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Class 11 RD Sharma Solutions – Chapter 13 Complex Numbers – Exercise 13.1

  • Last Updated : 17 Dec, 2020

Question 1: Evaluate the following:

(i) i457

(ii) i528

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(iii) 1/i58



(iv) i37 + 1/i67

(v) [i41 + 1/i257]9

(vi) (i77 + i70 + i87 + i414)3

(vii) i30 + i40 + i60

(viii) i49 + i68 + i89 + i110

Solution:

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) × i     [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

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

Solution:

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.

Question 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

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




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