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 $\displaystyle w = \sqrt[n]{1 + 0i}$
Then $\displaystyle w^n = 1 + 0i$
$\displaystyle w^n = e^{2\pi i}$.
This means by DeMoivre's Theorem, the focus root is
$\displaystyle w = e^{\frac{2\pi i}{n}}$.
But there must be exactly $\displaystyle n$ roots, all spaced evenly about a circle.
So that means any $\displaystyle w$ can be written as
$\displaystyle w = e^{\frac{2k\pi i}{n}}$, where $\displaystyle k$ is a positive integer no greater than $\displaystyle n$.
That means
$\displaystyle w^2 = e^{\frac{4k\pi i}{n}}$
$\displaystyle w^3 = e^{\frac{6k\pi i}{n}}$
$\displaystyle \vdots$
$\displaystyle w^n = e^{2k\pi i}$
But any integer multiple of $\displaystyle 2\pi$ gives the same result as $\displaystyle 2\pi$.
So that means $\displaystyle w^n = e^{2\pi i}$.
Now let's go back to the original problem...
We have $\displaystyle w + w^2 + w^3 + \dots + w^n$.
This is a geometric series with $\displaystyle a = w$ and $\displaystyle r = w$.
So the sum is
$\displaystyle \frac{a(1 - r^n)}{1 - r}$
$\displaystyle = \frac{w(1 - w^n)}{1 - w}$
$\displaystyle = \frac{e^{\frac{2k\pi i}{n}}\left(1 - e^{2\pi i}\right)}{1 - e^{\frac{2k\pi i}{n}}}$.
But $\displaystyle e^{2\pi i} = 1$.
So that means the sum is
$\displaystyle \frac{e^{\frac{2k\pi i}{n}}\left(1 - 1\right)}{1 - e^{\frac{2k\pi i}{n}}}$
$\displaystyle = \frac{0e^{\frac{2k\pi i}{n}}}{1 - e^{\frac{2k\pi i}{n}}}$
$\displaystyle = 0$.
Q.E.D.