prove that the next two affirmations are equivalent :
1) every non constant $f(x)\in \mathbb C[x]$ has all of its roots in $\mathbb C$
2)every non constant $f(x)\in \mathbb C[x]$ has at least one root in $\mathbb C$
the fisrt implication ($1\Rightarrow 2$) is trivial but I don´t know how to prove the second implication ($2\Rightarrow 1$)
I would really appreciate your help :)
If $f(x)$ has a root in $\mathbb{C}$ like $z_0$ then $f_1(x)=\frac{f(x)}{x-z_0}\in \mathbb{C}[x]$ too. So $f_1$ has a root in $\mathbb{C}$ like $z_1$. We can construct $f_{i+1}$ from $f_i$ and receive roots $z_i$ and $z_{i+1}$. Since roots of a polynomial are finite then there is $n\in \mathbb{N}$ such that $f_n$ is constant and $f=f_n (x-z_0)\ldots(x-z_{n-1})$.