The Residue of an Analytic Function at a Point

The Residue of an Analytic Function at a Point

Consider the annulus $A(z_0, 0, r)$ (where $r > 0$), and suppose that $f$ is analytic on $A$. Then the Laurent series expansion of $f$ on $A(z_0, 0, r)$ is:

(1)
\begin{align} \quad f(z) = \sum_{n=-\infty}^{\infty} a_n(z - z_0)^n \end{align}

Where for each $n \in \mathbb{Z}$, $\displaystyle{a_n = \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{(w - z_0)^{n+1}} \: dw}$ where $\gamma$ is any circle inside $A(z_0, 0, r) = D(z_0, r) \setminus \{ z_0 \}$. By The Deformation Theorem, each $a_n$ can be evaluated by instead considering $\gamma$ to be any positively oriented closed piecewise smooth curve inside $A(z_0, 0, r)$.

Now look specifically at the term $a_{-1}$. We have that:

(2)
\begin{align} \quad a_{-1} = \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{(w - z_0)^{-1 + 1}} \: dw = \frac{1}{2\pi i} \int_{\gamma} f(w) \: dw \end{align}

For any positively oriented closed piecewise smooth curve $\gamma$ in $A(z_0, 0, r)$. This coefficient is very important and given a name for that reason.

 Definition: If $f$ is analytic on $A(z_0, 0, r)$ and if $\displaystyle{\sum_{n=-\infty}^{\infty} a_n(z - z_0)^n}$ is the corresponding Laurent series expansion on $A(z_0, 0, r)$ then the Residue of $f$ at $z_0$ is $\mathrm{Res} (f, z_0) = a_{-1}$.

For example, suppose that we want to evaluate the following integral:

(3)
\begin{align} \quad \int_{\gamma} z \sin \left ( \frac{1}{z} \right ) \: dz \end{align}

Where $\gamma$ is any positively oriented closed piecewise smooth curve in $\mathbb{C} \setminus \{ 0 \}$.

Note that the function $\displaystyle{f(z) = z \sin \left ( \frac{1}{z} \right )}$ is analytic on $\mathbb{C} \setminus \{ 0 \}$. We've already found the Laurent series expansion for $\displaystyle{\sin \left ( \frac{1}{z} \right )}$ on the Laurent's Theorem for Analytic Complex Functions page to be:

(4)
\begin{align} \quad \sin \left ( \frac{1}{z} \right ) = \sum_{n=0}^{\infty} \frac{(-1)^n}{z^{2n+1} (2n+1)!} = \frac{1}{z} - \frac{1}{3! z^3} + \frac{1}{5! z^5} - ... \end{align}

Therefore the Laurent series expansion for $\displaystyle{z \sin \left ( \frac{1}{z} \right )}$ is:

(5)
\begin{align} \quad z\sin \left ( \frac{1}{z} \right ) = \sum_{n=0}^{\infty} \frac{(-1)^n}{z^{2n} (2n+1)!} = 1 - \frac{1}{3! z^2} + \frac{1}{5! z^4} - ... \end{align}

Note that $\mathrm{Res} (f, 0) = a^{-1}$ is the coefficient attached to $\frac{1}{z}$ in the expansion above. But this means that $\mathrm{Res} (f, 0) = 0$. In other words:

(6)
\begin{align} \quad \mathrm{Res} (f, 0) = \frac{1}{2\pi i } \int_{\gamma} z \sin \left ( \frac{1}{z} \right ) \: dz = 0 \end{align}

For any positively-oriented closed piecewise smooth curve in $\mathbb{C} \setminus \{ 0 \}$. Hence $\displaystyle{\int_{\gamma} z \sin \left ( \frac{1}{z} \right ) \: dz = 0}$ for any of such $\gamma$.

We will now state an extremely simple result regarding the residues of an analytic function $f$ at removable singularities $z_0$.

 Proposition 1: If $z_0$ is a removable singularity of $f$ then $\mathrm{Res} (f, z_0) = 0$.
• Proof: By the Theorem on the Laurent's Theorem for Analytic Complex Functions page, if $z_0$ is a removable singularity then $a_n = 0$ for all $n < 0$ in the Laurent series expansion of $f$ on $A(z_0, 0, r)$ ($r > 0$). Therefore $a_{-1} = \mathrm{Res} (f, 0) = 0$. $\blacksquare$