In Exercise 13.1 of Chapter 13, we shall discuss problems based on the integral powers of IOTA (i) and also study what the need for complex numbers is. The RD Sharma Class 11 Solutions for Maths are 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 are available in PDF format, from the links given below.
RD Sharma Solutions for Class 11 Maths Exercise 13.1 Chapter 13 – Complex Numbers
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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.
i457 = i (456 + 1)
= i 4(114) × i
= (1)114 × i
= i [since i4 = 1]
(ii) i 528
Let us simplify.
i 528 = i 4(132)
= (1)132
= 1 [since i4 = 1]
(iii) 1/ i58
Let us simplify.
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.
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.
[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.
(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.
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.
i 49 + i 68 + i 89 + i 110 = i(48 + 1) + i68 + i(88 + 1) + i(116 + 2)
= (i4)12×i + (i4)17 + (i4)22×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.
i49 + i68 + i89 + i110 = i (48 + 1) + i68 + i(88 + 1) + 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) i30 + i80 + i120
Let us simplify.
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.
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.
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.
[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.
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.
(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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