The Residue of an Analytic Function at a Pole Singularity

Recall from The Residue of an Analytic Function at a Point page that if $f$ is analytic on the annulus $A(z_0, 0, r)$ and if $\displaystyle{\sum_{n=-\infty}^{\infty} a_n(z - z_0)^n}$ is the Laurent series expansion of $f$ on $A(z_0, 0, r)$ then the residue of $f$ at $z_0$ is defined as:

(1)

\begin{align} \quad \mathrm{Res} (f, z_0) = a_{-1} \end{align}

We proved in a simple proposition that if $z_0$ is a removable singularity of $f$ then $\mathrm{Res} (f, z_0) = 0$. We will now look at another proposition which characterizes the residues $\mathrm{Res} (f, z_0)$ for which $z_0$ is a pole singularity of order $k$.

Proposition 1: If $z_0$ is a pole singularity of order $k$ of $f$ then $\displaystyle{\mathrm{Res} (f, z_0) = \lim_{z \to z_0} \frac{[(z - z_0)^k f(z)]^{(k-1)}}{(k-1)!}}$.

• Proof: Let $z_0$ be a pole singularity of order $k$ of $f$. Then there exists a function, $g$, which is analytic, $g(z_0) \neq 0$, such that:

(2)

\begin{align} \quad f(z) = \frac{1}{(z - z_0)^k} g(z) \quad (*) \end{align}

• Hence for any positively oriented closed piecewise smooth curve $\gamma$ in $A(z_0, 0, r)$ ($r > 0$) we have that:

(3)

\begin{align} \quad \mathrm{Res} (f, z_0) &= \frac{1}{2\pi i} \int_{\gamma} f(z) \: dz \\ &= \int_{\gamma} \frac{1}{2\pi i} \frac{g(z)}{(z - z_0)^k} \: dz \\ \end{align}

• By reversing Cauchy's Integral Formula for Derivatives we have that:

(4)

\begin{align} \quad \mathrm{Res} (f, z_0) &= \frac{1}{(k - 1)!} g^{(k-1)}(z_0) \end{align}

• From $(*)$ we see that:

(5)

\begin{align} \quad g^{(k-1)}(z_0) = [f(z)(z - z_0)^k]^{(k-1)} \biggr \lvert_{z=z_0} \end{align}

• Thereforefore:

(6)

\begin{align} \quad \mathrm{Res} (f, z_0) &= \lim_{z \to z_0} \frac{[f(z) (z - z_0)^k]^{(k-1)}}{(k-1)!} \quad \blacksquare \end{align}

For example, consider the following function:

(7)

\begin{align} \quad f(z) = \frac{e^{z^2} - 1}{z^3} \end{align}

Suppose that we want to find $\mathrm{Res} (f, 0)$. Note that $z_0 = 0$ is an isolated singularity of $f$. If we can show that $z_0 = 0$ is a pole singularity of order $k$ of $f$ then we can apply the theorem above to compute $\mathrm{Res} (f, 0)$.

We have that:

(8)

\begin{align} \quad e^{z^2} - 1 = \sum_{n=0}^{\infty} \frac{(z^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{z^{2n}}{n!} = [1 + z^2 + \frac{z^4}{2!} + \frac{z^6}{3!} + ...] - 1 \end{align}

Therefore:

(9)

\begin{align} \quad \frac{e^{z^2} - 1}{z^3} = \frac{1}{z^3} \left [ z^2 + \frac{z^4}{2!} + \frac{z^6}{3!} + ... \right ] = \frac{1}{z} + \frac{z}{2!} + \frac{z^3}{3!} + ... \end{align}

Hence $z_0 = 0$ is a pole singularity of order $k = 1$ of $f$. We also see from the expansion above that $\mathrm{Res} (f, 0) = 1$ (the coefficient attached to the term, $\displaystyle{\frac{1}{z}}$, but let's use the formula in the previous proposition to compute this. Since $z_0 = 0$ is a pole of order $k = 1$ we have that:

(10)

\begin{align} \quad \mathrm{Res} (f, 0) = \lim_{z \to z_0} \frac{[f(z)(z - 0)]^{(1-1)}}{(1-1)!} = \lim_{z \to 0} f(z) \cdot z = \lim_{z \to 0} \frac{e^{z^2} - 1}{z^3} \cdot z = \lim_{z \to 0} \frac{e^{z^2}-1}{z^2} \end{align}

By L'Hospital's rule:

(11)

\begin{align} \quad \mathrm{Res} (f, 0) = \lim_{z \to 0} \frac{2ze^{z^2}}{2z} = \lim_{z \to 0} e^{z^2} = 1 \end{align}