InterviewSolution
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InterviewSolution
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Show that `{i^(23)+((1)/(i))^(29)}^(2) = -4`. |
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Answer» We have `i^(23)=i^((4xx5+3))=(i^(4))^(5)xx i^(3) = 1 xx (-i) = -i." "[because i^(4) = 1 and i^(3) = -i]` `((1)/(i))^(29) = (1)/(i^(29))=(1)/(i^(29))xx(i^(3))/(i^(3))=(i^(3))/(i^(32))=(-i)/(1) =-i." "[because i^(3) =-i and i^(32) =1]` `therefore" "{i^(23)+((1)/(i))^(29)}^(2)=(-i-i)^(2)=(-2i)^(2)=4i^(2)=4xx(-1)=-4`. |
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