# Equation satisfied by nth roots of unity

• Benny
In summary, the conversation discusses the relationship between the nth roots of 1 and the polynomial p(z) = zn - 1. It is shown that these roots also satisfy another polynomial q(z) = (z - w0)...(z - wn-1) and thus are the same roots. The second question asks to show that the product of the omegas is (-1)^n, which can be proven by comparing coefficients.
Benny
Q. Fix n>= 1. If the nth roots of 1 are w_0,...,w_(n-1), show that they satisfy:

$$\left( {z - \omega _0 } \right)\left( {z - \omega _1 } \right)...\left( {z - \omega _{n - 1} } \right) = z^n - 1$$

I tried considering z^n = 1.

$$z^n = e^{i2\pi + 2k\pi i} \Rightarrow z = e^{\frac{{i2\pi }}{n} + \frac{{2k\pi i}}{n}}$$ with k = 0,1,2...n-1.

Also how would I do the following? The relevant information is given in the stem of the previous question I posted.

Q. Show that the omegas satisfy: $$\omega _0 \omega _1 ...\omega _{n - 1} = \left( { - 1} \right)^{n - 1}$$

Again I haven't really gotten anywhere in my attempts.

$$\omega _0 \omega _1 ...\omega _{n - 1} = \left( {e^{\frac{{i2\pi }}{n} + \frac{{2\left( 0 \right)\pi i}}{n}} } \right)\left( {e^{\frac{{i2\pi }}{n} + \frac{{2\left( 1 \right)\pi i}}{n}} } \right)...\left( {e^{\frac{{i2\pi }}{n} + \frac{{2\left( {n - 1} \right)\pi i}}{n}} } \right)$$

$$= e^{\frac{{i2\pi n}}{n} + \frac{{2\left( {0 + 1 + ...\left( {n - 1} \right)} \right)\pi i}}{n}} = e^{i2\pi } e^{\frac{{2\left( {\sum\limits_{j = 0}^{n - 1} k } \right)\pi i}}{n}} = e^{\frac{{2\left( {\sum\limits_{j = 0}^{n - 1} k } \right)\pi i}}{n}}$$

$$= e^{\frac{{2\left( {\sum\limits_{j = 0}^{n - 1} k } \right)\pi i}}{n}} = e^{\frac{{2\left( {\sum\limits_{j = 1}^n {\left( {k - 1} \right)} } \right)\pi i}}{n}}$$

So that's all I've been able to do. Some help would be appreciated.

Last edited:
Do you know the fundamental theorem of algebra? You're essentially given the roots of the polynomial p defined by p(z) = zn - 1. Now it's easy to see by inspection that those roots are the very same roots of the polynomial q defined by q(z) = (z - w0)...(z - wn-1). You have two polynomials with the same roots, and it's easy to check that they have the same leading coefficient, namely 1. So they're the same.

As for the second question, it's hard to tell what you're doing. The sums appear to be indexed by j, but I only see k appearing in the summand. I suppose you meant those two to be the same. Well go ahead and compute the sum. You should be able to see quite easily after doing this that the desired result follows.

Yeah there's an error in the index of the summation...j should be k. Thanks for the help.

Benny for 2) as well simply compare coefficents. The product of omegeas is $$(-1)^n$$, the right hand side is -1

## 1. What is the equation satisfied by the nth roots of unity?

The equation satisfied by the nth roots of unity is x^n - 1 = 0. This means that when you raise each nth root of unity to the nth power, the result will be 1.

## 2. How many solutions does the equation have?

The equation has n solutions, where n is the degree of the equation. In this case, the degree is n, so there are n solutions.

## 3. What are the solutions to the equation?

The solutions to the equation are the nth roots of unity, which are complex numbers that satisfy the equation x^n - 1 = 0. These solutions are written in the form e^(2πik/n), where k is an integer from 0 to n-1.

## 4. What is the significance of the nth roots of unity?

The nth roots of unity have many applications in mathematics and science, including complex number theory, number theory, and signal processing. They also have connections to symmetries in geometry and physics.

## 5. How can the equation satisfied by the nth roots of unity be used in practical applications?

The equation can be used in practical applications such as calculating the sum of an infinite geometric series and finding the roots of a polynomial equation. It also has applications in fields such as computer graphics, cryptography, and engineering.

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