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anonymous

  • 5 years ago

how do you solve i to the negative 7 power?

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  1. cherrilyn
    • 5 years ago
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    the answer is i

  2. anonymous
    • 5 years ago
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    but how do you solve it? i to the negative 7 power

  3. cherrilyn
    • 5 years ago
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    i 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

  4. cherrilyn
    • 5 years ago
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    Anything 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

  5. anonymous
    • 5 years ago
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    rewrite 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

  6. cherrilyn
    • 5 years ago
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    1/-i is i.

  7. anonymous
    • 5 years ago
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    thanks for your help

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