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

In Exercise 13.1 of Chapter 13, we shall discuss problems based on the integral powers of IOTA (i), and also study what is the need for complex numbers. The RD Sharma Class 11 Solutions for Maths is designed for students who aim to score high marks in their board exams. By referring to these solutions, students can follow the correct methodology for solving problems and can also clear their doubts pertaining to any question. Students can easily download RD Sharma Class 11 solutions, which is available in the pdf format, from the links given below.

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1. Evaluate the following:

(i) i ⁴⁵⁷

(ii) i ⁵²⁸

(iii) 1/ i⁵⁸

(iv) i ³⁷ + 1/i ⁶⁷

(v) [i ⁴¹ + 1/i ²⁵⁷]

(vi) (i ⁷⁷ + i ⁷⁰ + i ⁸⁷ + i ⁴¹⁴)³

(vii) i ³⁰ + i ⁴⁰ + i ⁶⁰

(viii) i ⁴⁹ + i ⁶⁸ + i ⁸⁹ + i ¹¹⁰

Solution:

(i) i ⁴⁵⁷

Let us simplify we get,

i⁴⁵⁷ = i ^{(456 + 1)}

= i ⁴⁽¹¹⁴⁾ × i

= (1)¹¹⁴ × i

= i [since i⁴ = 1]

(ii) i ⁵²⁸ 

Let us simplify we get,

i ⁵²⁸ = i ⁴⁽¹³²⁾

= (1)¹³² 

= 1 [since i⁴ = 1]

(iii) 1/ i⁵⁸

Let us simplify we get,

1/ i⁵⁸ = 1/ i ⁵⁶⁺²

= 1/ i ⁵⁶ × i²

= 1/ (i⁴)¹⁴ × i²

= 1/ i² [since, i⁴ = 1]

= 1/-1 [since, i² = -1]

= -1

(iv) i ³⁷ + 1/i ⁶⁷

Let us simplify we get,

i ³⁷ + 1/i ⁶⁷ = i³⁶⁺¹ + 1/ i⁶⁴⁺³

= i + 1/i³ [since, i⁴ = 1]

= i + i/i⁴

= i + i

= 2i

(v) [i ⁴¹ + 1/i ²⁵⁷]

Let us simplify we get,

= [i + 1/i]⁹ [since, 1/i = -1]

= [i – i]

= 0

(vi) (i ⁷⁷ + i ⁷⁰ + i ⁸⁷ + i ⁴¹⁴)³

Let us simplify we get,

(i ⁷⁷ + i ⁷⁰ + i ⁸⁷ + i ⁴¹⁴)³ = (i^{(76 + 1)} + i^{(68 + 2)} + i^{(84 + 3)} + i^{(412 + 2)} ) ³

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

= (i + (– 1) + (– i) + (– 1)) ³ 

= (– 2)³

= – 8

(vii) i ³⁰ + i ⁴⁰ + i ⁶⁰

Let us simplify we get,

i ³⁰ + i ⁴⁰ + i ⁶⁰ = i^{(28 + 2) +} i⁴⁰ + i⁶⁰

= (i⁴)⁷ i² + (i⁴)¹⁰ + (i⁴)¹⁵

= i² + 1¹⁰ + 1¹⁵ 

= – 1 + 1 + 1

= 1

(viii) i ⁴⁹ + i ⁶⁸ + i ⁸⁹ + i ¹¹⁰

Let us simplify we get,

i ⁴⁹ + i ⁶⁸ + i ⁸⁹ + i ¹¹⁰ = i^{(48 + 1)} + i⁶⁸ + i^{(88 + 1)} + i^{(116 + 2)}

= (i⁴)¹²×i + (i⁴)¹⁷ + (i⁴)¹¹×i + (i⁴)²⁹×i²

= i + 1 + i – 1

= 2i

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

Solution:

Given:

1 + i¹⁰ + i²⁰ + i³⁰ = 1 + i^{(8 + 2) +} i²⁰ + i^{(28 + 2)}

= 1 + (i⁴)² × i² + (i⁴)⁵ + (i⁴)⁷ × i²

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

= 0

Hence , 1 + i¹⁰ + i²⁰ + i³⁰ is a real number.

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

Let us simplify we get,

i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = i ^{(48 + 1)} + i⁶⁸ + i^{(88 + 1)} + i^{(108 + 2)}

= (i⁴)¹² × i + (i⁴)¹⁷ + (i⁴)¹¹ × i + (i⁴)²⁷ × i²

= i + 1 + i – 1 [since i⁴ = 1, i² = – 1]

= 2i

∴ i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = 2i

(ii) i³⁰ + i⁸⁰ + i¹²⁰

Let us simplify we get,

i³⁰ + i⁸⁰ + i¹²⁰ = i^{(28 + 2)} + i⁸⁰ + i¹²⁰

= (i⁴)⁷ × i² + (i⁴)²⁰ + (i⁴)³⁰

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

= 1

∴ i³⁰ + i⁸⁰ + i¹²⁰ = 1

(iii) i + i² + i³ + i⁴

Let us simplify we get,

i + i² + i³ + i⁴ = i + i² + i²×i + i⁴

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

= i – 1 – i + 1

= 0

∴ i + i² + i³ + i⁴ = 0

(iv) i⁵ + i¹⁰ + i¹⁵

Let us simplify we get,

i⁵ + i¹⁰ + i¹⁵ = i^{(4 + 1)} + i^{(8 + 2)} + i^{(12 + 3)}

= (i⁴)¹×i + (i⁴)²×i² + (i⁴)³×i³

= (i⁴)¹×i + (i⁴)²×i² + (i⁴)³×i²×i

= 1×i + 1 × (– 1) + 1 × (– 1)×i

= i – 1 – i

= – 1

∴ i⁵ + i¹⁰ + i¹⁵ = -1

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

Let us simplify we get,

= [i¹⁰ (i⁵⁸² + i⁵⁸⁰ + i⁵⁷⁸ + i⁵⁷⁶ + i⁵⁷⁴) / (i⁵⁸² + i⁵⁸⁰ + i⁵⁷⁸ + i⁵⁷⁶ + i⁵⁷⁴)]

= i¹⁰

= i⁸ i²

= (i⁴)² i²

= (1)² (-1) [since i⁴ = 1, i² = -1]           

                             = -1 

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

(vi) 1 + i² + i⁴ + i⁶ + i⁸ + … + i²⁰

Let us simplify we get,

1 + i² + i⁴ + i⁶ + i⁸ + … + i²⁰ = 1 + (– 1) + 1 + (– 1) + 1 + … + 1

= 1

∴ 1 + i² + i⁴ + i⁶ + i⁸ + … + i²⁰ = 1

(vii) (1 + i)⁶ + (1 – i)³

Let us simplify we get,

(1 + i)⁶ + (1 – i)³ = {(1 + i)² }³ + (1 – i)² (1 – i)

= {1 + i² + 2i}³ + (1 + i^{2 –} 2i)(1 – i)

= {1 – 1 + 2i}³ + (1 – 1 – 2i)(1 – i)

= (2i)³ + (– 2i)(1 – i)

= 8i³ + (– 2i) + 2i²

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

= – 10 i – 2

= – 2(1 + 5i)

= – 2 – 10i

∴ (1 + i)⁶ + (1 – i)³ = – 2 – 10i