anonymous
  • anonymous
how do you solve i to the negative 7 power?
Mathematics
schrodinger
  • schrodinger
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cherrilyn
  • cherrilyn
the answer is i
anonymous
  • anonymous
but how do you solve it? i to the negative 7 power
cherrilyn
  • cherrilyn
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

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cherrilyn
  • cherrilyn
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
anonymous
  • anonymous
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
cherrilyn
  • cherrilyn
1/-i is i.
anonymous
  • anonymous
thanks for your help

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