# Equation satisfied by nth roots of unity

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.

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