Lecture 9: Singularities of Complex Functions

We classify in this lecture singularities of complex functions. Let us recall that a point $z_0$ where the function $f(z)$ is not analytic is called a singular point of $f(z)$ .

Definition 1. Let $z_0$ be a singular point of $f(z)$ . If $f$ is analytic in $0<|z-z_0|<R$ , for some $R>0$ , we say $z_0$ is an isolated singular point of $f(z)$ .

Example 2. Let $f(z)=\dfrac{1}{1-z}$ , then $z=1$ is an isolated singular point of $f(z)$ since $f(z)$ is analytic besides at the point $z=1$ .

Example 3. Let $f(z)=\dfrac{e^z}{(z-i)^2}$ , then $z=i$ is an isolated singular point of $f(z)$ since $f(z)$ is analytic besides at the point $z=i$ .

Example 4. Let $f(z)=\dfrac{1}{\sin z}$ , then $z=k\pi$ ( $k\in\mathbb Z$ ) are all isolated singular points of $f(z)$ .

Example 5. Let $f(z)=e^{1/z}$ , then $z=0$ is an isolated singular points of $f(z)$ .

Removable singularities

Definition 6. Let $z_0$ be an isolated singular point of $f(z)$ . We say that $z_0$ is a removable singular point of $f(z)$ if $\displaystyle\lim_{z\to z_0} f(z) = c_0$ for some $c_0\in\mathbb C$ .

Here are two classical examples of removable singular points.

Example 7. Let $f(z)=\dfrac{\sin z}{z}$ , then $z=0$ is an isolated singular points of $f(z)$ (this is easy to check since $f'(z)=\dfrac{z\cos z-\sin z}{z^2}$ which blows up at $z=0$ ). Moreover, $z=0$ is a removable singular point since $\displaystyle\lim_{z\to 0}f(z)=1$ .

Example 8. Let $f(z)=\dfrac{e^z-1}{z}$ , then $z=0$ is an isolated singular points of $f(z)$ (since $f'(z)=\dfrac{(z-1)e^z}{z^2}$ which blows up at $z=0$ ). Moreover, $z=0$ is a removable singular point since $\displaystyle\lim_{z\to 0} f(z)=1$ .

Poles of different orders

Definition 9. Let $z_0$ be an isolated singular point of $f(z)$ . We say that $z_0$ is an $n$ -th order pole of $f(z)$ if there exists a positive integer $n$ and an analytic function $g(z)$ (in a neighborhood of $z_0$ ) such that $f(z)=\dfrac{g(z)}{(z-z_0)^n} \qquad \mbox{and} \qquad g(z_0)\neq 0.$ If $n=1$ , we say $z_0$ is a simple pole.

Example 10. Let $f(z)=\dfrac{1}{(z-1)(z-2)}$ . Then $z=1$ and $z=2$ are both isolated singular points. Let $g(z)= \dfrac{1}{(z-1)}$ . Then $g(z)$ is analytic when $|z-2|\le 0.5$ and $f(z)=\dfrac{g(z)}{z-2}$ . Therefore, $z=2$ is a simple pole. For the same reason, $z=1$ is also a simple pole.

Example 11. Let $f(z)=\dfrac{3z+1}{(z-1)^3(z-2)(z-5)^2}$ . Then $z=1$ is a 3rd order pole, $z=5$ is a second order pole and $z=1$ is a simple pole.

Example 12. Let $f(z)=\dfrac{3z+1}{z\sin z}$ . Then it is clear that $z=0$ is an isolated singular point, but is not removable. Using Taylor series of $\sin z$ at $z=0$ , we have $\begin{gathered} f(z)=\dfrac{3z+1}{z\sin z} = \dfrac{3z+1}{z(z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots)}=\dfrac{3z+1}{z^2(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots)} \\ =\dfrac{3z+1}{z^2}\left(1+(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots)+(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots)^2+\cdots\right) \end{gathered}$ Therefore, $z=0$ is a second order pole.

Remark 13 ([Warning!]). The requirement that $g(z_0)\neq 0$ is essential in the definition of poles. For instance, let $f(z)=\dfrac{\sin z}{z-1}$ , then $z=1$ is a simple pole. However, define the analytic function $g(z)=\sin z(z-1)^3$ , we have $f(z)=\dfrac{\sin z}{z-1}=\dfrac{\sin z(z-1)^3}{(z-1)^4} =\dfrac{g(z)}{(z-1)^4}.$ Therefore, it looks like that $z=1$ is a fourth order pole. This is NOT the case since $g(1)=0$ .

Essential singular points

Definition 14. Let $z_0$ be an isolated singular point of $f(z)$ . We call $z_0$ an essential singular point of $f(z)$ if it is neither a removable singular point nor a pole.

Example 15. Let $f(z)=e^{1/z}$ . Then $z=0$ is an isolated singular point. However, $z=0$ is not removable. To see why $z=0$ is not a pole, we observe that we can not have a finite $n$ such that $z^n e^{1/z}$ is analytic.

Singularities at Infinity

Using the transform $z\to 1/z$ , we can study singularities of functions at $z=\infty$ : the type of singularity of $f(z)$ at $z=\infty$ is the same as that of $f(1/z)$ at $z=0$ .

Example 16. The function $f(z)=z^2$ has a second order pole at $z=\infty$ . This is because $f(1/z)=1/z^2$ has a second order pole at $z=0$ .

Example 17. The function $f(z)=e^z$ has an essential singularity at $z=\infty$ . This is because $f(1/z)=e^{1/z}$ has an essential singularity at $z=0$ .

Guided review and additional examples

Learning goals

Classify isolated singularities as removable, poles, or essential.
Use limits and Laurent coefficients to recognize each type.
Determine the order of a pole.
Extend a function across a removable singularity.

Classification checklist

Near an isolated singularity $z_0$ , inspect the Laurent series. No negative powers means the singularity is removable. Finitely many negative powers means a pole, whose order is the largest negative exponent. Infinitely many negative powers means an essential singularity.

Worked example 1: remove a singularity

For $f(z)=\sin z/z$ at $z=0$ ,

$\frac{\sin z}{z}=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots.$

There are no negative powers, and $\lim_{z\to0}\sin z/z=1$ . Defining $f(0)=1$ extends $f$ to an entire function.

Worked example 2: compare a pole and an essential singularity

The function $1/z^3$ has a pole of order $3$ at zero because $z^3(1/z^3)=1$ . By contrast,

$e^{1/z}=1+\frac1z+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\cdots$

has infinitely many negative powers, so zero is an essential singularity. Multiplying by any fixed power of $z$ cannot make it analytic at zero.

Check your understanding

What kind of singularity does $(1-\cos z)/z^2$ have at $0$ , and what value removes it?

Show the answer

Since $1-\cos z=z^2/2-z^4/4!+\cdots$ , the quotient tends to $1/2$ . The singularity is removable, and defining the value at zero to be $1/2$ gives an analytic extension.