Question

How do you simplify `i^17`?

Answer 1

You should write `i^17` as `i^16*i` this is same as `(i^2)^8 * i` simplifying this you would get the answer as `i`

Explanation:

Any problem of type `i^n`

Case 1: when `n` is even

Rewrite it as `(i^2)^(n/2)` use the rule `i^2=-1` and then you would get `-1` when `n/2` is odd and `+1` when `n/2` is even
Example to understand this case

`i^6 = (i^2)^3 = (-1)^(3) = -1` here `n" is even and`n/2# is odd

`i^8 = (i^2)^4 = (-1)^4 = 1` here `n` is even and `n/2` is even.

Case 2: when `n` is odd

Rewrite `i^n` as `i^(n-1)*i^1`
Apply the case(1) rules this time using `n-1` instead of `n` the final answer would be `-i` or `i` depending on `i^(n-1)`

Answer 2

`i^17 = i`

Explanation:

First note that `i^4 = i^2*i^2 = -1*-1 = 1`

In general note that if `k` and `c` are integers and `n = 4k+c` then:

`i^n = i^(4k+c) = i^(4k)i^c = (i^4)^k i^c = 1^k i^c = i^c`

In our case we can take `k = 4` and `c = 1` to find:

`i^17=i^(16+1)=i^16 i^1 = (i^4)^4 i = 1^4 i = i`