Show the nth root of unity are the vertice of regular polygon and a formula for the perimeter... 1) Show the nth root of unity are the vertice of regular polygon
2) Find the formula for the perimeter of that polygon called "ln" and prove
$lim_{n\rightarrow \infty }l_n=2\pi$
My attempt
Let $z=1$such that $z\in \mathbb{C}$. We need the nth root of the unity.
Let $w\in \mathbb{C}$. such that $w^{\frac{1}{n}}=z$
Then 
$w_k=cos(\frac{\theta+2k\pi}{n})+isin(\frac{\theta+2k\pi}{n})\,\,\,\,(1)$ for $k=0,1,...,n-1$
As $z=1$ then $\theta=0$. Replacing in $(1)$ we have:
$w_k=cos(\frac{2k\pi}{n})+isin(\frac{2k\pi}{n})=e^{i\frac{2k\pi}{n}}\,\,\,\,(2)$
Here, i'm stuck. 
I make a graphich representation for $k=4$ and is a polygin of four vertices. But for $n$ i'm stuck.
For the 2) question i don't have idea. Can someone help me?
 A: You want to show that the angle between $e^{2\pi k i /n}$ and $e^{2\pi (k+1)i/n}$ is constant.  
Note that if you divide these two complex numbers you get the resulting angle of rotation between the two. $$ \frac {e^{2\pi (k+1)i/n}}{e^{2\pi ki /n}}= e^{2\pi i/n}$$ which is the same for all $k$, that is they are vertices of a regular polygon, considering that they all have unit length.
For the side-length of the polygon you need to find the norm of the difference of two consecutive roots,  for example $$|1-e^{2\pi i/n}|$$ 
Multiply the result by n and let n goes to $\infty$ to get your $2\pi$ 
A: In De Moivre's formula, you see that the argument of two consecutive roots differ of $2\pi/n$. Since the module of every root is equal to $1$, the points are distributed lie on the circumference $|z|=1$ and the angular distance between two consecutive points is $2\pi/n$. Therefore the points are vertex of a regular $n$-agon. By the chord theorem, the length of the $n$-agon is $2\sin(\pi/n)$. Therefore $$ l_n=2n\sin(\pi/n)=2\pi\cdot\frac{\sin(\pi/n)}{\pi/n}\overset{n\to \infty}{\to} 2\pi. $$ 
