# Isolated singularities.

The function has an isolated singularity at if it is analytic on a deleted open neighborhood of , but is not analytic at .

Example 8.3.1
1. has an isolated singularity at 0.
2. has an isolated singularity at 2.

Definition 8.3.2   Suppose that has an isolated singularity at .
1. If there exists a function , analytic on an open neighborhood of and such that , then the singularity of at is called removable.
2. Suppose that for every , , where verify the following conditions:
• and are analytic at ;
• and ;
Then the function has a pole at . If is a zero of order of , we say that has a pole of order at .
3. If is neither a pole nor a removable singularity of , we say that has an essential singularity at .

Proposition 8.3.3 (Riemann)   Suppose that has an isolated singularity at . If , then the singularity is removable.

Example 8.3.4   Let . This function has a singular point at 1, but:

Therefore

and the singularity is removable.

Example 8.3.5   Let . This function has a singularity at 2. But:

and the singularity is not removable.

Recall that in Calculus, you would have proven that the real valued function of the real variable given by has two infinite one-sided limits at 2.

Proposition 8.3.6   Suppose that has an isolated singularity at . If there exists a natural number such that and , then has a pole of order at .

Proposition 8.3.7   Suppose that has a singular point at . Expand as a Laurent series on a pointed neighborhood of , say

1. If there exists a negative integer such that for every integer and , then has a pole of order at .
2. If the Laurent series expansion does contain negative powers of , but there is no such , the point is an essential singularity of .

Example 8.3.8   has a pole of order 1 at 0 as a Laurent series expansion of about 0 is precisely .

Example 8.3.9   has a pole of order 2 at 3. To prove this according to Proposition 3.7, we need some work:

A Taylor expansion of about 3 is:

Thus

Example 8.3.10   has an essential singularity at 0. The MacLaurin expansion of is

By substitution we get a Laurent expansion of about 0:

The negative powers are not bounded below.

Noah Dana-Picard 2007-12-24