Which is equivalent to i^18

How would you find the i to the power of anything first of all? And how do you know if it's 1, or i, or -i, or -1?? Step by step solutions would be helpful :D

remember, i^1 is just i, i^2 is -1, i^3 is -1, and i^4 is 1, so you divide 18 by 4 and you get a remainder of 2, and so it's (1)(1)(1)(1)(i)(i), which we established i^2 is -1, so the answer is -1

The value i raised to an even power will either be -1 or 1. If the power is divisible by 4, the result is 1. If it's only divisible by 2 and not 4, the result is -1.

i^2 = -1 (i * i = -1)

i^4 = 1 (i * i * i * i = -1 * -1 = 1)

i^6 = -1 (i * i * i * i * i * i = -1 * -1 * -1 = -1)

Since 18 is divisible by 2 but not by 4, i^18 = -1.

(Odd powers of i result in +i and -i following a similar pattern.)

Last year I had an oldschool math teacher who made us memorize the patterns based upon factors and all that fun stuff.
This year my teacher, who's also funny , young, and my varsity coach just told us to type it into our graphing calcs.
if you have Texas instruments it should be MATH > NUM > iPart( (Hit Enter) > i^18
If you want to do it RIGHT NOW go to http://mathway.com/problem.aspx?p=algebr…
If you "evaluate" it should be -1
-1
-1
-1
-1
-1

i=√-1
i^2 = √-1√-1 = -1
i^3 = √-1√-1√-1 = -1i = -i
i^4 = √-1√-1√-1√-1 = (-1)(-1) = 1
i^5 = i
i^6 = -1
i^7 = -i
i^8 = 1
See the pattern? Therefore i^16 (16 being a multiple of 4) would equal 1. Then i^18 would equal -1.

Answer: -1

-1. Every 4 i's is a cycle. 18/4 has a remainder of 2. Since i=√-1, i^2=-1 (squaring the square root of a number just leaves you with that number)