What are the "roots of unity"?

Question

A question is asking me to "find the sixth roots of unity and represent them on an Argand diagram".

I don't need you to do the problem for me, I'd rather attempt it myself. However, I don't understand what it's asking me to do. What is unity? Is it $1$, i.e. $\cos(0+2\pi k)$ where $k$ is an integer?

Answer 1

A sixth root of unity is any complex number $z$ such that $z^6=1$. "Unity" is an old-fashioned term for "one."

You can use the De Moivre formula to express the solutions of the above equation in terms of sines and cosines.

Answer 2

Yes, unity represents $1$. So there are six complex roots of unity $z_i,$ such that $$z_i^6 = 1,\;\;\;1 \leq i \leq 6$$

From De Moivre's formula (valid for all real $x$ and integers $n$), we have

$$(\cos x + i \sin x)^n = \cos nx + i \sin nx.$$

Setting $x = 2π/n$ gives an $\color{blue}{\bf \text{nth root of unity}}$:

$$\left(\color{blue}{\bf \cos\frac{2\pi}{n} + i \sin\frac{2\pi}{n}}\right)^n = \cos 2\pi + i \sin 2\pi = 1,$$ and so for $k = 1, 2, ⋯ , n − 1,$

$$\left(\cos\frac{2\pi}{n} + i \sin\frac{2\pi}{n}\right)^k= \cos\frac{2k\pi}{n} + i \sin\frac{2k\pi}{n} \neq 1$$

Answer 3

Yes, unity is $1=\operatorname{cis}(2\pi \rm{i})=\operatorname{cis}(4\pi \rm{i})=\operatorname{cis}(-2\pi \rm{i})=\operatorname{cis}(2k\pi \rm{i}), \ k\in\mathbb Z$.

A sixth root of unity is a complex number $z\in\mathbb C$ s.t. $z^6=1$.
To find the sixth roots of unity use the following:

• Each complex number $z\in\mathbb C$ can be written in the form $z=r\operatorname{cis}(\theta \rm{i})$ with $r\geq0.$
• If $z=r\operatorname{cis}(\theta \rm{i})$ then $z^k=r^k\operatorname{cis}(k\theta \rm{i}), \ k\in\mathbb Z$.
• There are only six sixth roots of unity.

So, you must find six compex numbers $z_i=r_i\operatorname{cis}(\theta_i \rm{i})$ such that $z_i^6=r^6\operatorname{cis}(6\theta_i \rm{i})=\operatorname{cis}(2k\pi \rm{i})$ for some $k\in\mathbb Z$ .