Question

How do you simplify `i^38`?

Answer 1

`i^38 = -1`

Explanation:

Let's see what happens if we compute any power of `i`:

`i = i`
`i^2 = -1`
`i^3 = i^2 * i = -1 * i = -i`
`i^4 = i^2 * i^2 = -1 * (-1) = 1`

`i^5 = i^4 * i = 1 * i = i`

... and so on, after that, the sequence `i`, `-1`, `-i` and `1` repeats itself.

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How can you determine which one it is for `i^38`?
Let's start with factorizing `38 = 4*9 + 2`:

`i^38 = i^(4*9+2) = i^(4*9) * i^2 = (i^4)^9 * i^2 = 1^9 * (-1) = -1`

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Let me also additionally show you a general way to determine `i^n` for any positive integer `n`.

There are four possibilites:

• if `n` can be divided by `4`, then `i^n = 1`
• if `n` can be divided by `2` (but not by `4`), then `i^n = -1`
• if `n` is an odd number but `n-1` can be divided by `4`, then `i^n = i`
• if `n` is an odd number but `n+1` can be divided by `4`, then `i^n = -i`

Described in a more formal way,

`i^n = {(1, " " n= 4k),(i, " " n = 4k + 1),(-1, " " n = 4k + 2),(-i, " " n= 4k + 3) :}`

for `k in NN_0`.