# Thread: Roots of unity Question

1. ## Roots of unity Question

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!

2. 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.

3. 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.