# Laurent Series and Singularity Proofs.

1. Apr 16, 2007

### moo5003

1. The problem statement, all variables and given/known data

Let D be a subset of C and D is open. Suppose a is in D and f:D\{a} -> C is analytic and injective. Prove the following statements:

a) f has in a, a non-essential singularity.
b) If f has a pole in a, then it is a pole of order 1.
c) If f has a removable singularity in a, then theanalytic extension of f to D is one to one too.

2. Relevant equations
Laurent Series
f(z) = h(1/z) + g(z) on the annulis.
a_n = 0 for all n<k for some k in Z if a is removable
a_n != 0 for infinite n<0 if a is essential

3. The attempt at a solution
a)I'm a little stumped at the moment. I've narrowed my goal down to showing that for some reason because f(z) is one to one that a_n goes to 0. My other guess was to assume that the sing. was essential but i'm not sure how to come up with a contradiction. Any hints you guys can provide would be helpful.

Last edited: Apr 16, 2007
2. Apr 16, 2007

### Dick

For essential singularities, look up the theorem known as "Little Picard". For pole singularities, the function near the pole behaves an awful lot like 1/(z-a)^n. Now remember a complex number has n n-th roots.

3. Apr 18, 2007

### moo5003

I think you meant to say use the Big Picard Theorem, but either way thanks for the tip it made the problem much more approachable. (Contradiction to injectivity arises since you can find two dotted disks that both map C or C\{b} for some b in C which implies that points map to the same thing =><= ). Thanks for your help