# Complex Polynomials and Minimal Values

Hello, I'm trying to prove the fundamental thereom of algebra using the minimal modulous principal. There is one step that I cannot make sense out of logically and I was wondering if one of you could explain it to me.

If the limit as |z| approaches infinity implies |P(z)| approaches infinity for some non-constant polynomial then there exists an a in C such that |P(a)| <or= |P(z)| for all z. Why does this follow?

## Answers and Replies

JasonRox
Homework Helper
Gold Member
Hello, I'm trying to prove the fundamental thereom of algebra using the minimal modulous principal. There is one step that I cannot make sense out of logically and I was wondering if one of you could explain it to me.

If the limit as |z| approaches infinity implies |P(z)| approaches infinity for some non-constant polynomial then there exists an a in C such that |P(a)| <or= |P(z)| for all z. Why does this follow?

I don't quite remember exactly how it goes, but I remember it using 1/|P(z)|.

I'd have to wait until I get home to find out how it went.

So, after reading wikipedia for awhile. It follows since you can find a closed disk around 0 with radius r where |f(z)|>or=|f(0)| where |z|>r since the limit goes to infinity. And since the disk is a compact set it must contain a minimum value of z_0. Thus |f(z_0)| <or= |f(x)| for all x in C.

Note: z_0 is on interior of the disk not the boundary.