# Complex Polynomials and Minimal Values

• moo5003
In summary, the fundamental theorem of algebra can be proven using the minimal modulus principle. If the limit as the absolute value of z approaches infinity implies the absolute value of P(z) approaches infinity for some non-constant polynomial, then there exists a complex number a such that the absolute value of P(a) is less than or equal to the absolute value of P(z) for all z. This can be shown by finding a closed disk around 0 with a radius where the absolute value of the polynomial is greater than or equal to the absolute value at 0, and since the disk is a compact set, it must contain a minimum value of z_0. This means that the absolute value of P(z_0) is less than or

#### moo5003

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?

moo5003 said:
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.