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 $n$?
Best Answer
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.]