find the polynomial equation

V

Vali

Junior Member
Joined
Feb 27, 2018
Messages
87
I have \(\displaystyle z_{k}=cos\frac{2k\pi}{5}+isin\frac{2k\pi}{5}, k=1,2,3,4\).

I need to find the polynomial equation for the roots $z_k(k=1,2,3,4)$

The right answer is \(\displaystyle x^4+x^3+x^2+x+1=0\).I tried to replace k with 1,2,3,4 to find the roots and then to use \(\displaystyle a(x-x_1)(x-x_2)(x-x3)(x-x_4)\) but I didn't get too far.
 
Vali said:
I have \(\displaystyle z_{k}=cos\frac{2k\pi}{5}+isin\frac{2k\pi}{5}, k=1,2,3,4\).
I need to find the polynomial equation for the roots $z_k(k=1,2,3,4)$
The right answer is \(\displaystyle x^4+x^3+x^2+x+1=0\).I tried to replace k with 1,2,3,4 to find the roots and then to use \(\displaystyle a(x-x_1)(x-x_2)(x-x3)(x-x_4)\) but I didn't get too far.
Click to expand...
First some notation. \(\displaystyle z_{k}=cos\frac{2k\pi}{5}+isin\frac{2k\pi}{5}={\exp \left( {\frac{{2K\pi i}}{5}} \right)}\)
The \(\displaystyle {\exp \left(i\theta \right)}=\cos(\theta)+i\sin(\theta)\) is for ease of computation.
There are five fifth roots of unity: \(\displaystyle z^5=1\) has five roots; \(\displaystyle z=\exp \left( \frac{2K\pi i}{5} \right),~K=0,1,\cdots,4\).
Each of the four given numbers are fifth roots of unity.
You say "The right answer is \(\displaystyle x^4+x^3+x^2+x+1=0\).
So you need show each of the given numbers is a root of that polynomial.
To that end SEE HERE You see usings the first of the numbers, \(\displaystyle K=1\) is a root, gives \(\displaystyle 0\).

 
But let's say I don't know the answer.How I get that answer without involving the answer ?
 
As pka pointed out, we know that the given roots are all but one of the 5th roots of unity, and we should observe that:

[MATH]x^5-1=(x-1)(x^4+x^3+x^2+x+1)[/MATH]
Now, since \(k=0\) is not included, which is associated with the root \(x=1\), we are then left with:

[MATH]f(x)=x^4+x^3+x^2+x+1[/MATH]
 
MarkFL said:
As pka pointed out, we know that the given roots are all but one of the 5th roots of unity, and we should observe that:

[MATH]x^5-1=(x-1)(x^4+x^3+x^2+x+1)[/MATH]
Now, since \(k=0\) is not included, which is associated with the root \(x=1\), we are then left with:

[MATH]f(x)=x^4+x^3+x^2+x+1[/MATH]
Click to expand...

How to know this thing?I mean, what should I notice?What should I verify to notice this?
 
Vali said:
How to know this thing?I mean, what should I notice?What should I verify to notice this?
Click to expand...
How do you expect any of us to know an answer to that?
What course are you studying? Have you studied the roots of unity?
If you have then you should have recognized those given numbers as fifth roots of unity.
 
Look at the given numbers: \(\displaystyle z_k= \cos\left(\frac{2k\pi}{5}\right)+ isin\left(\frac{2k\pi}{5}\right)\). That is of the form x+iy with \(\displaystyle x^2+ y^2= 1\). Those five points lie along a circle of radius 1 centered at the origin and equally spaced. That should be enough to tell you that they are "cyclotomic" points. n points, equally spaced along a circle of radius 1 with center at the origin of the complex plane satisfy the polynomial equation \(\displaystyle x^n+ x^{n-1}+ \cdot\cdot\cdot+ x+ 1= 0\).
 
Thank you @HallsofIvy I get it.
@pka I studied roots of unity at school but I never met roots of unity with complex numbers.At school we usually know the equation and we recognice the roots of unity.I didn't recognize them because I'm not used to use roots of unity in exercises with complex numbers.
Thank you for your help!
 
":I studied roots of unity at school but I never met roots of unity with complex numbers."

Are you sure of that? The only "roots of unity" that are real numbers are 1 and -1!
 
You must log in or register to reply here.
Top