Prove if $w=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$ then the $n$ roots can be written as $1,w,w^2,...,w^{n-1}$

Question

The roots of $z^n=1$ are called root of the unity.

Prove if $w=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$ then the $n$ roots can be written as $1,w,w^2,...,w^{n-1}$

My attempt:

Let $z,w\in\mathbb{C}$ such that $w=z^n$ then $w^{1/n}=z$, by hypothesis we know that: $w=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$ then,$

Answer 1

By exponential form and Euler's identity we have that

$$1=e^{i(2\pi k)}=\cos (2\pi k)+i\sin(2\pi k)$$

therefore

$$\sqrt[n]1=e^{i\frac{2\pi}n k}=\cos \left(\frac{2\pi}n k\right)+i\sin\left(\frac{2\pi}n k\right)=w^k$$

for $k=0,1,\ldots,n-1$.

Answer 2

Okay, so notice that $z^n=1$ means you can let $z^n=e^0=e^{2\pi i}=e^{4\pi i}=e^{2(n-1)\pi i}$.

Now, just take the $n$-th root of each possibility to get $n$ different solutions, the roots of unity!

Edit: Oh, it would be a crime to leave out $e^{i\theta}=\cos(\theta)+i\sin(\theta)$!

Answer 3

If $0\le k\le n-1$, by de Moivre's theorem $(w^k)^n=\cos2\pi k+i\sin2\pi k=1$. This exhausts the $n$ roots of $z^n-1$.