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We are asked to evaluate the following, applying the definition that
\n$\\large i^2$ That is: $ \\sqrt{-1} \\times \\sqrt{-1} $ or $i \\times i$. This one is straightforward from the definition of $i$;
\n$\\large i^2=-1$
\n\n
$\\large i^3$ That is: $ \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} $
\nNotice that the first two could be written as $-1$ so this becomes $-1 \\times i$
\n$\\large i^3=-i$
\n\n
$\\large i^4$ That is: $ \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} $
\nRemember that \"pairs\" become $-1$ and this becomes $-1 \\times -1$
\n$\\large i^4=1$
\n\n
$\\large i^5$ That is: $ \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} $
\nFrom the previous example, this becomes $1 \\times i$
\n$\\large i^5=i$
\n\n
$\\large i^6$ That is: $ \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1}$
\nAgain, thinking in \"pairs\" we have $-1 \\times -1 \\times -1$
\n$\\large i^6=-1$
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\n(Hint: use previous answers to inform later ones)
\n\n$\\large i^2$ That is: $ \\sqrt{-1} \\times \\sqrt{-1} $
\n$\\large i^2=$ [[0]]
\n\n
$\\large i^3$ That is: $ \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} $
\n$\\large i^3=$ [[1]]
\n\n$\\large i^4$ That is: $ \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} $
\n$\\large i^4=$ [[2]]
\n\n
$\\large i^5$ That is: $ \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} $
\n$\\large i^5=$ [[3]]
\n\n
$\\large i^6$ That is: $ \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1} \\times \\sqrt{-1}$
\n$\\large i^6=$ [[4]]
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\nEvaluate:
\n$ \\large i^{\\var{P1}}=$ [[0]]
\n\n
Evaluate:
\n$ \\large i^{\\var{P2}}=$ [[1]]
\n\n
Evaluate:
\n$ \\large i^{\\var{P3}}=$ [[2]]
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