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Math Help - Complex Analysis

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    MHF Contributor chiph588@'s Avatar
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    Complex Analysis

    Suppose that  f is an entire functionand that there are positive constants  A and  m with  |f(z)| \leq A|z|^{m} if  |z| \geq R_{0} . Show that  f is a polynomial of degree  m or less.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Suppose that  f is an entire functionand that there are positive constants  A and  m with  |f(z)| \leq A|z|^{m} if  |z| \geq R_{0} . Show that  f is a polynomial of degree  m or less.
    isn't this just a slight twist on the (extended) Liouville's theorem?
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    Quote Originally Posted by chiph588@ View Post
    Suppose that  f is an entire functionand that there are positive constants  A and  m with  |f(z)| \leq A|z|^{m} if  |z| \geq R_{0} . Show that  f is a polynomial of degree  m or less.
    Let R>R_0.

    If f is an entire function then f(z) = \sum_{n=0}^{\infty}a_n z^n and a_n = \frac{1}{2\pi i}\oint \limits_{|z|=R} \frac{f(z)}{z^{n+1}} dz.

    If |f| \leq M on |z|=R then \left| a_n \right| = \left| \frac{1}{2\pi i}\oint \limits_{|z|=R} \frac{f(z)}{z^{n+1}}dz \right| \leq \frac{M}{R^n}

    By hypothesis |f(z)| \leq A |z|^m for |z| \geq R_0. Therefore, if R > R_0 then on |z| = R we have |f(z)| \leq A |z|^m = A\cdot R^m

    Thus, |a_n| \leq \frac{A \cdot R^m}{R^n}.
    Thus, if n>m then by making R sufficiently large we have |a_n| = 0 \implies a_n = 0.

    Thus, f is a polynomial of degree at most m.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    That's the proof I used, does that prove it for the disk  |z| < R_{0} also?
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