# Fundamental Theorem of Algebra Optimisation Proof

I've recently been looking into optimisation and have found it quite fascinating. During my readings, I came across a proof for the fundamental theorem of algebra.

I follow the proof for the most part, but I get lost at one step. The proof takes a polynomial of degree $n$, denoted as $p(z)$ where $z$ is complex and shows that $\left|p(z)\right|$ is coercive and thus at least one minimiser $\hat{x}$ exists.

What I do not follow is the proof then saying that we can assume $\hat{x}=0$ without loss of generality and thus show that $p(\hat{x})=0$. Could someone please explain the justification behind this step?

• If $\hat{x} \neq 0$, consider the polynomial $p(z- \hat{x})$. $0$ is a minimizer for $|p(z- \hat{x})|$, and so if you prove that $0$ is a zero of $p(z- \hat{x})$, then $\hat{x}$ is a zero of $p$. – Crostul Aug 13 '16 at 13:12

If $\hat{x}$ is a non-zero number $a$, then the polynomial $q(z):=p(z+a)$ reaches the minimum at $0$ (thus $\hat{x}=0$ for $q$).
Now, if you prove that $q(0)=0$, then you have $p(a)=0$ and replacing $p$ by $q$ doesn't change the nature of the problem.