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simplify i^13 i^53 i^5
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divide the exponent by 4 and take the integer remainder
so first one is \[i^{13}=i^1=i\]
\[i^p=i^{4 \cdot q+r}\] just divide p each time by 4 and the remainder will be number that you really want to look at
second one is \[i^{53}=i^1=i\]
i i i
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and the last one is... \[i^5=i^1=i\] the i's have it
since \[i^p=i^{4 \cdot q+r}=(i^4)^qi^r=1^qi^r=1i^r=i^r\]
ooh fancy division algoriddim!
i i i
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