Need help figuring out what this Maths question means

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The discussion centers on understanding the concept of "nth roots of 1" in the context of a polynomial equation. It explains that the nth roots of 1 are complex numbers that satisfy the equation z^n = 1, and there are exactly n distinct roots. The roots can be expressed using De Moivre's theorem, where each root is represented as cis(2(k/n)π) for integer values of k. This means that for any positive integer n, there are multiple solutions to the equation, similar to how there are two square roots of one or four fourth roots of one. The clarification helps in grasping the relationship between roots and polynomial equations.
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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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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
 
Square root of one: \sqrt{1} = 1^{\frac{1}{2}}

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

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etc.

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

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

z^n = 1

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?
 
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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