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Question -

Find the values of the following expressions:

(i) i49┬а+ i68┬а+ i89┬а+ i110

(ii) i30┬а+ i80┬а+ i120

(iii) i + i2┬а+ i3┬а+ i4

(iv) i5┬а+ i10┬а+ i15

(v) [i592┬а+ i590┬а+ i588┬а+i586┬а+ i584] / [i582┬а+ i580┬а+i578┬а+ i576┬а+ i574]

(vi) 1 + i2┬а+ i4┬а+ i6┬а+ i8┬а+тАж + i20

(vii) (1 + i)6┬а+ (1 тАУ i)3



Answer -

(i)┬аi49┬а+i68┬а+ i89┬а+ i110

Let us simplify weget,

i49┬а+i68┬а+ i89┬а+ i110┬а= i┬а(48+ 1)┬а+ i68┬а+ i(88 + 1)┬а+ i(108+ 2)

= (i4)12┬а├Чi + (i4)17┬а+ (i4)22┬а├Ч i +(i4)27┬а├Ч i2

= i + 1 + i тАУ 1 [sincei4┬а= 1, i2┬а= тАУ 1]

= 2i

тИ┤ i49┬а+i68┬а+ i89┬а+ i110┬а= 2i

(ii)┬аi30┬а+i80┬а+ i120

Let us simplify weget,

i30┬а+i80┬а+ i120┬а= i(28 + 2)┬а+ i80┬а+i120

= (i4)7┬а├Чi2┬а+ (i4)20┬а+ (i4)30

= тАУ 1 + 1 + 1 [since i4┬а=1, i2┬а= тАУ 1]

= 1

тИ┤ i30┬а+i80┬а+ i120┬а= 1

(iii)┬аi + i2┬а+i3┬а+ i4

Let us simplify weget,

i + i2┬а+i3┬а+ i4┬а= i + i2┬а+ i2├Чi+ i4

= i тАУ 1 + (тАУ 1) ├Ч i +1 [since i4┬а= 1, i2┬а= тАУ 1]

= i тАУ 1 тАУ i + 1

= 0

тИ┤ i + i2┬а+i3┬а+ i4┬а= 0

(iv)┬аi5┬а+ i10┬а+i15

Let us simplify weget,

i5┬а+ i10┬а+i15┬а= i(4 + 1)┬а+ i(8 + 2)┬а+ i(12+ 3)

= (i4)1├Чi+ (i4)2├Чi2┬а+ (i4)3├Чi3

= (i4)1├Чi+ (i4)2├Чi2┬а+ (i4)3├Чi2├Чi

= 1├Чi + 1 ├Ч (тАУ 1) + 1├Ч (тАУ 1)├Чi

= i тАУ 1 тАУ i

= тАУ 1

тИ┤ i5┬а+i10┬а+ i15┬а= -1

(v)┬а[i592┬а+i590┬а+ i588┬а+ i586┬а+ i584]/ [i582┬а+ i580┬а+ i578┬а+ i576┬а+i574]

Let us simplify weget,

[i592┬а+i590┬а+ i588┬а+ i586┬а+ i584]/ [i582┬а+ i580┬а+ i578┬а+ i576┬а+i574]

= [i10┬а(i582┬а+i580┬а+ i578┬а+ i576┬а+ i574)/ (i582┬а+ i580┬а+ i578┬а+ i576┬а+i574)]

= i10

= i8┬аi2

= (i4)2┬аi2

= (1)2┬а(-1) [since i4┬а= 1, i2┬а=-1]

= -1┬а

тИ┤ [i592┬а+i590┬а+ i588┬а+ i586┬а+ i584]/ [i582┬а+ i580┬а+ i578┬а+ i576┬а+i574] = -1

(vi)┬а1 + i2┬а+i4┬а+ i6┬а+ i8┬а+ тАж + i20

Let us simplify weget,

1 + i2┬а+i4┬а+ i6┬а+ i8┬а+ тАж + i20┬а=1 + (тАУ 1) + 1 + (тАУ 1) + 1 + тАж + 1

= 1

тИ┤ 1 + i2┬а+i4┬а+ i6┬а+ i8┬а+ тАж + i20┬а=1

(vii)┬а(1 + i)6┬а+(1 тАУ i)3

Let us simplify weget,

(1 + i)6┬а+(1 тАУ i)3┬а= {(1 + i)2┬а}3┬а+ (1 тАУi)2┬а(1 тАУ i)

= {1 + i2┬а+2i}3┬а+ (1 + i2 тАУ┬а2i)(1 тАУ i)

= {1 тАУ 1 + 2i}3┬а+(1 тАУ 1 тАУ 2i)(1 тАУ i)

= (2i)3┬а+(тАУ 2i)(1 тАУ i)

= 8i3┬а+(тАУ 2i) + 2i2

= тАУ 8i тАУ 2i тАУ 2 [sincei3┬а= тАУ i, i2┬а= тАУ 1]

= тАУ 10 i тАУ 2

= тАУ 2(1 + 5i)

= тАУ 2 тАУ 10i

тИ┤ (1 + i)6┬а+(1 тАУ i)3┬а= тАУ 2 тАУ 10i

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