Dale Tate

2022-06-27

I have given the following nonconstant complex polynomial $h(x)={x}^{n}+{a}_{n-1}{x}^{n-1}+\dots +{a}_{1}x+{a}_{0}$. In the lecture our Prof. told us that using the minimal principle one could find $z\in \mathbb{C}$ such that $h(z)=0$

Hadley Cunningham

Beginner2022-06-28Added 20 answers

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.

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.

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