I am trying to show that 1 has n distinct roots of degree n, or in other word that the equations $$z^n=1$$ has n different roots over the complex field.
I know that the fundamental theorem of Algebra ensures me that there are n, but it does not ensure that they are different, so how can I do it?
Thanks
P.S. it's not homework :)
If $z_1, \ldots, z_n$ are pairwise distinct roots of an $n$-th degree polynomial $p(z)$ then they all have multiplicity 1 and there are no other roots. This I hope is clear, and if not I can explain it further.
The numbers $z_1, \ldots, z_n$ defined by $$z_k = \cos(2 \pi k / n) + i \sin (2 \pi k / n)$$ are all roots of $z^n - 1$ by Moivre's formula. They are also pairwise distinct. Therefore, they are the only roots of $z^n - 1$.
In short, we guess $n$ pairwise distinct roots of $z^n = 1$, and then we conclude that there cannot be any others.
A multiple root of $f(x)$ is also a root of $f'(x)$ (simply because the derivative of $f(x)=(x-a)^2g(x)$ is $f'(x)=2(x-a)g(x)+(x-a)^2g'(x)$). You can compute $\gcd(f(x),f'(x))$ with the euklidean algorithm. In this case simply $\gcd(f(x),f'(x)) = \gcd(x^n-1,nx^{n-1})=1$ because $\frac 1nxf'(x)-f(x)=1$.
