# The root of polynomial

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• Apr 24th 2013, 05:50 PM
Prove It
Re: The root of polynomial
Quote:

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

$\displaystyle z^5-i=0$ 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 $\displaystyle \displaystyle \frac{2\pi}{5}$

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

\displaystyle \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 \displaystyle \begin{align*} 2\pi \end{align*} will also work.
• Apr 25th 2013, 07:32 AM
topsquark
Re: The root of polynomial
Quote:

Originally Posted by M670
I got the point of $\displaystyle Z^5-i=0$$\displaystyle z^5=i$ thus$\displaystyle i$ is one root
but finding the other with the roots I don't understand how $\displaystyle z^5 = i = e^{\pi / 2 + 2 \pi k}$ 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 $\displaystyle z^5 = i = e^{i \pi / 2 + 2 i \pi k}$

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

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

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

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

where do you go from here?

-Dan
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