Question

How do you simplify `i^15`?

Answer 1

`i^15 = -i`

Explanation:

Remember that `i^2 = -1`.

Thus,

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

Also, remember the power rule

`a^m * a^n = a^(m+n)`

Thus, you have

`i^15 = i^(4 + 4 + 4 + 3) = i^4 * i^4 * i^4 * i^3 = 1 * 1 * 1 * i^3 = i^3 = i^2 * i = -1 * i = -i`

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Also, I'd like to offer you a more general solution for `i^n`, with `n` being any positive integer.

Try to recognize the pattern:

`i = i`
`i^2 = -1`
`i^3 = i^2 * i = -1 * i = -i`
`i^4 = i^3 * i = -i * i = -i^2 = 1`
`i^5 = i^4 * i = 1 * i = i`
`i^6 = i^4 * i^2 = -1`
...

So, basically, the power of `i` is always `i`,`-1`, `-i`, `1`, and then repeat.

Thus, to compute `i^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`.