Using Maximum Modulus Principle

[d2j2003]

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]

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

 

[d2j2003]

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

 

[Dick]

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.

 

[d2j2003]

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?

 

[d2j2003]

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

 

[Dick]

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.

 

[d2j2003]

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]

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

Of course M is real. Saying |p(z)|<=M doesn't make any sense if M is complex. The whole point is that if |f(z)|<=M on the boundary of an analytic domain then |f(z)|<=M on the interior of that domain. That's the maximum principle.

 

[d2j2003]

ok i thought so, just seemed like a weird variable for a real number.. but I think i understand it now, thank you so much