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anonymous
 5 years ago
how do you solve i to the negative 7 power?
anonymous
 5 years ago
how do you solve i to the negative 7 power?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but how do you solve it? i to the negative 7 power

cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0i is defined as √(1) Therefore, it's in cycles of 4: i = √(1) i² = 1 i³ =  √(1) (or just i) i^4 = 1 So first, let's get the negative exponent out of there: i^(7) is the same as 1/i^7

cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0Anything to the negative power is the same as putting it in the denominator and using the positive exponent: i^0 = 1 i^1 = i i^2 = 1 i^3 = i Then it repeats this pattern for every four powers: i^4 = 1 i^5 = i i^6 = 1 i^7 = i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0rewrite i in polar coordinates, exp(j*pi/2) and raise that to the 7th which becomes exp(j7pi/2), then using Euler's Identity rewrite it as cos(7pi/2)  i*sin(7pi/2), which gives you +i
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