How do you simplify #i^27#?

1 Answer
sente
Nov 9, 2015

#i^27 = -i#

Explanation:

#i^27 = i^(24+3)=i^24*i^3=i^(4*6)*i^(2+1)=(i^4)^6*i^2*i#

Now, we know that #i^2 = -1# and so #i^4 = (i^2)^2 = 1#

Substituting that in gives us

#i^27 = 1^6*(-1)*i = -i#

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The above is a formal way of going about it, but for quick mental calculations, remembering that #i^4=1# allows you to "pull out" the greatest multiple of 4 possible from the exponent.
This leaves you with an exponent of 0, 1, 2, or 3 (any greater and there is another multiple of 4 to remove).
Then it is easy to work out that #i^0 = 1#, #i^1 = i#, #i^2 = -1#, and #i^3 = -i#.

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