Cauchy residue theorem

Section 1.21 Cauchy residue theorem

Definition 1.21.1. Isolated singularity.

Let \(f\) be a complex function. An element \(z_0\in \C\) is an isolated singularity of \(f\) if \(f\) is analytic on an open punctured disc

\begin{equation*} B_R(z_0)-\{z_0\}=\{z\in \C\colon 0 < \abs{z-z_0}< R\} \end{equation*}

for some positive \(R\in \C\cup \{\infty\}\text{.}\)

Definition 1.21.2. Residue.

Let \(z_0\) be an isolated singularity of the complex function \(f\text{.}\) The residue of \(f\) at \(z_0\text{,}\) denoted \(\res_f(z_0)\text{,}\) is defined as \(\res_f(z_0)=c_{-1}\text{,}\) where

\begin{equation*} f(z)=\sum_{n=-\infty}^\infty c_{n}(z-z_0)^n\text{,} \end{equation*}

is a Laurent series representation of \(f\) on any open punctured ball \(U=B_R(z_0)-\{z_0\}\) on which \(f\) is analytic.

Remark 1.21.3. Residue.

Why is \(\res_f(z_0)\) well-defined for an isolated singularity \(z_0\) of \(f\text{?}\) If \(f\) is analytic on \(U=B_R(z_0)-\{z_0\}\) and \(V=B_S(z_0)-\{z_0\}\text{,}\) with \(R< S\text{,}\) then by Theorem 1.20.1 the Laurent series corresponding to the two open annular regions are identical, and in fact we have

\begin{equation*} c_{n}=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{(z-z_0)^{n+1}}\, dz\text{,} \end{equation*}

for all \(n\text{,}\) where \(\gamma=z_0+re^{it}\text{,}\) \(t\in [0,2\pi]\) for any \(r\in (0,R)\text{.}\)

Theorem 1.21.4. Cauchy residue theorem.

Assume \(U\) is open and that \(f\) is analytic on \(U-\{z_1,z_2,\dots, z_n\}\text{,}\) where for all \(1\leq k\leq n\text{,}\) \(z_k\) is an isolated singularity of \(f\text{.}\) If \(B\subseteq U\) is any closed ball containing \(\{z_1,z_2,\dots, z_n\}\) in its interior, then

\begin{equation} \int_\gamma f\, dz=2\pi i\sum_{k=1}^n\res_f(z_k)\text{,}\tag{1.80} \end{equation}

where \(\gamma\) is simple counterclockwise oriented parametrization of \(\partial B\text{.}\)

Proof.

Example 1.21.5. Cauchy residue theorem.

Compute \(\int_\gamma z^3e^{1/z}\text{,}\) where \(\gamma=Re^{it}\text{,}\) \(t\in [0,2\pi]\) for any \(R> 0\text{.}\)

Solution.

Since \(f(z)=z^3e^{1/z}\) is analytic on \(\C-\{0\}\text{,}\) we see that \(0\) is an isolated singularity of \(f\) and

\begin{equation*} \int_\gamma f\, dz=2\pi i\res_f(0)=2\pi i(1/24)=\frac{\pi i}{12}\text{,} \end{equation*}

where we make use of the Laurent expansion

\begin{equation*} f(z)= \cdots +\frac{1}{120z^2}+\frac{1}{24z}+\frac{1}{6}+\frac{1}{2}z+z^2+z^3 \end{equation*}

computed in Example 1.20.4.

Example 1.21.6. Cauchy residue theorem: two singularities.

Compute \(\int_\gamma \tfrac{e^z}{z^3+z^2}\, dz\text{,}\) where \(\gamma=Re^{it}\text{,}\) \(t\in [0,2\pi]\text{,}\) for any \(R> 1\text{.}\)

Solution.

Since \(f(z)=\tfrac{e^z}{z^3+z^2}\) is analytic on \(\C-\{0,-1\}\)

Theorem 1.21.7. Residue at simple pole.

Let \(f=p/q\text{,}\) where \(p\) and \(q\) are analytic on \(B_R(z_0)\text{.}\) If \(z_0\) is a zero of \(q\) of order 1 (i.e., \(\ord_q(z_0)=1\)), then \(z_0\) is an isolated singularity of \(f\) and

\begin{equation} \res_f(z_0)=\frac{p(z_0)}{q'(z_0)}\text{.}\tag{1.81} \end{equation}

Proof.

Since \(f\) is analytic on \(B_R(z_0)-\{z_0\}\text{,}\) \(z_0\) is an isolated singularity of \(f\text{.}\) Since \(\ord_q(z_0)=1\text{,}\) we have \(q(z)=(z-z_0)g(z)\) where \(g\) is analytic on \(B_R(z_0)\) and \(g(z_0)\ne 0\text{.}\) It follows that \(p/g\) is analytic on \(B_R(z_0)\) and we have

\begin{align*} f(z) \amp = \frac{1}{z-z_0}\cdot \frac{p(z)}{g(z)} \\ \amp = \frac{1}{z-z_0}\sum_{n=0}^\infty a_n(z-z_0)^n \\ \amp = \frac{1}{z-z_0}(p(0)/g(0)+a_1(z-z_0)+a_2(z-z_0)^2+\cdots)\\ \amp = \frac{p(0)/g(0)}{z}+a_1+a_2(z-z_0)+\cdots \text{.} \end{align*}

Thus

\begin{equation*} \ref_f(z_0)=\frac{p(0)}{g(0)}=\frac{p(0)}{q'(0)}\text{,} \end{equation*}

since \(q'(z)=g(z)+(z-z_0)g'(z)\text{.}\)

Example 1.21.8. Residue at simple pole.

Let \(f(z)=\tfrac{\tan z}{z}\text{.}\) Find all isolated singularities of \(f\) and compute their corresponding residues.

Solution.

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