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Question -

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



Answer -

(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

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