lack of understanding roots of unity

Question

I have tried looking for several videos and none of them really covered what I was looking for. The best way to learn for me is by doing. Sorry if this is hard to understand. I had no idea what was going on in class:

I changed the exponent from the homework question.find the six roots of unity for $z^6=1$

I think I am supposed to use the following formula: $z^n = e^{\Big(\frac{2 \pi k}{n}i \Big)^n} = e^{2 \pi k} = \cos{2 \pi k} + i\sin(2 \pi k)$ for $k = 0,1,...n-1$ but honestly I have no idea

Right off the bat aren't two answers 1,-1?

Answer 1

$$z^6 = 1 \implies z^3 = \pm1$$

• Case $1: z^3 =1 \implies (z^3-1) = (z-1)(z^2+z+1) = 0 \implies z = 1 $ or $ z = \frac{-1\pm i\sqrt{3}}{2} $

• Case 2: $z^3 = -1 \implies (z^3+1)=(z+1)(z^2-z+1) = 0 \implies z = -1 $ or $ z = \frac{1\pm i\sqrt3}{2}$

---

By using your method,

$z = e^{\frac{2\pi ki}{n}}=\cos\frac{2\pi k}{n}+ i\sin\frac{2\pi k}{n}$ and we have $n=6$

$\bullet k = 0 \implies z= \cos 0 = 1$

$\bullet k = 1\implies z= \cos \frac{\pi}{3}+i\sin\frac\pi3 = \frac{1+i\sqrt3}{2}$

$\bullet k = 2\implies z= \cos \frac{2\pi}{3}+i\sin\frac{2\pi}3 = \frac{-1+i\sqrt3}{2}$

$\bullet k = 3\implies z= \cos \pi+i\sin\pi = -1$

$\bullet k = 4\implies z= \cos \frac{4\pi}{3}+i\sin\frac{4\pi}3 = \frac{-1-i\sqrt3}{2}$

$\bullet k=5 \implies z = \cos \frac{5\pi}{3}+i\sin\frac{5\pi}{3}=\frac{1-i\sqrt3}{2}$

Answer 2

$z^6=1=e^{i×2kπ}, k=0,1,2,...,5$

$z=e^{i×\frac{kπ}3}=\cos{\frac{kπ}3}+i\sin{\frac{kπ}3}$

Insert each value of $k=0,1,2,3,4,5$ and you will get each value of the root.

$z_0=0$

$z_1=\frac{1+i\sqrt3}2$

$z_2=\frac{-1+i\sqrt3}2$

$z_3=-1$

$z_4=\frac{-1-i\sqrt3}2$

$z_5=\frac{1-i\sqrt3}2$

Answer 3

Yes you are right, right off the bat there are two roots, $\pm 1$. But there are also $4$ others.

So any complex number can be represented as $re^{ti}$ where $r$ is a positive numbers (or $0$) and $t$ is a real number, $0 \le t < 2 \pi$.

We also know that if you have the complex number $z = re^{ti}$, then it is true that $z^n = r^ne^{nti}$ (but notice that $nt$ may now be larger than $2 \pi$, so $z^n$ is not in the "standard" form; but this is ok, it is still a complex number, just not in the "standard" form).

Finally, if $x, y$ are real numbers, then $e^{xi} =e^{yi}$ if and only if $x, y$ differ by a multiple of $2 \pi$; that is, if and only if $y = x + 2 k \pi$ for some integer $k$.

So armed with these three things, we are able to find all complex numbers $z$ such that $z^6 = 1$.

Consider a complex number $z$ such that $z^6 = 1$. We may represent $z$ as $re^{ti}$ with $ 0 \le t < 2 \pi$. Now, if $z^6=1$, then $r^6e^{6ti} = 1 = 1e^{0i}$. Thereore $r^6 = 1$, and since $r$ is a positive real number, the only solution is $r=1$. Also we have that $e^{6ti} = e^{0i}$. So $0$ and $6t$ must differ by a multiple of $2 \pi$. That is, $6t = 0 + 2k \pi$ for some integer $k$. Now I encourage you to set $k = 0, 1, 2, 4, 5$, and solve for $t$. Then set $k$ equal to other integers, and you will see that you do not get any new solutions from other $k's$.

Answer 4

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

or you could do something like

$z^6 - 1 = (z^3 - 1)(z^3 + 1) = (z-1)(z^2 + z + 1)(z+1)(z^2 - z + 1)$

Use the quadratic formula were we don't factor nicely.