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Complex Analysis question

  1. Jun 13, 2006 #1
    So my professor threw in what he called an extra 'hard' question for a practice test. So naturally I have a question about it. It relates to the Maximum Modulus Principle:

    a) Let [tex]p(z) = a_0 + a_1 z + a_2 z^2 + ...[/tex]
    and let M = max |p(z)| on |z|=1.
    Show that [tex]|a_i|< M[/tex] for [tex]i = 0,1,2. [/tex]

    b) What is the order of the zero at infinity if f(z) is a rational function of the form

    [tex]f(z) = \frac {p(z)}{q(z)}[/tex]

    where both p(z) and q(z) are both polynomials and [tex] deg(p) < deg(q). [/tex]
     
    Last edited: Jun 13, 2006
  2. jcsd
  3. Jun 13, 2006 #2

    NateTG

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    As you have it, that's not necessarily true.
    Consider:
    [tex]p(z) \equiv 0[/tex]
    That is, the polynomial is constant zero.
    Then [tex]M=0[/tex],but
    [tex]a_i = M = 0 [/tex] which contradicts [tex]|a_i|<M[/tex]
     
  4. Jun 13, 2006 #3
    okay, it should be [tex]|a_i|[/tex] is less than or equal to M for [tex]i=0,1,2.[/tex]
     
  5. Jun 13, 2006 #4

    NateTG

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    So, have you tried anything?
     
  6. Jun 13, 2006 #5
    Well, for the first part, given that |p(z)| [tex]\leq[/tex]|M| for |z|=1,
    and with:
    p(0)=[tex]a_0[/tex]
    p'(0)=[tex]a_1[/tex]
    p''(0)=[tex]2a_2[/tex],

    then in general,

    [tex]|p^k (0)| \geq \frac{k!}{2\pi i} \int_{|z|=1} \frac{f(z)}{z^(k+1)} dz \Rightarrow \frac{k!}{2\pi i} \int_{0}^{2\pi} f( e^(it) ) dt \leq k!M[/tex].

    thats about as far for that one. the other one is

    [tex]\frac{p(z)}{q(z)} = \frac{a_0 + a_1 z +...+a_k z^k}{b_0 + b_1 z +...+b_l z^l}[/tex] for some l>k. After that, I'm still working.
     
    Last edited: Jun 13, 2006
  7. Jun 13, 2006 #6

    shmoe

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    a) check your inequalities a little more carefeully, specifically comparing the derivatives at 0 to the integral.

    b) What is the order of the zero of f(1/z) at z=0?
     
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