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How do you calculate i^8? I know there is some trick to it. Also i^25.

Guest Jun 8, 2017
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 #1
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i^8 = 1

LithianNerd  Jun 8, 2017
 #2
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Yes. Definitely there is some trick.

Remember that \(i^2=-1\),

so \(i^4=i^2\cdot i^2=-1\cdot -1=1\)

Therefore \(i^{4n}=(i^4)^n=1^n=1\), for any integer n.

\(i^8=(i^4)^2=1^2=1\)

\(i^{25}=i^{24}\times i=(i^4)^6\times i=1\times i = i\)

MaxWong  Jun 9, 2017

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