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Using Maximum Modulus Principle

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data

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

    Show that |p(z)|≤M|z|[itex]^{n}[/itex] if |z|≥1

    2. Relevant equations

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

    3. The attempt at a solution

    Book says to apply the maximum modulus principle to [itex]\frac{p(z)}{z^{n}}[/itex] on domain |z|>1 but I am unsure why?
     
  2. jcsd
  3. Apr 13, 2012 #2

    Dick

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    p(z)/z^n is 'analytic at infinity'. Has the book gone over what that means?
     
  4. Apr 13, 2012 #3
    I believe f(z) is analytic at infinity if f(1/z) is analytic at z=0
     
  5. Apr 13, 2012 #4

    Dick

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    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.
     
  6. Apr 13, 2012 #5
    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?
     
  7. Apr 13, 2012 #6
    I'm thinking now I have to show that |p(z)/z^n|≤M if |z|≥1 is this correct?
     
  8. Apr 13, 2012 #7

    Dick

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    Yes, so |p(z)/z^n|<=M on |z|>1. Otherwise it would have a local maximum.
     
  9. Apr 13, 2012 #8
    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
     
  10. Apr 13, 2012 #9

    Dick

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    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.
     
  11. Apr 13, 2012 #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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