If is analytic only in a punctured disk , then  is said to have an isolated singularity at Let A denote the set of all isolated singularities.  We say that the isolated singularity is an artificial (or removable) singularity if f can be extended to   so that the resulting function is holomorphic on .  An isolated singularity of  at  is called a pole of  if  where , is analytic in and . The integer is said to be the order of the pole and if , then we say that  is a simple pole of

We say that the isolated singularity is an essential singularity  if is neither an artificial singularity nor a pole.

A function  is called a meromorphic function if  is analytic for all values in C except at points where poles occur.


Theorem.  An isolated singularity of  at is a pole if and only if


Proof () This follows immediately from the definition of a pole.
() Let Then has an isolated singularity at is not identically and  has a removable singularity at  Then it is easy to show that there exists a positive integer  such that in a neighborhood of  where is analytic there with Thus is analytic in a neighborhood of  Now with  and the proof is complete.

We can now write down the following immediate-to-prove but practical:


Corollary.  has a pole of order  at  if and only if has a zero of order  at  


Examples.  1)  is an artificial singularity of since

2)      For each integer n, is a simple pole of

3)       is an essential singularity of since


Weierstrass Theorem.  Let  A be a singular subset of an open subset D of C and let f be holomorphic in D-A.  If is an essential singularity of f, then for every r>0 with D-A we have X.

Proof.  The proof is by contradiction.  Assume that X.  Then there exists X such that .  Then there exists such that

.  So .  Thus the function is holomorphic in and .  That is, g is bounded in .  By a previous proposition is an artificial (removable) singularity of g.  So we may regard g as a holomorphic function on  We now seek a contradiction in each of the following two exclusive cases:

Case 1. : There exists such that and therefore such that .  So .  Therefore by the triangle inequality.  That is, f is bounded on .  Moreover, if then f is bounded being continuous there.  As a result, f is bounded on .  By a previous proposition, is a removable singularity of f which is a contradiction.

Case 2. That is is a zero of g.  Let m denote the order of this zero.  Thus there exists a holomorphic function  on such that and .  We have   That is, or   Notice that h is holomorphic on  and .  This means that  is a pole of f of order m which is again a contradiction.