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cgonzalez31Best ResponseYou've already chosen the best response.0
yeah, assuming you mean i as the square root of 1, then i^0 = 1 (any # to the power of 0 is one) i ^1 = i i^2 = 1 (the square root goes away) i^3 = i (since the left hand side is equivalent to i^2 * i = i ) From then on the values loop, so i^4 = 1, i^5 = i, etc. You can also make them up from the values of previous terms to verify, i.e. i^5 = i^3 * i^2 = 1*i = i
 2 years ago

Wink_PinkBest ResponseYou've already chosen the best response.0
Is it Diff E problem?
 2 years ago
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