Question

How do you simplify `i^27`?

Answer 1

`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`.