# Roots of unity

• Dec 29th 2008, 07:30 PM
particlejohn
Roots of unity
In solving $z^n = 1$ we get the distinct solutions $z = e^{i(2k \pi/n)}$. We get this as follows:

Let $z = re^{i \theta}$. Then $(re^{i \theta})^{n} = 1$ or $r^{n}e^{i n \theta} = 1e^{i0}$. Thus $r = 1$ and $\theta = 2 k \pi/n$.

How do we draw the solutions? Why do we write $\omega_{n} = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n}$? Like the three cube roots of $1$ form an equilateral triangle inscribed in the unit circle. But does it matter how we draw it?
• Dec 29th 2008, 07:43 PM
Mush
Quote:

Originally Posted by particlejohn
In solving $z^n = 1$ we get the distinct solutions $z = e^{i(2k \pi/n)}$. We get this as follows:

Let $z = re^{i \theta}$. Then $(re^{i \theta})^{n} = 1$ or $r^{n}e^{i n \theta} = 1e^{i0}$. Thus $r = 1$ and $\theta = 2 k \pi/n$.

How do we draw the solutions? Why do we write $\omega_{n} = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n}$? Like the three cube roots of $1$ form an equilateral triangle inscribed in the unit circle. But does it matter how we draw it?

On the complex plane, the number $z = e^{i \theta}$ is defined as a rotation of $\theta$ radians from the point 1+0i.

So let n = 1,2,3,4,5, and find out what $\theta$ is at these points. Then simply rotate that amount about the unit circle from the point 1+0i.

Unless you're talking about drawing the general case... which is pretty meaningless.
• Dec 30th 2008, 01:18 AM
Complex roots of unity
Hello particlejohn
Quote:

Originally Posted by particlejohn
In solving $z^n = 1$ we get the distinct solutions $z = e^{i(2k \pi/n)}$. We get this as follows:

Let $z = re^{i \theta}$. Then $(re^{i \theta})^{n} = 1$ or $r^{n}e^{i n \theta} = 1e^{i0}$. Thus $r = 1$ and $\theta = 2 k \pi/n$.

How do we draw the solutions? Why do we write $\omega_{n} = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n}$? Like the three cube roots of $1$ form an equilateral triangle inscribed in the unit circle. But does it matter how we draw it?

Your working is OK up to and including

$r = 1$ and $\theta = 2 k \pi/n$

But then you've written

$\omega_{n} = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n}$

Which isn't quite right: you've missed the $k$ out. So let's put it right. Putting the values of $r$ and $\theta$ into $z = re^{i \theta}$ we get:

$z = 1e^{i \theta}=\cos \theta + i\sin\theta$

$=\cos \frac{2k\pi}{n} + i \sin \frac{2k\pi}{n}$

Now note two things:

• because $r = 1$, the points representing the values of $z$ lie on a circle, radius $1$, centre $(0,0)$
• distinct points are obtained by values of $k$ between $0$ and $n-1$ inclusive. When $k = 2n, 2n+1, \dots$ the points are repeated.

So, where do these points lie?

When $k=0$, $z$ is wholly real. This is simply the value $z=1$, and is represented by the point $(1,0)$.

The remaining values of $k$ divide the whole circle into $n$ equal parts, corresponding to angles at the centre of $\frac{\pi}{n}, \frac{2\pi}{n},\frac{3\pi}{n}, \dots$ The points representing these values of $z$, then, lie on the vertices of a regular $n$-sided polygon, with centre $(0,0)$, one vertex at $(1,0)$ and each side subtending an angle at the centre of $\frac{2\pi}{n}$.

I hope that clears it up.

• Dec 30th 2008, 03:28 PM
particlejohn
Actually, isn't my $\omega_n$ correct? Because then all the roots can be expressed as $1, \omega, \omega^{2}, \ldots, \omega^{n-1}$?
• Dec 30th 2008, 06:59 PM
mr fantastic
Quote:

Originally Posted by particlejohn
Actually, isn't my $\omega_n$ correct? Because then all the roots can be expressed as $1, \omega, \omega^{2}, \ldots, \omega^{n-1}$?

The other roots can indeed be constructed from $\omega_n$:

$1, \omega_{\color{red}n}, \omega_{\color{red}n}^{2}, \ldots, \omega_{\color{red}n}^{n-1}$

and you get the same values as Grandad.

But you never said this is what you meant in your original post (and your notation is not quite right).