# Thread: The root of polynomial

1. ## 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?

2. ## Re: The root of polynomial

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?
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

3. ## 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

4. ## Re: The root of polynomial

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?

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: $\displaystyle \rho\cdot\xi^k~~k=0,1,2,3,4~.$

I got the point of $\displaystyle Z^5-i=0 $$\displaystyle z^5=i thus\displaystyle i is one root but finding the other with the roots I don't understand how \displaystyle z^5 = i = e^{\pi / 2 + 2 \pi k} this is going to help me. The lightbulb didn't go off yet 6. ## Re: The root of polynomial Originally Posted by Plato Do you understand that \displaystyle r\exp(i\theta)=r(\cos(\theta)+i\sin(\theta))~? Now if \displaystyle \rho = \exp \left( {\frac{{i\pi }}{{10}}} \right) then you can see that \displaystyle \rho^5=i. Let \displaystyle \xi = \exp \left( {\frac{{2\pi i}}{5}} \right) Then the five roots you are looking to find are: \displaystyle \rho\cdot\xi^k~~k=0,1,2,3,4~. sorry what do yo mean by exp? exponent ? and yes the first part looks like DeMovire's Theorem 7. ## Re: The root of polynomial Originally Posted by M670 sorry what do yo mean by exp? exponent ? and yes the first part looks like DeMovire's Theorem What does "yo" mean? 8. ## Re: The root of polynomial Originally Posted by M670 I got the point of \displaystyle Z^5-i=0$$\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
Hint: Simply put: $\displaystyle \left ( z^5 \right ) ^{1/5} = z^1 = z$. How can you use the properties of exponents here?

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

-Dan

9. ## Re: The root of polynomial

Originally Posted by M670
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.

10. ## Re: The root of polynomial

Originally Posted by Plato
What does "yo" mean?
Sorry "yo" was supposed to be "you"

11. ## 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 ?

12. ## Re: The root of polynomial

Fyi I asked 2 math teachers and neither could explain how to solve this

13. ## Re: The root of polynomial

Originally Posted by M670
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 $\displaystyle \cos \left( {\frac{{\pi + 4k\pi }}{{10}}} \right) + i\sin \left( {\frac{{\pi + 4k\pi }}{{10}}} \right)~~k=0,1,2,3,4$

14. ## Re: The root of polynomial

Originally Posted by Plato
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.

15. ## Re: The root of polynomial

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

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