- #1

Benny

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Q. Fix n>= 1. If the nth roots of 1 are w_0,...,w_(n-1), show that they satisfy:

[tex]

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

[/tex]

I tried considering z^n = 1.

[tex]

z^n = e^{i2\pi + 2k\pi i} \Rightarrow z = e^{\frac{{i2\pi }}{n} + \frac{{2k\pi i}}{n}}

[/tex] 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: [tex]\omega _0 \omega _1 ...\omega _{n - 1} = \left( { - 1} \right)^{n - 1} [/tex]

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

[tex]\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)[/tex]

[tex]

= 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}}

[/tex]

[tex]

= 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}}

[/tex]

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

[tex]

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

[/tex]

I tried considering z^n = 1.

[tex]

z^n = e^{i2\pi + 2k\pi i} \Rightarrow z = e^{\frac{{i2\pi }}{n} + \frac{{2k\pi i}}{n}}

[/tex] 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: [tex]\omega _0 \omega _1 ...\omega _{n - 1} = \left( { - 1} \right)^{n - 1} [/tex]

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

[tex]\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)[/tex]

[tex]

= 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}}

[/tex]

[tex]

= 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}}

[/tex]

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

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