Question

How do you simplify `i^109`?

Answer 1

`i^100 = 1`

Explanation:

We will be using the following:

• `(x^a)^b = x^(ab)`

• `i^2 = -1`

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Because `i^2 = -1`, we know that `i^4 = (i^2)^2 = (-1)^2 = 1`

Using this, we can take `i` to a large integer power and quickly evaluate it by "pulling out" multiples of `4` from the exponent, like so:

`i^100 = i^4*25 = (i^4)^25 = 1^25 = 1`

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While the above suffices for the given problem, it is also applicable to integers which are much larger, or not multiples of `4`. For example, looking at `i^528533`

We can write

`528533 = 528500 + 32 + 1`
`= 100*5285 + 4*8 + 1`
`= 4*(25*5288) + 4*8 + 1`
`= 4(25*5288 + 8) + 1`

So, applying this to our problem:

`i^528533 = i^(1+4(25*5288))`
`= i^1*(i^4)^(25*5288)`
`=i * 1^(25*5288)`
`=i*1`
`= i`