Question #4c230

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Question #4c230

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Answer 1 · 2018-01-23T21:25:22

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Explanation:

I'm guessing this is $i^{235} + i^{29}$ $i^235+i^29$ .

First we need to know:

$i^{1} = i$ $i^1=i$ , $i^{2} = - 1$ $i^2=-1$ , $i^{3} = - i$ $i^3=-i$ , and $i^{4} = 1$ $i^4=1$ .

$i^{235} = i^{232 + 3} = i^{232} \cdot i^{3} = {(i^{4})}^{58} \cdot i^{3} = {(1)}^{58} \cdot (- i) = - i$ $i^235=i^(232+3) = i^232*i^3 = (i^4)^58*i^3 = (1)^58*(-i)=-i$

$i^{29} = i^{28 + 1} = i^{28} \cdot i = {(i^{4})}^{7} \cdot i = i$ $i^29= i^(28+1)=i^28*i=(i^4)^7*i=i$

So:

$i^{235} + i^{29} = - i + i = 0$ $i^235+i^29 = -i+i=0$

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Question #4c230

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