This is an exercise from Alhfors Complex Analysis book- to show that an analytic function with a nonessential singularity at infinity must be a polynomial. It seems like it should probably be pretty straight forward, but I must be missing something. If it has a removable singularity at infinity than it extends to an analytic function on the Riemann sphere, and so must be constant by Liouville's theorem. What if there is a pole at infinity though? This was homework some time ago, and I never finished it :/ but have been thinking about it again recently. Thanks :)
Analytic functions with nonessential singularity at infinity must be a polynomial
13
$\begingroup$
complex-analysis
2 Answers
5
Another hint: look at the function $f(\frac{1}{z})$ at z = 0, it has a nonessential singularity at 0...
2
Hint: consider the Laurent series in the annulus $0 < |z| < \infty$.
-
0Ahlfors book hasn't introduced Laurent series yet, so I was looking for a way along another path- hopefully. Thank you though. – 2011-03-29