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Math Help - Easy imaginary numbers problem?

  1. #1
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    Easy imaginary numbers problem?

    Simplify i^-19?

    Thanks!
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  2. #2
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    Re: Easy imaginary numbers problem?

    Quote Originally Posted by woahitzme View Post
    Simplify i^-19?

    Thanks!
    \displaystyle \begin{align*} i^{-19} &= \left(e^{\frac{\pi}{2}i}\right)^{-19} \\ &= e^{-\frac{19\pi}{2}i} \\ &= e^{\left(-4\pi - \frac{3\pi}{2}\right)i} \\ &= e^{-\frac{3\pi}{2}i} \\ &= e^{\frac{\pi}{2}i} \\ &= i  \end{align*}
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    Re: Easy imaginary numbers problem?

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} i^{-19} &= \left(e^{\frac{\pi}{2}i}\right)^{-19} \\ &= e^{-\frac{19\pi}{2}i} \\ &= e^{\left(-4\pi - \frac{3\pi}{2}\right)i} \\ &= e^{-\frac{3\pi}{2}i} \\ &= e^{\frac{\pi}{2}i} \\ &= i  \end{align*}

    Umm...

    i^{20} = 1

    i^{-19} = i^{-19}i^{20} = i
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    Re: Easy imaginary numbers problem?

    Quote Originally Posted by richard1234 View Post
    Umm...

    i^{20} = 1

    i^{-19} = i^{-19}i^{20} = i
    I preferred to show a method that allows you to exponentiate ANY complex number
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  5. #5
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    Re: Easy imaginary numbers problem?

    Good point...
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