What are the nth roots of unity?

[Hoplite]

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"?
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.

 

[davidY]

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

 

[cepheid]

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?

 

[Hoplite]

Oops. Thanks, I shouldn't have missed that. I'll put it down to being the beginning of the semester.