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    eulers formula

    z^3=-i that gives z= -i Give all roots in the form r*e^eiv
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  2. #2
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    Re: eulers formula

    Quote Originally Posted by kalemale View Post
    z^3=-i that gives z= -i Give all roots in the form r*e^eiv
    First of all, \displaystyle \begin{align*} (-i)^3 \neq -i \end{align*}, but \displaystyle \begin{align*} i^3 \end{align*} does...

    Anyway, if \displaystyle \begin{align*} z = r\,e^{i\theta} \end{align*}, then we have

    \displaystyle \begin{align*} z^3 &= -i \\ \left(r\,e^{i\theta}\right)^3 &= e^{-\frac{\pi}{2}i} \\ r^3e^{3i\theta} &= 1e^{-\frac{\pi}{2}i} \\ r^3 = 1\textrm{ and } 3i\theta = -\frac{\pi}{2}i \\ r = 1 \textrm{ and } \theta = -\frac{\pi}{6} \end{align*}

    All the roots are evenly spaced about a circle, so have the same radius and are separated by an angle of \displaystyle \begin{align*} \frac{2\pi}{3} \end{align*}, so the other angles are \displaystyle \begin{align*} -\frac{\pi}{6} - \frac{2\pi}{3} = -\frac{5\pi}{6} \end{align*} and \displaystyle \begin{align*} -\frac{\pi}{6} + \frac{2\pi}{3} =  \end{align*}

    So the three roots are

    \displaystyle \begin{align*} 1e^{-\frac{5\pi}{6}i}, 1e^{-\frac{\pi}{6}i}, 1e^{\frac{\pi}{2}i} \end{align*}
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