Roots of unity Question

**noobonastick** · April 7th 2010, 10:42 PM

Hello, can I get some help with the following?

Show that if w is an nth root of unity (w != 1 and n > 1) then
w + w^2 + ...+ w^n = 0

Thanks!

**Prove It** · April 7th 2010, 11:08 PM

Originally Posted by noobonastick

Hello, can I get some help with the following?

Show that if w is an nth root of unity (w != 1 and n > 1) then
w + w^2 + ...+ w^n = 0

Thanks!

If $w = \sqrt[n]{1 + 0i}$

Then $w^n = 1 + 0i$

$w^n = e^{2\pi i}$ .

This means by DeMoivre's Theorem, the focus root is

$w = e^{\frac{2\pi i}{n}}$ .

But there must be exactly $n$ roots, all spaced evenly about a circle.

So that means any $w$ can be written as

$w = e^{\frac{2k\pi i}{n}}$ , where $k$ is a positive integer no greater than $n$ .

That means

$w^2 = e^{\frac{4k\pi i}{n}}$

$w^3 = e^{\frac{6k\pi i}{n}}$

$\vdots$

$w^n = e^{2k\pi i}$

But any integer multiple of $2\pi$ gives the same result as $2\pi$ .

So that means $w^n = e^{2\pi i}$ .

Now let's go back to the original problem...

We have $w + w^2 + w^3 + \dots + w^n$ .

This is a geometric series with $a = w$ and $r = w$ .

So the sum is

$\frac{a(1 - r^n)}{1 - r}$

$= \frac{w(1 - w^n)}{1 - w}$

$= \frac{e^{\frac{2k\pi i}{n}}\left(1 - e^{2\pi i}\right)}{1 - e^{\frac{2k\pi i}{n}}}$ .

But $e^{2\pi i} = 1$ .

So that means the sum is

$\frac{e^{\frac{2k\pi i}{n}}\left(1 - 1\right)}{1 - e^{\frac{2k\pi i}{n}}}$

$= \frac{0e^{\frac{2k\pi i}{n}}}{1 - e^{\frac{2k\pi i}{n}}}$

$= 0$ .

Q.E.D.

**HallsofIvy** · April 8th 2010, 03:12 AM

Originally Posted by noobonastick

Hello, can I get some help with the following?

Show that if w is an nth root of unity (w != 1 and n > 1) then
w + w^2 + ...+ w^n = 0

Thanks!

I w is an nth root of unity then w^n= 1, of course, so that equation becomes 1+ w+ w^2+ ...+ w^{n-1}= 0.

Also, since w^n= 1, w^n- 1= (w- 1)(1+ w+ w^2+ ...+ w^{n-1})= 0.

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