SOLUTION: how do u solve problems like these: i^29 i^155

Question 6751: how do u solve problems like these:
i^29
i^155
Answer by prince_abubu(198) (Show Source): You can put this solution on YOUR website!
I take it that i is the imaginary number that is equivalent to .

There's a trick here that many teachers don't show when computing . We're gonna have to start from scratch here.
i^1 = i
i^2 = i * i = -1
i^3 = i^2 * i = -i
i^4 = i^3 * i = -i * i = 1

i^5 = i^4 * i = i
i^6 = i^5 * i = i * i = -1
i^7 = i^6 * i = -1 * i = -i
i^8 = i^7 * i = -i * i = 1
and so on.

If you noticed, it goes from i ---> -1 ----> -i ----> 1 and back again to i. It's like a cycle. Another thing that you probably noticed is that every power of i divisible by 4 equals 1, so i^4 = i^8 = ...= i^40 = ... = i^400 = 1.

What about the rest? Actually all you have to do is to take that power n, divide by 4, and pay attention to the remainder.
If the remainder is 0, then i^n = 1
If the remainder is 1, then i^n = -1
If the remainder is 2, then i^n = i
If the remainder is 3, then i^n = -i

So let's take i^29. n = 29. Divide that by 4. You'll get 7 remainder 1. Just because it's remainder 1, i^29 = -1. I think you can take it from here.
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