Given a polynomial in $\mathbb{C}$, I want to show that a polynomial of the form $$P_n(z)=\sum^n_{i=0} a_iz^i$$ can be decomposed into something like
$$P_n(z)=(z-z_0)\cdot Q_{n-1}(z),$$ where $Q$ is a polynomial of degree $n-1$ and $P(z_0)=0$.
I started doing the following:
$$P_n(z)=\sum^n_0 a_i((z-z_0)+z_0)^i=\sum^n_{i=0}a_i\sum^i_{k=0}{i\choose k}(z-z_0)^{i-k}z_0^k,$$ but now I don't know how to continue
From the Euclidean algorithm applied to polynomials we know that for any $z_0$
$P_n(z) = (z-z_0)Q_{n-1}(z) + R(z)$
where $R(z)$ has degree less than the degree of $z-z_0$ - so in fact $R(z)$ must be a constant $R$ and we have
$P_n(z) = (z-z_0)Q_{n-1}(z) + R$
But if you know that $z_0$ is a root of $P_n$ then $P_n(z_0) = 0$ and so $R=0$.
The difficult part is showing the $P_n$ must have at least one root in $\mathbb{C}$. If we replace $\mathbb{C}$ by $\mathbb{R}$ then this is not necessarily true because $\mathbb{R}$ is not algebraically closed.