Question

how do you simplify i^17?

Answer 1

when i is put to an odd exponent you always get -i.
i^3 = -i
i^5 = -i
i^7 = -i
and so on.

Answer 2

so it's just -i?

Answer 3

yep.

Answer 4

NO

Answer 5

\[i^{16}~\times i\]

Answer 6

divide the exponent by 4 and look at the remainder. take i to that power and you'll have your answer

Answer 7

oh oops
i^17 = i
forgot
i^16 = 1

Answer 8

it's i

Answer 9

Ok, when you have even numbers you always get -1. When you have odd you always get i. Therefore, i^16 * i = -1 * i

Answer 10

no

Answer 11

i^3 = -i
i^4 = 1
i^2 = -1

Answer 12

\[i^{4n}=1\]\[i^{4n+1}=i\]\[i^{4n+2}=-1\]\[i^{4n+3}=-i\]\[i^{4n+4}=i^{4n}=1\]YOU GET THE IDEA.

Answer 13

ok thanks :)

Answer 14

Snap! I messed up.

Solomon is right.. srry

Answer 15

Anytime!

Answer 16

i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, etc

The pattern repeats, grouping them into groups of four.

Divide the exponent by 4 and take the remainder. Raise i to the power of remainder.

Since 17 divide by 4 yields remainder of 1,

i^17 = i^1 = i

answer: i @kiergurl5