Why are roots of unity always distinct? As per the fundamental theorem of algebra, $z^n-1=0$ has $n$ complex solutions--but, for a general polynomial, some of those solutions may have multiplicity $> 1$, they may repeat.
Why is it that roots of unity never repeat, i.e., that the equation above always has $n$ distinct solutions for all $n$?
 A: If a polynomial $\phi$ has a repeated root $a$ then $\phi(X)=(X-a)^k \psi(X)$ for some polynomial $\psi$ and some $k\geqslant 2$. Then $\phi'(X)=k(X-a)^{k-1}\psi(X)+(X-a)^k \psi'(X)$, and $\phi'(a)=0$.
Apply this to $X^n-1$ whose derivative is $nX^{n-1}$: the only candidate for a repeated root is $0$ and it is not even a root.
[Note that over some fields, where the characteristic divides $n$, we do get repeated roots.]
A: Because we know explicitely its $n$ roots $$\exp(2i\pi k/n), \ \ \ k=0,1,...(n−1)$$
and that all of them are distinct.
A: Suppose a complex root $ \ \alpha \ $ had multiplicity 2 . Then $ \ (z \ - \ \alpha)^2 \ $ would generate a term $ \ -2 \alpha z \ $; similarly, higher multiplicities would produce terms which had powers of $ \ z \ $ other than $ \ x^n \ $ or the "constant term".
The coefficient of the $ \ z^{n-1} \ $ term is zero, which is the negative of the sum of the $ \ n \ $ zeroes of the polynomial.  In order to meet this condition and not produce "unwanted" terms, the zeroes must all be distinct.
We can in fact go further:  the zeroes must be paired in complex conjugates  $ \ a \ \pm \ bi \ $ .  If  $ \ n \ $ is even, then $ \ 1 \ $ and $ \ -1 \ $ are zeroes, and the other $ \ (n-2) \ $ zeroes (an even number) must have "conjugate pairs" with the $ \ a \ $'s in one pair balanced by $ \ -a \ $'s in another pair.  If  $ \ n \ $ is odd, then $ \ 1 \ $ is a zero, and  the other $ \ (n-1) \ $ zeroes (an even number) must have "conjugate pairs" with the $ \ a \ $'s in the pairs summing to $ \ -1 \ $.
