How do you simplify i^27?

1 Answer
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.