Using Maximum Modulus Principle

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Homework Help Overview

The discussion revolves around a problem involving a polynomial \( p(z) \) of degree \( n \) and the application of the Maximum Modulus Principle to show that \( |p(z)| \leq M |z|^n \) for \( |z| \geq 1 \). Participants are exploring the implications of the principle in the context of analytic functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the Maximum Modulus Principle to the function \( \frac{p(z)}{z^n} \) and question the reasoning behind this approach. There is exploration of the concept of a function being analytic at infinity and its implications for the domain \( |z| > 1 \). Some participants express uncertainty about the nature of the variable \( M \) and its implications for the problem.

Discussion Status

The discussion is active with participants examining the conditions under which the Maximum Modulus Principle applies. There is a recognition that \( \frac{p(z)}{z^n} \) is analytic on the specified domain, and some participants are clarifying the nature of \( M \) as a real number. The conversation reflects a mix of understanding and uncertainty, with no explicit consensus reached yet.

Contextual Notes

Participants note the lack of specification regarding the variable \( M \) and express curiosity about the implications of the boundary condition \( |z| = 1 \) in relation to the overall problem.

d2j2003
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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?
 
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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
 
d2j2003 said:
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?
 
d2j2003 said:
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
 
d2j2003 said:
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.
 
  • #10
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
 

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