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Math Help - Explain i^-5

  1. #1
    NME
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    Explain i^-5

    Could somebody explain to me why i^-5= -i and not 1/i

    Thanks
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  2. #2
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    Re: Explain i^-5

    i^{-5} = i^{-4} \cdot i^{-1} = i^{-1}

    However, if you multiply by i/i (which happens to be the conjugate) you get i^{-1} \cdot \dfrac{i}{i} = \dfrac{i}{i^2} = -i


    \dfrac{1}{i} is a correct intermediate step but we tend to want a real denominator so we multiply through by the complex conjugate which is i/i in this case
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  3. #3
    NME
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    Re: Explain i^-5

    I don't understand how i/iČ=-i
    Shouldn't that also be i^-1 since i^1/i^2= i^(1-2)
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    Re: Explain i^-5

    Quote Originally Posted by NME View Post
    I don't understand how i/iČ=-i
    Shouldn't that also be i^-1 since i^1/i^2= i^(1-2)
    Think back to the very basics of complex numbers and how we define i^2 (or i, depending on your textbook).

    In other words you can simplify i^2 into a real number
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  5. #5
    NME
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    Re: Explain i^-5

    Oh that's right xD! How could i even miss that!

    Well thank you very much for your help and patience !
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  6. #6
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    Re: Explain i^-5

    Quote Originally Posted by NME View Post
    Could somebody explain to me why i^-5= -i and not 1/i

    Thanks
    \displaystyle i^{-5} = \frac{1}{i^5} = \frac{1}{i^4\cdot i} = \frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i
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  7. #7
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    Re: Explain i^-5

    Another way of looking at it: i+ \frac{1}{i}= \frac{i^2}{i}+ \frac{1}{i}= \frac{-1}{i}+ \frac{1}{i}= 0

    That's why \frac{1}{i} is -i.
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  8. #8
    NME
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    Re: Explain i^-5

    Quote Originally Posted by HallsofIvy View Post
    Another way of looking at it: i+ \frac{1}{i}= \frac{i^2}{i}+ \frac{1}{i}= \frac{-1}{i}+ \frac{1}{i}= 0

    That's why \frac{1}{i} is -i.
    Thank you sir (:
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  9. #9
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    Re: Explain i^-5

    another way of looking at it is this:

    suppose we had some number (which we don't know what it is), let's call it x, and -x = 1/x.

    then (-x)(-x) = (1/x)(-x) (nothing funny going on here).

    but (-x)(-x) = (-1)(x)(-1)(x) = (-1)(-1)(x)(x) = x^2, while (1/x)(-x) = -x/x = (-1)(x/x) = (-1)(1) = -1.

    therefore, for such an x, x^2 = -1.

    so x is a square root of -1. this tells us not only that -i = 1/i, but also that -(-i) = 1/(-i), or: i = 1/(-i).

    in geometric terms: if you go 1/4 the way around a circle, the way back home is the same way you get 1/4 the way around the circle in the opposite direction.
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