Class 11 RD Sharma Solutions - Chapter 13 Complex Numbers - Exercise 13.1

Chapter 13 of RD Sharma's Class 11 Mathematics textbook delves into the fascinating world of complex numbers. Complex numbers are an extension of real numbers incorporating both the real part and an imaginary part. This chapter focuses on solving problems related to the fundamental properties and operations of complex numbers providing a solid foundation for further study in algebra and calculus.

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part. They are expressed in the form a+bi, where a is the real part the imaginary part and i is the imaginary unit with property i²=−1. Complex numbers enable us to solve equations that have no real solutions and are crucial in various fields including engineering, physics, and applied mathematics.

Question 1: Evaluate the following:

(i) i⁴⁵⁷

(ii) i⁵²⁸

(iii) 1/i⁵⁸

(iv) i³⁷ + 1/i⁶⁷

(v) [i⁴¹ + 1/i²⁵⁷]⁹

(vi) (i⁷⁷ + i⁷⁰ + i⁸⁷ + i⁴¹⁴)³

(vii) i³⁰ + i⁴⁰ + i⁶⁰

(viii) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰

Solution:

We know that i = √-1

i² = -1

i³ = -i

i⁴ = 1

(i) i⁴⁵⁷

To find iⁿ,

As n is greater than 4, so we divide 457 by 4, we get

When dividing 457 by 4 we get quotient (p) as 114 and remainder (q) as 1

Therefore, substituting the value of p and q in the equation iⁿ = i^4p+q we get.

i⁴⁵⁷ = i ^{4(114) + 1}

i⁴⁵⁷ = i ⁴⁽¹¹⁴⁾ × i

i⁴⁵⁷ = (1)¹¹⁴ × i    [As i⁴ = 1, therefore 1¹¹⁴ = 1]

i⁴⁵⁷ = i 

(ii) i⁵²⁸

To find iⁿ,

As n is greater than 4, so we divide 528 by 4, we get

When dividing 528 by 4 we get quotient (p) as 132 and the remainder (q) as 0

Therefore, substituting the value of p and q in the equation iⁿ = i^4p+q we get.

i⁵²⁸ = i⁴⁽¹³²⁾

i⁵²⁸= (1)¹³²    [As i⁴ = 1, therefore 1¹³² = 1]

i⁵²⁸= 1 

(iii) 1/ i⁵⁸

To find iⁿ,

As n is greater than 4, so we divide 58 by 4, we get

When dividing 58 by 4 we get quotient (p) as 14 and the remainder (q) as 2

Therefore, substituting the value of p and q in the equation iⁿ = i4p+q we get.

1/ i⁵⁸ = 1/ i ^{4(14) + 2}

1/ i⁵⁸ = 1/ i⁴⁽¹⁴⁾ × i²      [As i⁴ = 1, therefore 1¹⁴ = 1]

1/ i⁵⁸ = 1/ i² [since, i² = -1]

1/ i⁵⁸ = 1/-1

1/ i⁵⁸ = -1

(iv) i³⁷ + 1/i⁶⁷

To find iⁿ,

As n is greater than 4, so we divide 37 and 67 by 4, we get

When dividing 37 by 4 we get quotient (p) as 9 and the remainder (q) as 1

When dividing 67 by 4 we get quotient (p) as 16 and the remainder (q) as 3

Therefore, substituting the value of p and q in the equation iⁿ = i^4p+q we get.

i³⁷ + 1/i⁶⁷ = i⁴⁽⁹⁾⁺¹ + 1/ i^4(16)+3

 = i⁴⁽⁹⁾×i + 1/ i⁴⁽¹⁶⁾×i³

= i + 1/i³     [As, i⁴ = 1]

Multiplying numerator and denominator by i, we get

= i + i/i⁴

= i + i

i³⁷ + 1/i⁶⁷ = 2i

(v) [i⁴¹ + 1/i²⁵⁷]⁹

To find iⁿ,

As n is greater than 4, so we divide 41 and 257 by 4, we get

When dividing  by 4 we get quotient (p) as 10 and the remainder (q) as 1

When dividing 257 by 4 we get quotient (p) as 64 and the remainder (q) as 1 

Therefore, substituting the value of p and q in the equation iⁿ = i^4p+q we get,

[i⁴¹ + 1/i²⁵⁷] = [i^4(10)+1 + 1/ i^4(64)+1]⁹

 = [ i⁴⁽¹⁰⁾×i + 1/ i⁴⁽⁶⁴⁾×i ]⁹

= [i + 1/i]⁹ [As, i⁴ = 1 and 1/i = -1]

