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:

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

Therefore

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

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:

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

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

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

  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:

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

Thus

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

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

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

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

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

The negative powers are not bounded below.

Noah Dana-Picard 2007-12-24