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Math Help - The root of polynomial

  1. #16
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    Re: The root of polynomial

    Quote Originally Posted by M670 View Post
    I am really not doing well on this question

    z^5-i=0 find all the roots of the polynomial I know there is going to be 5 roots and 2 of them should be -i and +i but i can't figure out the rest?
    In one circle there will be five roots, all evenly spaced around a circle, so having the same magnitude and separated by the same angle \displaystyle \frac{2\pi}{5}

    \displaystyle \begin{align*} z^5 &= i \\ z^5 &= e^{\frac{\pi}{2}i} \\ z &= \left( e^{\frac{\pi}{2}i} \right) ^{\frac{1}{5}} \\ z &= e^{\frac{\pi}{10}i}  \end{align*}

    So the first of these roots is \displaystyle \begin{align*} z = e^{\frac{\pi}{10}i} \end{align*}, and the rest are separated by an angle of \displaystyle \begin{align*} \frac{2\pi}{5} \end{align*}, so in the region \displaystyle \begin{align*} \left( \pi , \pi \right] \end{align*} we have

    \displaystyle \begin{align*} z &= \left\{ e^{-\frac{7\pi}{10}i} , e^{-\frac{3\pi}{10}i} , e^{\frac{\pi}{10}i} , e^{\frac{\pi}{2}i}, e^{\frac{9\pi}{10}i} \right\} \end{align*}

    Of course, all multiples of \displaystyle \begin{align*} 2\pi \end{align*} will also work.
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  2. #17
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    Re: The root of polynomial

    Quote Originally Posted by M670 View Post
    I got the point of Z^5-i=0 z^5=i thus i is one root
    but finding the other with the roots I don't understand how z^5 = i = e^{\pi / 2 + 2 \pi k} this is going to help me.

    The lightbulb didn't go off yet
    Oops! I left the "i" out in my equation. Sorry about that. It should have been z^5 = i = e^{i \pi / 2 + 2 i \pi k}

    C'mon. How do you solve the equation x^5 = 32?
    x^5 = 32

    \left ( x^5 \right ) ^{1/5} = \left ( 32 \right ) ^{1/5} = \left ( 2^5 \right )^{1/5} = 2^{5 \cdot 1/5} = 2^1 = 2

    So when you have
    z^5 = i = e^{i \pi / 2 + 2 i \pi k}

    \left ( z^5 \right )^{1/5} = \left ( e^{i \pi / 2 + 2 i \pi k} \right ) ^{1/5}

    where do you go from here?

    -Dan
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