Isolated singularities.

The function $f$ has an isolated singularity at $z_0$ if it is analytic on a deleted open neighborhood of $z_0$ , but is not analytic at $z_0$ .

Example 8.3.1  

1. $f(z)=1/z$ has an isolated singularity at 0.
2. $f(z)= e^{\frac {1}{z-2}}$ has an isolated singularity at 2.

Definition 8.3.2   Suppose that $f$ has an isolated singularity at $z_0$ .

1. If there exists a function $g$ , analytic on an open neighborhood $V$ of $z_0$ and such that $\forall z \in V-{z_0}, \; g(z)=f(z)$ , then the singularity of $f$ at $z_0$ is called removable.
2. Suppose that for every $z \neq z_0$ , $f(z)=N(z)/D(z)$ , where $N,D$ verify the following conditions:
   • $N$ and $D$ are analytic at $z_0$ ;
   • $N(z_0) \neq 0$ and $D(z_0)=0$ ;
   Then the function $f$ has a pole at $z_0$ . If $z_0$ is a zero of order $n$ of $D$ , we say that $f$ has a pole of order $n$ at $z_0$ .
3. If $z_0$ is neither a pole nor a removable singularity of $f$ , we say that $f$ has an essential singularity at $z_0$ .

Proposition 8.3.3 (Riemann)   Suppose that $f$ has an isolated singularity at $z_0$ . If $\underset{z \rightarrow z_0}{\text{lim}} (z-z_0)f(z)=0$ , then the singularity is removable.

Example 8.3.4   Let $f(z) = \frac {z^2-(1+i)z+i}{z^2-3z+2}$ . This function has a singular point at 1, but:

$$(z-1)f(z)= (z-1) \frac {(z-1)(z-i)}{(z-1)(z-2)} = \frac {(z-1)(z-i)}{z-2}$$

Therefore

$$\underset{z \rightarrow 1}{\text{lim}} (z-1)f(z)=0$$

and the singularity is removable.

Example 8.3.5   Let $f(z) = \frac {z^3-1}{z-2}$ . This function has a singularity at 2. But:

$$\underset{z \rightarrow 2}{\text{lim}} (z-2)f(z)= \underset{z \rightarrow 2}{\text{lim}} z^3-1 = 7 \neq 0$$

and the singularity is not removable.

Recall that in Calculus, you would have proven that the real valued function of the real variable $x$ given by $f(x)=\frac {x^3-1}{x-2}$ has two infinite one-sided limits at 2.

Proposition 8.3.6   Suppose that $f$ has an isolated singularity at $z_0$ . If there exists a natural number $n$ such that $\underset{z \rightarrow z_0}{\text{lim}} (z-z_0)^nf(z) \neq 0$ and $\underset{z \rightarrow z_0}{\text{lim}} (z-z_0)^{n+1}f(z)=0$ , then $f$ has a pole of order $n$ at $z_0$ .

Proposition 8.3.7   Suppose that $f$ has a singular point at $z_0$ . Expand $f(z)$ as a Laurent series on a pointed neighborhood of $z_0$ , say

$$f(z)= \underset{n=- \infty}{\overset{+ \infty}{\sum}} c_n (z-z_0)^n$$

1. If there exists a negative integer $N$ such that $c_n=0$ for every integer $n<N$ and $c_N \neq 0$ , then $f$ has a pole of order $N$ at $z_0$ .
2. If the Laurent series expansion does contain negative powers of $z-z_0$ , but there is no such $N$ , the point $z_0$ is an essential singularity of $f$ .

Example 8.3.8   $f(z)=1/z$ has a pole of order 1 at 0 as a Laurent series expansion of $f$ about 0 is precisely $f(z)=z^{-1}$ .

Example 8.3.9   $f(z)=e^z/(z-3)^2$ has a pole of order 2 at 3. To prove this according to Proposition 3.7, we need some work:

A Taylor expansion of $e^z$ about 3 is:

$$e^z=e^3+e^3(z-3)+e^3\frac {(z-3)^2}{2!}+e^3\frac {(z-3)^3}{3!}+ \dots$$

Thus

$$\frac {e^z}{(z-3)^2} = e^3(z-3)^{-2}+e^3(z-3)^{-1}+e^3+e^3+e^3(z-3)+e^3\frac {(z-3)^2}{2!}+ \dots$$

Example 8.3.10   $f(z)=e^{1/z}$ has an essential singularity at 0. The MacLaurin expansion of $e^z$ is

$$e^z=1+z+\frac {z^2}{2!}+\frac {z^3}{3!}+\frac {z^4}{4!} + \dots$$

By substitution we get a Laurent expansion of $e^{\frac {1}{z}}$ about 0:

$$e^{\frac {1}{z}} = 1+z^{-1}+\frac {z^{-2}}{2!}+\frac {z^{-3}}{3!}+\frac {z^{-4}}{4!} + \dots$$

The negative powers are not bounded below.