Calculate i^21 + i^22 + i^23 + i^24?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to calculate i^21 + i^22 + i^23 + i^24

i is an imaginary number equal to sqrt (-1). i^2 = -1, i^3 = (-1)*i = -i and i^4 = (-1)^2 = 1.

i^21 + i^22 + i^23 + i^24

=> i*i^20 + i^2*i^20 + i^3*i^20 + i^4*i^20

=> i^20[i -1 - i + 1]

=> i^20*0

=> 0

The result of i^21 + i^22 + i^23 + i^24 = 0

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Based on the fact that i^2 = -1 and i^4 = 1, we'll get:

i^21 = i^(20+1) = i^20*i = [(i^4)^5]*i = 1*i = i

i^22 = i^(20+2) = i^20*i^2 = [(i^4)^5]*i^2 = 1*(-1) =-1

i^23 = i^(20+3) = i^20*i^3 = [(i^4)^5]*i^3 = 1*(-i) = -i

i^24 = i^(20+1) = i^20*i^4 = [(i^4)^5]*i^4 = 1*1 =1

i^21 + i^22 + i^23 + i^24 = i - 1 - i + 1=0

Therefore, the result of addition of powers of i is i^21 + i^22 + i^23 + i^24 = 0.

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