Equation satisfied by nth roots of unity

[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.

I haven't been able to get anywhere with this so can someone please help out?

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.

 

[AKG]

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 - w_n-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.

 

[Benny]

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

 

[heardie]

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