Could we find two complex numbers such as $z_1$ and $z_2$, where both of them are roots of unity (de Moivre numbers) and also when we define
$$s = z_1 + z_2$$
$s$ is also a root of unity?
Finding two roots of unity which their sum is also root of unity
complex numbersroots-of-unity
Related Solutions
Since $\omega=\exp\left(\frac{2\pi i}{n}\right)$, we have $\omega^{n/2}=\exp\left(\frac{2\pi i}{n}\frac{n}{2}\right)=\exp(\pi i)=-1$. This and the law $\omega^{a+b}=\omega^a\omega^b$ give the formula you ask about in your first question. In words: multiplying a complex number by $\omega$ rotates the vector in the complex plane corresponding to that number by $360^\circ/n$. Repeating $n$ times brings the vector back to its starting position; repeating only $n/2$ times rotates by $180^\circ$, which is the same as multiplying by $-1$.
As to your second question, when $n$ is odd, $\omega^0=1$ is the only real root; it's the root we already knew about before we learned about complex numbers; all other roots have an imaginary part. When $n$ is even, there are two real roots, $1$ and $-1=\omega^{n/2}$. Again, we knew both of these roots before we learned complex numbers. (E.g. the two real sixth roots of 1 are $1$ and $-1$.) All other roots have an imaginary part.
First, there are at most $n$ $n$th roots of unity, because $x^n-1$ can have at most $n$ roots (as a consequence of the Factor Theorem applied in $\mathbb{C}$).
Second, if $\omega$ is an $n$th root of unity, that means that $\omega^n = 1$. But then, for any integer $k$, we have $$(\omega^k)^n = \omega^{kn} = (\omega^n)^k = 1^k = 1,$$ so $\omega^k$ is also an $n$th root of unity.
So now the question is which ones are different? If $\omega$ is such that $\omega^n=1$ but $\omega^{\ell}\neq 1$ for any $0\lt \ell\lt n$, then $\omega^r=\omega^s$ if and only if $\omega^{r-s}=1$, if and only if $n|r-s$: indeed, using the division algorithm, we can write $r-s$ as $qn + t$, with $0\leq t\lt n$ (division with remainder). So then $$1=\omega^{r-s} = \omega^{qn+t} = \omega^{qn}\omega^t = (\omega^n)^q\omega^t = 1^q\omega^t = \omega^t.$$ But we are assuming that $\omega^t\neq 1$ if $0\lt t\lt n$; since $0\leq t\lt n$ and $\omega^t=1$, the only possibility left is that $t=0$; that is, that $n|r-s$.
Thus, if $\omega^{\ell}\neq 1$ for $0\lt \ell\lt n$ and $\omega^n=1$, then $\omega^r=\omega^s$ if and only if $n|r-s$, which is the same as saying $r\equiv s\pmod{n}$.
So it turns out that $\omega^0$, $\omega^1$, $\omega^2,\ldots,\omega^{n-1}$ are all different (take any two of the different exponents: the difference is not a multiple of $n$), and they are all roots of $x^n-1$, and so they are all the roots of $x^n-1$.
So it all comes down to finding an $\omega$ with the property that $\omega^k\neq 1$ for $0\lt k\lt n$, but $\omega^n=1$. And $$\large\omega = e^{2\pi i/n}$$ has that property.
Best Answer
If $|z_1|=|z_2|=1$ and $|z_1+z_2|=1$, then $$1=|z_1+z_2|^2=(z_1+z_2)(\bar{z}_1+\bar{z}_2)=z_1\bar{z}_1+z_2\bar{z}_2+z_1\bar{z}_2+\bar{z}_1z_2.$$ That is, $$1=|z_1|^2+|z_2|^2+2\Re(z_1\bar{z}_2)=1+1+2\Re(z_1\bar{z}_2).$$ Hence, $$\Re\left(\frac{z_1}{z_2}\right)=\Re(z_1\bar{z}_2)=-\frac12.$$ However, $$\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}=\frac{1}{1}=1.$$ That is, $$\frac{z_1}{z_2}=\frac{-1\pm\sqrt{3}i}{2}=\left(\frac{1\pm\sqrt{3}i}{2}\right)^2.$$ This means $$(z_1,z_2)=(u\alpha,u\alpha^{-1})$$ for some complex number $u$ such that $|u|=1$ and $\alpha=\frac{1\pm\sqrt{3}i}{2}$; in this case, $z_1+z_2=u$. If you want $z_1$ and $z_2$ be roots of unity, then it automatically follows that $u$ is also a root of unity.