# Using Maximum Modulus Principle

## Homework Statement

Suppose p is a polynomial of degree n and |p(z)|≤M if |z|=1

Show that |p(z)|≤M|z|$^{n}$ if |z|≥1

## Homework Equations

Maximum Modulus Principle: If f is a nonconstant analytic function on a domain D, then |f| can have no local maximum on D.

## The Attempt at a Solution

Book says to apply the maximum modulus principle to $\frac{p(z)}{z^{n}}$ on domain |z|>1 but I am unsure why?

Dick
Homework Helper
p(z)/z^n is 'analytic at infinity'. Has the book gone over what that means?

I believe f(z) is analytic at infinity if f(1/z) is analytic at z=0

Dick
Homework Helper
I believe f(z) is analytic at infinity if f(1/z) is analytic at z=0

That's it. And p(z)/z^n is one of those functions, yes? It's just a polynomial in 1/z. If that's the case you can treat the function as analytic on the domain |z|>1.

so we can say that since p(z)/z^n is analytic at infinity then it is analytic on the domain |z|>1 so according to the maximum modulus principle, p(z)/z^n can have no local maximum on domain (|z|>1) right?

I'm thinking now I have to show that |p(z)/z^n|≤M if |z|≥1 is this correct?

Dick
Homework Helper
so we can say that since p(z)/z^n is analytic at infinity then it is analytic on the domain |z|>1 so according to the maximum modulus principle, p(z)/z^n can have no local maximum on domain (|z|>1) right?

Yes, so |p(z)/z^n|<=M on |z|>1. Otherwise it would have a local maximum.

and M is just a real number? it is not specified anywhere... and what about |z|=1 since it says on |z|≥1 and we have looked at |z|>1

Dick