Roots of unity are distincts For every $n\in\Bbb N$ and $$z_{k}:= \cos(2\pi k /n)+i\sin(2\pi k /n), \qquad k = 0,\ldots,n-1$$
we have $z_k^n=1$. 

How to show, in a simple way, that $z_k\neq z_l$ for every $k\neq l$?

By simple I mean that I know roughly the following:


*

*I know a bit of trigonometry (high school level) and I just proved De Moivre's formula by induction.

*I just discovered complex numbers.

*I know what injectivity/surjectivity/bijectivity is but I never proved that $\sin:[-\pi/2,\pi/2]\to[-1,1]$ and $\cos:[0,\pi]\to[-1,1]$ are bijective. (maybe it's time for it).

*I don't know what is $e^{z}$ for $z\in\Bbb C$.

*I know what is a field and that $(\Bbb C,+,\cdot)$ is a field.

*I don't know what is $\Bbb R^n$ for $n>2$ (and know more or less what is $\Bbb R^2$). Moreover I don't know what is a vector space.
 A: Consider the regular polygon with $n$ vertices circumscribed in the unit circle with a vertex situated on the positive $x$-axis. One may draw line segments from the origin to each vertex and determine the angle that line segment makes with the one on the positive $x$-axis: the angles are given by $\theta=2\pi\frac{k}{n}$ for $k=0,1,\cdots,n-1$. One can then determine the coordinates of the vertices using the definition of sine and cosine (as base and height of a right triangle with unit-length hypotenuse): they are $(\cos2\pi\frac{k}{n},\sin\pi\frac{k}{n})$ for $k=0,1,\cdots,n-1$. Changing the representation of these points to the complex number form $a+bi$ doesn't change the geometric fact that they are $n$ distinct points in the plane.
A: It isn't true that $z_k \neq z_l$ for every $k \leq l$. In fact, basic trigonometry gives you that $z_k = z_{k+n}$ for all $k$. What is true is that $z_0,\ldots,z_{n-1}$ are all distinct.
Proving from scratch that $z_0,\ldots,z_{n-1}$ is somewhat tiring, but only makes use of the following facts:


*

*$\sin x$ is monotone increasing from $0$ to $\pi/2$, monotone decreasing from $\pi/2$ to $3\pi/2$, and then monotone increasing from $3\pi/2$ to $2\pi$.

*$\cos x$ is monotone decreasing from $0$ to $\pi$ and monotone increasing from $\pi$ to $2\pi$.

*$\sin x$ is positive for $0 < x < \pi$ and negative for $\pi < x < 2\pi$.

*$\cos x$ is positive for $0 < x < \pi/2$, negative for $\pi/2 < x < \pi$, positive for $\pi < x < 3\pi/2$, and negative for $3\pi/2 < x < 2\pi$.
Another, perhaps simpler approach, is to compute $z_k - z_\ell$ using known trigonometric formulas:
$$
\cos \frac{2\pi k}{n} - \cos \frac{2\pi \ell}{n} + i \sin \frac{2\pi k}{n} - i \sin \frac{2\pi \ell}{n} = 2\sin \frac{\pi (k-\ell)}{n} \left(-\sin \frac{\pi (k+\ell)}{n} + i\cos \frac{\pi (k+\ell)}{n}\right).
$$
This time you need to use your knowledge of zeroes of $\sin$ and $\cos$:


*

*If $\sin x = 0$ then $x = m\pi$ for an integer $m$.

*If $\cos x = 0$ then $x = (m + 1/2)\pi$ for an integer $m$.
There are probably other approaches.
A: If you look at the graphs of $\sin(x)$ and $\cos(x)$ for $0\leq x\leq 2\pi$. You'll see that they cannot both (simultaneously) take the same value for any such $x$.
In other words you cannot find $x,y∈[0,2\pi]$ such that $\sin(x)=\sin(y)$ and $\cos(x)=\cos(y)$.  
Another way to see it is to convince yourself that the $n$-th roots of unity are equally spaced around the unit circle in $\mathbb C$.  That follows from the way raising to the $n$-th power works on complex numbers.  It multiplies the angle by $n$ and raises the radius to the $n$-th power (which is just $1^n=1$ for points on the unit circle).  Thus raising to the $n$-th power just multiplies the angle by $n$. 
