I have given the following nonconstant complex polynomial $h(x)=x^n+a_{n-1}x^{n-1}+…+a_1x+a_0$. In the lecture our Prof. told us that using the minimal principle one could find $z\in \Bbb{C}$ such that $h(z)=0$.
So I mean $h$ is clearly an analytic function and $\Bbb{C}$ is open and path connected and thus also connected. Now I should find $z_0\in \Bbb{C}$ such that $$|f(z)|\geq |f(z_0)|>0$$ for all $z$ in the domain. Then the minimal principle sais that $f$ is constant.
My idea was now to find such a $z_0$ to apply the minimal principle and then to show that for some $w$ $f(w)=0$ Then we should also have $f(z)=0$.
But I‘m not sure if this works and therefore I wanted to ask if someone could help me and say if my idea is correct.
This would be very nice.
Best Answer
Suppose $h(z)$ is a polynomial of degree at least $1$ and $h(z)$ has no roots.
Consider $g(z) = \frac {1}{h(z)}.$ If $h$ has no roots, $g$ is analytic.
$|g(0)| > 0$ (or else we have found a root and have a contradiction.)
Over the domain $|z| \le R$ the maximum modulus theorem says the maximum of $|g(z)|$ must lie on the circle $|z| = R$
But if as |z| gets to be large, $|h(z)|$ goes to infinity.
Hence for large enough R, $|z| = R$ implies $0<|g(z)| < |g(0)|$ which is a contradiction.