问题1：评估以下内容：
(i)我457
(ii)我528
(iii)1 / i 58
(iv)i 37 + 1 / i 67
(v)[i 41 + 1 / i 257 ] 9
(vi)(i 77 + i 70 + i 87 + i 414 ) 3
(vii)我30 +我40 +我60
(viii)i 49 + i 68 + i 89 + i 110
解决方案：
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 = i4p+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 = i4p+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 = i4p+q we get.
i37 + 1/i67 = i4(9)+1 + 1/ i4(16)+3
= i4(9)×i + 1/ i4(16)×i3
= i + 1/i3 [As, i4 = 1]
Multiplying numarator 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 = i4p+q we get,
[i41 + 1/i257] = [i4(10)+1 + 1/ i4(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 = i4p+q we get,
(i77 + i70 + i87 + i414)3 = (i4(17)+ 1 + i4(21) + 2 + i4(21) + 3 + i4(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 = i4p+q we get,
i30 + i40 + i60 = i4(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 = i4p+q we get,
i49 + i68 + i89 + i110 = i4(12) + 1 + i4(17) + i4(22) + 1 + i4(27) + 2
= i4(12) × i + i4(17) + i4(22) × i + i4(27) × i2
= i + 1 + i – 1 [As, i4 = 1]
= 2i
问题2：证明1 + i 10 + i 20 + i 30是实数吗?
解决方案：
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 = i4p+q we get,
1 + i10 + i20 + i30 = 1 + i4(2) + 2 + i4(5) + i4(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.
问题3：查找以下表达式的值：
(i)我49 +我68 +我89 +我110
(ii)我30 +我80 +我120
(iii)我+我2 +我3 +我4
(iv)我5 +我10 +我15
(v)[i 592 + i 590 + i 588 + i 586 + i 584 ] / [i 582 + i 580 + i 578 + i 576 + i 574 ]
(vi)1 + i 2 + i 4 + i 6 + i 8 +…+ i 20
(vii)(1 + i) 6 +(1 – i) 3
解决方案：
(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 = i4p+q we get,
i49 + i68 + i89 + i110 = i4(12) + 1 + i4(17) + i4(22) + 1 + i4(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 = i4p+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 = i4p+q we get,
i5 + i10 + i15 = i4(1) + 1 + i4(2) + 2 + i4(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 = i4p+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