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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?

rcoder
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2 Answers2

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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 $2\pi$

Mohammad Riazi-Kermani
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3

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 $$ l_n=2n\sin(\pi/n)=2\pi\cdot\frac{\sin(\pi/n)}{\pi/n}\overset{n\to \infty}{\to} 2\pi. $$

Maurizio Moreschi
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