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Math Help - Pole

  1. #1
    Senior Member slevvio's Avatar
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    Pole

    Hello everyone I was wondering if anyone could give me a hand with this

    Let f be an entire function such that |f(z)| \rightarrow \infty as | z| \rightarrow \infty. Define g by

     <br />
g(z)=f(1/z)
    where  (z \not= 0)<br />

    Show that g has a pole at 0, and, by considering the Laurent series for g about 0, show that f is a polynomial function.

    Im not so worried about the 2nd part yet but i was wondering how to show that g has a pole, intuitvely i can see that it has one at zero

    i am trrying to express g as

    g(z) = \frac{h(z)}{z} where h(z) is analytic on some open disc around 0 and k(0) \not= 0
    but im not sure how to do this

    any help would be appreciated, thank you
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Leeds, UK
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    Quote Originally Posted by slevvio View Post
    Let f be an entire function such that |f(z)| \rightarrow \infty as | z| \rightarrow \infty. Define g by

     <br />
g(z)=f(1/z)
    where  (z \not= 0)<br />

    Show that g has a pole at 0.
    Let h(z) = 1/g(z)\ (z\ne0). Then h is analytic in some neighbourhood of 0 and tends to 0 as z\to0. So if we extend the domain of h by defining h(0)=0 then h is analytic at 0, and in fact has a zero of some order at 0. Therefore g has a pole of that same order at 0.
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  3. #3
    Senior Member slevvio's Avatar
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    great, thank you
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