= [i – i]⁹

= 0

(vi) (i⁷⁷ + i⁷⁰ + i⁸⁷ + i⁴¹⁴)³

To find iⁿ,

As n is greater than 4, so we divide 77, 70, 87 and 414 by 4, we get

When dividing 77 by 4 we get quotient (p) as 19 and the remainder (q) as 1.

When dividing 70 by 4 we get quotient (p) as 17 and the remainder (q) as 2.

When dividing 87 by 4 we get quotient (p) as 21 and the remainder (q) as 3.

When dividing 414 by 4 we get quotient (p) as 103 and the remainder (q) as 2 .

Therefore, substituting the value of p and q in the equation iⁿ = i^4p+q we get,

(i⁷⁷ + i⁷⁰ + i⁸⁷ + i⁴¹⁴)³ = (i^{4(17)+ 1} + i^{4(21) + 2} + i^{4(21) + 3} + i^{4(103) + 2} )³

= (i⁴⁽¹⁷⁾× i + i⁴⁽²¹⁾ × i² + i⁴⁽²¹⁾ × i³ + i⁴⁽¹⁰³⁾ × i² )³     [As, i⁴ = 1 ]

= (i + i² + i³ + i²)³ [As, i³ = – i, i² = – 1]

= (i + (– 1) + (– i) + (– 1))³

= (– 2)³

= – 8

(vii) i³⁰ + i⁴⁰ + i⁶⁰

To find iⁿ,

As n is greater than 4, so we divide 30, 40 and 60 by 4, we get

When dividing 30 by 4 we get quotient (p) as 7 and the remainder (q) as 2.

When dividing 40 by 4 we get quotient (p) as 10 and the remainder (q) as 0.

When dividing 60 by 4 we get quotient (p) as 15 and the remainder (q) as 0.

Therefore substituting the value of p and q in the equation iⁿ = i^4p+q we get,

i³⁰ + i⁴⁰ + i⁶⁰ = i^{4(7) + 2} + i⁴⁽¹⁰⁾ + i⁴⁽¹⁵⁾

= i⁴⁽⁷⁾ × i² + i⁴⁽¹⁰⁾ + i⁴⁽¹⁵⁾

= i² + 1¹⁰ + 1¹⁵      [As, i⁴ = 1 and i² = -1]

= – 1 + 1 + 1

= 1

(viii) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰

To find iⁿ,

As n is greater than 4, so we divide 49, 68, 89 and 110 by 4, we get

When dividing 49 by 4 we get quotient (p) as 12 and the remainder (q) as 1.

When dividing 68 by 4 we get quotient (p) as 17 and the remainder (q) as 0.

When dividing 89 by 4 we get quotient (p) as 22 and the remainder (q) as 1.

When dividing 110 by 4 we get quotient (p) as 27 and the remainder (q) as 2.

Therefore substituting the value of p and q in the equation iⁿ = i^4p+q we get,

i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = i^{4(12) + 1} + i⁴⁽¹⁷⁾ + i^{4(22) + 1} + i^{4(27) + 2}

= i⁴⁽¹²⁾ × i + i⁴⁽¹⁷⁾ + i⁴⁽²²⁾ × i + i⁴⁽²⁷⁾ × i²

= i + 1 + i – 1    [As, i4 = 1]

= 2i

Question 2: Show that 1 + i¹⁰ + i²⁰ + i³⁰ is a real number?

Solution:

Given: 1 + i¹⁰ + i²⁰ + i³⁰

To find iⁿ,

As n is greater than 4, so we divide 10, 20, and 30 by 4, we get

When dividing 10 by 4 we get quotient (p) as 2 and the remainder (q) as 2.

