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Thread: The root of polynomial

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    Member M670's Avatar
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    The root of polynomial

    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?
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    Forum Admin topsquark's Avatar
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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

    $\displaystyle z^5-i=0$ 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?
    Try this...

    $\displaystyle z^5 - i = 0$

    $\displaystyle z^5 = i$

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

    where k is any integer.

    -Dan
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    Member M670's Avatar
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    Re: The root of polynomial

    I am not familiar with that what you are showing me sorry, all my text talks about is conjugate zero theorem for solving this kind of question....
    which i am really having hard time applying to this question
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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
    $\displaystyle z^5-i=0$ 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?

    Do you understand that $\displaystyle r\exp(i\theta)=r(\cos(\theta)+i\sin(\theta))~?$

    Now if $\displaystyle \rho = \exp \left( {\frac{{i\pi }}{{10}}} \right)$ then you can see that $\displaystyle \rho^5=i$.

    Let $\displaystyle \xi = \exp \left( {\frac{{2\pi i}}{5}} \right)$

    Then the five roots you are looking to find are: $\displaystyle \rho\cdot\xi^k~~k=0,1,2,3,4~.$
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    Member M670's Avatar
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    Re: The root of polynomial

    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
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    Member M670's Avatar
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    Re: The root of polynomial

    Quote Originally Posted by Plato View Post
    Do you understand that $\displaystyle r\exp(i\theta)=r(\cos(\theta)+i\sin(\theta))~?$

    Now if $\displaystyle \rho = \exp \left( {\frac{{i\pi }}{{10}}} \right)$ then you can see that $\displaystyle \rho^5=i$.

    Let $\displaystyle \xi = \exp \left( {\frac{{2\pi i}}{5}} \right)$

    Then the five roots you are looking to find are: $\displaystyle \rho\cdot\xi^k~~k=0,1,2,3,4~.$
    sorry what do yo mean by exp? exponent ? and yes the first part looks like DeMovire's Theorem
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    Re: The root of polynomial

    Quote Originally Posted by M670 View Post
    sorry what do yo mean by exp? exponent ? and yes the first part looks like DeMovire's Theorem
    What does "yo" mean?
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    Forum Admin topsquark's Avatar
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    Re: The root of polynomial

    Quote Originally Posted by M670 View Post
    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
    Hint: Simply put: $\displaystyle \left ( z^5 \right ) ^{1/5} = z^1 = z$. How can you use the properties of exponents here?

    Quote Originally Posted by M670 View Post
    sorry what do yo mean by exp? exponent ? and yes the first part looks like DeMovire's Theorem
    $\displaystyle exp(x) = e^x$

    -Dan
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    Re: The root of polynomial

    Quote Originally Posted by M670 View Post
    sorry what do yo mean by exp? exponent ? and yes the first part looks like DeMovire's Theorem
    You want to look at this webpage.
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    Re: The root of polynomial

    Quote Originally Posted by Plato View Post
    What does "yo" mean?
    Sorry "yo" was supposed to be "you"
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    Re: The root of polynomial

    I think I am confused because you guys are telling me to use polar form to find my complex roots and oh wait light bulb ... So I am looking for my roots which are x intercepts and then they are the a out of a+bi which means i need to somehow expand my z^5 into a polynomial and wind up with imaginary numbers ?
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    Member M670's Avatar
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    Re: The root of polynomial

    Fyi I asked 2 math teachers and neither could explain how to solve this
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    Re: The root of polynomial

    Quote Originally Posted by M670 View Post
    Fyi I asked 2 math teachers and neither could explain how to solve this

    That is a real sad comment on the state mathematics education in your education authority.
    Why the heck don't you complain?

    Your roots are $\displaystyle \cos \left( {\frac{{\pi + 4k\pi }}{{10}}} \right) + i\sin \left( {\frac{{\pi + 4k\pi }}{{10}}} \right)~~k=0,1,2,3,4$
    Last edited by Plato; Apr 24th 2013 at 05:52 PM.
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    Member M670's Avatar
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    Re: The root of polynomial

    Quote Originally Posted by Plato View Post
    That is a real sad comment on the state mathematics education in your education authority.
    Why the heck don't you complain?

    Your roots are $\displaystyle \cos \left( {\frac{{\pi + 2k\pi }}{{10}}} \right) + i\sin \left( {\frac{{\pi + 2k\pi }}{{10}}} \right)~~k=0,1,2,3,4$
    See I live in montreal and the new gov't is cutting education funding and promoting language issues in an attempted to create a new country .... so we end up with phd math students as teachers and if its a low level course they don't really care because they are more interested in the research they are doing ...

    But I love it when you answer my questions, more importantly make me try and figure out the answer myself ...

    This was a question from last years final so i figured i should at least try and understand it.
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    Re: The root of polynomial

    I am looking for my nth root of unity in this question
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