Edwin Koh
Posted on Sunday, 16 November, 2003 - 07:25 am:
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.)
Edwin Koh
Posted on Sunday, 16 November, 2003 - 07:30 am:
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.
Edwin Koh
Posted on Sunday, 16 November, 2003 - 07:50 am:
P/S - f is only analytic in a neighbourhood about a.
David Loeffler
Posted on Sunday, 16 November, 2003 - 10:12 am:
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 =C∪{∞} 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