Need help figuring out what this Maths question means

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In summary, the conversation discusses finding the nth roots of 1 and how they satisfy a specific equation. The concept of nth roots and their relationship to polynomial roots is explored, and it is concluded that there are n roots for any given number, just as there are two square roots and four fourth roots.
  • #1
Hoplite
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Hi everybody, this is my first post here.

I got this question, but I don't know what it means:


Fix n ≥ 1. If the nth roots of 1 are w0, w1, w2, . . . , wn−1, show that they satisfy:

(z − w0)(z − w1)(z - w2) · · · (z − wn−1) = z^n − 1

(z and wn are all complex numbers)

What I don't understand is, what does it mean by "nth roots of 1"? :confused:
I think by "roots" it means polynomial roots, but what does it mean to have a root of a number in this context?

Any help would be appreciated.
 
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  • #2
try this:
let a^n=1
=cis(0+2k*pi) for k=integer
by de-moivres
a=cis(2(k/n)*pi)

there you have the roots, the bth root is probably k=b
 
  • #3
Square root of one: [tex] \sqrt{1} = 1^{\frac{1}{2}} [/tex]

Fourth root of one: [tex] \sqrt[4]{1} = 1^{\frac{1}{4}} [/tex]

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etc.

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nth root of one: [tex] \sqrt[n]{1} = 1^{\frac{1}{n}} [/tex]

There is more than one nth root i.e. more than one number (call them [itex] z [/itex]) that satisfies the equation:

[tex] z^n = 1 [/tex]

In fact there are exactly "n" of them, just as there are two square roots of one, and four fourth roots of one. Do you see why?
 
  • #4
Oops. Thanks, I shouldn't have missed that. I'll put it down to being the beginning of the semester. :redface:
 

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