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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Access answers to RD Sharma Solutions for Class 11 Maths Exercise 13.1 Chapter 13 – Complex Numbers

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 41 + 1/i 257] = [i40+1 + 1/ i256+1]

= [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,

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

= [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 + i2 – 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

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