SOLUTION: I have a question about the powers of "i". I am not understanding the facts about how "i" to the second power = blank or "i" to the 6th powers = blank and so fourth. Can you explai

Algebra ->  Algebra  -> Radicals -> SOLUTION: I have a question about the powers of "i". I am not understanding the facts about how "i" to the second power = blank or "i" to the 6th powers = blank and so fourth. Can you explai      Log On

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Question 241146: I have a question about the powers of "i". I am not understanding the facts about how "i" to the second power = blank or "i" to the 6th powers = blank and so fourth. Can you explain this procedure in detail and explain how it is used in the radical equation?

Answer by vleith(2517) About Me  (Show Source):
You can put this solution on YOUR website!
i is defined as the sqrt%28-1%29
So
i^1 = i
i^2 = -1
i^3 = -i
i^4 = (-1)(-1) = 1
And it repeats in a cycle of 4
i^5 = i^4*i = 1+i = i
i^6 = i^4*i^2 = 1*i^2 = 1 *(-1) = -1
i^7 = i^4*i^3 = 1*i^3 = -i
etc etc
how do you use in a radical? If the value under the radical is negative, then look at that as (-1)*aPositive
for instance sqrt%28-9%29+=+sqrt%28%28-1%29%2A9%29+=+sqrt%28-1%29%2Asqrt%289%29 = plus or minus 3i