⁸⁹ + i ¹¹⁰
(ii) i + i ² + i ³ + i ⁴
Solution:
(i) i ⁴⁹ + i ⁶⁸ + i ⁸⁹ + i ¹¹⁰
= (i ⁴ ) ¹² . i + (i ⁴ ) ¹⁷ + (i ⁴ ) ²² . i + (i ⁴ ) ²⁷ . i ²
= (1) ¹² . i + (1) ¹⁷ + (1) ²² . i + (1) ²⁷ (-1) ……[∵ i ⁴ = 1, i ² = -1]
= i + 1 + i – 1
= 2i

(ii) i + i ² + i ³ + i ⁴
= i + i ² + i ² . i + i ⁴
= i – 1 – i + 1 [∵ i ² = -1, i ⁴ = 1]
= 0

Question 7.
Find the value of 1 + i ² + i ⁴ + i ⁶ + i ⁸ + …… + i ²⁰ .
Solution:
1 + i ² + i ⁴ + i ⁶ + i ⁸ + ….. + i ²⁰
= 1 + (i ² + i ⁴ ) + (i ⁶ + i ⁸ ) + (i ¹⁰ + i ¹² ) + (i ¹⁴ + i ¹⁶ ) + (i ¹⁸ + i ²⁰ )
= 1 + [i ² + (i ² ) ² ] + [(i ² ) ³ + (i ² ) ⁴ ] + [(i ² ) ⁵ + (i ² ) ⁶ ] + [(i ² ) ⁷ + (i ² ) ⁸ ] + [(i ² ) ⁹ + (i ² ) ¹⁰ ]
= 1 + [-1 + (- 1) ² ] + [(-1) ³ + (-1) ⁴ ] + [(-1) ⁵ + (-1) ⁶ ] + [(-1) ⁷ + (-1) ⁸ ] + [(-1) ⁹ + (-1) ¹⁰ ] [∵ i ² = -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) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\)

(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.
Find the value of:
(i) x ³ – x ² + x + 46, if x = 2 + 3i
(ii) 2x ³ – 11x ² + 44x + 27, if x = \(\frac{25}{3-4 i}\)
Solution:
(i) 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)

∴ x ³ – x ² + x + 46 = (x ² – 4x + 13)(x + 3) + 7
= 0(x + 3) + 7 ……[From (i)]
= 7

(ii) x = \(\frac{25}{3-4 i}\)

∴ x = 3 + 4i
∴ x – 3 = 4i
∴ (x – 3) ² = 16i ²
∴ x ² – 6x + 9 = 16(-1) …….[∵ i ² = -1]
∴ x ² – 6x + 25 = 0 …….(i)

∴ 2x ³ – 11x ² + 44x + 27
= (x ² – 6x + 25) (2x + 1) + 2
= 0 . (2x + 1) + 2 ……[From (i)]
= 0 + 2
= 2