When dividing 20 by 4 we get quotient (p) as 5 and the remainder (q) as 0.

When dividing 30 by 4 we get quotient (p) as 7 and the remainder (q) as 2.

Therefore substituting the value of p and q in the equation iⁿ = i^4p+q we get,

1 + i¹⁰ + i²⁰ + i³⁰ = 1 + i^{4(2) + 2} + i⁴⁽⁵⁾ + i^{4(7) + 2}

= 1 + i⁴⁽²⁾ × i² + i⁴⁽⁵⁾ + i⁴⁽⁷⁾ × i²

= 1 – 1 + 1 – 1 [As, i⁴ = 1, i² = – 1]

= 0

Therefore , 1 + i¹⁰ + i²⁰ + i³⁰ =0, and 0 is a real number.

Question 3: Find the values of the following expressions:

(i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰

(ii) i³⁰ + i⁸⁰ + i¹²⁰

(iii) i + i² + i³ + i⁴

(iv) i⁵ + i¹⁰ + i¹⁵

(v) [i⁵⁹² + i⁵⁹⁰ + i⁵⁸⁸ + i⁵⁸⁶ + i⁵⁸⁴]/[i⁵⁸² + i⁵⁸⁰ + i⁵⁷⁸ + i⁵⁷⁶ + i⁵⁷⁴]

(vi) 1 + i² + i⁴ + i⁶ + i⁸ + … + i²⁰

(vii) (1 + i)⁶ + (1 – i)³

Solution:

(i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰

To find iⁿ,

As n is greater than 4, so we divide 49, 68, 89 and 110 by 4, we get

When dividing 49 by 4 we get quotient (p) as 12 and the remainder (q) as 1.

When dividing 68 by 4 we get quotient (p) as 17 and the remainder (q) as 0

When dividing 89 by 4 we get quotient (p) as 22 and the remainder (q) as 1.

When dividing 110 by 4 we get quotient (p) as 27 and the remainder (q) as 2.

Therefore substituting the value of p and q in the equation iⁿ = i^4p+q we get,

i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = i^{4(12) + 1} + i⁴⁽¹⁷⁾ + i^{4(22) + 1} + i^{4(27) + 2}

= i⁴⁽¹²⁾ × i + i⁴⁽¹⁷⁾ + i⁴⁽²²⁾ × i + i⁴⁽²⁷⁾ × i²

= i + 1 + i – 1    [As, i4 = 1]

= 2i

Therefore, i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = 2i

(ii) i³⁰ + i⁸⁰ + i¹²⁰

To find iⁿ,

As n is greater than 4, so we divide 30, 80, and 120 by 4, we get

When dividing 30 by 4 we get quotient (p) as 7 and the remainder (q) as 2.

When dividing 80 by 4 we get quotient (p) as 20 and the remainder (q) as 0.

When dividing 120 by 4 we get quotient (p) as 30 and the remainder (q) as 0.

Therefore substituting the value of p and q in the equation iⁿ = i^4p+q we get,

i³⁰ + i⁸⁰ + i¹²⁰ = i4(7) + 2 + i4(20) + i4(30)

= i⁴⁽⁷⁾ × i² + i⁴⁽²⁰⁾ + i⁴⁽³⁰⁾

= – 1 + 1 + 1 [As, i⁴ = 1, i² = – 1]

= 1

Therefore, i³⁰ + i⁸⁰ + i¹²⁰ = 1

(iii) i + i² + i³ + i⁴

i + i² + i³ + i⁴ = i + i² + i²⁺¹ + i⁴

= i + i² + i²×i + i4

= i – 1 + (– 1) × i + 1 [As, i⁴ = 1, i² = – 1]

= i – 1 – i + 1

= 0

Therefore, i + i² + i³ + i⁴ = 0

(iv) i⁵ + i¹⁰ + i¹⁵

To find iⁿ,

As n is greater than 4, so we divide 5, 10, and 10 by 4, we get

When dividing 5 by 4 we get quotient (p) as 1 and the remainder (q) as 1.

When dividing 10 by 4 we get quotient (p) as 2 and the remainder (q) as 1.

When dividing 15 by 4 we get quotient (p) as 3 and the remainder (q) as 3.

