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 i4Â = 1]
(ii) i 528
Let us simplify we get,
i 528Â = i 4(132)
= (1)132
= 1 [since i4Â = 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, i4Â = 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, i4Â = 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)Â + i(68 + 2)Â + i(84 + 3)Â + i(412 + 2)Â )Â 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) +Â i40Â + i60
= (i4)7Â i2Â + (i4)10Â + (i4)15
= i2Â + 110Â + 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)Â + i68Â + i(88 + 1)Â + i(116 + 2)
= (i4)12×i + (i4)17 + (i4)22×i + (i4)29×i2
= i + 1 + i – 1
= 2i
2. Show that 1 + i10Â + i20Â + i30Â is a real number?
Solution:
Given:
1 + i10Â + i20Â + i30Â = 1 + i(8 + 2) +Â i20Â + i(28 + 2)
= 1 + (i4)2 × i2 + (i4)5 + (i4)7 × i2
= 1 – 1 + 1 – 1 [since, i4 = 1, i2 = – 1]
= 0
Hence , 1 + i10Â + i20Â + i30Â is a real number.
3. Find the values of the following expressions:
(i) i49Â + i68Â + i89Â + i110
(ii) i30Â + i80Â + i120
(iii) i + i2Â + i3Â + i4
(iv) i5Â + i10Â + i15
(v) [i592 + i590 + i588 + i586 + i584] / [i582 + i580 + i578 + i576 + i574]
(vi) 1 + i2Â + i4Â + i6Â + i8Â + … + i20
(vii) (1 + i)6 + (1 – i)3
Solution:
(i) i49Â + i68Â + i89Â + i110
Let us simplify we get,
i49Â + i68Â + i89Â + i110 = i (48 + 1)Â + i68Â + i(88 + 1)Â + i(108 + 2)
= (i4)12 × i + (i4)17 + (i4)22 × i + (i4)27 × i2
= i + 1 + i – 1 [since i4 = 1, i2 = – 1]
= 2i
∴ i49 + i68 + i89 + i110 = 2i
(ii) i30Â + i80Â + i120
Let us simplify we get,
i30Â + i80Â + i120 = i(28 + 2)Â + i80Â + i120
= (i4)7 × i2 + (i4)20 + (i4)30
= – 1 + 1 + 1 [since i4 = 1, i2 = – 1]
= 1
∴ i30 + i80 + i120 = 1
(iii) i + i2Â + i3Â + 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) i5Â + i10Â + i15
Let us simplify we get,
i5Â + i10Â + i15 = i(4 + 1)Â + i(8 + 2)Â + 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)2Â i2
= (1)2Â (-1) [since i4Â = 1, i2Â = -1]
= -1Â
∴ [i592 + i590 + i588 + i586 + i584] / [i582 + i580 + i578 + i576 + i574] = -1
(vi) 1 + i2Â + i4Â + i6Â + i8Â + … + 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