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Simplify i31.
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HI!!
i bet this is \(i^{31}\) am i right?
yes:)))
ok her is the deal \[i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\] and so on
the pattern continues your answer has to be one of those four \(i,-1,-i,1\)
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take the integer remainder when you divide \(31\) by \(4\)
not a decimal, just the remainder so 4 goes in to 31 7 times, with a remainder of 3 (3 left over) so \[i^{31}=i^3=-i\]
it has to be one of those four, like i said, and that is how you pick it
oh okay,i never realized thats how you figured it out!
not too hard when you see it right?\[i^{202}=i^2=-1\] for example
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not at all! thank you :)
\[\color\magenta\heartsuit\]
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