Therefore substituting the value of p and q in the equation iⁿ = i^4p+q we get,

i⁵ + i¹⁰ + i¹⁵ = i^{4(1) + 1} + i^{4(2) + 2} + i^{4(3) + 3}

= i⁴×i + i⁴⁽²⁾×i² + i⁴⁽³⁾×i³

= i⁴×i + i⁴⁽²⁾×i² + i⁴⁽³⁾×i²×i     [As, i⁴ = 1, i² = – 1]

= 1×i + 1 × (– 1) + 1 × (– 1)×i

= i – 1 – i

= – 1

Therefore, i⁵ + i¹⁰ + i¹⁵ = -1

(v) [i⁵⁹² + i⁵⁹⁰ + i⁵⁸⁸ + i⁵⁸⁶ + i⁵⁸⁴] / [i⁵⁸² + i⁵⁸⁰ + i⁵⁷⁸ + i⁵⁷⁶ + i⁵⁷⁴]

[i⁵⁹² + i⁵⁹⁰ + i⁵⁸⁸ + i⁵⁸⁶ + i⁵⁸⁴] / [i⁵⁸² + i⁵⁸⁰ + i⁵⁷⁸ + i⁵⁷⁶ + i⁵⁷⁴]

= [i¹⁰ (i⁵⁸² + i⁵⁸⁰ + i⁵⁷⁸ + i⁵⁷⁶ + i⁵⁷⁴) / (i⁵⁸² + i⁵⁸⁰ + i⁵⁷⁸ + i⁵⁷⁶ + i⁵⁷⁴)]       [Taking i¹⁰ as common from numerator]

= i¹⁰

To find iⁿ,

As n is greater than 4, so 

When dividing 10 by 4 we get quotient (p) as 2 and the remainder (q) as 2.

Therefore substituting the value of p and q in the equation iⁿ = i^4p+q we get,

= i⁴⁽²⁾⁺²

= i⁴⁽²⁾ × i²

= -1 [As, i⁴ = 1, i² = -1]

= -1  

Therefore, [i⁵⁹² + i⁵⁹⁰ + i⁵⁸⁸ + i⁵⁸⁶ + i⁵⁸⁴] / [i⁵⁸² + i⁵⁸⁰ + i⁵⁷⁸ + i⁵⁷⁶ + i⁵⁷⁴] = -1

(vi) 1 + i² + i⁴ + i⁶ + i⁸ + … + i²⁰

When n is greater than 4, so we divide n by 4,

Here we will divide all the values greater than 4 i.e 6, 8, 10, 12, 14, 16, 18, and 20.

1 + i² + i⁴ + i⁶ + i⁸ + … + i²⁰ = 1 + i² + i⁴ + i⁴⁺² + i⁴⁺⁴   + ... + i⁴⁽⁵⁾

= 1 + (– 1) + 1 + (– 1) + 1 + … + 1   [As, i4 = 1, i2 = -1]

= 1

Therefore, 1 + i² + i⁴ + i⁶ + i⁸ + … + i²⁰ = 1

(vii) (1 + i)⁶ + (1 – i)³

(1 + i)⁶ + (1 – i)³ = [(1 + i)² ]³ + (1 – i)² (1 – i)

= [1 + i² + 2i]³ + (1 + i² – 2i)(1 – i)   [By using formula (a+b)² = a² + b²+ 2ab ]

= [1 – 1 + 2i]³ + (1 – 1 – 2i)(1 – i)

= (2i)³ + (– 2i)(1 – i)

= 8i³ + (– 2i) + 2i²

= – 8i – 2i – 2 [As, i³ = – i, i² = – 1]

= – 10i – 2

= – 2 – 10i

Therefore (1 + i)⁶ + (1 – i)³ = – 2 – 10i

Read More:

• Complex Numbers
• Complex Number Formula

Conclusion

Exercise 13.1 in Chapter 13 of RD Sharma’s Class 11 textbook provides the comprehensive set of the problems designed to the enhance the understanding of the complex numbers. By practicing these problems students can effectively grasp the basic operations and properties of the complex numbers laying the groundwork for the more advanced topics in the mathematics.