Why $x^{n}=1$'s solution is equal to regular polygon in complex plane? My major is Electrical Engineering.
Although I'm not majoring mathamatics, but I got one question.
In my circuit theory class, I just learned Euler's formula to understand R-L-C circuit's
analysis
But one day, another professor told me that you have to make regular polygon program unless
you got low score.
so, I have to make regular polygon drawing program.
my solution is let's find $x^{n}=1$ solution.
here is my $x^{n}=1$ solution (Let me skip my prove)
$x_k=e^{\frac{i(2k-2)\pi}{n}}$ (n, k is must be natural number) $(1\le k \le n, n\ge1)$
but why $x^n=1$ solution equals to drawing regular polygon??
I mean, if I solve  ${x^3=1}$ equation, I got regular triangle (in complex plane)
if I solve  ${x^4=1}$ equation, I got regular square (in complex plane)
if I solve  ${x^5=1}$ equation, I got regular pentagon (in complex plane)
but why?
Do I have to learn Galois theory?
Galois proved that over 5 dimension equation, there is no general quadratic formula.
but I found general equation in some cases.
why?
Some people mentioned that you have to learn group theory to understand
equation.
Do I have to learn group theory to understand it?
Thank you for reading my noob article.
 A: Very nice question. Here is the thing about complex multiplication: it works best in polar coordinates.
Every complex number $z$ can be written in the form $r e^{i \phi}$ where $r$ is a real number and $\phi$ an angle (so technically also a real number). When drawing $z$ in the complex plane the number $r$, called the modulus, is the distance from 0 to $z$ and $\phi$, called the argument is the angle that the line from 0 to $z$ makes with the positive real axis. In other words: $(r, \phi)$ are the location of $z$ in standard polar coordinates.
Now here is the thing you need to know:

If you multiply two complex numbers you multiply their moduli and add their arguments

So if $z = re^{i \phi}$ and $w = s e^{i \theta}$ then $zw = rse^{i (\theta + \phi)}$.
This has a very interesting consequence for numbers with modulus 1.

Multiplying an arbitrary complex number $w$ with $e^{i \phi}$ looks graphically just like rotating $w$ over angle $\phi$ around the origin (counter clockwise)

This simple result (much much much more simple than any part of Galois theory) is what you need to get your answer.
To find the $n$ points of a regular $n$-gon you start with the number 1. You rotate it $2\pi/n$ degrees to get to the next point on you n-gon, which corresponds to multiplying by $e^{i 2 \pi/n}$. Then you multiply it with that number again to get to the third point etc.
After $n$ steps you are back where you started: at the number 1. So apparently multiplying 1 $n$ times with $e^{i 2 \pi/n}$ yields $1$ again or in other words $(e^{i 2 \pi/n})^n = 1$ or, equivalently $e^{i 2 \pi/n}$ is a solution to $x^n = 1$.
Now in a completely similar fashion you can show that the other points on the $n$-gon are also solutions to that equation.
EDIT: what I showed above is that the points of a regular $n$-gon that
a) lie on a circle with radius 1 and
b) have number 1 as one of the points
all are solutions to $x^n = 1$. You asked the oposite question: why do the solutions of $x^n$ form a regular $n$-gon. What we know from the above is that at least some $n$-element subset of the set of solutions to $x^n = 1$ do form an $n$-gon, namely the $n$ points on the special $n$-gon discussed above.
So the only thing left two show is that the equation $x^n = 1$ has no other solutions besides the $n$ we just studied.
[END OF EDIT, original answer continues below]
There are two ways of doing that: you use the description of multiplication above to see that every other point in the complex plane has modulus either too small or too big or otherwise the wrong argument, or you use some abstract theory to show that no degree $n$ polynomial can have more than $n$ roots.
A: Your question leads to De Moivre’s formula and $n$-th roots, which says that for $n\in\mathbb{N}$ applies: $(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)$.
I recommend a nice book "A Friendly Approach to Complex Analysis" by Sara Maad Sasane and Amol Sasane, to which I refer for answering the OP's question.
Let us consider complex numbers that satisfy $x^n=z$. Then $z=r(\cos\theta+i\sin\theta)$ for non-negative radius $r\ge0$ and $\theta\in[0,2\pi)$. By taking the power $x^n=z$, where $x=\rho(\cos\alpha+i\sin\alpha)$ we obtain:
$$x^n=\rho^n(\cos(n\alpha)+i\sin(n\alpha))=r(\cos\theta+i\sin\theta)=z$$
Since $\rho^n=r$, we have $\rho=\sqrt[n]{r}$. The angle that $x^n$ makes with the positive real axis is $n\alpha$ and it is in the set $\{\ldots,\theta-4\pi,\theta-2\pi,\theta,\theta+2\pi,\theta+4\pi,\ldots\}$, since the
angle made by a nonzero $z$ with the positive real axis is unique only up to integral multiples of $2\pi$, that is, instead of $\theta$, we could just as well have used $\theta+2\pi\cdot k$ for any integer $k$. We get an angle $\alpha\in\left\{\frac{\theta}{n}+\frac{2\pi}{n}k~\middle|~ k\in\mathbb{Z}\right\}$ and distinct $x$ for $\alpha\in\left\{\frac{\theta}{n},\frac{\theta}{n}+\frac{2\pi}{n},\frac{\theta}{n}+2\frac{2\pi}{n},\ldots,\frac{\theta}{n}+(n-1)\frac{2\pi}{n}\right\}$. Let us choose $z=1$, then we obtain the $n$th roots of unity, which are located at the vertices of an $n$-sided regular polygon inscribed in a circle:

