1. ## Roots of unity

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

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

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

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

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

How do we draw the solutions? Why do we write $\displaystyle \omega_{n} = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n}$? Like the three cube roots of $\displaystyle 1$ form an equilateral triangle inscribed in the unit circle. But does it matter how we draw it?
On the complex plane, the number $\displaystyle z = e^{i \theta}$ is defined as a rotation of $\displaystyle \theta$ radians from the point 1+0i.

So let n = 1,2,3,4,5, and find out what $\displaystyle \theta$ 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.

3. ## Complex roots of unity

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

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

How do we draw the solutions? Why do we write $\displaystyle \omega_{n} = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n}$? Like the three cube roots of $\displaystyle 1$ 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

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

But then you've written

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

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

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

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

Now note two things:

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

So, where do these points lie?

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

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

I hope that clears it up.

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

5. Originally Posted by particlejohn
Actually, isn't my $\displaystyle \omega_n$ correct? Because then all the roots can be expressed as $\displaystyle 1, \omega, \omega^{2}, \ldots, \omega^{n-1}$?
The other roots can indeed be constructed from $\displaystyle \omega_n$:

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

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).