Singularities of Analytic Complex Functions

Like in elementary calculus, it is important to study the behaviour of singularities of functions to obtain a better understanding of the function itself. We begin by giving a definition of a singularity for an analytic complex function.

Definition: A Singularity of an analytic function $f$ is a point $z_0$ for which $f$ is not analytic at $z_0$. A singularity $z_0$ of $f$ is said to be an Isolated Singularity if there exists an open disk $D(z_0, r)$ for which $f$ is analytic on the punctured disk $D(z_0, r) \setminus \{ z_0 \}$.

For example, consider the function $\displaystyle{f(z) = \frac{z^3}{z^2 - 1}}$. Then $f$ is analytic everywhere it is defined but not analytic at points where $f$ is undefined. At $z = 1, -1$ $f$ is undefined and hence $f$ is not analytic at $z = 1, -1$. Therefore $1$ and $-1$ are singularities of $f$ - both of which are isolated as illustrated in the image below.

We will now look at the three main types of isolated singularities of complex analytic functions.

Definition: $z_0$ is a Removable Singularity of $f$ if $\displaystyle{\lim_{z \to z_0} f(z)}$ exists.

For example, consider the function $\displaystyle{f(z) = \frac{z^2}{z}}$. The point $z_0 = 0$ is an isolated singularity of $f$ and notice that:

(1)

\begin{align} \quad \lim_{z \to 0} \frac{z^2}{z} = \lim_{z \to 0} z = 0 \end{align}

Therefore $z_0 = 0$ is a removable singularity of $f$.

Definition: $z_0$ is a Pole Singularity of $f$ if $\displaystyle{\lim_{z \to z_0} \mid f(z) \mid = \infty}$.

For example, consider the function $\displaystyle{g(z) = \frac{1}{z}}$. The point $z_0 = 0$ is an isolated singularity of $g$ and notice that:

(2)

\begin{align} \quad \lim_{z \to 0} \mid g(z) \mid = \lim_{z \to 0} \frac{1}{\mid z \mid} = \infty \end{align}

Therefore $z_0 = 0$ is a pole singularity of $g$.

Definition: $z_0$ is an Essential Singularity of $f$ if $\displaystyle{\lim_{z \to z_0} \mid f(z) \mid \neq \infty}$ and does not exist.

For example, consider the function $\displaystyle{h(z) = e^{1/z}}$. The point $z_0 = 0$ is an isolated singularity of $h$ and notice that:

(3)

\begin{align} \quad \lim_{z \to 0} \mid h(z) \mid = \lim_{z \to 0} \mid e^{1/z} \mid = \mathrm{does \: not \: exist} \end{align}

To prove this, consider the limit of $h$ as $z \to 0$ along the positive $x$-axis:

(4)

\begin{align} \quad \lim_{z \to 0, y =0, x >0} \mid h(z) \mid = \lim_{x \to 0+} \mid e^{1/x} \mid = \infty \end{align}

And consider the limit of $h$ as $z \to 0$ along the negative $x$-axis:

(5)

\begin{align} \quad \lim_{z \to 0, y =0, x < 0} \mid h(z) \mid = \lim_{x \to 0-} \mid e^{1/x} \mid = 0 \end{align}

Since these limits are not equal, we conclude that $\displaystyle{\lim_{z \to 0} \mid h(z) \mid}$ does not exist (and does not equal $\infty$.