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9. Value of i4k + j4k+1 + i4k+2 + i4k+3 is​

Answer 1

$i^{4k}+i^{4k+1}+i^{4k+2}+i^{4k+3}=0$

Given :  $i^{4k}+i^{4k+1}+i^{4k+2}+i^{4k+3}$

( corrected Question )

To Find : Evaluate

Solution:

$i^{4k}+i^{4k+1}+i^{4k+2}+i^{4k+3}$

one Law of Exponent : xᵃ⁺ᵇ=xᵃ × xᵇ

i² = - 1

i³ = - i

i⁴ = 1

$i^{4k}+i^{4k+1}+i^{4k+2}+i^{4k+3}$

Taking $i^{4k}$ common

$i^{4k}$ (1 + i + i²  + i³)

= $i^{4k}$ (1 + i - 1 - i)

= $i^{4k}$ (0)

= 0

Hence $i^{4k}+i^{4k+1}+i^{4k+2}+i^{4k+3}=0$

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