Edwin
Koh
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| Posted on Sunday, 16
November, 2003 - 07:25 am: |
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Let f(z) be
an analytic function. If f has a zero of order n at a,
then f(z)=(z-a)n g(z) where g(z) is analytic
and g(a)=/= 0. Is there a similar representation for f if
a = infinity ? (An example of such a function f(z) would
be f(z)=p(z)/q(z) where p and q are polynomials of the
same degree.)
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Edwin
Koh
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| Posted on Sunday, 16
November, 2003 - 07:30 am: |
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Oops, a
correct example would be f(z) = p(z)/q(z) - c/d where c
and d are the leading coefficients of p(z) and q(z)
respectively.
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Edwin
Koh
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| Posted on Sunday, 16
November, 2003 - 07:50 am: |
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P/S - f is
only analytic in a neighbourhood about a.
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David
Loeffler
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| Posted on Sunday, 16
November, 2003 - 10:12 am: |
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Yes, just look at g(z)=f(1/z). If
f is 'analytic at ¥', then g will have a
removable singularity at 0 and we can apply all the usual
machinery. (The last sentence is true by definition.)
You can think about this in terms of extending the maps
to maps from S2® S2 where S2=CÈ{ ¥} is the Riemann sphere. Using the general
theory of Riemann surfaces one can show that any function
meromorphic on the Riemann sphere is in fact a rational
function (a quotient of polynomials).
David
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