• +91 9971497814
• info@interviewmaterial.com

# RD Chapter 13- Complex Numbers Ex-13.1 Interview Questions Answers

### Related Subjects

Question 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

(i) i 457

Let us simplify weget,

i457 =i (456 + 1)

= i 4(114) ×i

= (1)114 ×i

= i [since i4 =1]

(ii) i 528

Let us simplify weget,

528 =i 4(132)

= (1)132

= 1 [since i4 =1]

(iii) 1/ i58

Let us simplify weget,

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) 37 +1/i 67

Let us simplify weget,

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 weget,

[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 weget,

(i 77 +i 70 + i 87 + i 414)=(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) 30 +i 40 + i 60

Let us simplify weget,

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) 49 +i 68 + i 89 + i 110

Let us simplify weget,

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

Question 2 :

Show that 1 + i10 + i20 + i30 isa real number?

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.

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

(i) i49 +i68 + i89 + i110

Let us simplify weget,

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 [sincei4 = 1, i2 = – 1]

= 2i

i49 +i68 + i89 + i110 = 2i

(ii) i30 +i80 + i120

Let us simplify weget,

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 weget,

i + i2 +i3 + i= 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 weget,

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 weget,

[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 weget,

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 weget,

(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 [sincei3 = – i, i2 = – 1]

= – 10 i – 2

= – 2(1 + 5i)

= – 2 – 10i

(1 + i)6 +(1 – i)3 = – 2 – 10i

krishan