SOLUTION: tutor could you help me please: how do I simplify: i^23 and i^24

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Question 138781: tutor could you help me please: how do I simplify: i^23 and i^24
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
i^4 = (-1)(-1) = 1
This cycle repeats as you increase the exponent. Hence i^5 = i^(4+1) = i^4 * i^i = 1*i = i
All you need to so is divide the given exponent by 4 and then find the remainder. The remainder will be either 0, 1, 2, 3. Which will correspond to 1, i, -1, -i respectively.
For i^23, we get 23%2F4+=+5+remainder+of+3 i%5E23 = i%5E%2820%2B3%29 = 1%2Ai%5E3 = i%5E3 = -i
For i^24, the remainder is 0. So i%5E24 = 1